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Engineering Mechanics

DYNAMICS

INSTRUCTOR

SOLUTIONS

MANUAL

Fourteenth Edition

R. C. Hibbeler

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12 - 1 .

Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t — 6) m/s 2 , where t is in seconds. What
is the particle’s velocity when t = 6 s, and what is its position
when t= 11 s?

SOLUTION

a = 2t — 6

dv = a dt

6) dt

v = t 2 - 6t

ds = v dt

6 1 ) dt

s =

3 1 2

When t = 6 s,
v = 0

When t = 11s,
s = 80.7 m

Ans.

Ans.

Ans:

s = 80.7 m

1

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12 - 2 .

If a particle has an initial velocity of v 0 = 12 ft/s to the
right, at s 0 = 0, determine its position when t = 10 s, if
a = 2 ft/s 2 to the left.

SOLUTION

(±>) s = s 0 + v 0 t + |a c f 2

= 0 + 12(10) + i(-2)(10) 2

= 20 ft

Ans.

Ans:

s — 20 ft

2

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12 - 3 .

A particle travels along a straight line with a velocity
v = (12 — 3f 2 ) m/s, where t is in seconds. When t = Is,the
particle is located 10 m to the left of the origin. Determine
the acceleration when t = 4 s, the displacement from
t = 0 to t = 10 s, and the distance the particle travels during
this time period.

SOLUTION

v = 12 - 3t 2

dv ^ „. , ,

a = — = —6t\. = 4 = —24 m/s
* * '

r

ds = v dt = /

(12 -

- 3t 2 )dt

J -10 J 1 J 1

s +

10 = 12 1 - t 3 - 11

s =

12 1 — r 3 — 21

s t=

o = -21

s\t=

o

ON

1

II

o

As :

= -901 - (-21) =

-880

m

From Eq. (1):

V =

0 when t = 2s

s\t=

2 = 12(2) - (2) 3 -

21 =

-5

s T =

= (21 - 5) + (901 -

5) =

912 m

( 1 )

Ans.

t-o

Ans.

Ans.

Ans:

a = -24 m/s 2
As = -880 m
s T = 912 m

3

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* 12 - 4 .

A particle travels along a straight line with a constant
acceleration. When s = 4 ft, v = 3 ft/s and when s = 10 ft,
v = 8 ft/s. Determine the velocity as a function of position.

SOLUTION

Velocity: To determine the constant acceleration a c , set ,v 0 = 4 ft, v 0 = 3 ft/s,
s = 10 ft and v = 8 ft/s and apply Eq. 12-6.

( ^*) v 2 = vl + 2a c (s - s 0 )

8 2 = 3 2 + 2 a c (10 - 4)

a c = 4.583 ft/s 2

Using the result a c = 4.583 ft/s 2 , the velocity function can be obtained by applying
Eq. 12-6.

(■ i *) v 2 = vl + 2a c (s - s 0 )

v 2 = 3 2 + 2(4.583) (s - 4)

v = (V9.17s - 27.7) ft/s Ans.

Ans:

v = (V9.17s - 27.7) ft/s

4

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12 - 5 .

The velocity of a particle traveling in a straight line is given
by v = (6f — 3 1 2 ) m/s, where t is in seconds. If s = 0 when
t — 0, determine the particle’s deceleration and position
when t = 3 s. How far has the particle traveled during the
3-s time interval, and what is its average speed?

SOLUTION

v = 61 - 3 1 2
dv

a = —— = 6 — 6 1
dt

At t = 3 s
a = —12 m/s 2
ds = v dt

f ds = f ( 6 1 — 3 t 2 )dt

Jo Jo
s = 3t 2 - f 3
At t = 3 s
5 = 0

Since v = 0 = 6t — 3f 2 , when t = 0 and t = 2 s
when t = 2 s, s = 3(2) 2 - (2) 3 = 4 m
57- = 4 + 4 = 8m

(v SB ) = — = ~~ = 2.67 m/s

\ / avg ^ '

Ans:

St = 8 m
w avg = 2.67 m/s

Ans.

Ans.

Ans.

Ans.

5

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12 - 6 .

The position of a particle along a straight line is given by
s = (1.5< 3 — 13.5t 2 + 22.5 1) ft, where t is in seconds.
Determine the position of the particle when t = 6 s and the
total distance it travels during the 6-s time interval. Hint:
Plot the path to determine the total distance traveled.

SOLUTION

Position: The position of the particle when t = 6 s is

4 =6s = 1.5(6 3 ) - 13.5(6 2 ) + 22.5(6) = -27.0 ft Ans.

Total DistanceTraveled: The velocity of the particle can be determined by applying
Eq. 12-1.

v = — = 4.50t 2 - 21.Ot + 22.5
dt

The times when the particle stops are

4.50t 2 - 27.0f + 22.5 = 0
t = Is and t = 5 s
The position of the particle at t = 0 s, 1 s and 5 s are
s|r=0s= l-5(0 3 ) - 13.5(0 2 ) + 22.5(0) = 0
s| I= i s = 1.5(1 3 ) - 13.5(1 2 ) + 22.5(1) = 10.5 ft
s| (=5s = 1.5(5 3 ) - 13.5(5 2 ) + 22.5(5) = -37.5 ft

From the particle’s path, the total distance is

s t ot = 10-5 + 48.0 + 10.5 = 69.0 ft Ans.

■t-ss t«

-2Wh 5=0 5*/05fi

(eS t=o i.=!S

Ans:

■s| f =6s = -27.0 ft
Xtot = 69.0 ft

6

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12 - 7 .

A particle moves along a straight line such that its position
is defined by s = ( t 2 — 6f + 5) m. Determine the average
velocity, the average speed, and the acceleration of the
particle when t = 6 s.

SOLUTION

t 2 - 6t + 5

ds

— = 2t — 6
dt

dv

It ~ ~

■ 0 when t = 3

:0 = 5

=3 = -4

=6 = 5

o

II

O 1 so

II

>3 1 ^

<i |<a
ii

Ans.

\ _ s T _ 9 + 9 _ o __

’ avg Ar 6 3 m / s

Ans.

=6 = 2 m/s 2

Ans.

Ans:

^avg 0
(^sp)avg ^ m/s

a \t=6s = 2 m/s 2

7

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* 12 - 8 .

A particle is moving along a straight line such that its
position is defined by s = (10 1 2 + 20) mm, where t is in
seconds. Determine (a) the displacement of the particle
during the time interval from t = 1 s to t = 5 s, (b) the
average velocity of the particle during this time interval,
and (c) the acceleration when r = Is.

SOLUTION

s = 10r 2 + 20

(a) s|j s = 10(1) 2 + 20 = 30 mm
s| 5s = 10(5) 2 + 20 = 270 mm
As = 270 - 30 = 240 mm

(b) At = 5 — 1 = 4 s

v

avg

As _ 240
A t 4

60 mm/s

(c) a =

d 2 s
dt 2

20 mm/s 2

(for all t)

Ans.

Ans.

Ans.

Ans:

As = 240 mm
V avg = 60 mm/s
a = 20 mm/s 2

8

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12 - 9 .

The acceleration of a particle as it moves along a straight
line is given by a = (2t — 1) m/s 2 , where t is in seconds. If
s = lm and v = 2 m/s when t = 0, determine the
particle’s velocity and position when f = 6 s. Also,
determine the total distance the particle travels during this
time period.

SOLUTION

a = 2t — 1

dv = a dt

l)dt

v = t 2 - t + 2

dx = v dt

t + 2)dt

s = -1 3 - - 1 2 + It + 1
3 2

When t = 6 s

v = 32 m/s

s = 67 m

Since v 0 for 0 < t < 6 s, then
d = 67 — 1 = 66 m

Ans.

Ans.

Ans.

Ans:

v = 32 m/s
s — 67 m
d = 66 m

9

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12 - 10 .

A particle moves along a straight line with an acceleration
of a = 5/(3s 1 / J + s 5,/2 ) m/s 2 , where s is in meters.
Determine the particle’s velocity when v = 2 m, if it starts
from rest when s = 1 m. Use a numerical method to evaluate
the integral.

SOLUTION

5

a

a ds — v dv

1 9

0.8351 =

2

v = 1.29 m/s

Ans.

Ans:

v = 1.29 m/s

10

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12 - 11 .

A particle travels along a straight-line path such that in 4 s
it moves from an initial position s A = — 8 m to a position
s B = +3 m. Then in another 5 s it moves from s B to
s c = — 6 m. Determine the particle’s average velocity and
average speed during the 9-s time interval.

SOLUTION

Average Velocity: The displacement from A to C is As = s c — S A = — 6 — (—8)
= 2 m.

As 2

t^ave = — = ■-- = 0.222 m/s Ans.

6 Af 4 + 5 '

Average Speed: The distances traveled from A to 6 and B to C are s A _, B = 8 + 3
= 11.0 m and s B _, c = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled
is s To t = s A ^, B + s B _> c = 11.0 + 9.00 = 20.0 m.

, \ ^Toi 20.0

(*V)avg = = 2.22 m/s Ans.

Ans:

v aV g = 0.222 m/s
(«sp)avg = 2.22 m/s

L 8m

J 3 m

r

r> -v

C L >

y

j

>

n

c

W ^ -T-

J 1 ,

'-dm

i-

5 e>

11

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12 - 12 .

Traveling with an initial speed of 70 km/h, a car accelerates
at 6000 km/h 2 along a straight road. How long will it take to
reach a speed of 120 km/h? Also, through what distance
does the car travel during this time?

SOLUTION

v — Vi + a c t
120 = 70 + 6000(f)

t = 8.33(10~ 3 ) hr = 30 s Ans.

v 2 = vf + 2 a c (s — v x )

(120) 2 = 70 2 + 2(6000) (s - 0)

s = 0.792 km = 792 m Ans.

Ans:

f = 30 s
s = 792 m

12

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13

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12 - 14 .

The position of a particle along a straight-line path is
defined by s = (P — 6 1 2 — 15t + 7) ft, where t is in seconds.
Determine the total distance traveled when t = 10 s. What
are the particle’s average velocity, average speed, and the
instantaneous velocity and acceleration at this time?

SOLUTION

s = P - 6t 2 - 15 1 + 7

When t = 10 s,

a = 48 ft/s 2 Ans.

When v = 0,

0 = 3t 2 — 12 1 - 15
The positive root is
t = 5 s

When t = 0, s = 7 ft
When t = 5 s, s = —93 ft
When t = 10 s, .y = 257 ft
Total distance traveled

s T = 7 + 93 + 93 + 257 = 450 ft

Ans.

As 257 - 7

At~ 10-0 - 25 - 0ft / s

Ans.

w^irlH 5 - 0 */ 8

Ans.

Ans:

v = 165 ft/s
a = 48 ft/s 2
s T = 450 ft
u avg = 25.0 ft/s
(Psp)avg = 45.0 ft/s

14

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12 - 15 .

A particle is moving with a velocity of v 0 when 5 = 0 and
t = 0. If it is subjected to a deceleration of a = — kv 3 ,
where k is a constant, determine its velocity and position as
functions of time.

SOLUTION

Ans.

Ans.

Ans:

i

\ 1/2

v = 2 kt +

Vo

/

1

iy

i Y /2

.V = —

2 kt

+

k

IA

VqJ

15

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* 12 - 16 .

A particle is moving along a straight line with an initial
velocity of 6 m/s when it is subjected to a deceleration of
a = (—1.5V 1 / 2 ) m/s 2 , where v is in m/s. Determine how far it
travels before it stops. How much time does this take?

SOLUTION

Distance Traveled: The distance traveled by the particle can be determined by
applying Eq. 12-3.

ds —

vdv

ds= -y dv

lo J6 m/s — 1.5v 2

•s = / —0.6667 v 2 dv

J 6 m/s

= —0.4444v 2 + 6.532 m

When v = 0, s= -0.4444^0 2 ) + 6.532 = 6.53 m

Ans.

Time: The time required for the particle to stop can be determined by applying
Eq. 12-2.

dt =

dv

[*=-[

Jo J 6 m/s 1.5u 2

t = -1.333 u 2

\6 m/s

= 3.266 - 1.333v 2 s

When v = 0,

t = 3.266 - 1.333 0 2 = 3.27 s

Ans.

Ans:

x = 6.53 m
t = 3.27 s

16

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12 - 17 .

Car B is traveling a distance d ahead of car A. Both cars are
traveling at 60 ft/s when the driver of B suddenly applies the
brakes, causing his car to decelerate at 12 ft/s 2 . It takes the
driver of car A 0.75 s to react (this is the normal reaction
time for drivers). When he applies his brakes, he decelerates
at 15 ft/s 2 . Determine the minimum distance d be tween the
cars so as to avoid a collision.

A B

SOLUTION

For B :

() v = v 0 + a c t

v B = 60 — 12 t

, 1 ,

(■**) s = s 0 + v 0 t + -a c t

1 ,

s B = d + 60t - — (12) (1)

For A:

() v = v 0 + a c t

v A = 60 - 15(r - 0.75), [t > 0.75]

, 1 ,

(-** ) S = s 0 + V 0 t + - a c t

1 ,

= 60(0.75) + 60(r - 0.75) - - (15) (t - 0.75) 2 , [t > 0.74] ( 2 )

Require v A = v B the moment of closest approach.

60 - 12 1 = 60 - 15 (t - 0.75)
t = 3.75 s

Worst case without collision would occur when = s B .

At t = 3.75 s,from Eqs. (1) and (2):

60(0.75) + 60(3.75 - 0.75) - 7.5(3.75 - 0.75) 2 = d + 60(3.75) - 6(3.75) 2
157.5 = d + 140.625

d = 16.9 ft Ans.

Ans:

d = 16.9 ft

17

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12 - 18 .

The acceleration of a rocket traveling upward is given by
a = (6 + 0.02s) m/s 2 , where s is in meters. Determine the
time needed for the rocket to reach an altitude of
s = 100 m. Initially, v = 0 and s = 0 when t = 0.

SOLUTION

a ds = v dv

/ (6 + 0.02 s) ds = / v dv

1 0 Jo

6 s + 0.01 s 2 = —v 2
2

v

= Vl2 s + 0.02 s 2

ds = v dt

,.100

ds

= / dt

Jo Vl2 s + 0.02 s 2 Jo

Vao2 ln |_

Vl2s + 0.02s 2 + sVO02 +

12

-,100

2 Vao 2 j,

= t

t = 5.62 s

Ans.

|

ill

-

Ans:

t = 5.62 s

18

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12 - 19 .

A train starts from rest at station A and accelerates at
0.5 m/s 2 for 60 s. Afterwards it travels with a constant
velocity for 15 min. It then decelerates at 1 m/s 2 until it is
brought to rest at station B. Determine the distance
between the stations.

SOLUTION

Kinematics: For stage (1) motion, v (l = 0, .v (l = ()./ = 60 s, and a c = 0.5 m/s 2 . Thus,
( X ) s = 5 o + Vot + - aj 2

s 1 = 0 + 0 + ^(0.5)(60 2 ) = 900 m

( X ) v = v 0 + a c t

Vi = 0 + 0.5(60) = 30 m/s

For stage (2) motion, Vq = 30 m/s, s 0 = 900 m, a c = 0 and t = 15(60) = 900 s. Thus,

( X ) s = *0 + | a c t 2

s 2 = 900 + 30(900) + 0 = 27 900 m

For stage (3) motion, Vq = 30 m/s, v = 0, sq = 27 900 m and a c = — 1 m/s 2 . Thus,

( X ) v = v 0 + a c t

0 = 30 + (-1 )t
t = 30 s

s = s 0 + v 0 t + -a c r

s 3 = 27 900 + 30(30) + ^(-1)(30 2 )

= 28 350 m = 28.4 km Ans.

Ans:

s = 28.4 km

19

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12 - 20 .

The velocity of a particle traveling along a straight line is
v = (3 t 2 — 6 1 ) ft/s, where t is in seconds. If s = 4 ft when
t = 0, determine the position of the particle when t = 4 s.
What is the total distance traveled during the time interval
t = 0 to t = 4 s? Also, what is the acceleration when t = 2 s?

SOLUTION

Position: The position of the particle can be determined by integrating the kinematic
equation ds = v dt using the initial condition s = 4 ft when f = Os. Thus,

ds = v dt

f ds = f (3 1 2 — 6 t)dt
J 4 ft Jo

s = (t 3 - 3 1 2 )

4 ft 0

s = (t 3 - 3 1 2 + 4) ft

When t = 4 s,

-t=o$o * O; I - -5(H) Zo 4,5 = 4 3 - 3(4 2 ) + 4 = 20 ft Ans. The velocity of the particle changes direction at the instant when it is momentarily brought to rest. Thus, v = 3t 2 - 6f = 0 t(3t - 6) = 0 t = 0 and t — 2 s The position of the particle at t = 0 and 2 s is s| 0s = 0 - 3(0 2 ) + 4 = 4ft 4s = 23 - 3(2 2 ) + 4 = 0 Using the above result, the path of the particle shown in Fig. a is plotted. From this figure, s Xot = 4 + 20 = 24 ft Ans. Acceleration: a dv dt 61) a = (6t - 6) ft/s 2 When f = 2 s, < 4=2 S = 6(2) - 6 = 6 ft/s 2 —> Ans. Ans: s Tot = 24 ft « 1 1=2 s = 6 ft/J 2 —* 20 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 21 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 22 . A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m/s. If the bag is released with the same upward velocity of 6 m/s when t = 0 and hits the ground when f = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant. SOLUTION (+ 1 ) h = 0 + ( 6)(8) + | (9.81)(8) 2 = 265.92 m During t = 8 s, the balloon rises h' = vt = 6(8) = 48 m Altitude = h + h' = 265.92 + 48 = 314 m Ans. (+1) v = v 0 + a c t v 6 + 9.81(8) = 72.5 m/s Ans. Ans: h = 314 m v = 72.5 m/s 22 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 23 . A particle is moving along a straight line such that its acceleration is defined as a = (— 2v) m/s 2 , where v is in meters per second. If v = 20 m/s when s = 0 and t = 0, determine the particle’s position, velocity, and acceleration as functions of time. SOLUTION a = —2v v = (20e 2f ) m/s Ans. a = — = (-40e 2t ) m/s 2 dt v ' Ans. L ds = v dt = L (20 e~ 2 ‘)dt s s — 10e~ 2t | o = — 10(e~ 2t - 1) 10(l - e~ 2 ') m -2m _ Ans. Ans: v = (20e ~ 2f ) m/s a = (-40e ~ 2 ') m/s 2 5 = 10(1 - e~ 2t ) m 23 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 24 . The acceleration of a particle traveling along a straight line is a = ^-s 1/2 m/s 2 , where s is in meters. If v = 0, s = 1 m when t = 0, determine the particle’s velocity at s = 2 m. SOLUTION Velocity: v dv = a ds v dv ± V 2 o V s 1 ^ 2 ds 1 3/2 J_(^3/2 _ 1} l/2 m/s V3 When s = 2m,t = 0.781 m/s. Ans. Ans: v = 0.781 m/s 24 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 25 . If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9.81 [1 — v 2 (1CT 4 )] m/s 2 , where v is in m/s and the positive direction is downward. If the body is released from rest at a very high altitude , determine (a) the velocity when f = 5 s, and (b) the body's terminal or maximum attainable velocity (as t —» oo). SOLUTION Velocity: The velocity of the particle can be related to the time by applying Eq. 12-2. (+ 1 ) a r dv Jo 9.81 [1 - (0.01«) 2 ] 1 f r dv , r dv 9.81 [J 0 2(1 + O.Olv) 1 Jo 2(1 - 0.01 v) J „ , / 1 + O.Olv 9.81t = 501n - Vl - O.Olv 100(e° 1962r - 1) e ° 1962 ' + 1 ( 1 ) a) When t = 5 s, then, from Eq. (1) 100[e° 1962 ( 5 ) - 1] V ~ g0.1962(5) + x 45.5 m/s ^,0.19621 _ i b) °°> e 0 . 1962 , + x ^ 1- Then, from Eq. (1) «max = 100 m/s Ans. Ans. Ans: (a) v = 45.5 m/s (b) «max = 100m/s 25 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 26 . The acceleration of a particle along a straight line is defined by a = (2 1 — 9) m/s 2 , where t is in seconds. At t = 0, s = 1 m and v = 10 m/s. When f = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity. SOLUTION a = 2t — 9 J 10 civ = / (2 1 — 9) dt Jo V - 10 = t 2 - 9 t v = t 2 - 9 t + 10 ds = J (f — 9t + 10) dt s — 1 — —P — 4.5 t 2 + 10 t 3 s = — t 3 — 4.5 f 2 + 10 t + 1 3 Note when t; = t 2 -9r+10 = 0: t = 1.298 s and t = 7.701 s When t = 1.298 s, s = 7.13 m When t = 7.701 s, s = —36.63 m When t = 9 s, s = = —30.50 m (a) s = —30.5 m Ans. (b) ^To i , = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50) ^To i , = 56.0 m Ans. (c) V = 10 m/s Ans. H.lMs Ans: (a) x = —30.5 m (b) Sxot = 56.0 m (c) v = 10 m/s 26 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 27 . When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity v f. If this variation of the acceleration can be expressed as a = ( g/v 2 f)(v 2 f — v 2 ), determine the time needed for the velocity to become v = Vf/2. Initially the particle falls from rest. SOLUTION t = 0.549 Ans. Ans: t = 0.549 27 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 28 . Two particles A and B start from rest at the origin s = 0 and move along a straight line such that a A = ( 6 1 — 3) ft/s 2 and a B = (12f 2 — 8) ft/s 2 , where t is in seconds. Determine the distance between them when t = 4 s and the total distance each has traveled in t = 4 s. SOLUTION Velocity: The velocity of particles A and B can be determined using Eq. 12-2. dv A = a A dt rt / dv A = (61 — 3)dt Jo Jo v A = 3 1 2 — 3t dv B dv R = a B dt r* (12 1 1 - 8)dt JO JO v B = 4 f 3 — 8 1 The times when particle A stops are 3t 2 — 3t = 0 t = Os and = 1 s The times when particle B stops are 4f 3 — 8f = 0 t = Os and t = \fl. s Position: The position of particles A and B can be determined using Eq. 12-1. II 03 "^3 v A dt •s A II ^3 / (3f 2 Jo Sa = 3 3 ' u — r 2 els g — v B dt 'S B dsg = / (4f 3 Jo S B = f 4 - 4f 2 The positions of particle A at t = 1 s and 4 s are *aU« = l3 -|(l 2 ) = -0.500 ft U s = 4 3 -|(4 2 ) = 40.0ft Particle A has traveled d A = 2(0.5) + 40.0 = 41.0 ft The positions of particle B at t = \/7. s and 4 s are s B \t=Vi = C V2) 4 - 4( V2) 2 = -4 ft s b lt=4 = (4) 4 — 4(4) 2 = 192 ft Particle B has traveled d B = 2(4) + 192 = 200 ft At t = 4 s the distance beween A and B is A s AB = 192 - 40 = 152 ft Ans. Ans. Ans. • 0.5 ft 40.5 ft S A = - 0.5 ft 5^=0 S, = 40.0 ft t= 1s / = 0 s I = 4 s 4.0 ft fe H-h ] 96 ft S H = - J O ft 5 fl = 0 t = d 2 s 1 = 0 s -O -H ■+- S B = 192 t = 4 s Ans: d A = 41.0 ft d B = 200 ft ks AB = 152 ft 28 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 29 . A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m/s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m/s. Determine the height from the ground and the time at which they pass. SOLUTION Origin at roof: Ball A: (+T ) s = s 0 + v 0 t + -a c t 2 -s = 0 + 5t - j(9.81 )t 2 Ball B\ (+t) s = s 0 + v 0 t + -a c t 2 -s = -30 + 20 1 - ^(9.81 )t 2 Solving, t = 2 s ■s = 9.62 m Distance from ground, d = (30 - 9.62) = 20.4 m Also, origin at ground, 1 2 s = s 0 + v 0 t + - a c t s A = 30 + 5t + i(—9.81 )t 2 s B = 0 + 20t + |(-9.81)t 2 Require = S B 1 1 30 + 5 1 + -(—9.81 )t 2 = 20 1 + -(—9.81 )t 2 t = 2 s s B = 20.4 m Ans. Ans. Ans. Ans. Ans: h = 20.4 m t = 2 s 29 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 30 . A sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a deceleration of a = (—6 1) m/s 2 , where t is in seconds, determine the distance traveled before it stops. SOLUTION Velocity: v 0 = 27 m/s at f 0 = 0 s. Applying Eq. 12-2, we have (+1) dv = adt dv = I —6 tdt J21 JO v = (27 — 3t 2 ) m/s ( 1 ) At v = 0, from Eq. (1) 0 = 27 - 'it 1 t = 3.00 s Distance Traveled: ,v 0 = 0 m at l t] = 0 s. Using the result v = 27 — 3/ 2 and applying Eq. 12-1, we have 4 ) ds = vdt ds = [21 - 3t z ) dt Jo Jo s = [lit — f 3 ) m At t = 3.00 s, from Eq. (2) 5 = 27(3.00) - 3.00 3 = 54.0 m ( 2 ) Ans. Ans: s = 54.0 m 30 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 31 . The velocity of a particle traveling along a straight line is v = Vq - ks, where k is constant. If s = 0 when t = 0, determine the position and acceleration of the particle as a function of time. SOLUTION Position: ( ) dt = ds dt = ds Jo v 0 - ks t|o = ~j In (v 0 - ks) , = - e k ‘ = Vo — ks Vo Vq — ks V 0 s = — 1 - e Velocity: v = v = v 0 e Acceleration: dv ds d dt dt kt Vo / — 11 - e dt dt a = —kv 0 e^ k ‘ v 0 e Ans. Ans. Ans: s a «o k (1 -kv 0 e kt 31 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 32 . Ball A is thrown vertically upwards with a velocity of v 0 . Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t < 2vq /g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant. SOLUTION Kinematics: First, we will consider the motion of ball A with (v A ) {] = v 0 , = 0, sa = h,t A = t', and (a c ) A = -g. ( + T) s A = (s A ) 0 + (v A ) 0 t A + -( a c ) A t A 2 h = 0 + v 0 f' + |(-g)(t') 2 h = v 0 t' - 1 1 ' 2 ( 1 ) (+t) V A = (v A )o + (a c ) A t A Va = v 0 + (~g)(t') v A = v 0 - gt' (2) The motion of ball B requires (v B ) 0 = v 0 > ( s b)o = 0, s B = h, t B = t' — t , and K)b = ~g- ( + t) s B = (s B ) q + ( v B ) 0 t B + ~g{a c ) B t B 2 h = 0 + v 0 (f' - 0 + | (~g)(t' ~ t) 2 h = v 0 (t' - t) - | (f' - tf (3) ( + t) V B = (v B )o + {a c ) B t B V B = V 0 + (~gW ~ t) V B = V 0 - g(t' - t) ( 4 ) Solving Eqs. (1) and (3), V' - 2 f ~ v o( 1 ' ~ 0 “ j t' = 2v 0 + gf 2g Substituting this result into Eqs. (2) and (4), f 2v 0 + gf = V 0 - g[-^~ = -\ gt = 2 gti f 2v 0 + gt v B = v 0 - g^——-f Ans. Ans. Ans. Ans: 2vp + gt 2 g 1 , va = ygf i V B = ^gt \ 32 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 33 . As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = — g 0 [R 2 /(R + y) 2 ], where g 0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If go = 9-81 m/s 2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that v = 0 as y —* oo. SOLUTION v dv = a dy v 2 0 g 0 R 2 2 v R + y o v = V2g 0 R = V2(9.81)(6356)(10) 3 = 11167 m/s = 11.2 km/s Ans. Ans: v = 11.2 km/s 33 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 34 . Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12-36), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y Q from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y 0 = 500 km? Use the numerical data in Prob. 12-36. SOLUTION From Prob. 12-36, (+t) R 2 a ~ -go (R + yf Since a dy = v dv then ~g 0 R 2 g 0 R goR 2 [ Thus y dy , (R + y) 2 -y ~2 1 fo R + y _ 1 _ R + y R + v dv i? ~2 v = - R 2go (To - y) (R + y)(R + y 0 ) When y Q = 500 km, y = 0, v = —6356(10 3 ). 2(9.81)(500)(10 3 ) 6356(6356 + 500)(10 6 ) v — —3016 m/s = 3.02 km/s I Ans. Ans. Ans: v = —R 2go(yo ~ y) (R + y)(R + To) Wimp = 3-02 km/s 34 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 35 . A freight train starts from rest and travels with a constant acceleration of 0.5 ft/s 2 . After a time t' it maintains a constant speed so that when t = 160 s it has traveled 2000 ft. Determine the time t' and draw the v-t graph for the motion. SOLUTION Total Distance Traveled: The distance for part one of the motion can be related to time t = t' by applying Eq. 12-5 with s 0 = 0 and v 0 = 0. ( ^ ) s = s 0 + v 0 1 + - a c t 2 sj = 0 + 0 + | (0.5)(t') 2 = 0.25(f') 2 The velocity at time t can be obtained by applying Eq. 12-4 with Vq = 0. () v = Vq + a c t = 0 + 0.5f = 0.5t (1) The time for the second stage of motion is t 2 = 160 — t' and the train is traveling at a constant velocity of v = 0.5t' (Eq. (l)).Thus, the distance for this part of motion is ( ^* ) s 2 = vt 2 = 0.5f'(160 - t ') = 80 1' - 0.5(f') 2 If the total distance traveled is i Xot = 2000, then rck/i) ta) ^Tot — S 1 + S 2 2000 = 0.25(f') 2 + 80t' - 0.5(t') 2 0.25(t') 2 - 80 1' + 2000 = 0 Choose a root that is less than 160 s, then l' = 27.34 s = 27.3 s Ans. v — t Graph: The equation for the velocity is given by Eq. (l).When t = t' = 27.34 s, v = 0.5(27.34) = 13.7 ft/s. Ans: t' = 27.3 s. When r = 27.3 s, v = 13.7 ft/s. 35 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 36 . The s-t graph for a train has been experimentally determined. From the data, construct the v-t and a-t graphs for the motion; 0 < t < 40 s. For 0 < t < 30 s, the curve is s = (0.4f 2 ) m, and then it becomes straight for f > 30 s. SOLUTION 0 < t < 30: x = 0.4 1 2 ds v = — = 0.8f dt dv a = —— = 0.8 dt 30 < t < 40: (600 — 360V *- 360 = U^r) (? - 30) 5 = 24(1 - 30) + 360 ds v = — = 24 dt dv a = — = 0 dt r(m) v("/s) oa SSL. Is HO •«*) Ans: v a s dt dv dt 24 (f = 0.8t = 0.8 - 30) + 360 v a d- s ~dt~ 2A ? = o dt 36 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 37 . Two rockets start from rest at the same elevation. Rocket A accelerates vertically at 20 m/s 2 for 12 s and then maintains a constant speed. Rocket B accelerates at 15 m/s 2 until reaching a constant speed of 150 m/s. Construct the a-t, v-t, and s-t graphs for each rocket until t = 20 s. What is the distance between the rockets when t = 20 s? SOLUTION For rocket A For t < 12 s + ] v A = (v A )o + a A t v A = 0 + 20 t v A = 20 t 1 , + T Sa = (uOo + (v A )o t + -a 4 t L S A = 0 + o T |(20) t 2 S 4 = 10 t 2 When t = 12 s, v A = 240 m/s s A = 1440 m For t > 12 s v A = 240 m/s = 1440 + 240(r - 12) For rocket B For t < 10 s +1 v B = (v B )o + a B t v B = 0 + 15 t v B = 15 t + T % = (%)o + Ob)o t + \a B t 1 Sb = 0 + 0 + —( 15 ) t 2 s B = 7.5 t 2 When? = 10 s, v B = 150 m/s s B = 750 m For t > 10 s v B = 150 m/s s B = 750 + 150(f - 10) When t = 20 s, s A = 3360 m, s B = 2250 m As = 1110 m = 1.11km 30 -- Vo' m /s) u 3a r* 15 Ans. Ans: As = 1.11 km 37 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 38 . A particle starts from s = 0 and travels along a straight line with a velocity v = (t 2 — At + 3) m/s, where t is in seconds. Construct the v—t and a—t graphs for the time interval 0 < t < 4 s. SOLUTION a-t Graph: Thus, dv a = — dt a = (It d i dt ? ~ 4t + ^ 4) m/s 2 «lr=o = 2(0) — 4 = —4 m/s 2 4=2 = 0 4=4 s = 2(4) — 4 = 4 m/s 2 The a—t graph is shown in Fig. a. dv v-t Graph: The slope of the v—t graph is zero when a = — = 0. Thus, dt a = 2t — 4 = 0 t = 2 s The velocity of the particle at t = 0 s, 2 s, and 4 s are 4=o s = 0 2 — 4(0) + 3 = 3 m/s 4 = 2 s = 2 2 - 4(2) + 3 = -1 m/s 4=4 s — 4 2 - 4(4) + 3 = 3 m/s The v—t graph is shown in Fig. b. -tfc) Ans: a\ ,=o = —4 m/s 2 fl|r=2s = 0 «|f=4s = 4 m/s 2 w|f=o = 3 m/s v If= 2 s = “I m/s w| f =4s = 3 m/s 38 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 39 . If the position of a particle is defined by s = [2 sin (-77-/5 )t + 4] m, where t is in seconds, construct the s—t, v—t, and a—t graphs for 0 < t < 10 s. SOLUTION V (">/*) Ans: s = 2 sin 2tt v =* cos 2tt a = srn + 4 39 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 40 . An airplane starts from rest, travels 5000 ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. It then climbs in a straight line with a uniform acceleration of 3 ft/s 2 until it reaches a constant speed of 220 mi/h. Draw the s-t, v-t, and a-t graphs that describe the motion. SOLUTION Vi = 0 v 2 = 162- mi (lh) 5280 ft h (3600 s)(l mi) 237.6 ft/s = A + 2 a c (s 2 - Si) (237.6) 2 = 0 2 + 2(u c )(5000 - 0) a c = 5.64538 ft/s 2 v 2 — v i + a c t 237.6 = 0 + 5.64538 t t = 42.09 = 42.1 s t>3 = 220 mi (lh) 5280 ft h (3600 s)(l mi) + 2 a c (s 3 - s 2 ) 322.67 ft/s (322.67) 2 = (237.6) 2 + 2(3 )(s - 5000) s = 12 943.34 ft v 3 = v 2 + a c t 322.67 = 237.6 + 3 t t = 28.4 s SiS 3 Aljtyj 2.) 42--1 70.+ -icv Ans: s = 12943.34 ft v 3 = v 2 + a c t t = 28.4 s 40 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 41 . The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft/s 2 and then decelerate at 2 ft/s 2 . Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a-t, v-t, and s-t graphs for the motion. SOLUTION +1 v 2 = Vi + a c ti ^'max 0 + 5 +1 v 2 = v 2 + a c t 0 v max 2 t 2 Thus h ~ 0.4 t 2 , 1 , + T «2 - Si + v ih + 2° c ^ h = 0 + 0 + i(5)(t?) = 2.5 A + T40-/? = 0 + v max t 2 - |( 2 ) A + T v 1 = V\ +2 a c (s — ij) vLx = 0 + 2(5)(* - 0) 'Uniax 10/7 0 = iLx + 2(—2)(40 - h) vLax = 160 “ 4/1 Thus, 10 h = 160 - 4/7 h = 11.429 ft v max = 10.69 ft/s t r = 2.138 s t 2 = 5.345 s t = t\ + t 2 = 7.48 s When t = 2.145, v = v max = 10.7 ft/s and h = 11.4 ft. «• (Wi 1 ) z..* 7.48 0 l WO Ans. Ans: t = 7.48 s. When t = 2.14 s, V = Wax = 10.7 ft/s h = 11.4 ft 41 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 42 . The velocity of a car is plotted as shown. Determine the total distance the car moves until it stops (t = 80 s). Construct the a-t graph. r(m/s) 10 SOLUTION Distance Traveled: The total distance traveled can be obtained by computing the area under the v — t graph. 40 80 -f (s) x = 10(40) + j(10)(80 - 40) = 600 m Ans. dv a-t Graph: The acceleration in terms of time t can be obtained by applying a = —. For time interval 0 s < t < 40 s, t dv a = — = 0 dt For time interval 40 s < t < 80 s, —-— = —-—, v = ( —-t + 20 ) m/s. t - 40 80 - 40 V 4 1 ' dv 1 , , a = — = — = —0.250 m/s 2 dt 4 ! o -0-250- - 40 So -td) For 0 < t < 40 s, a — 0. For 40 s < t < 80, a = —0.250 m/s 2 . Ans: s = 600 m. For 0 < t < 40 s, a = 0. For 40 s < t < 80 s, a = —0.250 m/s 2 42 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-43. The motion of a jet plane just after landing on a runway is described by the a-t graph. Determine the time t' when the jet plane stops. Construct the v-t and s-t graphs for the motion. Here s = 0, and v = 300 ft/s when t = 0. SOLUTION a (m/s 2 ) v-t Graph. The v-t function can be determined by integrating dv = a dt. ForO < t < 10 s, a = 0. Using the initial condition v = 300 ft/s at t = 0, dv > 300 ft/s v — 300 — 0 v = 300 ft/s a — (— 20 ) For 10 s < t < 20 s, - t - 10 -10 - (- 20 ) 20 - 10 a = (t initial condition v = 300 ft/s at t = 10 s, Ans. 30) ft/s 2 . Using the v - 300 30) dt J 10 s - 30r + 550 t 10 s ft/s Ans. At t = 20 s. = —(20 2 ) - 30(20) + 550 = 150 ft/s 1=20 s 2 For 20 s < t < t' , a = —10 ft/s. Using the initial condition v = 150 ft/s at t = 20 s, dv = - 10 dt 150 ft/s I 20 s v — 150 = (-lot) v = (-lot + 350) ft/s It is required that at f — t', v = 0. Thus 0 = —10 f' + 350 t' = 35 s Ans. Ans. Using these results, the v-t graph shown in Fig. a can be plotted s-t Graph. The s-t function can be determined by integrating ds = v dt. For 0 < f < 10 s, the initial condition is s = 0 at t = 0. ds= 300 dt Jo Jo s = {300 t} ft Ans. At = 10 s, ^1 r=io s = 300(10) = 3000 ft 43 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-43. Continued For 10 s < t < 20 s, the initial condition is s = 3000 ft at t = 10 s. f ds = I (~f 2 ~ 30 r + 550)dt J 3000 ft J 10sV2 J I 3000 ft s - 3000 = ( ^t 3 - 15 1 2 + 550f 10 s s = | -t 3 - 15 1 2 + 550 1 - 11671 ft At t = 20 s, x = i (20 3 ) - 15(20 2 ) + 550(20) - 1167 = 5167 ft For 20 s < t < 35 s, the initial condition is s = 5167 ft at t = 20 s. [ ds = [ (— 10r + 350) dt J 5167 ft J 20 s t s - 5167 = (-5t 2 + 350r) 20 s 5 = { —5t 2 + 350r + 167} ft At t = 35 s, = —5(35 2 ) + 350(35) + 167 = 6292 ft t = 35 s using these results, the s-t graph shown in Fig. b can be plotted. 'vCftlO Ans. i(s) Ans: t’ = 35 s For 0 < t < 10 s, s = |300f) ft v = 300 ft/s For 10 s < t < 20 s, s = |if 3 - 15t 2 + 550f - 11671 ft v = — 30r + 55o| ft/s For 20 s < t < 35 s, s = { —5f 2 + 350f + 167} ft v = (-10f + 350) ft/s 44 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 44 . The v-t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude of 4 m/s 2 . If the plates are spaced 200 mm apart, determine the maximum velocity v max and the time t' for the particle to travel from one plate to the other. Also draw the s-t graph. When t = t'/ 2 the particle is at s = 100 mm. SOLUTION a c = 4 m/s 2 - = 100 mm = 0.1 m 2 p 2 = Vo + 2 a c (s - s 0 ) pL* = 0 + 2(4)(0.1 - 0) Vmax = 0.89442 m/s = 0.894 m/s v = v 0 + a c t' t' 0.89442 = 0 + 4(—) y 2 t ' = 0.44721 s = 0.447 s /-Ui.ix- <\ sAo Ans. 1 ^ S = So + v o t + - a c t s = 0 + 0 + | (4)(t) 2 s = 2? When t = 0.44721 2 = 0.2236 = 0.224 s. s = 0.1 m / ds= ~ J 0.894 JO. v = -4 r+1.788 r s pt 4 dt ds= (-41+1.788) dt J 0.1 J 0.2235 x = - 2 t 2 + 1.788 t - 0.2 When t = 0.447 s, s = 0.2 m Ans: t' = 0.447 s 5 — 0.2 m 45 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-45. The v-t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where t' = 0.2 s and w ma x= 10 m/s. Draw the s-t and a-t graphs for the particle. When t = t'/2 the particle is at s = 0.5 m. SOLUTION For 0 < t < 0.1 s, v = 100 t a dv dt 100 ds = v dt 100 t dt When t = 0.1 s, x = 0.5 m For 0.1 s < t < 0.2 s, v = -lOOt + 20 dv a = — = - 100 dt ds = v dt ds= (-100f + 20)dt J0.5 J 0.1 s - 0.5 = (-50 t 2 + 20 t - 1.5) s = — 50 t 2 + 20 t — 1 When t = 0.2 s, s = 1 m When t = 0.1 s, s = 0.5 m and a changes from 100 m/s 2 to —100 m/s 2 . When t = 0.2 s, s = 1 m. I 00 -\ov ox 0.\ 4 C$>

K*3

Ans:

When t = 0.1 s,
s = 0.5 m and a changes from
100 m/s 2 to -100 m/s 2 . When t = 0.2 s,
s = 1 m.

46

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12-46.

The a-s graph for a rocket moving along a straight track has
been experimentally determined. If the rocket starts at s = 0
when v = 0, determine its speed when it is at
s — 75 ft, and 125 ft, respectively. Use Simpson’s rule with
n = 100 to evaluate v at s = 125 ft.

SOLUTION

0 < s < 100

v = VlOs

At s = 75 ft, v = Vl50 = 27.4 ft/s
At x = 100 ft, v = 31.623

/»125

/ vdv= / [5 + 6(Vs - 10) 5 / 3 ] ds

J 31.623 J 100

2 V

= 201.0324

31.623

v = 37.4 ft/s

a (ft/s 2 )

o = 5 + 6(/T- 10) 5 '

100

i (ft)

Ans.

Ans.

Ans:

= 27.4 ft/s

= 37.4 ft/s

47

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12 - 47 .

A two-stage rocket is fired vertically from rest at s = 0 with
the acceleration as shown. After 30 s the first stage, A, burns
out and the second stage, B , ignites. Plot the v-t and s-t
graphs which describe the motion of the second stage for

5 < t < 60 s.

a (m/s 2 )

SOLUTION

v-t Graph. The v—t function can be determined by integrating dv = a dt.

12 ( 2 \ , ,

For 0 < t < 30 s, a = — t = I —t J m/s . Using the initial condition v = 0 at t = 0,

f V f ' 2

/ dv = t —tdt

Jo Jo 5

v = | 1 m/s

V(mlsj

At t = 30 s.

= —(30 2 ) = 180 m/s
t = 30 s 5

For 30 < t < 60 s, a = 24 m/s 2 . Using the initial condition v =
f v dv = f 24 dt

J 180 m/s J 30 s

u - 180 = 24 t

30 s

v = {24t — 540) m/s
At t = 60 s.

5Cm)

■t(&)

48

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12-47. Continued

For 30 s < t < 60 s, the initial condition is s = 1800 m at t = 30 s.

ds = / (24 1 - 540 )dt

1 1800 m

s - 1800 = (lit 2 - 540f)

30 s

5 = [lit 2 - 540r + 7200} m
At t = 60 s,

t = 60 s

= 12(60 2 ) - 540(60) + 7200 = 18000 m

Using these results, the s—t graph in Fig. b can be plotted.

Ans:

For 0 < t < 30 s,

For 30 < t < 60 s,
v = {24f — 540} m/s
s = {12t 2 - 540f + 7200} m

49

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*12-48.

The race car starts from rest and travels along a straight
road until it reaches a speed of 26 m/s in 8 s as shown on the
v-t graph. The flat part of the graph is caused by shifting
gears. Draw the a-t graph and determine the maximum
acceleration of the car.

SOLUTION

For 0 < t < 4 s

a

A v

TT

14

T

3.5 m/s 2

For 4 s < t < 5 s

A v

a = —— = 0
A t

For 5 s < t < 8 s

_ A v _ 26 - 14
“ T7 “ 8-5

4 m/s 2

a max = 4.00 m/s 2

>1

a. 5

* S i

-**(*)

Ans.

Ans:

flmax = 4.00 m/s 2

50

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12-49.

The jet car is originally traveling at a velocity of 10 m/s o (m/s 2 )

when it is subjected to the acceleration shown. Determine
the car’s maximum velocity and the time t' when it stops.

When t = 0, s = 0.

6

SOLUTION

v-t Function. The v—t function can be determined by integrating dv = a dt. For
0<f<15s,a = 6 m/s 2 . Using the initial condition v = 10 m/s at t = 0,

v — 10 = 6t

v = [61 + 10) m/s

The maximum velocity occurs when t = 15 s. Then

v mm = 6(15) + 10 = lOOm/s Ans.

For 15 s < t < f', a = —4 m/s, Using the initial condition v = 100 m/s at t = 15 s,

dv =

- 4dt

15 s

J 100 m/s

v - 100 = (—40

15 s

v = {— 4t + 160} m/s
It is required that v = 0 at t = t'.Then
0 = -4 1' + 160 t' = 40 s

Ans.

15

f'

t( s)

Ans:

Wiax = 100m/s
t' = 40 s

51

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12-50.

The car starts from rest at s = 0 and is subjected to an
acceleration shown by the as graph. Draw the vs graph
and determine the time needed to travel 200 ft.

a (ft/s 2 )

SOLUTION

For s < 300 ft

a ds = v dv

12 ds =

12 s

1

2

v

2

v = 4.90 s 1 / 2

At v = 300 ft, v = 84.85 ft/s
For 300 ft < s < 450 ft

a ds = v dv

(24 — 0.04 s) ds = / v dv

24 x - 0.02 x 2 - 5400 = 0.5 v 2 - 3600
v = (—0.04 s 2 + 48 s - 3600) 1 / 2
At s = 450 ft, v = 99.5 ft/s
v = 4.90 s 1 / 2
ds

dt

= 4.90 s 1/2

/»200

s ds= 4.90 dt

2 ^

= 4.90 t

o

t = 5.77 s

12

6

300 450

.(ft)

Ans.

Ans:

For 0 < s < 300 ft,
v = { 4.90 S' 1 / 2 } m/s.

For 300 ft < s < 450 ft,
v = {(—0.04s 2 + 48. - 3600) 1 / 2 } m/s.
s = 200 ft when t = 5.77 s.

52

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12-51.

The v-t graph for a train has been experimentally v (m/s)

ds =

1

20 '

When t = 60 s.

oVlO

t ]dt

Ans.

■S' I r - 60 s = ( 6 ° 2 ) = 180 m

For 60 s < t < 120 s, v = 6 m/s. Using the initial condition 5 = 180 m at t = 60 s,

ds =

6 dt

s - 180 = 6 t

60s

s = {6t — 180} m
At t = 120 s,

s|,-i20s = 6(120) - 180 = 540 m
v - 6 10-6

Ans.

For 120 s < t < 180 s,

t - 120 180 - 120

condition s = 540 m at t = 120 s,

\v = ) — t - 2 } m/s. Using the initial

ds = I ( — t — 2 | dt

s - 540 = ( — t 2 - 21
V 30

s = ( ^-f 2 — 2 1 + 300 } m

Ans.

At f = 180 s,

s|<=i80s = ^( 18 ° 2 ) - 2(180) + 300 = 1020m
Using these results, s-t graph shown in Fig. a can be plotted.

53

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12-51. Continued

a-t Graph. The a-t function can be determined using a =

dv

dt'

For 0 < t < 60 s,

toO

dt

= 0.1 m/s 2

d(6)

For 60 s < t < 120 s, a = -= 0

dt

For 120 s < t < 180 s,

^(l5 r “)

dt

= 0.0667 m/s 2

Using these results, a-t graph shown in Fig. b can be plotted.

Ans.

Ans.

Ans.

Ans:

For 0 < t < 60 s,

a = 0.1 m/s 2 .

For 60 s < t < 120 s,
s = {6t — 180} m,
a = 0. For 120 s < t < 180 s,

s = — 2 1 + 30o| m,

a = 0.0667 m/s 2 .

54

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*12-52.

A motorcycle starts from rest at s = 0 and travels along a
straight road with the speed shown by the v-t graph.
Determine the total distance the motorcycle travels until it
stops when t = 15 s. Also plot the a-t and s-t graphs.

SOLUTION

For t < 4 s

dv i

a = —— = 1.25
dt

ds = 1.25 t dt

Jo Jo

s = 0.625 t 2

When t = 4 s, s = 10 m

For 4 s < t < 10 s
dv

a= * = °

[ ds = [5 dt

J 10 J 4

x = 5 t - 10

When t = 10 s, x = 40 m

For 10 s < t < 15 s
dv

a = — = -1
dt

ds = / (15 — t) dt

MO

'10

x — 15 t 0.5 t 2 60
When t = 15 s, s = 52.5 m

HTs*)

10 / is

T - >*($) Ans. Ans: When t = 15 s, s = 52.5 m 55 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-53. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v-t graph. Determine the motorcycle’s acceleration and position when t = 8 s and t = 12 s. SOLUTION At t = 8 s A s = f v dt s-0 = |(4)(5) + (8 - 4)(5) = 30 j = 30 m At t = 12 s dv —5 dt 5 -1 m/s 2 A s = f v dt s-0 = |(4)(5) + (10 - 4)(5) + |(15 - 10)(5) - |(|)(5)(|)(5) s = 48 m Ans. Ans. Ans. Ans. Ans: At t = 8 s, a = 0 and s = 30 m. At t = 12 s, a = — 1 m/s 2 and s = 48 m. 56 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-54. The v-t graph for the motion of a car as it moves along a straight road is shown. Draw the s-t and a-t graphs. Also determine the average speed and the distance traveled for the 15-s time interval. When t = 0, s = 0. v (m/s) 15 SOLUTION s-t Graph. The s-t function can be determined by integrating ds = v dt. For 0 < t < 5 s, v = 0.6 1 2 . Using the initial condition s = 0 at t = 0, s = { 0.2f 3 } m Ans. At t = 5 s, s|t=5s = 0.2(5 3 ) = 25 m For 5 s < t < 15 s, v — 15 t - 5 0-15 15-5 s = 25 m at t = 5 s, 3 t)dt 45 3 s - 25 = —t - -t 2 - 93.75 2 4 s = |i(90f - 3 1 2 - 275) | m At t = 15 s, 3 1). Using the initial condition Ans. 5 = j [ 90(15) - 3(15 2 ) - 275] = 100 m Ans. Thus the average speed is s T 100 m , rw = — = -= 6.67 m/s Ans. g t 15 s using these results, the s-t graph shown in Fig. a can be plotted. 57 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-54. Continued a-t Graph. The a-t function can be determined using a = dv dt' d{ 0.6 r 2 ) For 0 < t < 5 s, a = = {1.2 f} m/s 2 at Ans. At t = 5 s, a = 1.2(5) = 6 m/s 2 Ans. d[ |(45 -3t)l For 5s<t<15s, a = -= —1.5 m/s 2 at Ans. j$(rr>)

For 0 < t < 5 s,
s = { 0.2f 3 } m
a = {1.21 ) m/s z
For 5 s < t < 15 s,

s = |^(90t - 3 1 2 - 275)| m

a = —1.5 m/s 2
At t = 15 s,
s = 100 m
t>avg = 6.67 m/s

58

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12-55.

An airplane lands on the straight runway, originally traveling
at 110 ft/s when s = 0. If it is subjected to the decelerations
shown, determine the time t' needed to stop the plane and
construct the s-t graph for the motion.

SOLUTION

Vo = 110 ft/s
A v = J a dt

0 - 110 = -3(15 - 5) - 8(20 - 15) - 3(f' - 20)
t' = 33.3 s

Ans.

*\t = 5s

= 550 ft

‘ f = 15s

= 1500 ft

= 1800 ft

= 2067 ft

Ans:

t' = 33.3 s
*\t=Ss = 550 ft
s I r=i5 s = 1500 ft

■S | ,=20 s = 1800 ft

s 1 1 = 33.3 s = 2067 ft

59

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*12-56.

Starting from rest at s = 0, a boat travels in a straight line
with the acceleration shown by the as graph. Determine
the boat’s speed when s = 50 ft, 100 ft, and 150 ft.

a (ft/s 2 )

SOLUTION

vs Function. The v—s function can be determined by integrating v dv = a ds.

a - 8 6-8 fl 1 , ,, .

, a = { —— s + 8 > ft/s . Using the initial

For 0 < s < 100 ft,

5-0 100-0

condition v— 0 at s = 0,

v dv = / ( —— i + 8 | ds

Jo J 0 V 50

50

v

2

= ( —— s 2 + 8 5
100

1 ,

- =85-5“

2 100

50

At s = 50 ft,

v I s = 50 ft =

At 5 = 100 ft,

(800 s - s z ) > ft/s

50

[800(50) - 50 2 ] = 27.39 ft/s = 27.4 ft/s

v I s=ioo ft = [800 (100) - 100 2 ] = 37.42 ft/s = 37.4 ft/s

Ans.

Ans.

For 100 ft < s < 150 ft.

a — 0

7^150 = 100^150 ;fl = {“I* + 18 Using the

initial condition v = 37.42 ft/s at s = 100 ft,

' 37.42 ft/s

vdv= I-s T 18 I ds

J 100 ft V 25

v

2

37.42 ft/s

= >- 50 * +18i '

= ||\/-3s 2 + 900s - 250001 ft/s

At s = 150 ft

i= i 50 ft = |V-3(150 2 ) + 900(150) - 25000 = 41.23 ft/s = 41.2 ft/s Ans.

100

150

-s(ft)

Ans:

v | s = 50 ft = 27.4 ft/s
v | s = 100 ft = 37.4 ft/s
v\s = 150 ft = 41.2 ft/s

60

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12-57.

Starting from rest at s = 0, a boat travels in a straight line
with the acceleration shown by the as graph. Construct the
vs graph.

a (ft/s 2 )

SOLUTION

vs Graph. The v—s function can be determined by integrating v dv = a ds. For

100

150

■ s (ft)

0 < s < 100 ft,

a — 8 6 — 8 (1 • , ■■■,

-= -, a = <- s + 8 > ft / s using the initial

s - 0 100 - 0 1 50 ' ' 6

condition v = 0 at v = 0,

/ v dv = / —— s + 8 ) ds

A) Jo V 50

1

=-s 2 + 8 s

100

1 9

— = 8 s- s

2 100

50

(800 s — s 2 ) f ft/s

At j = 25 ft, 50 ft, 75 ft and 100 ft

vI,=25it = ^[800 (25) -25 2 ] = 19.69 ft/s
vI,=soft = oj ] 5() [800 (50) -50 2 ] = 27.39 ft/s
v|, =75 ft = ^[800 (75) -75 2 ] = 32.98 ft/s
vI,=iooft = yjjfi [800 (100) —100 2 ] = 37.42 ft/s

For 100 ft < s < 150 ft,

a — 0 6 — 0 (3 i 9

-= -; a = { - s + 18 / ft/s 2 using the

5 - 150 100 - 150 I 25 ' ' B

initial condition v = 37.42 ft/s at v = 100 ft,

3

/ vdv= I — — ,s +18 | ds

' 37.42 ft/s J 100 ft V 25

37.42 ft/s

= , -5T +18 *

100 ft

= ||\/-3j 2 + 900v - 250001 ft/s

At j = 125 ft and s = 150 ft
1

Ans:

For 0 < s < 100 ft,

t 2 Is=i25ft = ^-V-3(l25 2 ) +900 (125) -25000 = 40.31 ft/s
v 1 5 =i 5 oft = |\/-3(150 2 ) +900 (150) -25000 = 41.23 ft/s

V =

I sj ^( 80Qy ” ■ ?2 )} ft /s

For 100 ft < x < 150 ft,
v = | V-3s 2 + 900.S- - 25000 \ ft/s

61

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12-58.

A two-stage rocket is fired vertically from rest with the
acceleration shown. After 15 s the first stage A burns out and
the second stage B ignites. Plot the v-t and s-t graphs which
describe the motion of the second stage for 0 < £ < 40 s.

SOLUTION

For 0 < t < 15

VI- fu,

Jo Jo

1 2

V = ~t
2

v = 112.5 when £ = 15 s
f ds =

f 0

1 3

s = - r

<o

— f 2 dt
2

s = 562.5 when £ = 15 s
For 15 < t < 40
a = 20

[ dv = [ 20 dt
J 112.5 J 1.5

v = 201 - 187.5

v = 612.5 when t = 40 s

[ ds = / (20 t - 187.5) dt

J 562.5 J 15

s = 10 £ 2 — 187.5 t + 1125
j = 9625 when t = 40 s

Ans:

For 0 < t < 15 s,

v = jVf 2 jm/s

For 15 s < £ < 40 s,
v = {20f — 187.5 m/sj
s = {10f 2 - 187.5f + 1125} m

62

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12-59.

The speed of a train during the first minute has been
recorded as follows:

t (s) 0 20 40 60

v (m/s) 0 16 21 24

Plot the v—t graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.

SOLUTION

The total distance traveled is equal to the area under the graph.

1 1 1

s T = 2 (20)(16) + - (40 - 20)(16 + 21) + - (60 - 40)(21 + 24) = 980 m Ans.

'* 0*0

to

Ans:

s T = 980 m

63

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* 12 - 60 .

A man riding upward in a freight elevator accidentally
drops a package off the elevator when it is 100 ft from the
ground. If the elevator maintains a constant upward speed
of 4 ft/s, determine how high the elevator is from the
ground the instant the package hits the ground. Draw the
v-t curve for the package during the time it is in motion.
Assume that the package was released with the same
upward speed as the elevator.

SOLUTION

For package:

(+ T) v 2 = + 2a c (s 2 ~ s 0 )

v 2 = (4) 2 + 2(—32.2)(0 - 100)
v = 80.35 ft/s i

(+ T) v = v 0 + a c t
-80.35 = 4 + (-32.2 )t
t = 2.620 s

For elevator:

(+T) s 2 = S 0 + vt

s = 100 + 4(2.620)
s = 110 ft

V

i

Ans.

Ans:

s = 110 ft

64

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12 - 61 .

Two cars start from rest side by side and travel along a
straight road. Car A accelerates at 4 m/s 2 for 10 s and then
maintains a constant speed. Car B accelerates at 5m/s 2
until reaching a constant speed of 25 m/s and then
maintains this speed. Construct the a-t, v-t, and s-t graphs
for each car until t = 15 s. What is the distance between the
two cars when t = 15 s?

SOLUTION

Car A :

v = v 0 + a c t
v A = 0 + 4?

At t = 10 s, v A = 40 m/s

1

s = s 0 + v 0 t + ~a c r
s A = 0 + 0 + |(4)t 2 = 21 2

At t = 10 s, = 200 m

t > 10 s, ds = v dt

r s A r’

ds = / 40 dt

) 200

/to

= 40f — 200

At t = 15 s, s A = 400 m
CsltB:

v = v 0 + a c t
Vb = 0 + 5t
25

When v B = 25 m/s, t = — = 5 s

1 n.

S = S 0 + v 0 t + —a c r
s B = 0 + 0 + | (5 )t 2 = 2.5 1 2

When t = 10 s, v A = (z>A)max = 40 m/s and s A = 200 m.
When t = 5 s, s B = 62.5 m.

When t = 15 s, s A = 400 m and s B = 312.5 m.

ID 1

IrCS)

I

If

tc S')

65

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66

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12 - 62 .

If the position of a particle is defined as s = (5 1 — 3 1 2 ) ft,
where t is in seconds, construct the s-t, v-t, and a-t graphs
for 0 < t < 10 s.

SOLUTION

v(K/0

Ans:

v = {5 - 6fj ft/s
a = — 6 ft/s 2

67

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12-63.

From experimental data, the motion of a jet plane while
traveling along a runway is defined by the v — / graph.
Construct the s — t and a — t graphs for the motion. When
t = 0, s = 0.

SOLUTION

V (m/s)
60-

20

s — t Graph: The position in terms of time / can be obtained by applying

ds 20

v = —. For time interval 0 s < / < 5 s, v = — t = (4f) m/s.

dt

When t — 5 s,

For time interval 5 s < t < 20 s,

When / = 20 s,

ds = vdt

ds = Atdt
Jo Jo

s = (2 f 2 ) m
s = 2(5 2 ) = 50 m,

ds = vdt

f ds = [ 20 dt

J 50 m J 5 a

s = (20 1 — 50) m
s = 20(20) - 50 = 350 m

20

30

For time interval 20 s < t < 30 s, y—yy = -yy, v - (At — 60) m/s.

ds — vdt

/ ds= (At — 60) dt

J 350 m J 20 a

s = (2f 2 - 60/ + 750) m

When / = 30 s, s = 2(30 2 ) - 60(30) + 750 = 750 m

a — t Graph: The acceleration function in terms of time / can be obtained by
dv

applying a = For time interval I) s £ / < 5 s, 5 s < ( < 20 s and
dt

dv a dv dv a

20 s < f £ 30 s, a = — = 4.00 m/s 2 , a = — = 0 and a = — = 4.00 m/s 2 ,
.. . dt dt dt '

respectively.

-1 (s)

i(s)

■tCS)

Ans:

For 0 < t < 5 s,

x = {2/ 2 } mandfl = 4 m/s 2

For 5 s < / < 20 s,

x = {20/ — 50} m and a = 0.

For 20 s < / < 30 s,

x = {2/ 2 — 60/ + 750} m

and a = A m/s 2 .

68

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*12-64.

The motion of a train is described by the a-s graph shown.
Draw the v-s graph if v = 0 at s = 0.

SOLUTION

V(mls)

-SCm)

P I 5=300 m = ^(300) = 30m / s

a — 3

0-3

For 300 m < s < 600 m, - ,

s — 300 600 - 300

initial condition v = 30 m/s at s = 300 m,

v dv = / I —— X + 6 ) rfs
1 30 m/s J 300 m V 100

:a = \ ~

100 '

s + 6 > m/s z , using the

30 m/s

200

s 2 + 6 s

300 m

V 2 1

— — 450 = 6 s -—1350

2 200

v =

At s = 600 m,

12s-s 2 - 1800 !• m/s

100 1 '

Ans.

v = ^12 (600) - ^ (600 2 ) -1800 = 42.43 m/s
Using these results, the v—s graph shown in Fig. a can be plotted.

Ans:

u = {^s} m/ s

69

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12-65.

The jet plane starts from rest at s = 0 and is subjected to the
acceleration shown. Determine the speed of the plane when
it has traveled 1000 ft. Also, how much time is required for
it to travel 1000 ft?

a (ft/s 2 )

75 —
50-

^s—

tar

a = 75 - 0.025s

--s (ft)

SOLUTION iooo

v-s Function. Here, --— = —-—; a = (75 — 0.025s] ft/s 2 . The function vi,\

s - 0 1000 - 0 1 1 ' w

can be determined by integrating v dv = a ds. Using the initial condition v = 0 at

s = 0,

v dv = / (75 - 0.025 s) ds

— = 75 s - 0.0125 s 2
2

v = { Vl50s - 0.025 s 2 } ft/s
At s = 1000 ft,

v = Vl50(1000) - 0.025(l000 2 )
= 353.55 ft/s = 354 ft/s

Ans.

ds

Time, t as a function of s can be determined by integrating dt = —. Using the initial
condition s = 0 at t = 0; v

dt =

t =

ds

/o Vl50s - 0.025 s 2

1 . Y150 - 0.05 .

- , sin -

VO025 V 150

t =

1

At s = 1000 ft,

t =

Vao25

l f t

— sin i

150 - 0.05 s
V 150

150 - 0.05(1000)

,-) _ all

VO025 1 2
= 5.319 s = 5.32 s

150

Ans.

Ans:

v = 354 ft/s
t = 5.32 s

70

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12 - 66 .

The boat travels along a straight line with the speed
described by the graph. Construct the s-t and a—s graphs.

Also, determine the time required for the boat to travel a
distance s — 400 m if s = 0 when t — 0.

SOLUTION

s—t Graph: For 0 < ,v < 100 m, the initial condition is .v = 0 when t = 0 s.

When s = 100 m,

100 = t 2 t = 10 s

For 100 m < s £ 400 m, the initial condition is s = 100 m when t = 10 s.

' J*'

. ds
dt = —
v

[ dt =

J 10 s

t - 10 = 51n
t

5 ~ 2 = ln 100

r ds

100 m 0.2s
S

100

s

j/5

S

~ 100
s

loo

s = (l3.53e'/ s ):

When s = 400 m,

400 = 13.53e (/s

t = 16.93 s = 16.9 s Ans.

The s-t graph is shown in Fig. a.
a—s Graph: For 0m<s< 100 m,

a = v— = (2s 1 / 2 )(s -1 / 2 ) = 2 m/s 2

For 100 m < s £ 400 m,
dv

a = v— = (0.2s)(0.2) = 0.04s
ds

When s = 100 m and 400 m,

«| s =ioom = 0.04(100) = 4 m/s 2
fl| s=4 oom = 0.04(400) = 16 m/s 2
The a-s graph is shown in Fig. b.

v (m/s)

SCjn)

atrrth*)

SCm)

0>)

Ans:

When s = 100 m,
t = 10 s.

When s = 400 m,
t = 16.9 s.

«| s =ioom = 4 m/s 2
^ I j=400 m = 16m/s 2

71

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12-67.

The v—s graph of a cyclist traveling along a straight road is
shown. Construct the a—s graph.

SOLUTION

as Graph: For 0 < s < 100 ft,

a = v V = (o.ls + 5)(0.l) = (0.01s + 0.5) ft/s 2 ... . .

ds [ All ) OjtftfsO

SCH)

a| s=100 f t = 0.0016(l00) - 0.76 = -0.6 ft/s 2
a| J= 350 f t = 0.0016(350) - 0.76 = -0.2 ft/s 2

The a—s graph is shown in Fig. a.

Thus at s = 0 and 100 ft

a| s=0 = 0.0l(0) + 0.5 = 0.5 ft/s 2

aUtooft = 0.0l(l00) + 0.5 = 1.5 ft/s 2
At s = 100 ft, a changes from a max = 1.5 ft/s 2 to a min = —0.6 ft/s 2 .

Ans:

At s = 100 s,

a changes from a max =1.5 ft/s 2
to o min = —0.6 ft/s 2 .

72

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* 12 - 68 .

The v-s graph for a test vehicle is shown. Determine its
acceleration when s = 100 m and when s = 175 m.

SOLUTION

0 < x < 150m:

1

V ~ 3 S ’

dv = — ds
3

v dv = a ds

3 S (

'1 \

—ds = a ds

,3 J

1

a — —s

9

At s = 100 m.

a = |(100)

150 < s < 200 m;

v = 200

dv = — ds
v dv — a ds

(200 — s)( — ds) = a ds

a = s — 200

At s = 175 m, a = 175 - 200 = - 25 m/s 2

v (m/s)

Ans.

Ans.

Ans:

At s = 100 s, a = 11.1 m/s 2
At s = 175 m, a = —25 m/s 2

73

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12-69.

If the velocity of a particle is defined as v(f) = (0.8r 2 i +
+ 5k} m/s, determine the magnitude and coordinate
direction angles a, p, y of the particle’s acceleration when
t = 2 s.

SOLUTION

v(f) = 0.8? 2 i + 12fV 2 j + 5k

a = — = 1.6i + 6r 1//2 j
dt J

When t = 2 s, a = 3.2i + 4.243j

a = V(3.2) 2 + (4.243) 2 = 5.31 m/s 2

u„ = - = 0.6022i + 0.7984j
a

a = cos ^ (0.6022) = 53.0°

P = cos 1 (0.7984) = 37.0°
y = cos ’(0) = 90.0°

Ans.

Ans.

Ans.

Ans.

Ans:

a = 5.31 m/s 2
a = 53.0°

P = 37.0°
y = 90.0°

74

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12-70.

The velocity of a particle is v = {3i + (6 — 2f)j} m/s, where t
is in seconds. If r = 0 when l = 0. determine the
displacement of the particle during the time interval
r = lstof = 3s.

SOLUTION

Position: The position r of the particle can be determined by integrating the
kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the
integration limit. Thus,

dr = vdt

dr =

3i +

(6 - 2t)]]d

t

r =

3ti + (6 1

4i

m

When t = Is and 3 s,

4=1 s = 3(l)i + [6(1) - l 2 ]j = [3i + 5j] m/s

4=3 8 = 3 ( 3 )> + [ 6 ( 3 ) ” 32 ]j = [ 9i + 9 J] m / s

Thus, the displacement of the particle is

Ar = 4=3 s “ 4=1 s

= (9i + 9j) - (3i + 5j)

= (6i + 4j) m Ans.

Ans:

Ar = { 6i + 4j } m

75

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12-71.

A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft),
is subjected to an acceleration of a = {6fi + 12t 2 k} ft/s 2 .
Determine the particle’s position ( x, y, z) at t = 1 s.

SOLUTION

Velocity: The velocity expressed in Cartesian vector form can be obtained by
applying Eq. 12-9.

dv = a dt

dv =

'o

v = {3/4 + 4/^} ft/s

Position: The position expressed in Cartesian vector form can be obtained by
applying Eq. 12-7.

dr = vdt

r — (3i + 2j + 5k) = f4 + f 4 k
r = {(t 3 + 3) i + 2j + (f 4 + 5)kj ft

When t = 1 s, r = (l 3 + 3)i + 2j + (l 4 + 5)k = {4i + 2j + 6k} ft.
The coordinates of the particle are

(4 ft, 2 ft, 6 ft)

Ans.

Ans:

(4 ft, 2 ft, 6 ft)

76

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*12-72.

The velocity of a particle is given by
v = {16r 2 i + 4f 3 j + (5 1 + 2)k} m/s, where t is in seconds. If
the particle is at the origin when t = 0, determine the
magnitude of the particle’s acceleration when t — 2 s. Also,
what is the x, y, z coordinate position of the particle at this
instant?

SOLUTION

Acceleration: The acceleration expressed in Cartesian vector form can be obtained
by applying Eq. 12-9.

a = — = {32fi + 12f 2 j + 5k) m/s 2

When t = 2 s, a = 32(2)i + 12^2 2 )j + 5k = {641 + 48j + 5k} m/s 2 . The magnitude
of the acceleration is

a = \/fl 2 + a 2 y + a\ = \/64 2 + 48 2 + 5 2 = 80.2 m/s 2

Ans.

Position: The position expressed in Cartesian vector form can be obtained by
applying Eq. 12-7.

dr = v dt

When f = 2 s,

Thus, the coordinate of the particle is

(42.7,16.0,14.0) m

Ans.

Ans:

(42.7, 16.0,14.0) m

77

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12-73.

The water sprinkler, positioned at the base of a hill, releases
a stream of water with a velocity of 15 ft/s as shown.
Determine the point B(x, y) where the water strikes the
ground on the hill. Assume that the hill is defined by the
equation y = (0.05x 2 ) ft and neglect the size of the sprinkler.

SOLUTION

v x = 15 cos 60° = 7.5 ft/s v y = 15 sin 60° = 12.99 ft/s

(i) s = v 0 t
x = 7.5 t

( + T) s = s 0 + v D t + ^a c t 2

y = 0 + 12.991 + ^ (—32.2)t 2
y = 1.732* - 0.286x 2

Since y = 0.05x 2 ,

0.05x 2 = 1.732* - 0.286* 2
x(0.336x - 1.732) = 0
x = 5.15 ft

y = 0.05(5.15) 2 = 1.33 ft
Also,

(i) x = v 0 t

x = 15 cos 60°t
( +1) 5 = s 0 + v 0 t + - a c t 2

y = 0 + 15 sin 60 °t + j (-32.2 )t 2
Since y = 0.05x 2

12.99r - 16.lt 2 = 2.8125t 2 t = 0.6869 s
So that,

x = 15 cos 60° (0.6868) = 5.15 ft
y = 0.05(5.15) 2 = 1.33 ft

Ans.

Ans.

Ans.

Ans.

Ans:

(5.15 ft, 1.33 ft)

78

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12-74.

A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft),
is subjected to an acceleration a = [6ti + 12t 2 k) ft/s 2 .
Determine the particle’s position (x, y, z) when t = 2 s.

SOLUTION

a = 6/i + 12f z k

I dv = I (6fi + 12f 2 k) dt

Jo Jo

v = 3r 2 i + 4r 3 k

{ dt = f (3t 2 i + 4r 3 k) dt
J r 0 Jo

r — (3i + 2j + 5k) = f 3 i + f 4 k

When t = 2 s
r = (lli + 2j + 21k) ft

Ans.

Ans:

r = (lli + 2j + 21k) ft

79

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12-75.

A particle travels along the curve from A to B in
2 s. It takes 4 s for it to go from B to C and then 3 s to go
from C to D. Determine its average speed when it goes from
A to D.

SOLUTION

s T = i(277)(10)) + 15 + i(2ir(5)) = 38.56
s T 38.56

v sn = — = ~,-- = 4.28 m/s

sp t, 2 + 4 + 3 '

Ans.

Ans:

Osp)avg = 4.28 m/s

80

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*12-76.

A particle travels along the curve from A to B in 5 s. It
takes 8 s for it to go from B to C and then 10 s to go from
C to A. Determine its average speed when it goes around
the closed path.

SOLUTION

The total distance traveled is
^Tot = S A b + S BC + S C a

= 20^/ + V20 2 + 30 2 + (30 + 20)
= 117.47 m

The total time taken is

f Tot — l AB + IflC + f CA

= 5 + 8 + 10
= 23 s

Thus, the average speed is

, ^ 5 Tot 117.47 m _ C11 ,

(Psp)avg = ' — = —^-= 5.107 m/s = 5.11 m/s Ans.

^Tot $Ans: (^sp)avg 5.11 m/s 81 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-77. The position of a crate sliding down a ramp is given by x = (0.25f 3 ) m, y = (1.5t 2 ) m, z = (6 — 0.75f 5 / 2 ) m, where t is in seconds. Determine the magnitude of the crate’s velocity and acceleration when t = 2 s. SOLUTION Velocity: By taking the time derivative of x, y, and z, we obtain the x, y, and z components of the crate’s velocity. v x = x = — (o.25t 3 ) = (o.75t 2 ) m/s v y = y = Jt i 1 - 5 * 2 ) = ( 3f ) m / s v z = z = (6 - 0.75t 5 / 2 ) = (-1.875t 3 / 2 ) m/s When t = 2 s, v x = 0.75(2 2 ) = 3 m/s v y = 3(2) = 6 m/s v z = —1.875(2) 3 / 2 = —5.303 m/s Thus, the magnitude of the crate’s velocity is v = V v x + v y 2 + v, 2 = \/3 2 + 6 2 + (—5.303) 2 = 8.551 ft/s = 8.55 ft Ans. Acceleration: The x,y, and z components of the crate’s acceleration can be obtained by taking the time derivative of the results of v x , v y , and v z , respectively. ax = v x = ^ (o.75f 2 ) = (1.50 m/s 2 a y = Vy = f (30 = 3 m/s 2 a z = i) z = -^(-1.8751 3/2 ) = (-2.8150/ 2 ) m/s 2 When f = 2 s, fl t =1.5(2) = 3m/s 2 fl_ v = 3 m/s 2 fl z = -2.8125(2 1 / 2 ) = -3.977 m/s 2 Thus, the magnitude of the crate’s acceleration is a = \/a x 2 + fly 2 + a. 2 = \/3 2 + 3 2 + (—3.977) 2 = 5.815 m/s 2 = 5.82 m/s Ans. Ans: v = 8.55 ft/s a = 5.82 m/s 2 82 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-78. A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y 2 = [120(10 3 )x] m. If the x component of acceleration is a x m/s 2 , where t is in seconds, determine the magnitude of the rocket’s velocity and acceleration when t = 10 s. SOLUTION Position: The parameter equation of x can be determined by integrating a x twice with respect to t. Substituting the result of .r into the equation of the path, y 2 = 120(10 3 )(^ 4 ) y = (50t 2 ) m Velocity: v y = y = ^(501 2 ) = (lOOr) m/s When t = 10 s, v x = ^(l0 3 ) = 83.33 m/s v y = 100(10) = 1000 m/s Thus, the magnitude of the rocket’s velocity is v = Vv x 2 + v y 2 = V83.33 2 + 1000 2 = 1003 m/s Ans. Acceleration: a y = v y = ^ (100r) = 100 m/s 2 When t = 10 s, a x = ^(!0 2 ) = 25 m /s 2 Thus, the magnitude of the rocket’s acceleration is a = \/a 2 + a 2 = "\/25 2 + 100 2 = 103 m/s 2 Ans. Ans: v = 1003 m/s a = 103 m/s 2 83 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-79. The particle travels along the path defined by the parabola y = 0.5x 2 . If the component of velocity along the x axis is v x = (5 1) ft/s, where f is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = Is. When t = 0, x = 0, y = 0. SOLUTION dx Position: The x position of the particle can be obtained by applying the v x = —. dx = v x dt x = (2.50 1 2 ) ft Thus, y = 0.5(2.50r 2 ) 2 = (3.125t 4 ) ft. At f = Is, x = 2.5(l 2 ) = 2.50 ft and y = 3.125(l 4 ) = 3.125 ft.The particle’s distance from the origin at this moment is d = V(2.50 - 0) 2 + (3.125 - 0) 2 = 4.00 ft Ans. Acceleration: Taking the first derivative of the path y = ().5jc 2 , we have y = xx. The second derivative of the path gives y = x 2 + xx (1) However, x = v x . When t = 1 s, v x Eq. (2) x = a x and y = a y . Thus, Eq. (1) becomes a y ~ v 2 + xa x (2) dv r = 5(1) = 5 ft/s a x = —— = 5 ft/s 2 , and x = 2.50 ft. Then, from dt a y = 5 2 + 2.50(5) = 37.5 ft/s 2 Also, a = \/a 2 + a 2 = \/5 2 + 37.5 2 = 37.8 ft/s 2 Ans. Ans: d = 4.00 ft a = 37.8 ft/s 2 84 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 80 . The motorcycle travels with constant speed v 0 along the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve. SOLUTION Vo y = c sin x ) \ c -- L -- - - L -- y = c sin \ — x y = — cl cos — x]x 77/77 Vy = — c v x ( cos — X VQ = 1?y + 1% Va = v l x V x = V 0 1+1 L C C 0 S “lZ X 1 + l z c ) cos 2 (z x V 0 TTC ( TT V v = - cos — X y L V L 1 + — c cos" I — X L ) \L Ans. Ans. Ans: v x = v 0 1+1 L C ) COS \l X VqTTC 7 T D y - L ( COS-X 1 + 1 z c j cos2 (z x 85 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 81 . A particle travels along the circular path from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C. SOLUTION Position: The coordinates for points B and C are [30 sin 45°, 30 — 30 cos 45°] and [30 sin 75°, 30 - 30 cos 75°]. Thus, r B = (30 sin 45° - 0)i + [(30 - 30 cos 45°) - 30]j = [21.21i - 21.21j) m r c = (30 sin 75° - 0)i + [(30 - 30 cos 75°) - 30]j = [28.98i - 7.765j[ m Average Velocity: The displacement from point B to C is Ar BC = r c — r B = (28.98i - 7.765j) - (21.21i - 21.21j) = [7.765i + 13.45j) m. Ar sc 7.765i + 13.45i (Vijc)avg = -= {3.88i + 6.72j[ m/s Ans. y Ans: (v B c)avg = {3.88i + 6.72jj m/s 86 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 82 . The roller coaster car travels down the helical path at constant speed such that the parametric equations that define its position are x = c sin kt, y = c cos kt,z = h - bt, where c, h, and b are constants. Determine the magnitudes of its velocity and acceleration. SOLUTION X = c sin kt X = ck cos kt x = —ck 2 sin kt y = c cos kt y = —ck sin kt y = —ck 2 cos kt z = h — bt z = -b z = 0 v = \^(ck cos kt) 2 + (—ck sin kt) 2 + (~b) 2 = V c 2 k 2 + b 2 a = \/(—ck 2 sin kt) 2 + (— ck 2 cos kt) 2 + 0 = ck 2 Ans. Ans. Ans: v = Vc 2 k 2 + b 2 a = ck 2 87 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-83. Pegs A and 3 are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10 m/s, determine the magnitude of the velocity and acceleration of peg A when r = lm. SOLUTION Velocity: The x and y components of the peg’s velocity can be related by taking the first time derivative of the path’s equation. i (2xx) + 2yy = 0 — xx + 2 yy — 0 y D v = 10 m/s 1 -xv x + 2 yv y = 0 At x = 1 m, (l ) 2 4 1 ( 1 ) Here, v x = 10 m/s and x = 1. Substituting these values into Eq. (1), 1 ( 1 )( 10 ) + 2 (V3 V 2 I v y = 0 = -2.887 m/s = 2.887 m/s i Thus, the magnitude of the peg’s velocity is v = Vv x 2 + v y 2 = VlO 2 + 2.887 2 = 10.4 m/s Ans. Acceleration: The x and y components of the peg’s acceleration can be related by taking the second time derivative of the path’s equation. 2 (xx + xx) + 2 (yy + yy) = 0 |(i 2 + xx) + 2 (y 2 + yy) = 0 z(v x 2 + xa x ) + 2(v y 2 + ya y ) = 0 ( 2 ) V3 Since v x is constant, a x — 0. When r = lm, y = —— m, v x = 10 m/s, and v y = —2.887 m/s. Substituting these values into Eq. (2), V3 (—2.887) z + — a. = 0 2 ( 1Q2 + 0 ) + 2 a y = -38.49 m/s 2 = 38.49 m/s 2 i Thus, the magnitude of the peg’s acceleration is a = \/a 2 + <jy 2 = Vo 2 + (—38.49) 2 = 38.5 m/s 2 Ans. Ans: v = 10.4 m/s a = 38.5 m/s 2 88 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12-84. The van travels over the hill described by y = (-1.5(10 -3 ) x 2 + 15) ft. If it has a constant speed of 75 ft/s, determine the x and y components of the van’s velocity and acceleration when x = 50 ft. SOLUTION Velocity: The x and y components of the van’s velocity can be related by taking the first time derivative of the path’s equation using the chain rule. y = — 1.5(l0 _3 )jc 2 + 15 y = — 3(l0~ 3 )xx u, = —3 10 3 xii. When x = 50 ft, = -3(l0~ 3 )(50)v x = —0.15V* The magnitude of the van’s velocity is v = Vu* 2 + v y 2 Substituting v = 75 ft/s and Eq. (1) into Eq. (2), 75 = VV + (-0.15v x ) 2 v x = 74.2 ft/s <— Substituting the result of v x into Eq. (1), we obtain v y = —0.15(—74.17) = 11.12 ft/s = 11.1 ft/s T ( 1 ) ( 2 ) Ans. Ans. a = -(16.504 + 0.15a*) (3) Since the van travels with a constant speed along the path, its acceleration along the tangent of the path is equal to zero. Here, the angle that the tangent makes with the horizontal at x = 50 ft is 6 = tan 1 ( / ) = tan 1 —3(l0~ 3 )x = tan _1 (-0.15) = -8.531°. \dx J x=50 ft J x=50 ft Thus, from the diagram shown in Fig. a, a r cos 8.531° — a„ sin 8.531° = 0 (4) Ans: Solving Eqs. (3) and (4) yields v x = 74.2 ft/s <— -2.42 ft/s = 2.42 ft/s 2 <— Ans. v y = 11.1 ft/s t a x = 2.42 ft/s 2 <— Oy ~ “ -16.1 ft/s = 16.1 ft/s z l Ans. a y = 16.1 ft/s 2 1 89 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-85. The flight path of the helicopter as it takes off from A is defined by the parametric equations x = (2 1 1 ) m and y = (0.04f 3 ) m, where t is the time in seconds. Determine the distance the helicopter is from point A and the magnitudes of its velocity and acceleration when t = 10 s. SOLUTION X = 2 1 2 y = 0.04 1 3 At t = 10 s, x = 200 m y = 40 m d = V(200) 2 + (40) 2 = 204 m a X 4 dy It 0.12t 2 dVy a v = —— = 0.24 1 y dt At t = 10 s, v = V(40) 2 + (12) 2 = 41.8 m/s a = V(4) 2 + (2.4) 2 = 4.66 m/s 2 Ans. Ans. Ans. Ans: d = 204 m v = 41.8 m/s a = 4.66 m/s 2 90 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 86 . Determine the minimum initial velocity v 0 and the corresponding angle 6 0 at which the ball must be kicked in order for it to just cross over the 3-m high fence. SOLUTION t = Po = 58.86 ( 1 ) Coordinate System: The x—y coordinate system will be set so that its origin coincides with the ball’s initial position. x-Motion: Here, (v 0 ) x = Vq cos 8, x 0 = 0, and x = 6 m. Thus, ( X ) X = x 0 + ( v 0 )x‘ 6 = 0 + (v 0 cos 8)t 6 Vq cos 6 y-Motion: Here, (v 0 ) x — v () sin 9. a y = -g = — 9.81 m/s 2 , and y 0 = O.Thus, ( + t) y = To + (v 0 ) y t + ^a y t 2 3 = 0 + v 0 (s'md)t + —(—9.81)f 2 3 = Vq (sin 0) t — 4.905I 2 Substituting Eq. (1) into Eq. (2) yields sin 26 — cos 2 6 ( 2 ) (3) From Eq. (3), we notice that v 0 is minimum when f(6) — sin 2 6 — cos 2 8 is df(6) maximum. This requires ^ = 0 dm = 2 cos 26 + sin 28 — 0 d6 tan 2 6 = —2 26 = 116.57° 8 = 58.28° = 58.3° Ans. Substituting the result of 6 into Eq. (2), we have 58.86 ( v o)min ~ sin 116.57° - cos 2 58.28' ,-= 9.76 m/s l CO TOO ' Ans. Ans: 8 = 58.3° (Po)min = 9.76 m/s 91 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-87. The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity at which it was launched, the angle of release 8 , and the height h. SOLUTION (i) s = v 0 t 18 = v A cos 0(1.5) (1) ( + 1 ) V 2 = v 0 2 + 2 a c (s - J 0 ) 0 = (v A sin 6) 2 + 2(-32.2 )(h - 3.5) ( +1) v = v 0 + a c t 0 = v A sin 8 — 32.2(1.5) (2) To solve, first divide Eq. (2) by Eq. (1) to get 8. Then 8 = 76.0° Ans. II )D bo Ans. h = 39.7 ft Ans. Ans: 8 = 76.0° v A = 49.8 ft/s h = 39.7 ft 92 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 88 . Neglecting the size of the ball, determine the magnitude v A of the basketball’s initial velocity and its velocity when it passes through the basket. SOLUTION ¥ Also, (*-) (i) Mx = ( V A )x = V A COS 30° ( 2 ) Vertical Motion. Here, ( v A ) y = u^sin 30 o |, (s A ) y = 0, (s B ) y = 3 — 2=1 m| and a y = 9.81 m/s 2 I ( +1) (s B ), = ('Ja)v + My t + ^a y t 2 1=0 + v A sin 30° t + i(-9.81)f 2 4.905? 2 - 0.5 v A t + 1 = 0 (3) Also ( + T) My = ( v A ) y + a y t (v B ) y = u^sin30° + (—9.81)? My = 0.5 v A - 9.81 1 (4) Solving Eq. (1) and (3) v A = 11.705 m/s = 11.7 m/s Ans. t = 0.9865 s Substitute these results into Eq. (2) and (4) Mx = 11-705 cos 30° = 10.14 m/s —» My = 0.5(11.705) - 9.81(0.9865) = -3.825 m/s = 3.825 m/s i Thus, the magnitude of \ B is v B = V(u B ) 2 + (v B ) y = \/l0.14 2 + 3.825 2 = 10.83 m/s = 10.8 m/s Ans. And its direction is defined by 0 R = tan My . Mx = tan 1 f3.825\ _ 10.14 = 20.67° = 20.7° Ans. Ans: v A = 11.7 m/s v B = 10.8 m/s 0 = 20.7° ^ 93 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-89. The girl at A can throw a ball at v A = 10 m/s. Calculate the maximum possible range R = R max and the associated angle 8 at which it should be thrown. Assume the ball is caught at B at the same elevation from which it is thrown. SOLUTION (i) s = s 0 + v 0 t R = 0 + (10 cos 8)t ( +1) v = v 0 + a c t -10 sin 6 = 10 sin 8 - 9.81r 20 . t = -sin 8 9.81 Thus, R = -sin 8 cos 8 9.81 R = 100 sin 28 Require, dR = 0 de 100 9.81 cos 28 = 0 cos 28(2) = 0 8 = 45° 100 , R = — (sin 90°) = 10.2 m ( 1 ) Ans. Ans. Ans: Rmax = 10-2 m 8 = 45° 94 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-90. Show that the girl at A can throw the ball to the boy at B by launching it at equal angles measured up or down from a 45° inclination. If v A = 10 m/s, determine the range R if this value is 15°, i.e., 8 1 = 45° - 15° = 30° and 8 2 = 45° + 15° = 60°. Assume the ball is caught at the same elevation from which it is thrown. R SOLUTION (i) S = S 0 + v 0 t R = 0 + (10 cos 8)t ( + 1 ) v = v 0 + a c t -10 sin 8 = 10 sin 8 - 9.81t 20 . t = -sin 8 9.81 ^ 200 . Thus, R = -sin 8 cos 8 9.81 R = —— sin 28 (1) 9.81 Since the function y = sin 28 is symmetric with respect to 8 = 45° as indicated, Eq. (1) will be satisfied if | </>] | = \4> 2 \ Choosing <(. = 15° or 8y = 45° - 15° = 30° and d 2 = 45° + 15° = 60°, and substituting into Eq. (1) yields R = 8.83 m Ans. Ans: R = 8.83 m 95 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-91. The ball at A is kicked with a speed v A = 80 ft/s and at an angle d A = 30°. Determine the point ( x, -y ) where it strikes the ground. Assume the ground has the shape of a parabola as shown. SOLUTION ( v A ) x = 80 cos 30° = 69.28 ft/s (v A ) y = 80 sin 30° = 40 ft/s (i)s = s 0 + v 0 t x = 0 + 69.28 1 ( +1) S = So + v 0 t + - a c t 2 —y = 0 + 40f + (—32.2 )t 2 y = —0.04x 2 From Eqs. (1) and (2): -y = 0.5774x - 0.003354x 2 0.04x 2 = 0.5774.x - 0.003354x 2 0.04335x 2 = 0.5774x x = 13.3 ft Thus y = -0.04 (13.3) 2 = -7.09 ft (1) ( 2 ) Ans. Ans. Ans: (13.3 ft, -7.09 ft) 96 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12-92. The ball at A is kicked such that d A = 30°. If it strikes the ground at B having coordinates jc = 15 ft, y — —9 ft, determine the speed at which it is kicked and the speed at which it strikes the ground. y x SOLUTION —0.04x 2 (i) s = Sq + V 0 1 15 = 0 + v A cos 30° t v A = 16.5 ft/s Ans. t = 1.047 s (i) (v B ) x = 16.54 cos 30° = 14.32 ft/s ( +1) v = v 0 + a c t (v B ) y = 16.54 sin 30° + (-32.2)(1.047) = -25.45 ft/s v B = V(14.32) 2 + (—25.45) 2 = 29.2 ft/s Ans. Ans: v A = 16.5 ft/s t = 1.047 s v B = 29.2 ft/s 97 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-93. A golf ball is struck with a velocity of 80 ft/s as shown. Determine the distance d to where it will land. SOLUTION (i) s = j 0 + A/ d cos 10° = 0 + 80 cos 55° t ( + t)s = s 0 + «of+^ a c t 2 d sin 10° = 0 + 80 sin 55 °( - * (32.2) (t 2 ) Solving t = 3.568 s d = 166 ft Ans. Ans: d = 166 ft 98 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-94. A golf ball is struck with a velocity of 80 ft/s as shown. Determine the speed at which it strikes the ground at B and the time of flight from A to B. SOLUTION (v A ) x = 80 cos 55° = 44.886 ( v A ) y = 80 sin 55° = 65.532 (i) J = so + v 0 1 d cos 10° = 0 + 45.886f (+T) * = so + v 0 t + -a c t 2 d sin 10° = 0 + 65.532 (r) + | (-32.2)(t 2 ) d = 166 ft t = 3.568 = 3.57 s (Vb)x = ( v a)x = 45.886 ( +1) V = v 0 + a c t {v B ) y = 65.532 - 32.2(3.568) (v B ) y = -49.357 v B = V(45.886) 2 + (—49.357) 2 v B = 67.4 ft/s Ans. Ans. Ans: t = 3.57 s v B = 67.4 ft/s 99 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 95 . The basketball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Neglecting the size of the ball, determine the magnitude v A of its initial velocity and the height h of the ball when it passes over player B. SOLUTION ( ) S = So + Vat 30 = 0 + v A cos 30° t AC (+T) s = s 0 + vot +-a/ 10 = 7 + v A sin 30° t AC ~ |(32.2)(& c ) Solving v A = 36.73 = 36.7 ft/s Ans. btc ~ 0.943 s ( ) s = s 0 + v 0 t 25 = 0 + 36.73 cos 30° t AB (+T) s = s 0 + v 0 t + -«/ h = 7 + 36.73 sin 30° t AB - |(32.2)(^ B ) Solving t AB = 0.786 s h = 11.5 ft Ans. Ans: v A = 36.7 ft/s h = 11.5 ft 100 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 96 . It is observed that the skier leaves the ramp A at an angle d A = 25° with the horizontal. If he strikes the ground at B, determine his initial speed v A and the time of flight t AB . SOLUTION () S = v 0 t 10C)(|) = v A cos 25 °t AB ( + T) s = s 0 + v 0 t + ^a c t 2 -4 - lOoQj = 0 + v A sin 25°t AB + ^(-9.81)^ Solving, v A = 19.4 m/s t AB = 4.54 s Ans: v A = 19.4 m/s t AB = 4.54 s V 101 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 97 . It is observed that the skier leaves the ramp A at an angle 0 A = 25° with the horizontal. If he strikes the ground at B , determine his initial speed v A and the speed at which he strikes the ground. SOLUTION Coordinate System: x—y coordinate system will be set with its origin to coincide with point A as shown in Fig. a. x-motion: Here, x A = 0, x B = 100 ( — ) = 80 m and ( v A ) x = i^cos 25°. ( X ) X B = *4 + (v A ) x t 80 = 0 + ( v A cos 25 °)t 80 t = v A cos 25° ( 1 ) y-motion: Here, y A = 0, y B = — [4 + 100^— J] = —64 m and ( v A ) y = v A sin 25° and a y = — g = —9.81 m/s 2 . (+T) y B = yA + (v A ) y t + -a y t 2 —64 = 0 + n^sin25°f + — (—9.81)t 2 4.905t 2 - v A sin 25° f = 64 Substitute Eq. (1) into (2) yields ( 2 ) 80 » 4.9051 -— = v A sin 25 c v A cos 25 c 80 v A cos 25° 6 ^; = 64 80 v A cos 25 c 80 = 20.65 = 4.545 v A cos 25° v A = 19.42 m/s = 19.4 m/s Substitute this result into Eq. (1), 80 Ans. t = 19.42 cos 25' = 4.54465 102 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-97. Continued Using this result, (+T) (v B )y = (v A )y + a y t = 19.42 sin 25° + (-9.81)(4.5446) = —36.37 m/s = 36.37 m/s i And ( ) ( v B ) x = ( v A ) x = v A cos 25° = 19.42 cos 25° = 17.60 m/s —> Thus, v B = V (> B ) 2 + (v B f y = V36.37 2 + 17.60 2 = 40.4 m/s Ans. Ans: v A = v B = 19.4 m/s 40.4 m/s 103 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-98. Determine the horizontal velocity v A of a tennis ball at A so that it just clears the net at B. Also, find the distance s where the ball strikes the ground. SOLUTION Vertical Motion: The vertical component of initial velocity is (?; 0 ) v = 0. For the ball to travel from A to B, the initial and final vertical positions are (s 0 ) v = 7.5 ft and s y — 3 ft, respectively. ( +1) s y = (s 0 ), + (v 0 ) y t + - (a c ) y t 2 1 , 3 = 7.5 + 0 + - (-32.2)1? ti = 0.5287 s For the ball to travel from A to C, the initial and final vertical positions are (s 0 ) y = 7.5 ft and s y = 0, respectively. ( + 1) Sy = (so), + My t + - (, a c ) y t 2 0 = 7.5 + 0 + i(-32.2 )t\ t 2 = 0.6825 s Horizontal Motion: The horizontal component of velocity is (v () ) x = v A . For the ball to travel from A to B , the initial and final horizontal positions are (s 0 ) x = 0 and s x = 21 ft, respectively. The time is t = t 1 = 0.5287 s. ( ^ ) s x = (s 0 ) x + (v 0 ) x t 21 = 0 + v A (0.5287) v A = 39.72 ft/s = 39.7 ft/s Ans. For the ball to travel from A to C, the initial and final horizontal positions are (so)x = 0 an d s x = (21 + s) ft, respectively. The time is t = t 2 = 0.6825 s. (^) s x = (s 0 ) x + (v 0 ) x t 21 + s = 0 + 39.72(0.6825) s = 6.11 ft Ans. Ans: v A = 39.7 ft/s s = 6.11 ft 104 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-99. The missile at A takes off from rest and rises vertically to B, where its fuel runs out in 8 s. If the acceleration varies with time as shown, determine the missile’s height h B and speed v B . If by internal controls the missile is then suddenly pointed 45° as shown, and allowed to travel in free flight, determine the maximum height attained, h c , and the range R to where it crashes at D. SOLUTION 40 « - 8 r - dv = a dt 51 v = 2.5t 2 When t = 8 s, v B = 2.5(8) 2 ds = v dt 2.5 t 2 dt 160 m/s x - 2.5 r 3 2.5 (8) 3 = 426.67 = 427 m (v B ) x = 160 sin 45° = 113.14 m/s (v B ) y = 160 cos 45° = 113.14 m/s ( + 1 ) V 2 = vl + 2a c 0 - s 0 ) 0 2 = (113.14) 2 + 2(—9.81) (s c - 426.67) h c = 1079.1 m = 1.08 km f(s) Ans. Ans. Ans. (i) 5 = *0 + V 0 t R = 0 + 113.14? (+1) s = s 0 + v 0 t + - a c t 2 0 = 426.67 + 113.14? + y(-9.81)f 2 Solving for the positive root, t = 26.36 s Then, R = 113.14 (26.36) = 2983.0 = 2.98 km Ans. Ans: v B = 160 m/s h B = 427 m h c = 1.08 km R = 2.98 km 105 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 100 . The projectile is launched with a velocity v 0 . Determine the range R, the maximum height h attained, and the time of flight. Express the results in terms of the angle 8 and v 0 . The acceleration due to gravity is g. SOLUTION ( X ) s = s 0 + v 0 t R = 0 + (r> 0 cos 8)t (+T) s = s 0 + v 0 t + - a c t 2 0 = 0 + Oq sin 8) t + —(—g)f 0 = v 0 sin 8 - i(g)( ——- ) 2 yv 0 cos 8 J Po R = — sin 28 g -t) t = R Vq (2 sin 8 cos 8) v 0 g cos 8 v 0 cos 8 2v ° ■ a -sin 8 g v 2 = vl+ 2a c (s - s 0 ) 0 = (v 0 sin 8) 2 + 2(—g)(h - 0) Vq , h = — sin 2 8 2 g y Ans. Ans. Ans. Ans: R = Vo . — sin g 2vq , t = -sin 8 g h = -sin 2 # 2 g 106 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 101 . The drinking fountain is designed such that the nozzle is located from the edge of the basin as shown. Determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C. SOLUTION Horizontal Motion: (i) 5 = V 0 t Tt R = v A sin 40° t t = Vertical Motion: R v A sin 40° ( + T) s = s 0 + v 0 t + - 0.05 = 0 + v A cos 40°t + —(—9.81)r Substituting Eq.(l) into (2) yields: -0.05 = v A cos 40° (-—- \v A sin 40 + 2 (— 9 - 81 ) R v A sin 40 c At = 4.905 R 2 sin 40° (R cos 40° + 0.05 sin 40°) At point B, R = 0.1 m. Anin A4 4.905 (0.1) 2 sin 40° (0.1 cos 40° + 0.05 sin 40°) At point C, R = 0.35 m. 4.905 (0.35)2 = 0.838 m/s Anax At sin 40° (0.35 cos 40° + 0.05 sin 40°) = 1.76 m/s ( 1 ) ( 2 ) Ans. Ans. Ans: Aiin = 0.838 m/s Anax = 1-76 m/s 107 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 102 . If the dart is thrown with a speed of 10 m/s, determine the shortest possible time before it strikes the target. Also, what is the corresponding angle 8 A at which it should be thrown, and what is the velocity of the dart when it strikes the target? SOLUTION Vertical Motion. Here, (v A ) y = 10 sin 0 A \, = ( s B ) y = 0 and a y = 9.81 m/s 2 i ( + ! ) (? B )y = (s A )y + (v A ) y t + ~ a y t 2 0 = 0 + (10 sin 0 A ) t + | (—9.81 )t 2 4.905 1 2 — (10 sin 0 A ) t = 0 t (4.905t — 10 sin 8 A ) = 0 Since t ¥= 0, then 4.905r — 10 sin d A = 0 Also ( + t) (v B )y = (v A )y + 2 a y [(5 B )j, - (s A ) v ] (■ v B f y = (10 sin 0 A ) 2 + 2 (-9.81) (0 - 0) (■ D B )y = —10 sin S A = 10 sin 0 A i Substitute Eq. (1) into (3) 4.905 (---) - 10 sin 0 A = 0 \ 10 cos 6 a J 1.962 — 10 sin d A cos d A = 0 (3) (4) Using the trigonometry identity sin 2 8 A = 2 sin d A cos 0 A , this equation becomes 1.962 — 5 sin 28 A = 0 sin 28 A = 0.3924 2 8 a = 23.10° and 28 A = 156.90° 8 A = 11.55° and 8 A = 78.45° 108 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-102. Continued Since the shorter time is required, Eq. (1) indicates that smaller d A must be choosen. Thus e A = 11.55° = 11.6° Ans. and t = - = 0.4083 s = 0.408 s 10 cos 11.55° Substitute the result of d A into Eq. (2) and (4) (v B )x = 10 cos 11.55° = 9.7974 m/s - Ans. (v B ) y = 10 sin 11.55° = 2.0026 m/si Thus, the magnitude of v B is v B = V(v b ) 2 x + (v B ) 2 y = V9.7974 2 + 2.0026 2 = 10 m/s And its direction is defined by e B J K) . iy B ) x tan 1 /2.0026 \ V 9.7974/ 11.55° = 11.6° ^5 Ans. Ans. Ans: e A = n. 6 ° t = 0.408 s e B = 11.6° ^ 109 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-103. If the dart is thrown with a speed of 10 m/s, determine the longest possible time when it strikes the target. Also, what is the corresponding angle d A at which it should be thrown, and what is the velocity of the dart when it strikes the target? SOLUTION Vertical Motion. Here, (v A ),, = 10 sin 0 A /. (.v^ ) v = (s B ) y = 0 and a y = —9.81 m/s 2 l. ( + t ) (S B )y = (S A )y + (V A )y t + ^a y t 2 0 = 0+ (10 sin 6 A ) t + — (—9.81) t 2 4.905 1 2 - (10 sin 0 A ) t = 0 t (4.905f — 10 sin d A ) = 0 Since t ^ 0, then 4.905 t — 10 sin d A = 0 (3) Also, iPafy = (v A )y + 2 a y [(s B ) y - (s A ) v ] (v B ) 2 = (10 sin 0 A ) 2 + 2 (-9.81) (0 - 0) ( v B ) y = —10 sin 6 a = 10sine A i (4) Substitute Eq. (1) into (3) 4.905 (---) - 10 sin d A = 0 \ 10 cos 6 a ) 1.962 — 10 sin d A cos d A = 0 Using the trigonometry identity sin 2d A = 2 sin d A cos 0 A , this equation becomes 1.962 — 5 sin 26 A = 0 sin 26 A = 0.3924 2 6 A = 23.10° and 26 A = 156.90° 6 A = 11.55° and d A = 78.44° 110 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-103. Continued Since the longer time is required, Eq. (1) indicates that larger d A must be choosen. Thus, e A = 78.44° = 78.4° Ans. and t = = 1.9974 s = 2.00 s 10 cos 78.44° Substitute the result of d A into Eq. (2) and (4) (■ v B ) x = 10 cos 78.44° = 2.0026 m/s- Ans. (v B ) y = 10 sin 78.44° = 9.7974 m/s i Thus, the magnitude of v B is v B = V(v b ) 2 x + {v B f y = V2.0026 2 + 9.7974 2 = 10 m/s And its direction is defined by 4 r (vb) . (v B ) x tan 1 / 9/7974 \ V2.0026/ 78.44° = 78.4° Ans. Ans. Ans: q a = 78.4° t = 2.00 s 6 b = 78.4° ^ 111 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12-104. The man at A wishes to throw two darts at the target at B so that they arrive at the same time. If each dart is thrown with a speed of 10 m/s, determine the angles 8 C and d D at which they should be thrown and the time between each throw. Note that the first dart must be thrown at 0 C ( >d D ), then the second dart is thrown at S D . SOLUTION () s = So + V 0 t 5 = 0+ (10 cos 8 ) t (+T) V = v 0 + a c t — 10 sin 8 = 10 sin 8 — 9.81 1 2(10 sin 6) 9.81 2.039 sin 8 From Eq. (1), 5 = 20.39 sin 8 cos 8 Since sin 28 = 2 sin 8 cos 8 sin 28 = 0.4905 The two roots are 8 D = 14.7° 8 C = 75.3° From Eq. (1): t D = 0.517 s k = 1-97 s So that A t = tc — t D — 1.45 s ( 1 ) Ans. Ans. Ans. Ans: d D = 14.7° 8 C = 75.3° A t = t c — t D = 1.45 s 112 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-105. The velocity of the water jet d ischarging from the orifice can be obtained from v = v2gh, where h = 2 m is the depth of the orifice from the free water surface. Determine the time for a particle of water leaving the orifice to reach point B and the horizontal distance x where it hits the surface. SOLUTION Coordinate System: The x-y coordinate system will be set so that its origin coincides with point A. The speed of the water that the jet discharges from A is v A = V2(9.81)(2) = 6.264 m/s jr-Motion: Here, (v A ) x = v A = 6.264 m/s, x A = 0. x B — x, and t = Du Thus, () x B = x A + (v A ) x t x = 0 + 6.2Mt A ( 1 ) y-Motion: Here, ( v A ) y = 0, a y = — g = —9.81 m/s 2 , = 0m, = —1.5 m, and t = t^.Thus, ( +1) y B = yA + (v A ) y t + - a y t 2 -1.5 = 0 + 0 + ^(-9.81X4 2 t A = 0.553 s Ans. Thus, x = 0 + 6.264(0.553) = 3.46 m Ans. Ans: t A = 0.553 s x = 3.46 m 113 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 106 . The snowmobile is traveling at 10 m/s when it leaves the embankment at A. Determine the time of flight from A to B and the range R of the trajectory. SOLUTION (i.) s B = s A + v A t R = 0 + 10 cos 40° t (+t) s B - s A + v A t + -a c t 2 = 0 + 10 sin 40°f - ^(9.81) f2 Solving: R = 19.0 m t = 2.48 s Ans. Ans. Ans: R = 19.0 m t = 2.48 s 114 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-107. The fireman wishes to direct the flow of water from his hose to the fire at B. Determine two possible angles 6i and 0 2 at which this can be done. Water flows from the hose at v A = 80 ft/s. SOLUTION -T) S = So + v 0 t 35 = 0 + (80)(cos 0)t 1 2 s = Sq + 1 H— a c t~ Thus, -20 = 0 - 80 (sin 6)t + - (-32.2 )t 2 . 0.4375 „ / 0.1914 20 = 80 sin 6 - t + 16.1 cos 6 cos 2 6 20 cos 2 6 = 17.5 sin 28 + 3.0816 Solving, 8i = 24.9° (below the horizontal) 0 2 = 85.2° (above the horizontal) Ans. Ans. Ans: 0 y = 24 . 9 °^ 0 2 = 85 . 2 °^ 115 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 108 . The baseball player A hits the baseball at v A — 40 ft/s and 0 A = 60° from the horizontal. When the ball is directly overhead of player B he begins to run under it. Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the ball was hit. i C SOLUTION Vertical Motion: The vertical component of initial velocity for the football is ( v o)y = 40 sin 60° = 34.64 ft/s. The initial and final vertical positions are (s 0 ) y = 0 and s y = 0, respectively. 0 = 0 + 34.64f + |(-32.2 ) t 2 t = 2.152 s Horizontal Motion: The horizontal component of velocity for the baseball is ( vq) x = 40 cos 60° = 20.0 ft/s. The initial and final horizontal positions are (Jo)* = 0 and s x = R, respectively. ( ) ** = (so)* + Oo)* t R = 0 + 20.0(2.152) = 43.03 ft The distance for which player B must travel in order to catch the baseball is d = R - 15 = 43.03 - 15 = 28.0 ft Ans. Player B is required to run at a same speed as the horizontal component of velocity of the baseball in order to catch it. v B = 40 cos 60° = 20.0 ft/s Ans. Ans: d = 28.0 ft v B = 20.0 ft/s 116 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-109. The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity at which it was launched, the angle of release 6 , and the height h. SOLUTION () s = Vot 18 = v A cos 0(1.5) (1) (+t) v 2 = vl + 2a c (s - s 0 ) 0 = (y A sin d) 2 + 2(-32.2)(h - 3.5) (+t) V = Vo + a c t 0 = v A sin 0 — 32.2(1.5) (2) To solve, first divide Eq. (2) by Eq. (1), to get fl.Then 6 = 76.0° Ans. v A = 49.8 ft/s Ans. h = 39.7 ft Ans. Ans: 6 = 76.0° v A = 49.8 ft/s h = 39.7 ft 117 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 110 . An automobile is traveling on a curve having a radius of 800 ft. If the acceleration of the automobile is 5 ft/s 2 , determine the constant speed at which the automobile is traveling. SOLUTION Acceleration: Since the automobile is traveling at a constant speed, a t v 2 Thus, a n = a = 5 ft/s . Applying Eq. 12-20, a n = —, we have P v V800(5) 63.2 ft/s = 0 . Ans. Ans: v = 63.2 ft/s 118 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 111 . Determine the maximum constant speed a race car can have if the acceleration of the car cannot exceed 7.5 m/s 2 while rounding a track having a radius of curvature of 200 m. SOLUTION Acceleration: Since the speed of the race car is constant, its tangential component of acceleration is zero, i.e., a, = 0. Thus, a ci n P 7.5 n 2 200 v = 38.7 m/s Ans. Ans: v = 38.7 m/s 119 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 112 . A boat has an initial speed of 16 ft/s. If it then increases its speed along a circular path of radius p = 80 ft at the rate of v = (1.5s) ft/s, where s is in feet, determine the time needed for the boat to travel s = 50 ft. SOLUTION a, = 1.5s 1.5s ds = v dv 0.75 s 2 = 0.5 v 2 - 128 v = = V256 + 1.5 s 2 dt In (s + Vs 2 + 170.7) I* = 1.225f In (s + Vs 2 + 170.7) - 2.570 = 1.225r At s = 50 ft, t = 1.68 s Ans. Ans: t = 1.68 s 120 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-113. The position of a particle is defined by r = (4 (t — sin f)i + (2 1 2 — 3)j| m, where t is in seconds and the argument for the sine is in radians. Determine the speed of the particle and its normal and tangential components of acceleration when f = Is. SOLUTION r = 4(t — sin f) i + (2 t 2 - 3)j di v = — = 4(1 - cos f)i + (4 t) j v| t=\ = 1.83879i + 4j v = V(1.83879) 2 + (4) 2 = 4.40 m/s e = tan 1 (— 4 -) = 65.312° \1.83879 / a = 4 sin ri + 4j a| t=1 = 3.3659i + 4j a = V(3.3659) 2 + (4) 2 = 5.22773 m/s 2 5 = 0 -$ = 15.392°

a 2 = 5.22773 cos 15.392° = 5.04 m/s 2

a n = 5.22773 sin 15.392° = 1.39 m/s 2

Ans.

Ans.

Ans.

Ans:

v = 4.40 m/s
a, = 5.04 m/s 2
a n = 1.39 m/s 2

121

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12-114.

The automobile has a speed of 80 ft/s at point A and an
acceleration a having a magnitude of 10ft/s 2 , acting in
the direction shown. Determine the radius of curvature
of the path at point A and the tangential component of
acceleration.

SOLUTION

Acceleration: The tangential acceleration is

a, = a cos 30° = 10 cos 30° = 8.66 ft/s 2 Ans.

and the normal acceleration is a n = a sin 30° = 10 sin 30° = 5.00 ft/s 2 . Applying

v 2

Eq. 12-20, a n = —, we have
P

v 2 80 2

p = — = -— = 1280 ft Ans.

p a„ 5.00

Ans:

a t = 8.66 ft/s 2
p = 1280 ft

122

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12-115.

The automobile is originally at rest at s = 0. If its speed is
increased by v = (0.05 1 2 ) ft/s 2 , where t is in seconds,
determine the magnitudes of its velocity and acceleration
when t = 18 s.

SOLUTION

a, = 0.05/ 2

dv =

Jo

05 t 2 dt

v = 0.0167 t 3
[ ds = f 0.0167 t 3 dt

Jo Jo
s = 4.167(10~ 3 ) t 4

When t = 18 s, v = 437.4 ft

Therefore the car is on a curved path.

v = 0.0167(18 3 ) = 97.2 ft/s

(97.2) 2 ,

a n = 2 -- = 39.37 ft/s 2

a, = 0.05(18 2 ) = 16.2 ft/s 2
a = V(39.37) 2 + (16.2) 2
a = 42.6 ft/s 2

Ans.

Ans.

Ans:

v = 97.2 ft/s
a = 42.6 ft/s 2

123

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* 12 - 116 .

The automobile is originally at rest s = 0. If it then starts to
increase its speed at v = (0.05t 2 ) ft/s 2 , where t is in seconds,
determine the magnitudes of its velocity and acceleration at
j = 550 ft.

SOLUTION

The car is on the curved path.
a t = 0.05 t 2

I dv = I 0.05 t 2 dt
Jo Jo

v = 0.0167 t 3

[ ds = / 0.0167 t 3 dt
Jo Jo

s = 4.167(10~ 3 ) f 4

550 = 4.167(10~ 3 ) t 4

t = 19.06 s

So that

v = 0.0167(19.06) 3 = 115.4
v = 115 ft/s
(115.4) 2

240

= 55.51 ft/s 2

a, = 0.05(19.06) 2 = 18.17 ft/s 2
a = V(55.51) 2 + (18.17) 2 = 58.4 ft/s 2

Ans.

Ans.

Ans:

v = 115 ft/s
a = 58.4 ft/s 2

124

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12 - 117 .

The two cars A and B travel along the circular path at
constant speeds v A = 80 ft/s and v B = 100 ft/s, respectively.
If they are at the positions shown when t = 0, determine the
time when the cars are side by side, and the time when they
are 90° apart.

SOLUTION

a) Referring to Fig. a , when cars A and B are side by side, the relation between
their angular displacements is

0 B = 0 A + tt ( 1 )

Here, s A = v A t = 801 and s B = v B t

100 t. Apply the formula s = rd or d = - Then

0b

0a

Sb _ lOOf _ 10
T b ~ ”390” ~~ 39 f

s A _ 80 1 _ 1

~r^~ 400 “ 5 ?

Substitute these results into Eq. (1)

10
— t
39

1

5

t + 7T

t = 55.69 s = 55.7 s

Ans.

(b) Referring to Fig. a, when cars A and B are 90° apart, the relation between their
angular displacements is

Here, s A
Then

0 B

= v A t

0 B

0 a

77

+ ^ = 0 A + TT

0 B = 0 A + — ( 2 )

$= 80rand% = v B t = 100 t. Applying the formulas = rd or 8 = r _ s B _ 100 t _ 10 ~V B ~ “ 39 f _ s A _ 80 1 _ 1 ~ V A ~ 400 ~ 5 X Substitute these results into Eq. (2) 10 1 77 — t — — t T — 39 5 2 t = 27.84 s = 27.8 s Ans. va Ans: When cars A and B are side by side, t = 55.7 s. When cars A and B are 90° apart, t = 27.8 s. 125 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 118 . Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 60 ft/s and is increasing its speed at the rate of 15 ft/s 2 until it travels for a distance of 10077 ft, after which it maintains a constant speed. Car B has a speed of 120 ft/s and is decreasing its speed at 15 ft/s 2 until it travels a distance of 6577 ft, after which it maintains a constant speed. Determine the time when they come side by side. SOLUTION Referring to Fig. a , when cars A and B are side by side, the relation between their angular displacements is 0 A = 0ft + 77 (1) The constant speed achieved by cars A and B can be determined from (v A )c = (Va)q + 2 (a A ), [s A ~ (so)a] (v A ) 2 = 60 2 + 2(15) (IOOtt - 0) (v A ) c = 114.13 ft/s ( v b)c = ( v b)o + 2 (a B ) t [s,b ~ C s 'o)fl] (v B ) 2 = 120 2 + 2 (-15) (6577 - 0) (v H )c = 90.96 ft/s The time taken to achieve these constant speeds can be determined from (v A )c = (v A ) 0 + (, a A ) t (t A ) i 114.13 = 60 + 15(C0! (? A )i = 3.6084 s (v B )c = (v B )o + (a B ) t (t B )i 90.96 = 120 + (-15) (t B ) i (t B ) i = 1-9359 s Let t be the time taken for cars A and B to be side by side. Then, the times at which cars A and B travel with constant speed are (Ut) 2 = t — ( t A \ = t — 3.6084 and (1 B ) 2 = t - (fa)! = t - 1.9359. Here, (s A )i = IOOtt, (s 4 ) 2 = (v A ) c (t A ) 2 = 114.13 (t - 3.6084), (i B )j = 6577 and (s B ) 2 = (v B ) c (t B ) 2 = 90.96 (t - 1.9359). Using, the formula s = rO or 6 = - , 0 A ( 0 A ) 1 + ( 0 A )2 feOi (Mh = IOOtt 114.13 (t - 3.6084) r A r A ~ 400 400 = 0.2853 t - 0.24414 ~ (0b )i + (0b)i (s b )i (%)2 _ 65 77 90.96 (t - 1.9359) r B + r B 390 + 390 = 0.2332 t + 0.07207 v u Substitute these results into Eq. (1), 0.2853 t - 0.24414 = 0.2332 t + 0.07207 + t t t = 66.39 s = 66.4 s Ans. Ans: t = 66.4 s 126 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 119 . The satellite S travels around the earth in a circular path with a constant speed of 20 Mm/h. If the acceleration is 2.5 m/s 2 , determine the altitude /;. Assume the earth’s diameter to be 12 713 km. SOLUTION 20 ( 10 6 ) v = 20 Mm/h = ' = 5.56(10 3 ) m/s 3600 dv Since a, = — = 0, then, dt ci — cifi — 2.3 — P (5.56(10 3 )) 2 P =- 25 -= 12.35(10°) m The radius of the earth is 12 713(10 3 ) 2 Hence, = 6.36(10°) m h = 12.35(10°) - 6.36(10°) = 5.99(10°) m = 5.99 Mm Ans. Ans: h = 5.99 Mm 127 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 120 . The car travels along the circular path such that its speed is increased by a t = (0.5e ( ) m/s 2 , where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s — 18 m starting from rest. Neglect the size of the car. SOLUTION dv = 0.5 e‘dt Jo Jo v = 0.5(e r — 1) /»18 r*t / ds = 0.5 / (e 1 — 1 )dt Jo Jo 18 = 0.5(e' - t - 1) Solving, t = 3.7064 s v = 0.5(e 3 ™ 64 - 1) = 19.85 m/s = 19.9 m/s a, = v = 0.5e'| f=37064s = 20.35 m/s 2 v 2 19.85 2 a n P 30 = 13.14 m/s 2 = Vof Ta 2 = V20.35 2 + 13.14 2 = 24.2 m/s 2 Ans. Ans. Ans: v = 19.9 m/s a = 24.2 m/s 2 128 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 121 . The car passes point A with a speed of 25 m/s after which its speed is defined by v = (25 — 0.15s) m/s. Determine the magnitude of the car’s acceleration when it reaches point B , where s = 51.5 m and x = 50 m. 16 m SOLUTION Velocity: The speed of the car at B is v B = [25 - 0.15(51.5)] = 17.28 m/s Radius of Curvature: 1 ^ y = 16 - — x 2 <h_ dx dx 2 625 ' = -3.2(l0^ 3 )x = —3.2( 10~ 3 ) 1 + dy'' 2 dx 3/2 1 + ( -3.2(l0^ 3 )x 3/2 d 2 y —3.2( 10 3 ) dx 2 V / = 324.58 m *=50 m Acceleration: < = 17,28/ = 0.9194 m/s 2 " p 324.58 ’ dv ds = (25 - 0.15s)(—0.15) = (o.225s - 3.75) m/s 2 When the car is at B (s = 51.5 m) a, = [0.225(51.5) - 3.75] = -2.591 m/s 2 Thus, the magnitude of the car’s acceleration at B is a = Vaf + a 2 n = V(-2.591) 2 + 0.9194 2 = 2.75 m/s 2 Ans. Ans: a = 2.75 m/s 2 129 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 122 . If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of a, = 0.5 m/s 2 , determine the magnitude of the car’s acceleration when j = 101.68 m and x = 0. SOLUTION Velocity: The speed of the car at C is % 2 = v A 2 + 2 a t (s c - s A ) v c 2 = 20 2 + 2(0.5)(100 - 0) Vc — 22.361 m/s Radius of Curvature: 625 — = —3.2(l0~ 3 ) dx v ' dx 2 = —3.2 10“ 1 + * N N) 1 _ ] 3/2 , 1 + ( -3.2 (io- 3 ) 01 3/2 d 2 y dx 2 -3.2 ;io- 3 ) 312.5 m Acceleration: 1 1 = i) = 0.5 m/s % 22.361 2 = 1.60 m/s 2 " p 312.5 The magnitude of the car’s acceleration at C is a = \/a 2 + a 2 = \/o.5 2 + 1.60 2 = 1.68 m/s 2 Ans. Ans: a = 1.68 m/s 2 130 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 123 . The motorcycle is traveling at 1 m/s when it is at A. If the speed is then increased at v = 0.1 m/s 2 , determine its speed and acceleration at the instant t — 5 s. SOLUTION a t = v = 0.1 1 ^ s - s 0 + v 0 t + -a c r s = 0 + 1(5) + ^(0.1)(5) 2 = 6.25 m y = 0.5x 2 dy dx d 2 y dx~ = 1 6.25 = + x 2 dx 6.25 + x 2 + In tVT In x + VT + x 2 = 12.5 X 0 Solving, x = 3.184 m y A .♦i f 2 dx [1 + X 2 p d 2 y = 37.17 m x=3.184 dx 2 V = Vq Cl c t = 1 + 0.1(5) = 1.5 m/s v 2 (1.5) 2 , a n = — = = °- 0605 m / S a = V(0.1) 2 + (0.0605) 2 = 0.117 m/s 2 Ans. Ans. Ans: v = 1.5 m/s a = 0.117 m/s 2 131 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 124 . The box of negligible size is sliding down along a curved path defined by the parabola y = QAx 2 . When it is at A ( x A = 2 m, y A = 1.6 m), the speed is v = 8 m/s and the increase in speed is dv/dt = 4m/s? Determine the magnitude of the acceleration of the box at this instant. SOLUTION y = 0.4 x 2 dy_ dx x—2 m 0.8x = 1.6 x—2 m d*l dx 2 = 0.8 x—2 m [l +$

)2j3/2

[l + (1.6) 2 ] 3 / 2

d 2 y

|0.8|

dx 2

x—2 m

8.396 m

v b 1 8 2 2

a„ =-=-- = 7.622 m/s

" p 8.396 '

a = Vaf+4 = ”\/(4) 2 + (7.622) 2 = 8.61 m/s 2

y

Ans.

Ans:

a = 8.61 m/s 2

132

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12 - 125 .

The car travels around the circular track having a radius of r— 300 m
such that when it is at point A it has a velocity of 5 m/s, which is
increasing at the rate of v = (0.06r) m/s 2 , where t is in seconds.
Determine the magnitudes of its velocity and acceleration when it
has traveled one-third the way around the track.

SOLUTION

a, = v = 0.06 1
dv = a t dt

dv =

0.06 1 dt

v = 0.03t 2 + 5
ds = v dt

ds =

(0.03t 2 + 5) dt

s = O.Olt 3 + 5 1
s = | (2 tt(300)) = 628.3185

O.Olf 3 + 5 1 - 628.3185 = 0
Solve for the positive root,
t = 35.58 s

v = 0.03(35.58) 2 + 5 = 42.978 m/s = 43.0 m/s
v 2 (42.978) 2

a n ~ ~

P

300

= 6.157 m/s 2

a, = 0.06(35.58) = 2.135 m/s 2
a = V(6.157) 2 + (2.135) 2 = 6.52 m/s 2

y

Ans.

Ans.

Ans:

v = 43.0 m/s
a = 6.52 m/s 2

133

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12 - 126 .

The car travels around the portion of a circular track having
a radius of r — 500 ft such that when it is at point A it has a
velocity of 2 ft/s, which is increasing at the rate of
v = (0.002r) ft/s 2 , where t is in seconds. Determine the
magnitudes of its velocity and acceleration when it has
traveled three-fourths the way around the track.

SOLUTION

a t = 0.002 s
a t ds = v dv

0.002v ds = v dv

o J 2

O.OOLs 2 = jv 2 - i(2) 2
v 2 = 0.002s 2 + 4

s = - [2tt( 500)] = 2356.194 ft

v 2 = 0.002(2356.194) 2 + 4
v = 105.39 ft/s = 105 ft/s
v 2 (105.39) 2

a n = — =

P

500

= 22.21 ft/s 2

a, = 0.002(2356.194) = 4.712 ft/s 2
a = V(22.21) 2 + (4.712) 2 = 22.7 ft/s 2

y

Ans.

Ans.

Ans:

v = 105 ft/s
a = 22.7 ft/s 2

134

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12 - 127 .

At a given instant the train engine at E has a speed of
20 m/s and an acceleration of 14 m/s 2 acting in the
direction shown. Determine the rate of increase in the
train’s speed and the radius of curvature p of the path.

SOLUTION

a t = 14 cos 75° = 3.62 m/s 2
a n = 14 sin 75°

(20) 2

p = 29.6 m

Ans:

a t = 3.62 m/s 2
p = 29.6 m

135

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* 12 - 128 .

The car has an initial speed v 0 = 20 m/s. If it increases its
speed along the circular track at s = 0, a, = (0.8s) m/s 2 ,
where s is in meters, determine the time needed for the car
to travel s = 25 m.

SOLUTION

The distance traveled by the car along the circular track can be determined by
integrating v dv = a t ds. Using the initial condition v = 20 m/s at s = 0,

[ v r 5

/ v dv = / 0.8 s ds

' 20 m/s J 0

= 0.4 s 2

20 m/s

= | Vo.8 (s 2 + 500) | m/s

ds

The time can be determined by integrating dt = — with the initial condition
s = 0 at t = 0. v

r25 m

ds

' o Vo.8 (s 2 + 500)

t =

1

V08

[ln(s + Vs 2 + 500)]

25 m
0

= 1.076 s = 1.08 s

Ans.

Ans:

t = 1.08 s

136

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12 - 129 .

The car starts from rest at s = 0 and increases its speed at
a t = 4 m/s 2 . Determine the time when the magnitude of
acceleration becomes 20 m/s 2 . At what position s does this
occur?

SOLUTION

Acceleration. The normal component of the acceleration can be determined from

_ v 2 _ v 2

9r ~J' Ur ~ 40

From the magnitude of the acceleration

a = \/a 2 + a 2 \ 20 = yj 4 2 + v = 28.00 m/s

Velocity. Since the car has a constant tangential accelaration of a,

4 m/s 2 ,

v = v 0 + a, t; 28.00 = 0 + 4f

t = 6.999 s = 7.00 s Ans.

v 2 = i>o + 2 a t s; 28.00 2 = 0 2 + 2(4) s

i = 97.98 m = 98.0 m Ans.

Ans:

t = 7.00 s
x = 98.0 m

137

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12 - 130 .

A boat is traveling along a circular curve having a radius of
100 ft. If its speed at t — 0 is 15 ft/s and is increasing at
v = ( 0 . 8 1 ) ft/s 2 , determine the magnitude of its acceleration
at the instant t = 5 s.

SOLUTION

v = 25 ft/s

a

n

p

25 2

Too

At t = 5 s,

6.25 ft/s 2

a t = v = 0.8(5) = 4 ft/s 2
a = Va 2 + a 2 = V4 2 + 6.25 2 = 7.42 ft/s 2

Ans.

Ans:

a = 7.42 ft/s 2

138

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12 - 131 .

A boat is traveling along a circular path having a radius of
20 m. Determine the magnitude of the boat’s acceleration
when the speed is it = 5 m/s and the rate of increase in the
speed is v — 2 m/s 2 .

SOLUTION

a, = 2 m/s 2
v 2 5 2

a„ = — = ™ = 1-25 m/s 2

p Zl)

a = Va/ + a 2 = \/2 2 + 1.25 2 = 2.36 m/s 2 Ans.

Ans:

a = 2.36 m/s 2

139

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* 12 - 132 .

Starting from rest, a bicyclist travels around a horizontal circular
path, p = 10 m, at a speed of v = (0.09f 2 + O.lt) m/s,
where t is in seconds. Determine the magnitudes of his velocity
and acceleration when he has traveled s = 3 m.

SOLUTION

[ ds = [ (0.09I 2 + 0.1 t)dt
Jo Jo

s = 0.031 3 + 0.051 2

When i = 3 m, 3 = 0.03 1 3 + 0.05t 2

Solving,
t = 4.147 s

ds 'j

v = — = 0.09r + O.lf

dt

v = 0.09(4.147) 2 + 0.1(4.147) = 1.96 m/s

Ans.

a t — —— — 0.18? + 0.1
dt

= 0.8465 m/s 2

1=4.147 s

a

Vaj + a 2 „ = V(0.8465) 2 + (0.3852) 2 = 0.930 m/s 2

Ans.

Ans:

v = 1.96 m/s
a = 0.930 m/s 2

140

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12 - 133 .

A particle travels around a circular path having a radius of
50 m. If it is initially traveling with a speed of 10 m/s and its
speed then increases at a rate of v = (0.05 v) m/s 2 ,
determine the magnitude of the particle’s acceleraton four
seconds later.

SOLUTION

Velocity: Using the initial condition v = 10 m/s at t = 0 s,
dv

dt =

dt =

r dv
ho m/s 0.05?;

v

t = 20 In -
10

v = (10e^ 20 ) m/s

When t — 4 s,

v = 10e 4/2 ° = 12.214 m/s

Acceleration: When v = 12.214 m/s (t = 4 s),

a, = 0.05(12.214) = 0.6107 m/s 2
?; 2 (12.214) 2

a n ~ ~

P

50

= 2.984 m/s 2

Thus, the magnitude of the particle’s acceleration is

a = Va f 2 + a„ 2 = V0.6107 2 + 2.984 2 = 3.05 m/s 2

Ans.

Ans:

a = 3.05 m/s 2

141

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142

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12 - 135 .

When t = 0, the train has a speed of 8 m/s, which is increasing
at 0.5 m/s 2 . Determine the magnitude of the acceleration of
the engine when it reaches point A, at t = 20 s. Here the radius
of curvature of the tracks is p A = 400 m.

t, = 8m/s

SOLUTION

Velocity. The velocity of the train along the track can be determined by integrating
dv = a, dt with initial condition v = 8 m/s at f = 0.

0.5 dt

< o

v - 8 = 0.5 t
v = {0.5 t + 8} m/s

At t = 20 s,

v \t = 20s = 0.5(20) + 8 = 18m/s

Acceleration. Here, the tangential component is a t = 0.5 m/s 2 . The normal
component can be determined from

v 2 18 2 ,

a n = — =-= 0.81 m/s 2

" p 400 '

Thus, the magnitude of the acceleration is

a = \/fl 2 + a 2

= Vo.5 2 + 0.81 2

= 0.9519 m/s 2 = 0.952 m/s 2 Ans.

Ans:

a = 0.952 m/s 2

143

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* 12 - 136 .

At a given instant the jet plane has a speed of
550 m/s and an acceleration of 50 m/s 2 acting in the
direction shown. Determine the rate of increase in the
plane’s speed, and also the radius of curvature p of the path.

SOLUTION

Acceleration. With respect to the n-t coordinate established as shown in Fig. a, the
tangential and normal components of the acceleration are

a, = 50 cos 70° = 17.10 m/s 2 = 17.1 m/s 2 Ans.

a n = 50 sin 70° = 46.98 m/s 2

However,

46.98

P

_ 55tf
P

= 6438.28 m = 6.44 km

Ans.

550 m/s

C<X)

Ans:

a, = 17.1 m/s 2
a„ = 46.98 m/s 2
p = 6.44 km

144

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145

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12 - 138 .

The motorcycle is traveling at 40 m/s when it is at A. If the
speed is then decreased at v = — (0.05 s) m/s 2 , where s is in
meters measured from A , determine its speed and
acceleration when it reaches B.

SOLUTION

Velocity. The velocity of the motorcycle along the circular track can be determined
by integrating vdv = a d s with the initial condition v = 40 m/s at s = 0. Here,
a, = —0.05s.

rv r

/ vdv = /

J 40 m/s J 0

—0.05 sds

1 40 m/s

n 2
2

= -0.025 s 2

40 m/s

v = { Vl600 - 0.05 s 2 } m/s

At B, s = rO = 150^y J = 50t t m. Thus

v B = n| J=50Tm = \/l600 — 0.05(50 tt) 2 = 19.14 m/s = 19.1 m/s Ans.

Acceleration. At B, the tangential and normal components are
a, = 0.05(50-77) = 2.5-77 m/s 2
19.14 2

v 2 b

150

= 2.4420 m/s 2

Thus, the magnitude of the acceleration is

a = Vn 2 + a 2 = \/(2.5-7r) 2 + 2.4420 2 = 8.2249 m/s 2 = 8.22 m/s Ans.

And its direction is defined by angle <fi measured from the negative t-axis, Fig. a.

_ _ /M _ _ 1 /2.4420
2.5tt

= 17.3° Ans.

= 17.27°

(&)

Ans:

v B = 19.1 m/s
a = 8.22 m/s 2
</> = 17.3°

up from negative —t axis

146

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12 - 139 .

Cars move around the “traffic circle” which is in the shape of
an ellipse. If the speed limit is posted at 60 km/h, determine
the minimum acceleration experienced by the passengers.

SOLUTION

x 2 y 2

H-~ — 1

a 2 b 2

b 2 x 2 + a 2 y 2 = a 2 b 2

b 2 (2x

) + a 2 (2y)

dy

b 2 x

dx

a 2 y

dy

— b 2 x

dx y -

a 2

d 2 y

fdy\ 2

+ \dx)

d 2 y

-b 2

a 2

d 2 y

-b 2

a 2

d 2 y

-b 2

a 2

d 2 y

-b 2

a 2

d 2 y

-b*

dx 2 "

«v

h 4 b 2
——— + —

1 +

-b 2 x

a 2 y

3/2

-b 2

2 3

ay

At x = 0, y = h.

y

147

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12-139. Continued

Thus
a, = 0

thnin 9

P fl

¥

Set a = 60 m, b = 40 m,
60(10) 3

v 2 b

3600

(16.67) 2 (40)

(60)

= 16.67 m/s

= 3.09 m/s 2

Ans.

Ans:

"min = 3.09 m/s 2

148

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* 12 - 140 .

Cars move around the “traffic circle” which is in the shape of
an ellipse. If the speed limit is posted at 60 km/h, determine
the maximum acceleration experienced by the passengers.

SOLUTION

b 2 x 2 +

c^y 2 = a 2 b 2

b 2 (2x) + a 2 (2y)

dy

Tx = 0

dy

b 2 x

dx

a 2 y

dy

— b 2 x

dx y =

a 2

d 2 y ^

/ dy\ 2

-b 2

^ y +

\dx J

a 2

d 2 y

-b 2

f-b 2 x\

^1

II

a 2

V a 2 y )

d 2 y

-h 4

dx 2

a 2 y 3

l +

-b z x

a 2 y

2„\2

3/2

~b A

2 3

ay

At x — a, y 33 0,
b 2

P = —
a

Then

a t = 0

Lt max Ll n

P 6 / b*

n

60(10 3 )

Set a = 60 m, b = 40 m, v = —— = 16.67 m/s

(16.67) 2 (60) ,,

«max = - (4()) .,-= 10.4 m/s

y

Ans.

Ans:

flmax = 10-4 m/s 2

149

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12 - 141 .

A package is dropped from the plane which is flying with a
constant horizontal velocity of v A — 150 ft/s. Determine
the normal and tangential components of acceleration and
the radius of curvature of the path of motion (a) at the
moment the package is released at A, where it has a
horizontal velocity of v A = 150 ft/s, and (b) just before it
strikes the ground at B.

SOLUTION

Initially (Point A)\
Ma = g = 32.2 ft/s 2

{a,) A = 0

Ma

va

Pa

32.2

(150) 2

Pa

p A = 698.8 ft

(y B )x = (y A ) x = iso ft/s

(+1) V 2 = Vo + 2a c (s - s 0 )

( v B f y = 0 + 2(32.2)(1500 - 0)
(v B ) y = 310.8 ft/s

v B = V(150) 2 + (310.8) 2 = 345.1 ft/s

6 = tan 1

tan

310.8/
150 J

64.23°

( a n ) B — g cos 8 = 32.2 cos 64.24° = 14.0 ft/s 2
(a,) B = gsin# = 32.2 sin 64.24° = 29.0 ft/s 2
(345.1) 2

vb

Mb = — l 14.0 =
Pb

Pb

, 3 \

p B = 8509.8 ft = 8.51(10 J ) ft

Ans.

Ans.

Ans:

Ma = g = 32.2 ft/s 2

(«/),4 = 0

p A = 699 ft
Mb = 14-0 ft/s 2
(a t ) B = 29.0 ft/s 2
p B = 8.5l(l0 3 ) ft

150

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12 - 142 .

The race car has an initial speed v A = 15 m/s at A. If it
increases its speed along the circular track at the rate
a, = (0.4s) m/s 2 , where s is in meters, determine the time
needed for the car to travel 20 m. Take p = 150 m.

SOLUTION

v dv

a, = 0.4s =
a ds = v dv
[ 0.4s ds =

ds

v dv

Jo

0.4s 2

715

0 z 15

225

2 ~~ ~2 ~2
v 2 = 0.4s 2 + 225

v = — = V0.4s 2 + 225
dt

ds

= / dt

Jo V0.4s 2 + 225 Jo

f s ds

Jo Vs 2 + 562.5
In (s + Vs 2 + 562.5)

= 0.632 456?

S

= 0.632 456?
o

In (s + Vs 2 + 562.5) - 3.166 196 = 0.632 456?

At s = 20m,
? = 1.21 s

Ans.

Ans:

? = 1.21 s

151

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152

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153

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154

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12 - 146 .

Particles A and B are traveling around a circular track at a
speed of 8 m/s at the instant shown. If the speed of B is
increasing by ( a ,) s = 4m/ s 2 , and at the same instant A has an
increase in speed of ( a,) A = 0.8f m/s 2 , determine how long it
takes for a collision to occur. What is the magnitude of the
acceleration of each particle just before the collision occurs?

SOLUTION

Distance Traveled: Initially the distance between the two particles is d 0 — p6
( 120 ° \

= 5 - 77 = 10.47 m. Since particle B travels with a constant acceleration,

V180° J F

distance can be obtained by applying equation

Sb = Mb + Oo )b( + -a c t 2
s B = 0 + 8f + — (4) t 2 = (8f + 21 2 ) m

The distance traveled by particle A can be obtained as follows.

dvA — ci a dt

dvA =

0.8 tdt

J 8 m/s JO

v A ~ (o.4t 2 + 8) m/s

dsA = va dt

ds a

[ (o.4r 2 + 8) dt
Jo

= 0.1333t 3 + 8 1

In order for the collision to occur

$a + do = s B 0.1333I 3 + 8 1 + 10.47 = 8t + It 2 Solving by trial and error t = 2.5074 s = 2.51 s '240 Note: If particle A strikes B then, = 5 t = 14.6 s > 2.51 s. 180° [ 1 ] [ 2 ] Ans. 77 I + s B . This equation will result in Acceleration: The tangential acceleration for particle A and B when t = 2.5074 are (a,) A = 0.8t = 0.8(2.5074) = 2.006 m/s 2 and ( a t ) B = 4 m/s 2 , respectively. When t = 2.5074 s,fromEq. [l],u^ = 0.4(2.5074 2 ) + 8 = 10.51 m/s and v B = ( v 0 ) B + a c t = 8 + 4(2.5074) = 18.03 m/s. To determine the normal acceleration, apply Eq. 12-20. Ma = — P , . _ V B ( a nJB ~ P 10.51 2 5 18.03 2 5 22.11 m/s 2 65.01 m/s 2 The magnitude of the acceleration for particles A and B just before collision are a A = V (a t )A + ( a„) 2 A = V / 2.006 2 + 22.II 2 = 22.2 m/s 2 Ans. a B = V \a t ) 2 B + (a„) 2 B = \/4 2 + 65.01 2 = 65.1 m/s 2 Ans. Ans: t = 2.51 s a a = 22.2 m/s 2 a B = 65.1 m/s 2 155 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 147 . The jet plane is traveling with a speed of 120 m/s which is decreasing at 40 m/s 2 when it reaches point A. Determine the magnitude of its acceleration when it is at this point. Also, specify the direction of flight, measured from the x axis. SOLUTION dy dx 15 x x = 80 m 0.1875 y d 2 y dx 2 15 -0.002344 P x = 80 m ' fdy' \ 2 ] 3/2 1 + \dx; 1 d 2 y dx 2 a " p 449.4 [1 + (0.1875) 2 ] 3 / 2 | —0.002344| (120) 2 = 449.4 m = 32.04 m/s 2 a n = —40 m/s 2 a = V(-40) 2 + (32.04) 2 = 51.3 m/s 2 Since — = tan 6 = 0.1875 dx 6 = 10 . 6 ° Ans. Ans. Ans: e = io.6° 156 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 148 . The jet plane is traveling with a constant speed of 110 m/s along the curved path. Determine the magnitude of the acceleration of the plane at the instant it reaches point A(y = 0). SOLUTION ^ = 151n fe dy _ 15 dx x d 2 y dx 2 = 0.1875 x = 80 m 15 "7T = -0.002344 x = 80 m 1 + far \dx )1 3/2 d 2 y dx 2 x = 80 m [l + (0.1875) 2 ] 3 / 2 | —0.002344| = 449.4 m v 2 (110) 2 , , fl " = 7 = Zt49J = 26 - 9m/S Since the plane travels with a constant speed, a t = 0. Hence a = a„ = 26.9 m/s 2 y Ans. Ans: a = 26.9 m/s 2 157 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 149 . The train passes point B with a speed of 20 m/s which is decreasing at a, = -0.5 m/s 2 . Determine the magnitude of acceleration of the train at this point. SOLUTION Radius of Curvature: y = 200ei“0 ^ = 200(—-— leiooo = 0.2e 1000 dx V1000 j = 0.2^—-— leiooo = 0.2( 10~ 3 dx 2 VI000/ ( * \ 2 fi + m] 3/2 1+ 0.2 eiooo \dxj \ / 3/2 d 2 y dx 2 Acceleration: a, = v = —0.5 m/s 2 0 . 2 (l 0 ^ 3 )eiooo _ t/ _ 20 2 a ” ~ p ~ 3808.96 = 0.1050 m/s 2 = 3808.96 m The magnitude of the train's acceleration at B is a = Va 2 + a 2 = V(-0.5) 2 + 0.1050 2 = 0.511 m/s 2 Ans. Ans: a = 0.511 m/s 2 158 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 150 . The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of a, = -0.25 m/s 2 . Determine the magnitude of the acceleration of the train when it reaches point B, where s AB = 412 m. SOLUTION Velocity: The speed of the train at B can be determined from v B 2 = v A 2 + 2 a t (s B - s A ) v B 2 = 30 2 + 2(—0.25)(412 - 0) Vg = 26.34 m/s Radius of Curvature: y = 200e™ dy — = 0.2e™“ dx d 2 y , x = 0.2 10^ 3 )em dx 2 V ’ 1+ Cdy\ \dx J 2' 3/2 1 + ( ■ T 0.2e lom 3/2 d 2 y dx 2 0.2 (l0 _3 )e 1 ® 5 = 3808.96 m Acceleration: a t = v The magnitude of the train’s acceleration at B is a = Vo 2 + a 2 = V(-0.5) 2 + 0.1822 2 = 0-309 m/s 2 -0.25 m/s 2 26.34 2 3808.96 = 0.1822 m/s 2 Ans. Ans: a = 0.309 m/s 2 159 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 151 . The particle travels with a constant speed of 300 mm/s along the curve. Determine the particle’s acceleration when it is located at point (200 mm, 100 mm) and sketch this vector on the curve. SOLUTION v = 300 mm/s dv a < = ht = ° 20 ( 10 3 ) dy dx d 2 y dx 2 20 ( 10 3 ) = -0.5 =200 *=200 40(10 3 ) = 5(10~ 3 ) [l + ffiW [l+(-0.5fl< <£l dx 1 5(10~ 3 ) = 279.5 mm v 2 (300) 2 2 a„ = — = _ = 322 mm/s n p 279.5 ' a — V a 2 + a 2 = V(0) 2 + (322) 2 = 322 mm/s 2 dy Since — = —0.5, dx 0 = tan _1 (-0.5) = 26.6° If y (mm) x (mm) Ans. Ans. Ans: a = 322 mm/s 2 0 = 26.6° If 160 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 152 . A particle P travels along an elliptical spiral path such that its position vector r is defined by r = {2 cos(0.1t)i + 1.5 sin(0.1t)j + (2f)k| m, where t is in seconds and the arguments for the sine and cosine are given in radians. When t = 8 s, determine the coordinate direction angles a, /3, and y, which the binormal axis to the osculating plane makes with the x, y, and z axes. Hint: Solve for the velocity \ P and acceleration a F of the particle in terms of their i, j, k components. The binormal is parallel to \ P X a F . Why? SOLUTION r P = 2 cos (0.1t)i + 1.5 sin (O.lrjj + 2rk Vp = r = —0.2 sin (0.1t)i + 0.15 cos (0.1t)j + 2k a F = r = —0.02 cos (0.1f)i — 0.015 sin (0.1f)j When t = 8 s, \ P = -0.2 sin (0.8 rad)i + 0.15 cos (0.8 rad)j + 2k = -0.143 47i + 0.104 51j + 2k a P = -0.02 cos (0.8 rad)i - 0.015 sin (0.8 rad)j = -0.013 934i - 0.010 76j Since the binormal vector is perpendicular to the plane containing the n-t axis, and a F and \ p are in this plane, then by the definition of the cross product, b = v F X a F = i J k -0.14 347 0.104 51 2 -0.013 934 -0.010 76 0 b = V(0.02152) 2 + (—0.027868) 2 + (0.003) 2 u b = 0.608 991 - 0.788 62j + 0.085k a = cos _1 (0.608 99) = 52.5° 13 = cos -1 (—0.788 62) = 142° 0.021 521 - 0.027 868j + 0.003k 0.035 338 Ans. Ans. y = cos _1 (0.085) = 85.1° Ans. Note: The direction of the binormal axis may also be specified by the unit vector Uj' = — Uj, which is obtained from b' = a p X \ p . For this case, a = 128°, /3 = 37.9°, y = 94.9° Ans. z Ans: a = 52.5° /3 = 142° y = 85.1° a = 128°, j8 = 37.9°, y = 94.9° 161 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 153 . The motion of a particle is defined by the equations x = (2 1 + t 2 ) m and y = ( t 2 ) m, where t is in seconds. Determine the normal and tangential components of the particle’s velocity and acceleration when t = 2 s. SOLUTION Velocity: Here, r = {(2f + f 2 ) i + t 2 j} m. To determine the velocity v, apply Eq. 12-7. dr v = — = {(2 + 2f) i + 2<j } m/s When f = 2 s, v = [2 + 2(2)]i + 2(2)j = {6i + 4j) m/s. Then v = V6 2 + 4 2 = 7.21 m/s. Since the velocity is always directed tangent to the path, v„ = 0 and v t = 7.21 m/s 4 The velocity v makes an angle d = tan -1 — = 33.69° with the x axis. Acceleration: To determine the acceleration a, apply Eq. 12-9. d\ , , a = — = {2i + 2j) m/s 2 Then a = V2 2 + 2 2 = 2.828 m/s 2 Ans. —1 ^ The acceleration a makes an angle 4> = tan — = 45.0° with the x axis. From the figure, a = 45° - 33.69 = 11.31°. Therefore, a n = a sin a = 2.828 sin 11.31° = 0.555 m/s 2 Ans. a t = a cos a. = 2.828 cos 11.31° = 2.77 m/s 2 Ans. Ans: v n = 0 v, = 7.21 m/s a„ = 0.555 m/s 2 a, = 2.77 m/s 2 162 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 154 . If the speed of the crate at A is 15 ft/s, which is increasing at a rate v = 3 ft/s 2 , determine the magnitude of the acceleration of the crate at this instant. SOLUTION Radius of Curvature: y = dy dx 1 cPy = l dx 2 8 Thus, 1 + fdy_ Y 3/2 \dx ) \ d 2 y dx 2 1 + 3/2 1 8 jc= 10 ft 32.82 ft Acceleration: a, = v = 3ft/s 2 a n p 15 2 32.82 6.856 ft/s 2 The magnitude of the crate’s acceleration at A is a = Vo, 2 + a„ 2 = \/3 2 + 6.856 2 = 7.48 ft/s 2 Ans. Ans: a = 7.48 ft/s 2 163 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 155 . A particle is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by 8 = cos 2f, where 8 is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when 6 = 30°. SOLUTION When 8 = f rad, dd 6 = — = —2 sin 2f dt d 2 0 8 = —~ = —4 cos 2 1 dt 2 f = cos 2 1 t = 0.5099 s = -1.7039 rad/s »s = -2.0944 rad/s 2 f=0.5099 s r = 4 r = 0 r = 0 a r = r - r8 2 = 0 - 4(-1.7039) 2 = -11.6135 in./s 2 a e = r8 + 2rd = 4(-2.0944) + 0 = -8.3776 in./s 2 a = Va 2 r + a 2 e = V(-11.6135) 2 + (-8.3776) 2 = 14.3 in./s 2 Ans. Ans: a = 14.3 in./s 2 164 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 156 . For a short time a rocket travels up and to the right at a constant speed of 800 m/s along the parabolic path y = 600 — 35x 2 . Determine the radial and transverse components of velocity of the rocket at the instant 6 = 60°, where 6 is measured counterclockwise from the x axis. SOLUTION y = 600 — 35x 2 y = —70xx dy , = -70x dx tan 60° = — x y = 1.73205lx 1.73205lx = 600 - 35x 2 x 2 + 0.049487x - 17.142857 = 0 Solving for the positive root, x = 4.1157 m dy tan 6 = — = —288.1 dx 6' = 89.8011° = 180° - 89.8011° - 60° = 30.1989° v r = 800 cos 30.1989° = 691 m/s v e = 800 sin 30.1989° = 402 m/s Soo'Wf Ans. Ans. Ans: v r = 691 m/s v e = 402 m/s 165 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 157 . A particle moves along a path defined by polar coordinates r = (2e‘) ft and 8 = (8f 2 ) rad, where t is in seconds. Determine the components of its velocity and acceleration when t = 1 s. SOLUTION When t = 1 s, r = 2e‘ = 5.4366 r = 2e‘ = 5.4366 r = 2e' = 5.4366 8 = 8f 2 8 = 16t = 16 8 = 16 v r = r = 5.44 ft/s Ans. v e = rd = 5.4366(16) = 87.0 ft/s Ans. a r = r — r(8) 2 = 5.4366 — 5.4366(16) 2 = —1386 ft/s 2 Ans. a 0 = r8 + 2 rd = 5.4366(16) + 2(5.4366)(16) = 261 ft/s 2 Ans. Ans: v r = v e = a r = a e = 5.44 ft/s 87.0 ft/s -1386 ft/s 2 261 ft/s 2 166 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 158 . An airplane is flying in a straight line with a velocity of 200 mi/h and an acceleration of 3 mi/h 2 . If the propeller has a diameter of 6 ft and is rotating at an angular rate of 120 rad/s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller. SOLUTION ^200 mi V 5280 ft V lh v P i = a P i - V h 1 mi 3600 s = 293.3 ft/s 3 mi \ / 5280 ftV lh V = 0.001 22 ft/s 2 h 2 )\ 1 mi y y 3600 s, v Pr = 120(3) = 360 ft/s v = Vvp, + vp r = V(293.3) 2 + (360) 2 = 464 ft/s (360) 2 a Pr ~ Vp r = 43 200 ft/s 2 P 3 a = Vah + a 2 Pr = V(0.001 22) 2 + (43 200) 2 = 43.2(10 3 ) ft/s 2 Ans. Ans. Ans: v = 464ft/s a = 43.2(l0 3 ) ft/s 2 167 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 159 . The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m/s and its acceleration is 7 m/s 2 . Express the velocity and acceleration of the washer at this point in terms of its cylindrical components. SOLUTION The position of the washer can be defined using the cylindrical coordinate system (r, 6 and z) as shown in Fig. a. Since 6 is constant, there will be no transverse com¬ ponent for v and a. The velocity and acceleration expressed as Cartesian vectors are v = v\ - r A o *AO r A o r A o = 28 = 7 (0 - 2)i + (0 - 3)j + (0 - 6)k - V(0 - 2) 2 + (0 - 3) 2 + (0 - 6) 2 - (0 - 2)i + (0 - 3)j + (0 - 6)k - V(0 - 2) 2 + (0 - 3) 2 + (0 - 6) 2 - ; j— 8i — 12j — 24kjm/s {—2i - 3j - 6k} m 2 /s u r ob _ 2i + 3j r OB V 2 2 + 3 2 u 7 = k Vl3 Vl3 Using vector dot product »,,v-»,-(-8i-12j-24k).(A s i + JEj) v z = vu z = (—81 — 12 j — 24 k) • (k) = —24.0m/s a z = a • u z = (—2 i — 3j — 6k)-k = — 6.00m/s 2 £ -3.606 m/s 2 Thus, in vector form v = {-14.2 u r - 24.0 Uj.) m/s Ans. a = {—3.61 u r — 6.00 u z ) m/s 2 Ans. These components can also be determined using trigonometry by first obtain angle <f> shown in Fig. a. OA = V 2 2 + 3 2 + 6 2 = 7 m OB = V 2 2 + 3 2 = VT3 Thus, sin (f> v r = v z = a r = «z = = — and cos 4> = v cos 4> = —28 ■v sin <fi = —28 a cos 4> = —7 a sin <p = —7 Vl3 Vl3 . Then J = -14.42 m/s = —24.0 m/s = -3.606 m/s 2 -6.00 m/s 2 Ans: v = {-14.2u,. - 24.OU;,} m/s a = {—3.61u r — 6.00U;,} m/s 2 168 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 160 . A radar gun at O rotates with the angular velocity of 6 — 0.1 rad/s and angular acceleration of 6 — 0.025 rad/s 2 , at the instant 6 — 45°, as it follows the motion of the car traveling along the circular road having a radius of r = 200 m. Determine the magnitudes of velocity and acceleration of the car at this instant. SOLUTION Time Derivatives: Since r is constant, r = r = 0 Velocity: v r = r = 0 Vq — r() = 200(0.1) = 20 m/s Thus, the magnitude of the car’s velocity is v = VV + v e 2 = VO 2 + 20 2 = 20 m/s Acceleration: Ans. a r = r-r8 2 = 0 - 200(0.1 2 ) = -2 m/s 2 a e = rd + 2rd = 200(0.025) + 0 = 5 m/s 2 Thus, the magnitude of the car’s acceleration is a = V / a r 2 + a/ = \/(—2) 2 + 5 2 = 5.39 m/s 2 Ans. Ans: v = 20 m/s a = 5.39 m/s 2 169 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 161 . If a particle moves along a path such that r = (2 cos t ) ft and d = (t/2) rad, where t is in seconds, plot the path r = f(d) and determine the particle’s radial and transverse components of velocity and acceleration. SOLUTION r = 2 cos t r = — 2 sin t r = —2 cost t 1 e =- e=- o=o 2 2 v r = r = — 2 sin t v e = rO = (2 cosf)^i^ = cos f a,.r- r(t 2 = —2 cos t - (2cos o(^) - -f cos , = rd + 2f(9 = 2 cos f(0) + 2(—2 sin = — 2 sin f Ans. Ans. Ans. Ans. Ans: v r = —2 sin t v g = cos t 5 a= — cos t r 2 a g = —2 sin t 170 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 162 . If a particle moves along a path such that r = (e fl ') m and d — t, where t is in seconds, plot the path r =/((?), and determine the particle’s radial and transverse components of velocity and acceleration. SOLUTION r = e at r = ae at r = a 2 e et 6 = t 6 = 1 (9 = 0 v r = r = ae at Ans. v a = rO = e at (l) = e at Ans. a r = r - rd 2 = a 2 e a ‘ - e a '(l) 2 = e a \a 2 - 1) Ans. a e = rd + 2r0 = e a ‘(0) + 2(ae a ')(l) = 2ae“‘ Ans. Ans: v r = ae at v g = e at a r = e at (a 2 — l) a g = 2 ae at 171 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 163 . The car travels along the circular curve having a radius r = 400 ft. At the instant shown, its angular rate of rotation is 0 = 0.025 rad/s, which is decreasing at the rate 0 = —0.008 rad/s 2 . Determine the radial and transverse components of the car’s velocity and acceleration at this instant and sketch these components on the curve. SOLUTION r = 400 r = 0 r = 0 0 = 0.025 8 = -0.008 v r — r = 0 v e = r8 = 400(0.025) = 10 ft/s a r = r - rd 2 = 0 - 400(0.025) 2 = -0.25 ft/s 2 a e = r8 + 2 r'e = 400(-0.008) + 0 = -3.20 ft/s 2 Ans: v r = 0 v e = 10 ft/s a r = —0.25 ft/s 2 % = —3.20 ft/s 2 r = 400 ft r Ans. Ans. Ans. Ans. 172 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 164 . The car travels along the circular curve of radius r = 400 ft with a constant speed of v = 30 ft/s. Determine the angular rate of rotation 8 of the radial line r and the magnitude of the car’s acceleration. t si \ SOLUTION v = V(0) 2 + ^400 8 j = 30 8 = 0.075 rad/s Ans. 0 = 0 a r = r - rd 2 = 0 - 400(0.075) 2 = -2.25 ft/s 2 a e = rd + 2 r8 = 400(0) + 2(0)(0.075) = 0 a = V(-2.25) 2 + (0) 2 = 2.25 ft/s 2 Ans. r = 400 ft Ans: 8 = 0.075 rad/s a = 2.25 ft/s 2 173 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 165 . The time rate of change of acceleration is referred to as the jerk , which is often used as a means of measuring passenger discomfort. Calculate this vector, a, in terms of its cylindrical components, using Eq. 12-32. SOLUTION a = I r—rd 2 )u,. + I rd + 2rd lu fl + z u a = (T - rd 2 - 2rddju r + \ Y - rb 2 Ju r + (^rd + r6 + 2 rd + 2rd ju e + ( rd + 2'rd )u fl + zu r + zu ; But, u,. = du 0 u e = -du r u z = 0 Substituting and combining terms yields a = fr - 3rd 2 ~ 3rdd^Ju r + ^3 rd + rd + 3rd - rd 3 )u fl + f 'z)u z Ans. Ans: a = (r — 3rd 2 - 3rdd)u r + (3rd + rd + 3rd - rd 3 ) u e + (z)u z 174 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 166 . A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by 9 = sin 3 1, where 0 is in radians, the argument for the sine are in radians, and t is in seconds. Determine the acceleration of the particle at 8 = 30°. The particle starts from rest at 9 = 0°. SOLUTION r = 6 in., r = 0, r = 0 6 — sin 3 1 9=3 cos3f 9 = -9 sin 3 1 At 9 = 30°, t = 10.525 s Thus, 9 = 2.5559 rad/s 9 = -4.7124 rad/s 2 a r = r— rO 2 = 0 - 6(2.5559) 2 = -39.196 a e = r8 + 2r9 = 6(- 4.7124) + 0 = - 28.274 a = V(- 39.196) 2 + (- 28.274) 2 = 48.3 in./s 2 Ans. Ans: a = 48.3 in./s 2 175 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 167 . The slotted link is pinned at O, and as a result of the constant angular velocity 6 = 3 rad/s it drives the peg P for a short distance along the spiral guide r = (0.4 6) m, where 6 is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant 6 = tt/3 rad. SOLUTION 6 = 3 rad/s r = 0.4 6 r = 0.4 6 r = 0.4 6 At 6 = j, r = 0.4189 r = 0.4(3) = 1.20 r = 0.4(0) = 0 v = r = 1.20 m/s v e = rO = 0.4189(3) = 1.26 m/s a r = r - r0 2 = 0 - 0.4189(3) 2 = -3.77 m/s 2 a 0 = rO + 2 rd = 0 + 2(1.20)(3) = 7.20 m/s 2 Ans: v r = 1.20 m/s v e = 1.26 m/s a r = —3.77 m/s 2 a g = 7.20 m/s 2 Ans. Ans. Ans. Ans. 176 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 168 . For a short time the bucket of the backhoe traces the path of the cardioid r =25(1 - cos 0) ft. Determine the magnitudes of the velocity and acceleration of the bucket when 6 = 120° if the boom is rotating with an angular velocity of 0 = 2 rad/s and an angular acceleration of 0 = 0.2 rad/s 2 at the instant shown. SOLUTION r = 25(1 - cos 0) = 25(1 - cos 120°) = 37.5 ft r = 25 sin 00 = 25 sin 120°(2) = 43.30 ft/s r = 25[cos 00 2 + sin 00] = 25[cos 120°(2) 2 + sin 120°(0.2)] = —45.67 ft/s 2 v r = r = 43.30 ft/s v e = rO = 37.5(2) = 75 ft/s v = Vv 2 r + v 2 e = V43.30 2 + 75 2 = 86.6 ft/s Ans. a r = r - r6 2 = -45.67 - 37.5(2) 2 = -195.67 ft/s 2 a e = rd + 2 rd = 37.5(0.2) + 2(43.30)(2) = 180.71 ft/s 2 a = \/a 2 + ag = \/(—195.67) 2 + 180.71 2 = 266 ft/s 2 Ans. Ans: v = 86.6 ft/s a = 266 ft/s 2 177 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 169 . The slotted link is pinned at O, and as a result of the constant angular velocity 0 = 3 rad/s it drives the peg P for a short distance along the spiral guide r = (0.4 0) m, where 0 is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = 0.5 m. SOLUTION r = 0.4 0 r = 0.4 0 r = 0.4 6 0 = 3 0 = 0 At r = 0.5 m, 0 = — = 1.25 rad 0.4 r = 1.20 r = 0 v r = r = 1.20 m/s Vo = r 0 = 0.5(3) = 1.50 m/s a r = r - r(d) 2 = 0 - 0.5(3) 2 = -4.50 m/s 2 a 0 = rd + 2 rO = 0 + 2(1.20)(3) = 7.20 m/s 2 Ans. Ans. Ans. Ans. Ans: v r = 1.20 m/s v e = 1.50 m/s a r = —4.50 m/s 2 a g = 7.20 m/s 2 178 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 170 . A particle moves in the x-y plane such that its position is defined by r = {2/i + At 1 ]} ft, where t is in seconds. Determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s. SOLUTION r = 2ti + 4 t 2 j\ t=2 = 4i + 16j a = 8 ft/s 2 <l> ~ 0 = 6.9112° v r = 16.1245 cos 6.9112° = 16.0 ft/s Ans. v e = 16.1245 sin 6.9112° = 1.94 ft/s Ans. S = 90° - 0 = 14.036° a r — 8 cos 14.036° = 7.76 ft/s 2 Ans. a g = 8 sin 14.036° = 1.94 ft/s 2 Ans. Ans: v r = 16.0 ft/s v g = 1.94 ft/s a r = 7.76 ft/s 2 a e = 1.94 ft/s 2 179 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 171 . At the instant shown, the man is twirling a hose over his head with an angular velocity 6 — 2 rad/s and an angular acceleration 6 = 3 rad/s 2 . If it is assumed that the hose lies in a horizontal plane, and water is flowing through it at a constant rate of 3 m/s, determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r = 1.5 m. SOLUTION r = 1.5 r = 3 r = 0 6 = 2 6 = 3 v r = r = 3 v e = rd = 1.5(2) = 3 v = V(3) 2 + (3) 2 = 4.24 m/s a r = r - r{6) 2 = 0 - 1.5(2) 2 = 6 a„ = rd + 2 rd = 1.5(3) + 2(3)(2) = 16.5 a = V(6) 2 + (16.5) 2 = 17.6 m/s 2 Ans. Ans. Ans: v = 4.24 m/s a = 17.6 m/s 2 180 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 172 . The rod OA rotates clockwise with a constant angular velocity of 6 rad/s. Two pin-connected slider blocks, located at B , move freely on OA and the curved rod whose shape is a lima§on described by the equation r = 200(2 - cos 8) mm. Determine the speed of the slider blocks at the instant 8 = 150°. SOLUTION Velocity. Using the chain rule, the first and second time derivatives of r can be determined. r = 200(2 — cos 8) r = 200 (sin 8) 8 = {200 (sin 8) 8} mm/s r = { 200[(cos 8)8 2 + (sin 0)0]} mm/s 2 The radial and transverse components of the velocity are v r = r = {200 (sin 8)8} mm/s v e = r() = {200(2 — cos 8)8} mm/s Since 8 is in the opposite sense to that of positive 8,8 = —6 rad/s. Thus, at 8 = 150°, v r = 200(sin 150°)(—6) = —600 mm/s v e = 200(2 - cos 150°)(-6) = -3439.23 mm/s Thus, the magnitude of the velocity is v = Vv 2 r + vl V(-600) 2 + (—3439.23) 2 = 3491 mm/s = 3.49 m/s Ans. These components are shown in Fig. a Ans: v = 3.49 m/s 181 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 173 . Determine the magnitude of the acceleration of the slider blocks in Prob. 12-172 when 8 = 150°. SOLUTION Acceleration. Using the chain rule, the first and second time derivatives of r can be determined r = 200(2 — cos 8) r = 200 (sin 8)8 = { 200 (sin 8)8 } mm/s r = {200[(cos 8)8 2 + (sin 6)0] } mm/s 2 Here, since 8 is constant, 8 = 0. Since 8 is in the opposite sense to that of positive 8 , 8 = -6 rad/s. Thus, at 8 = 150° r = 200(2 - cos 150°) = 573.21 mm r = 200(sin 150°)(—6) = -600 mm/s r = 200[ (cos 150°)(—6) 2 + sinl50°(0)] = —6235.38 mm/s 2 The radial and transverse components of the acceleration are a r = r - r8 2 = -6235.38 - 573.21 (-6) 2 = -26870.77 mm/s 2 = -26.87 m/s 2 a e = rd + 2 r8 = 573.21(0) + 2(-600)(-6) = 7200 mm/s 2 = 7.20 m/s 2 Thus, the magnitude of the acceleration is a = Va 2 + aj = V(— 26.87) 2 + 7.20 2 = 27.82 m/s 2 = 27.8 m/s 2 Ans. These components are shown in Fig. a. Ans: a = 27.8 m/s 2 182 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 174 . A double collar C is pin connected together such that one collar slides over a fixed rod and the other slides over a rotating rod. If the geometry of the fixed rod for a short distance can be defined by a lemniscate, r 2 = (4 cos 26) ft 2 , determine the collar’s radial and transverse components of velocity and acceleration at the instant d — 0° as shown. Rod OA is rotating at a constant rate of 6 = 6 rad/s. SOLUTION r 2 = 4 cos 26 rr = —4 sin 26 6 rr = r 2 = -4 sin 26 6 - 8 cos 26 6 2 when 6 = 0, 0 = 6, 6 = 0 r = 2,r = 0,r = -144 v r = r = 0 v e = r6 = 2(6) = 12 ft/s a r = r — r6 2 = -144 - 2(6) 2 = -216 ft/s 2 a e = r6 + 2 r6 = 2(0) + 2(0)(6) = 0 Ans: v r = 0 v e = 12 ft/s a r = —216 ft/s 2 cift — 0 r 2 = 4 cos 2 6 Ans. Ans. Ans. Ans. 183 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 175 . A block moves outward along the slot in the platform with a speed of r = (4 1) m/s, where t is in seconds. The platform rotates at a constant rate of 6 rad/s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = 1 s. SOLUTION r = 4f| r=1 = 4 r = 4 <9 = 6 <9 = 0 / dr = [ At dt Jo Jo r = 2t 2 ]o = 2 m v = V / (r) 2 + (rd) 2 = V (4) 2 + [2(6)] 2 = 12.6 m/s a = V(r - rd 2 ) 2 + [r0 + 2 rO) 2 = V[4 - 2(6) 2 ] 2 + [0 + 2(4)(6)] 2 = 83.2 m/s 2 Ans. Ans. Ans: v = 12.6 m/s a = 83.2 m/s 2 184 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 176 . The car travels around the circular track with a constant speed of 20 m/s. Determine the car’s radial and transverse components of velocity and acceleration at the instant 6 = tt /4 rad. SOLUTION v = 20 m/s r = 400 cos 0 r = -400 sin 0 0 r = -400(cos 6(d) 2 + sin 6 6) V 2 = (r) 2 + ( rd) 2 0 = rr + rd(rd + rd) Thus r = 282.84 (20) 2 = [-400 sin 45° 6] 2 + [282.84 6] 2 6 = 0.05 r = -14.14 0 = —14.14[—400(cos 45°)(0.05) 2 + sin 45° 0} + 282.84(0.05)[—14.14(0.05) + 282.846] 0 = 0 r = -0.707 v r = r = —14.1 m/s Ans. v g = rd = 282.84(0.05) = 14.1 m/s Ans. a r = r - r (0) 2 = -0.707 - 282.84(0.05) 2 = -1.41 m/s 2 Ans. % = rd + 2 rd = d + 2(—14.14)(0.05) = -1.41 m/s 2 Ans. Ans: v r = —14.1 m/s v 0 = 14.1 m/s a r = —1.41 m/s 2 u s = —1.41 m/s 2 185 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 177 . The car travels around the circular track such that its transverse component is 8 = (0.006f 2 ) rad, where t is in seconds. Determine the car’s radial and transverse components of velocity and acceleration at the instant f = 4 s. SOLUTION 0 = 0.006 f 2 | (=4 = 0.096 rad = 5.50° 8 = 0.012 1 1 (=4 = 0.048 rad/s 0 = 0.012 rad/s 2 r = 400 cos 8 r = -400 sin 8 8 r = -400(cos 8 (8) 2 + sin# 8) At 8 = 0.096 rad r = 398.158 m r = -1.84037 m/s r = -1.377449 m/s 2 v r = r = —1.84 m/s v e = rd = 398.158(0.048) = 19.1 m/s a r = r — r(df = -1.377449 - 398.158(0.048) 2 = -2.29 m/s 2 a B = r 8 = 2r8 = 398.158 (0.012) + 2(-1.84037)(0.048) = 4.60m/s 2 Ans. Ans. Ans. Ans. Ans: v r = -1.84 m/s v e = 19.1 m/s a r = —2.29 m/s 2 a e = 4.60 m/s 2 186 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 178 . The car travels along a road which for a short distance is defined by r = (200/(9) ft, where 6 is in radians. If it maintains a constant speed of v = 35 ft/s, determine the radial and transverse components of its velocity when 0 = tt/3 rad. SOLUTION 200 _ 600 = 7r/3rad 77 ’ 200 • r = - 1800 • — it -e 6 = tt/3 rad 1800 ■ 600 • v r = r = -r- 6 v e = r0 = - 0 ir = v r + Vg 1800 • 35 = (- r e | + 600 • 6 = 0.1325 rad/s 1800 (0.1325) = -24.2 ft/s v e = —(0.1325) = 25.3 ft/s 7 T Ans. Ans. Ans: v r = —24.2 ft/s v g = 25.3 ft/s 187 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 179 . A horse on the merry-go-round moves according to the equations r = 8 ft, 0 = (0.6 1 ) rad, and z = (1.5 sin 0) ft, where t is in seconds. Determine the cylindrical components of the velocity and acceleration of the horse when t = 4 s. z SOLUTION r = 8 6 = 0.6 t r = 0 6 = 0.6 r = 0 0 = 0 z = 1.5 sin 0 z = 1.5 cos 0 0 z = —1.5 sin 0 (0) 2 + 1.5 cos 0 0 At r = 4 s 0 = 2.4 z = -0.6637 z = -0.3648 v r = 0 Ans. v e = 4.80 ft/s Ans. v z = —0.664 ft/s Ans. a, = 0 - 8(0.6) 2 = -2.88 ft/s 2 Ans. a e = 0 + 0 = 0 Ans. a z = —0.365 ft/s 2 Ans. Ans: v r = 0 v e = 4.80 ft/s v z = -0.664 ft/s a r = -2.88 ft/s 2 = 0 a z = -0.365 ft/s 2 188 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 180 . A horse on the merry-go-round moves according to the equations r = 8 ft, = 2 rad/s and z = (1.5 sin 6) ft, where t is in seconds. Determine the maximum and minimum magnitudes of the velocity and acceleration of the horse during the motion. SOLUTION r = 8 r = 0 0 = 2 r = 0 6 = 0 z = 1.5 sin 6 z = 1.5 cos 6 0 z = —1.5 sin 6 (6) 2 + 1.5 cos 6 0 v r = r = 0 Vff = r 6 = 8(2) = 16 ft/s (Pr)max = Z = 1.5(cOS 0°)(2) = 3 ft/s 0>z)min = z = 1.5(cos 90°)(2) = 0 «max = V(16) 2 + (3) 2 = 16.3 ft/s «min= V(16) 2 + (0) 2 = I6ft/S a r = r — r(d) 2 = 0 - 8(2) 2 = -32 ft/s 2 a e = r 6 + 2 rO = 0 + 0 = 0 (a z ) max = z = —1.5(sin 90°)(2) 2 = -6 (a z )min = z = —1.5(sin 0°)(2) 2 = 0 « max = V(-32) 2 + (0) 2 + (-6) 2 = 32.6 ft/s 2 « min = V(-32) 2 + (0) 2 + (0) 2 = 32 ft/s 2 Ans. Ans. Ans. Ans. Ans: t^max 16.3 ft/s «min = 16 ft/s tfmax = 32.6 ft/s 2 «min = 32 ft/s 2 189 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 181 . If the slotted arm AB rotates counterclockwise with a constant angular velocity of 8 = 2 rad/s, determine the magnitudes of the velocity and acceleration of peg P at 8 = 30°. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. SOLUTION Time Derivatives: r — 4 sec 8 r = (4 sec0(tanfl)fi) ft/s 6 = 2 rad/s r = 4[sec0(tan0)0 + 6(sec 6(sec 2 d)8 + tanfl sec@(tan 8)8)] 8 = 0 = 4[sec0(tan0)(9 + 8 2 (sec38 + tan 2 0 sec#)] ft/s 2 When 8 = 30°, 4 =30 o = 4 sec30° = 4.619ft r 1^=30° = (4 sec30° tan30°)(2) = 5.333 ft/s r | e=30 ° = 4[0 + 2 2 (sec 3 30° + tan 2 30° sec 30°)] = 30.79 ft/s 2 Velocity: v r = r = 5.333 ft/s v e = r8 = 4.619(2) = 9.238ft/s Thus, the magnitude of the peg’s velocity is v = Vv 2 + v e 2 = V5.333 2 + 9.238 2 = 10.7 ft/s Ans. Acceleration: a r = r - rd 2 = 30.79 - 4.619(2 2 ) = 12.32 ft/s 2 a e = rd + 2rd = 0 + 2(5.333)(2) = 21.23 ft/s 2 Thus, the magnitude of the peg’s acceleration is a = Va 2 + a e 2 = Vl2.32 2 + 21.23 2 = 24.6 ft/s 2 Ans. Ans: v = 10.7 ft/s a = 24.6 ft/s 2 D 190 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 182 . The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When 6 = 30°, the angular velocity and angular acceleration of arm AB are 0 = 2 rad/s and d = 3 rad/s 2 , respectively. Determine the magnitudes of the velocity and acceleration of the peg P at this instant. SOLUTION Time Derivatives: r = 4 see# r = (4sec0(tan<9)0) ft/s 0 = 2 rad/s r = 4[sec0(tan0)fl + d (secdsec 2 68 + tan0sec0(tan0)fi)] 6 = 3 rad/s 2 = 4[sec0(tan0)6> + 8 2 (sec 3 6° + tan 2 0°sec0°)] ft/s 2 When 6 = 30°, H»= 30 ° = 4 sec 30° = 4.619 ft 4=30° = (4 sec30°tan30°)(2) = 5.333 ft/s 4=30° = 4[(sec30° tan30°)(3) + 2 2 (sec 3 30° + tan 2 30° sec30°)] = 38.79ft/s 2 Velocity: v r = r = 5.333 ft/s v e = rd = 4.619(2) = 9.238 ft/s Thus, the magnitude of the peg’s velocity is v = VV+ v e 2 = V5.333 2 + 9.238 2 = 10.7 ft/s Ans. Acceleration: a r = r - rd 2 = 38.79 - 4.619(2 2 ) = 20.32 ft/s 2 a e = rQ + 2rd = 4.619(3) + 2(5.333)(2) = 35.19 ft/s 2 Thus, the magnitude of the peg’s acceleration is a = VV+ a e 2 = V20.32 2 + 35.19 2 = 40.6 ft/s 2 Ans. Ans: v = 10.7 ft/s a = 40.6 ft/s 2 D 191 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 183 . A truck is traveling along the horizontal circular curve of radius r = 60 m with a constant speed v = 20 m/ s. Determine the angular rate of rotation 0 of the radial line r and the magnitude of the truck’s acceleration. SOLUTION r = 60 r = 0 r = 0 v = 20 v r = r = 0 v e = r <9 = 60 6 v = V(v r ) 2 + (v e f 20 = 60 6 0 = 0.333 rad/s Ans. a r = r - r(ti) 1 = 0 - 60(0.333) 2 = - 6.67 m/s 2 a e = rd + 2'rO = 60 6 Since v = rO v = 'rd + r6 0 = 0 + 60 e (9 = 0 Thus, = 0 a a r = 6.67 m/s 2 Ans. Ans: 0 = 0.333 rad/s a = 6.67 m/s 2 192 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 184 . A truck is traveling along the horizontal circular curve of radius r = 60 m with a speed of 20 m/s which is increasing at 3m/s 2 . Determine the truck’s radial and transverse components of acceleration. SOLUTION r = 60 a t = 3 m/ s 2 a n r (20) 2 60 6.67 m/s 2 a r — —a n = —6.67 m/s 2 Ans. a e — a t = 3 m/s 2 Ans. Ans: a r = —6.67 m/s 2 a e = 3 m/s 2 193 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 185 . The rod OA rotates counterclockwise with a constant angular velocity of 0 = 5 rad/s. Two pin-connected slider blocks, located at B , move freely on OA and the curved rod whose shape is a limatjon described by the equation r = 100(2 - cos 0) mm. Determine the speed of the slider blocks at the instant 0 = 120°. SOLUTION 0 = 5 r = 100(2 — cos 0) r = 100 sin 00 = 500 sin 0 r = 500 cos 00 = 2500 cos 0 At 0 = 120°, v r = r = 500 sin 120° = 433.013 v e = r0 = 100 (2 - cos 120°)(5) = 1250 v = \/(433.013) 2 + (1250) 2 = 1322.9 mm/s = 1.32 m/s x cos 9) mm Ans. Ans: v = 1.32 m/s 194 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 186 . Determine the magnitude of the acceleration of the slider blocks in Prob. 12-185 when 6 = 120°. SOLUTION 6 = 5 0 = 0 r = 100(2 — cos 6) r = 100 sin 60 = 500 sin 6 r = 500 cos 69 = 2500 cos 6 a r = r - r0 2 = 2500 cos 6 — 100(2 — cos 6)( 5) 2 = 5000(cos 120° — 1) a s = rd + 2 r6 = 0 + 2(500 sin 6)( 5) = 5000 sin 120° = 4330.1 mm/s 2 a = V(-7500) 2 + (4330.1) 2 = 8660.3 mm/s 2 = 8.66 m/s 2 = -7500 mm/s 2 Ans. Ans: a = 8.66 m/s 2 195 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 187 . The searchlight on the boat anchored 2000 ft from shore is turned on the automobile, which is traveling along the straight road at a constant speed of 80 ft/s. Determine the angular rate of rotation of the light when the automobile is r = 3000 ft from the boat. SOLUTION r = 2000 esc 8 r = —2000 esc 8 ctn 8 At r = 3000 ft, 8 = 41.8103° r = -3354.102 8 v = V(r) 2 + (r 8) 2 (80) 2 = [(—3354.102) 2 + (3000) 2 ](6») 2 8 = 0.0177778 = 0.0178 rad/s Ans: 8 = 0.0178 rad/s Ans. 196 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 188 . If the car in Prob. 12-187 is accelerating at 15 ft/s 2 and has a velocity of 80 ft/s at the instant r = 3000 ft, determine the required angular acceleration 6 of the light at this instant. SOLUTION r = 2000 esc 6 r = — 2000 esc dctn8 6 At r= 3000 ft, 6 = 41.8103° r = -3354.102 0 a e = r d + 2 rd a„ = 3000 6 + 2(—3354.102)(0.0177778) 2 Since a 0 = 15 sin 41.8103° = 10 m/s Then, 6 = 0.00404 rad/s 2 Ans. 4 l.. 3 V0^ & Ans: 8 = 0.00404 rad/s 2 197 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 189 . A particle moves along an Archimedean spiral r = (89) ft, where 6 is given in radians. If 6 = 4 rad/s (constant), determine the radial and transverse components of the particle’s velocity and acceleration at the instant 9 = tt/ 2 rad. Sketch the curve and show the components on the curve. SOLUTION Time Derivatives: Since 9 is constant, 9 = 0. y r = 89 = = 4tt ft r = 89 = 8(4) = 32.0 ft/s Velocity: Applying Eq. 12-25, we have v r = r = 32.0 ft/s v g = r9 = 4tt (4) = 50.3 ft/s Acceleration: Applying Eq. 12-29, we have a r = r - r9 2 = 0 - 4tt( 4 2 ) = -201 ft/s 2 a e = rO + 2r9 = 0 + 2(32.0)(4) = 256 ft/s 2 Ans: a r = —201 ft/s 2 a e = 256 ft / s 2 198 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 190 . Solve Prob. 12-189 if the particle has an angular acceleration 0 = 5 rad/s 2 when 0 = 4 rad/s at 0 = 7r/2 rad. SOLUTION Time Derivatives: Here, r = 80 = 8^j = 4tt ft r = 80 = 8(4) = 32.0 ft/s r = 80 = 8(5) = 40 ft/s 2 Velocity: Applying Eq. 12-25, we have v r = r = 32.0 ft/s v 0 = r0 = 4ir(4) = 50.3 ft/s Acceleration: Applying Eq. 12-29, we have a r = r - rd 2 = 40 - 4tt( 4 2 ) = -161 ft/s 2 a g = rd + 2r0 = 4rr(5) + 2(32.0)(4) = 319 ft/s 2 Ans: v r = 32.0 ft/s v e = 50.3 ft/s a r = —161 ft/s 2 a e = 319 ft/s 2 199 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 191 . The arm of the robot moves so that r = 3 ft is constant, and its grip A moves along the path z = (3 sin 4d) ft, where 6 is in radians. If 9 = (0.5 1) rad, where t is in seconds, determine the magnitudes of the grip’s velocity and acceleration when t = 3 s. SOLUTION 0 = 0.5 t r = 3 z — 3 sin 2 1 e = 0.5 r = o II 6 cos 2 1 o = 0 r = 0 z = — 12 sin 2 1 At t = 3 s, z = -0.8382 z = 5.761 z = 3.353 V r = 0 Vg = : 3(0.5) = 1.5 W* = : 5.761 V = V(0) 2 + (1.5) : ' + (5.761) 2 = 5.95 ft/s a r = 0 - 3(0.5) 2 = -0.75 = 0 + 0 = 0 a z — ^ 3.353 a = V(-0.75) 2 + i (0) 2 + (3.353) : ! = 3.44 ft/s Ans. Ans. Ans: v = 5.95 ft/s a = 3.44 ft/s 2 200 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 192 . For a short time the arm of the robot is extending at a constant rate such that r = 1.5 ft/s when r = 3 ft, z = (4 1 2 ) ft, and 0 = 0.5r rad, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the grip A when t = 3 s. SOLUTION 0 = 0.5 t rad r = 3 ft z = 4 t 2 ft 6 = 0.5 rad/s r = 1.5 ft/s z = 8 t ft/s (9 = 0 r — 0 z = 8 ft/s 2 At f = 3 s, (9 = 1.5 r = 3 z = 36 0 = 0.5 r = 1.5 z = 24 0 = 0 r — 0 z = 8 v r = = 1.5 v g = = 3(0.5) = -- 1.5 v z = = 24 V = V(1.5) 2 + (1 ■5) 2 + (24) 2 = 24. a r = 0 — 3(0.5) 2 = -0.75 a„ = 0 + 2(1.5)(0.5) = 1.5 a z = 8 a = V(-0.75) 2 + (1.5) 2 + (8) 2 = 8.17 ft/s 2 Ans. Ans. Ans: v = 24.1 ft/s a = 8.17 ft/s 2 201 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 193 . The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod AB. If the angular velocity of AB is given as d = ( e o.5 1 ) rac j/s, where t is in seconds, and the path defined by the fixed rod is r = |(0.4 sin 8 + 0.2)| m, determine the radial and transverse components of the collar’s velocity and acceleration when f = Is. When t = 0, 8 = 0. Use Simpson’s rule with n = 50 to determine 8 at t = 1 s. SOLUTION 0 = e a5,2 | r=1 = 1.649 rad/s 6 = e°- 5,2 f| f=1 = 1.649 rad/s 2 8 = [ e°- 5f dt = 1.195 rad = 68.47° do r = 0.4 sin 8 + 0.2 r = 0.4 cos 8 8 r = -0.4 sin 8 8 2 + 0.4 cos 8 8 At t = Is, r = 0.5721 r = 0.2421 r = -0.7697 v r = r = 0.242 m/s v e = r 0 = 0.5721(1.649) = 0.943 m/s a r = r - rd 2 = -0.7697 - 0.5721(1.649) 2 a T = —2.33 m/s 2 a e = rd + 2 rd = 0.5721(1.649) + 2(0.2421)(1.649) a e = 1.74 m/s 2 Ans. Ans. Ans. Ans. Ans: v r = 0.242 m/s v e = 0.943 m/s a r = —2.33 m/s 2 ci e = 1.74 m/s 2 202 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 194 . The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod AB. If the mechanism is to be designed so that the largest speed given to the collar is 6 m/s, determine the required constant angular velocity 8 of rod AB. The path defined by the fixed rod is r = (0.4 sin 8 + 0.2) m. SOLUTION r = 0.4 sin 0 + 0.2 r = 0.4 cos 8 8 v r = r = 0.4 cos 0 6 v g = rd = (0.4 sin 8 + 0.2) 8 v 2 = v 2 r + v„ (6) 2 = [(0.4 cos 8) 2 + (0.4 sin 8 + 0.2) 2 ](0) 2 36 = [0.2 + 0.16 sin 8](8) 2 The greatest speed occurs when 8 = 90°. 8 = 10.0 rad/s Ans: 8 = 10.0 rad/s Ans. 203 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 204 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 196 . The motor at C pulls in the cable with an acceleration a c = (3 t 2 ) m/s 2 , where t is in seconds. The motor at D draws in its cable at a D = 5 m/s 2 . If both motors start at the same instant from rest when d = 3 m, determine (a) the time needed for d = 0, and (b) the velocities of blocks A and B when this occurs. SOLUTION For A: SA + (Ut ~ S C ) = l 2v a = v c 2 a A = a c = —3 1 2 a A = -1.5t 2 = 1.5f 2 -> v A — 0.5 1 3 —> s A = 0.125 1 4 -> For B: a B = 5 m/s 2 «— v B = 5t «- s B = 2.51 2 <— Require s A + s B = d 0.125 f 4 + 2.5 1 2 = 3 Set u = t 2 0.125 u 2 + 2.5m = 3 The positive root is u = 1.1355. Thus, t = 1.0656 = 1.07 s Ans. v A = 0.5(1.0656) 3 = 0.6050 v B = 5(1.0656) = 5.3281 m/s ?a = + v A/B 0.6050i = —5.3281i + v A / B i v A /b = 5.93 m/s Ans. Ans: t = 1.07 s v A / B ~ 5.93 ms/s —» 205 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 197 . The pulley arrangement shown is designed for hoisting materials. If SC remains fixed while the plunger P is pushed downward with a speed of 4 ft/s, determine the speed of the load at A. SOLUTION 5s B + (s B - s A ) = l 6 s B - s A = l 6 v B - v A = 0 6(4) = v A v A = 24 ft/s Ans. Ans: v = 24 ft/s 206 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 198 . If the end of the cable at A is pulled down with a speed of 5 m/s, determine the speed at which block B rises. SOLUTION Position Coordinate. The positions of pulley B and point A are specified by position coordinates s B and S 4 , respectively, as shown in Fig. a. This is a single-cord pulley system. Thus, s B + 2 (s B — a) + s A = 1 3 s B + s A = 1 + 2a (1) Time Derivative. Taking the time derivative of Eq. (1), 3v b + v A = 0 (2) Here v A = +5 m/s, since it is directed toward the positive sense of s A . Thus, 3v b + 5 = 0 v B = —1.667 m/s = 1.67 m/sf Ans. The negative sign indicates that v B is directed toward the negative sense of s B . Ans: v B = 1.67 m/s 207 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 199 . Determine the displacement of the log if the truck at C pulls the cable 4 ft to the right. SOLUTION 2s b + (s B - s c ) = l 3 s B - s c = / 3Asg — A s c = 0 Since As c = —4, then 3As B —4 A s B = -1.33 ft = 1.33 ft —> Ans. Ans: A s B = 1.33 ft -» 208 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 200 . Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5 s. SOLUTION v B =^ = 4m/st s B + (% “ %) = h % + C*c - *d) = h s a + 2 s D = h Thus, 2 s B — s c = l x 2% ~ s d = h s a + 2 Sd = h 2v a = v c 2v c = v D v A = ~2 v d 2(2v b ) = v D v A = -2(4 v B ) v A = ~8v b v A = —8(—4) = 32 m/s i Ans. Ans: v A = 32 m/s I 209 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 201 . Starting from rest, the cable can be wound onto the drum of the motor at a rate of v A = (3f 2 ) m/s, where t is in seconds. Determine the time needed to lift the load 7 m. SOLUTION v B = = 4 m/s f Sb + Ob ~ s C ) = h s c + Oc ~ s D ) = l 2 s A + 2 s D ) = / 3 Thus, 2 s B — s c = 1 1 2 sc ~ s D = l 2 s a + 2 s D = h 2v b = v c 2v c = v D v A = —2 v d v A = -8 v b 3 t 2 = —8 v b “ 3 2 v B = ^t 2 t = 3.83 s Ans. Ans: t = 3.83 s 210 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 202 . If the end A of the cable is moving at v A = 3 m/s, determine the speed of block B. v A = 3 m/s C D, SOLUTION Position Coordinates. The positions of pulley B, D and point A are specified by position coordinates s B , s D and respectively as shown in Fig. a. The pulley system consists of two cords which give 2 s B + s D = lx (1) and 0 a ~ s D ) + (b ~ s D ) = l 2 Ut - 2 s„ 4 b (2) Time Derivative. Taking the time derivatives of Eqs. (1) and (2), we get 2 v B + v D = 0 (3) v A - 2v d = 0 ( 4 ) Eliminate v 0 from Eqs. (3) and (4), v A + 4 v b = 0 ( 5 ) Here v A = +3 m/s since it is directed toward the positive sense of s A . Thus 3 + Av b = 0 v B = —0.75 m/s = 0.75 m/s <— Ans. The negative sign indicates that \ D is directed toward the negative sense of s B . Ans: v B = 0.75 m/s 211 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 203 . Determine the time needed for the load at B to attain a speed of 10 m/s, starting from rest, if the cable is drawn into the motor with an acceleration of 3 m/s 2 . SOLUTION Position Coordinates. The position of pulleys B , C and point A are specified by position coordinates s B , s c and s A respectively as shown in Fig. a. The pulley system consists of two cords which gives s B + 2(s B - s c ) = /, 3 s B - 2s c = h (1) And % + = h (2) Time Derivative. Taking the time derivative twice of Eqs. (1) and (2), 3 a B — 2 ac = 0 (3) And a c + a A = 0 (4) Eliminate a c from Eqs. (3) and (4) 3 a B + 2a A = 0 Elere, a A = +3 m/s 2 since it is directed toward the positive sense of s A . Thus, 3 a B + 2(3) = 0 a B = —2 m/s 2 = 2 m/s 2 | The negative sign indicates that a B is directed toward the negative sense of s B . Applying kinematic equation of constant acceleration, +! v B = (v B ) 0 + a B t 10 = 0 + 2 t t = 5.00 s Ans. Ans: t = 5.00 s 212 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 204 . The cable at A is being drawn toward the motor at v A = 8 m/s. Determine the velocity of the block. SOLUTION Position Coordinates. The position of pulleys B , C and point A are specified by position coordinates s B , s c and s A respectively as shown in Fig. a. The pulley system consists of two cords which give s B + 2(s b - s c ) = h 3 s B — 2 s c = /j (1) And s c + 5 .4 — h Time Derivative. Taking the time derivatives of Eqs. (1) and (2), we get 3v b - 2v c = 0 And v c + v A = 0 ( 2 ) ( 3 ) ( 4 ) Eliminate v c from Eqs. (3) and (4), 3v b + 2v a = 0 Flere v A = +8 m/s since it is directed toward the positive sense of S 4 . Thus, 3v b + 2(8) =0 v B = —5.33 m/s = 5.33 m/s f Ans. The negative sign indicates that \ B is directed toward the negative sense of s B . v B = 5.33 m/s'j' 213 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 205 . If block A of the pulley system is moving downward at 6 ft/s while block C is moving down at 18 ft/s, determine the relative velocity of block B with respect to C. SOLUTION s a “f 2 s B + 2 sc — / v A + 2v B + 2 Vq = 0 6 + 2v b + 2(18) = 0 v B = -21 ft/s = 21 ft/s | + i v B = v c + v B/c -21 = 18 + v B / C v B/c = -39 ft/s = 39 ft/s t Ans: v B /c = 39 ft/s t Ans. 214 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 206 . Determine the speed of the block at B. 6 m/s SOLUTION Position Coordinate. The positions of pulley B and point A are specified by position coordinates s B and respectively as shown in Fig. a. This is a single cord pulley system. Thus, s B + 2(jg — a — b) + ( s B — ci) + Sa = l 4s B + ^ = / + 3a + 2b (1) Time Derivative. Taking the time derivative of Eq. (1), 4 v B + v A = 0 (2) Here, Va = + 6 m/s since it is directed toward the positive sense of s^.Thus, 4v b + 6 = 0 v B = — 1.50m/s = 1.50m/s<^ Ans. The negative sign indicates that x B is directed towards negative sense of s B . Ans: v B = 1.50 m/s 215 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 207 . Determine the speed of block A if the end of the rope is pulled down with a speed of 4 m/s. SOLUTION Position Coordinates: By referring to Fig. a, the length of the cord written in terms of the position coordinates s A and ,v ;j is s B + s A + 2 (s A — a) = l s B + 3x^4 = / + 2 a Time Derivative: Taking the time derivative of the above equation, (+1) v B + 3v a = 0 Flere, v B = 4 m/s. Thus, 4 + 3v A = 0 v A = —133 m/s = 1.33 m/s f Ans. c Ans: v A = 1.33 m/s 216 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 208 . The motor draws in the cable at C with a constant velocity of v c = 4 m/s. The motor draws in the cable at D with a constant acceleration of a D = 8 m/s 2 . If v D = 0 when t = 0, determine (a) the time needed for block A to rise 3 m, and (b) the relative velocity of block A with respect to block B when this occurs. SOLUTION (a) a D = 8 m/s 2 Vd = 8 1 s D = 41 2 s D + 2 s A = 1 \s D = — A Sa = ~2 t 2 -3 = —2 t 2 t = 1.2247 = 1.22 s (b) v A = s A = -4t = -4(1.2247) = -4.90 m/s = 4.90 m/sf SB + ( s b ~ s C ) = 1 2v b = v c = -4 v B = —2m/s = 2 m/s | (+ 1 ) \ A = \ B + y A/B 4.90 = -2 + v A / B v A / B = —2.90 m/s = 2.90 m/s f Ans. Ans: v A/B = 2.90 m/s | 217 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 218 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 210 . The 16-ft-long cord is attached to the pin at C and passes over the two pulleys at A and D. The pulley at A is attached to the smooth collar that travels along the vertical rod. When s B = 6 ft, the end of the cord at B is pulled downwards with a velocity of 4 ft/s and is given an acceleration of 3 ft/s 2 . Determine the velocity and acceleration of the collar at this instant. SOLUTION 2 Vs 2 a + 3 2 + s B = / 2 Q)cd + + S B =0 SB 2sa (4 + 9)5 4 = -24(4 + 9) 5 - 244 1(4 + 9) 2 - I 2s a s a )(4 + 9) 2s a s a S B ~ 2(4 + s A s A ) 2(s a s a ) 2 (4 + 9) 2 (4 + 9) 2 At Sg = 6 ft, s B = 4 ft/s, s B =3 ft/s 2 2V4 + 3 2 + 6 = 16 *a = 4 ft 4 2(4)(4) (4 2 + 9)5 v A — 4 = —2.5 ft/s = 2.5 ft/sT Ans. ^ = _ 2[(-2.5) 2 + 4(4)] 2[4(—2.5)] 2 (4 2 + 9)5 (4 2 + 9)5 a a = 4 = -2.4375 = 2.44 ft/s 2 T Ans. Ans: v A = 2.5 ft/s t a A = 2.44 ft/s 2 t 219 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-211. The roller at A is moving with a velocity of v A = 4 m/s and has an acceleration of a A = 2m/s 2 when x A = 3 m. Determine the velocity and acceleration of block B at this instant. SOLUTION Position Coordinates. The position of roller A and block B are specified by position coordinates x A and y B respectively as shown in Fig. a. We can relate these two position coordinates by considering the length of the cable, which is constant Vxi + 4 2 + y B = l y B = i - + 16 Velocity. Taking the time derivative of Eq. (1) using the chain rule, dye l/o \ i dx A -^- = 0—-(xi + 16)-2(Zx A ) — ( 1 ) dy B dt x A dx A V x A + 16 dt dy B dx A However,—— = v B and —— = r^.Then dt dt X A V B = - / , V A Vx A + 16 ( 2 ) At x A = 3 m, v A = +4m/s since is directed toward the positive sense of x A . Then Eq. (2) give 3 v B =- (4) = —2.40 m/s = 2.40 m/s f Ans. V3 2 + 16 The negative sign indicates that v B is directed toward the negative sense of y B . Acceleration. Taking the time derivative of Eq. (2), dx A « ,dx a dv B dt x a(-\)(x a + 16) 3/2 + ( x a + 16 ) 1/2 TrT dt v A ~ x a (x 2 a + 16) - 1 / 2 ' dt dv B dv A dx A However, —— = a B , —— = a A and —— = v A . Then dt dt dt a B = 2 2 X A V A vl x A a A (xi + 16) 3 / 2 (xi + 16)V2 (xi + 16) 1 / 2 16 v A + a A x A [x A + 16) ° B= (xi + 16 ) 3 ^ At x A = 3 m, v A = +4 m/s, a A = +2 m/s 2 since and a A are directed toward the positive sense of x A . 16(4 2 ) + 2(3) (3 2 + 16) a B — = —3.248 m/s 2 = 3.25m/s 2 | Ans. (3 2 + 16) 3 / 2 The negative sign indicates that a B is directed toward the negative sense of y B . »j = 4m/s Ans: v B = 2.40 m/s | a B = 3.25 m/s 2 \ 220 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 212 . The girl at C stands near the edge of the pier and pulls in the rope horizontally at a constant speed of 6 ft/s. Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft. SOLUTION The length / of cord is \/ (8) - + x\ + xq — l Taking the time derivative: ^[( 8) 2 + x\Y m 2 x b x b + x c = 0 Xq — 6 ft/s When AB = 50 ft, x B = V(50) 2 - (8) 2 = 49.356 ft From Eq. (1) i[(8) 2 + (49.356) 2 ] 1/2 2(49.356)(ig) + 6 = 0 x B = - 6.0783 = 6.08 ft/s <- Ans: x B = 6.08 ft/s <— 6 ft/s ( 1 ) Ans. 221 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 213 . If the hydraulic cylinder H draws in rod fiC at 2 ft/s, determine the speed of slider A. A SOLUTION 2s u + s A = l 2v h = —v A 2(2) = -v A v A = -4 ft/s = 4 ft/s <— 7k/um Ans: v A = 4 ft/s 222 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 214 . At the instant shown, the car at A is traveling at 10 m/s around the curve while increasing its speed at 5 m/s 2 . The car at B is traveling at 18.5 m/s along the straightaway and increasing its speed at 2 m/s 2 . Determine the relative velocity and relative acceleration of A with respect to B at this instant. SOLUTION v A = 10 cos 45°i - 10 sin 45°j = { 7.071i - 7.071j } m/s v B = {18.5i} m/s v a/b = v A ~ v B = (7.071i - 7.071j) - 18.5i = {-11.429i - 7.071j } m/s v A/B = V(-11.429) 2 + (-7.071) 2 = 13.4 m/s 0 = tan 7.071 11.429 = 31.7° F' v\ 10 2 (fl ^ = 7 = Too = lm/s2 (a A ) t = 5m/s 2 a A = (5 cos 45° — 1 cos 45°)i + (—1 sin 45° — 5 sin 45°)j = {2.828i - 4.243j) m/s 2 a b = {2i} m /s 2 = *a ~ u B = (2.828i - 4.243j) - 2i = (0.828i - 4.24jj m/s 2 a A/B = V0.828 2 + (—4.243) 2 = 4.32 m/s d = tan 4.243 0.828 = 79.0° v B = 18.5 m/s A ns. Ans. Ans. Ans. Ans: V A /B = 13 - 4 m / S e v = 31.7° F' a A / B = 4.32 m/s 2 0 a = 79.0° ^ 223 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12-215. The motor draws in the cord at B with an acceleration of a B = 2 m/s 2 . When s A = 1.5 m, v B = 6 m/s. Determine the velocity and acceleration of the collar at this instant. SOLUTION Position Coordinates. The position of collar A and point B are specified by s A and s c respectively as shown in Fig. a. We can relate these two position coordinates by considering the length of the cable, which is constant. s B + v/.vy + 2 2 — / s B = l — VjJ + 4 Velocity. Taking the time derivative of Eq. (1), f-„-iW + 4 1-" Mr in ds ft dt ds a Vsi + 4 dt ds /j ds A However, —— = v B and = v A . Then this equation becomes v B = dt S A -v A ( 2 ) Vsl + 4 At the instant s A = 1.5 m, v B = +6 m/s. v B is positive since it is directed toward the positive sense of s B . 1.5 6 = - :V A Vl.5 2 + 4 v A = — 10.0 m/s = 10.0 m/s • The negative sign indicates that v A is directed toward the negative sense of s A . Acceleration. Taking the time derivative of Eq. (2), dv B dt ■M~f)( s i + 4 ) 3/2 ds A dt + (si + 4)-V2 dt V A - s A (sj + 4) V 2 dv A dt dv B dv A ds A However,—— = a B , —— = a A and —— = w^.Then dt dt dt a B 2 2 SaVa vi a A s A (si + 4 ) 3 / 2 (sX + 4 ) 1 / 2 (sX + 4 ) 1 / 2 4 v} + a A s A (s A + 4) ^ (sX + 4 ) 3 / 2 At the instant s A = 1.5 m, a B = +2 m/s 2 . a B is positive since it is directed toward the positive sense of s B . Also, v A = —10.0 m/s. Then |" 4(—10.0 ) 2 + a j 4 (1.5)(1.5 2 + 4)' 2 _ [ (1.5 2 + 4 ) 3 / 2 a A = —46.0 m/s 2 = 46.0 m/s 2 <— Ans. The negative sign indicates that a A is directed toward the negative sense of S 4 . Ans: v A = 10.0 m/s < cij 1 = 46.0 m/s 2 224 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 216 . If block B is moving down with a velocity v B and has an acceleration a B , determine the velocity and acceleration of block A in terms of the parameters shown. SOLUTION l = s B + \/s | + h 2 0 = 4 + ^(4 + h 2 ) 1/2 2 s A s A v A = s A = - s b (s 2 a + h 2 ) 1 ' 2 Sa v A =~v B (l + Ans. a A = v A = - v B (l + ) 1/2 - v B (^ ^)(1 + (J^j ) 1/2 (/z 2 )( - 2)(s A ) 3 s a a A =-a* 1 + (A)> 2+ ^!(1 + (A)Vp 2 \w 4 W/ Ans: v A = -v B [\ + aA = ~a B ( 1+ 225 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 217 . The crate C is being lifted by moving the roller at A downward with a constant speed of v A = 2 m/s along the guide. Determine the velocity and acceleration of the crate at the instant s = lm. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hint: Relate the coordinates x c and x A using the problem geometry, then take the first and second time derivatives. SOLUTION x c + Vjc^ + (4) 2 = / x c + + 16y L/2 (2x A )(x A ) = 0 x c ~ + 16 Y m (2x 2 A ){xjt) + 0*d + 16) 1/2 (*a) 2 + (x A + 16 y 1/2 (x A )(x A ) = 0 / = 8 m, and when i = lm, x c = 3 m x A = 3 m v a ~ x A = 2 m/s l i A = x A = 0 Thus, v c + [(3) 2 + 16] 1/2 (3)(2) = 0 v c = —1.2 m/s = 1.2 m/s f Ans. a c ~[( 3) 2 + 16] —3/2 (3) 2 (2) 2 + [(3) 2 + 16]^ 1/2 (2) 2 + 0 = 0 a c = —0.512 m/s 2 = 0.512 m/s 2 f Ans. Ans: v c = 1.2 m/s] a c = 0.512 m/s 2 ] 226 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 218 . Two planes, A and B , are flying at the same altitude. If their velocities are v A = 500 km/h and v B = 700 km/h such that the angle between their straight-line courses is 8 = 60°, determine the velocity of plane B with respect to plane A. SOLUTION Relative Velocity. Express \ A and \ B in Cartesian vector form, \ A = {—500 j } km/h \ B = { 700 sin 60°i + 700cos60°j} km/h = {350\/3i +350J } km/h Applying the relative velocity equation. = V A + \ B/A 350\/3i + 350j = -500j + \ B / A \ B/A = {35oV3i + 850j } km/h Thus, the magnitude of \ B / A is \ B/A = V(350V3) 2 + 850 2 = 1044.03 km/h = 1044 km/h Ans. And its direction is defined by angle 8 , Fig. a. 8 = tan -1 (— j = 54.50° = 54.5° Ans. V350V3/ v A = 500 km/h 6 *-; Ans: Vb/a = 1044 km/h 8 = 54.5°^2 227 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 228 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 220 . The boat can travel with a speed of 16 km/h in still water. The point of destination is located along the dashed line. If the water is moving at 4 km/h, determine the bearing angle 8 at which the boat must travel to stay on course. SOLUTION \ B = V W + V B/W v B cos 70 °i + v B sin 70°j = (±>) v B cos 70° = 0 + (+T) v B sin 70° = -4 2.748 sin 8 - cos Solving, 8 = 15.1° — 4j + 16 sin 0i + 16 cos 8 j 16 sin 8 + 16 cos 8 8 + 0.25 = 0 Ans. Ans: 8 = 15.1° 229 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 221 . Two boats leave the pier P at the same time and travel in the directions shown. If v A = 40 ft/s and v B = 30 ft/s, determine the velocity of boat A relative to boat B. How long after leaving the pier will the boats be 1500 ft apart? SOLUTION Relative Velocity: + Va/b 40 sin 30°i + 40 cos 30°j = 30 cos 45 °i + 30 sin 45 °j + v A / B v A/B = {—1.213i + 13.43j} ft/s Thus, the magnitude of the relative velocity v A / B is v A/B = V(-1.213) 2 + 13.43 2 = 13.48 ft/s = 13.5 ft/s Ans. And its direction is 0 = tan 1 13.43 1.213 84.8° Ans. One can obtained the time t required for boats A and B to be 1500 ft apart by noting that boat B is at rest and boat A travels at the relative speed v A / B = 13.48 ft/s for a distance of 1500 ft. Thus 1500 _ 1500 v A /b 13.48 111.26 s = 1.85 min Ans. Ans: v B = 13.5 ft/s 0 = 84.8° t = 1.85 min 230 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 222 . A car is traveling north along a straight road at 50 km/h. An instrument in the car indicates that the wind is coming from the east. If the car’s speed is 80 km/h, the instrument indicates that the wind is coming from the northeast. Deter¬ mine the speed and direction of the wind. SOLUTION Solution I Vector Analysis: For the first case, the velocity of the car and the velocity of the wind relative to the car expressed in Cartesian vector form are v c = [50j] km/h and y w/c = ( v w/ch i. Applying the relative velocity equation, we have Vw = v c + \ w/c v w = 50j + (v w/c \ i Vw = {v w/c \i + 50j (1) For the second case, Vc — [80j] km/h and v^/c = ( v w/c )2 cos 45°i + (v W / C )2 sin 45° j. Applying the relative velocity equation, we have = v c + \ w/c = 80j + (v w/c ) 2 cos 45°i + (v w/c ) 2 sin 45° j Vw = ( v w/c )2 cos 45° i + [80 + (v w/c ) 2 sin 45°]j (2) Equating Eqs. (1) and (2) and then the i and j components. Ml = (Vw/c )2 cos 45° (3) 50 = 80 + (v w / c )2 sin 45° ( 4 ) Solving Eqs. (3) and (4) yields (v w /ch = -42.43 km/h (v w / c h = “30 km/h Substituting the result of ( v w / c )j into Eq. (1), \ w = [—30i + 50j] km/h Thus, the magnitude of \ w is v w = \/(—30) 2 + 50 2 = 58.3 km/h Ans. and the directional angle 9 that \ w makes with the x axis is 50 9 = tan -1 — = 59.0° 30 Ans. Ans: v w = 58.3 km/h 9 = 59.0° 231 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 223 . Two boats leave the shore at the same time and travel in the directions shown. If v A = 10 m/s and v B = 15 m/s, determine the velocity of boat A with respect to boat B. How long after leaving the shore will the boats be 600 m apart? SOLUTION Relative Velocity. The velocity triangle shown in Fig. a is drawn based on the relative velocity equation \ A = \ B + v A / B - Using the cosine law, v A /b = VlO 2 + 15 2 — 2 ( 10) (15) cos 75° = 15.73 m/s = 15.7 m/s Ans. Then, the sine law gives sin cj> _ sin 75° 10 “ 15.73 0 = 37.89° The direction of \ A / B is defined by d = 45° - </> = 45° - 37.89° = 7.11° ^ Alternatively, we can express and \ B in Cartesian vector form V 4 = {—10 sin 30°i + 10 cos 30°j } m/s = {—5.00i + 5V3j} m/s \ B = {15 cos 45°i + 15sin45°j} m/s = { 7.5'V / 2i + 7.5 V^j } m/s. Applying the relative velocity equation Va = + \ A/B —500i + 5\/3j = 7.5 V 2 i + 7.5V2j + \ A/B VA/B = { 15.611 - 1.946j } m/s Thus the magnitude of \ A / B is v A j B = V(-15.61) 2 + (-1.946) 2 = 15.73 m/s = 15.7 m/s And its direction is defined by angle 0, Fig. b, (i 946\ d = tan 1 ) = 7.1088° = 7.11° ^ Heres^/g = 600 m. Thus t = , 15.61, S A/B Va/b 600 15.73 = 38.15 s = 38.1 s Ans. Ans: v A/B = 15.7 m/s d = 7.11° IF" t = 38.1 s 232 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 224 . At the instant shown, car A has a speed of 20 km/h, which is being increased at the rate of 300 km/h 2 as the car enters an expressway. At the same instant, car B is decelerating at 250 km/h 2 while traveling forward at 100 km/h. Determine the velocity and acceleration of A with respect to B. SOLUTION v A = {— 20 j) km/h \ B = ( 100 j) km/h Va/b = - v b = (—20j - lOOj) = {—120j} km/h v a/b ~ 120 km/h I Ans. v A 20 2 , , (a A ) n ~ — = —r = 4000 km/h” (a At ~ 300 km/h - p 0.1 a A = —4000i + (—300j) = {—4000i - 300jj km/h 2 a B = j—250j} km/h 2 a A/B ~ a A ~ a B = (—4000i - 300j) - (—250j) = j-4000i - 50jj km/h 2 a A/B = V(-4000) 2 + (-50) 2 = 4000 km/h 2 Ans. 6 = tan -1 = 0.716° IF' Ans. 4000 Ans: v a/b = 120 km/h J. a A/B = 4000 km/h 2 6 = 0.716° IF 233 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 225 . Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 90 ft/s and is increasing its speed at the rate of 15 ft/s 2 , whereas B has a speed of 105 ft/s and is decreasing its speed at 25 ft/s 2 . Determine the relative velocity and relative acceleration of car A with respect to car B at this instant. SOLUTION Dt = + \ A/B —90i = —105 sin 30° i + 105 cos30°j + \ A / B y A /B = {—37.51 - 90.93j} ft/s v AjB = V(-37.5) 2 + (-90.93) 2 = 98.4 ft/s 9 = tan y 90.93 37.5 = 67.6° ip' v A Ans. Ans. a A ~ a B + a A/B (90) 2 —151-T/j/pj = 25 cos 60°i — 25 sin 60°j — 44.1 sin 60°i — 44.1 cos 60°j + a A / B a a/b = {10.691 + 16.70j) ft/s 2 aA/B = V(10.69) 2 + (16.70) 2 = 19.8 ft/s 2 Ans. '(l0.6 9 ) ' Ans. Ans: v A /B = 98.4 ft/s 0 V = 67.6° IP' CIa/b = 19.8 ft/s 2 e„ = 51A°^L 234 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 226 . A man walks at 5 km/h in the direction of a 20-km/h wind. If raindrops fall vertically at 7 km/h in still air , determine the direction in which the drops appear to fall with respect to the man. SOLUTION Relative Velocity: The velocity of the rain must be determined first. Applying Eq. 12-34 gives V- = + V r/lv = 20 i + (-7 j) = { 20 i - 7 j } km/h Thus, the relatives velocity of the rain with respect to the man is v r = + v r/m 20 i - 7 j = 5 i + v r/m U/m = { 15 i - 7 j } km/h The magnitude of the relative velocity \ r / m is given by v r / m = Vl5 2 + (— l) 1 = 16.6 km/h Ans. And its direction is given by 6 tan 1 7_ 15 25.0° ^ Ans. Ans: v r / m = 16.6 km/h e = 25.0° ^ 235 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 227 . At the instant shown, cars A and B are traveling at velocities of 40 m/s and 30 m/s, respectively. If B is increasing its velocity by 2 m/s 2 , while A maintains a constant velocity, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is p B = 200 m. SOLUTION Relative velocity. Express v /( and \ B as Cartesian vectors. \ A = {40j } m/s \ B = {— 30 sin 30°i + 30 cos 30°j } m/s = {— 15i + 15V3j} m/s Applying the relative velocity equation, Hi = H* + \ B/A -15i + 15V3j = 40j + \ B/A Vb/a = { — 15i - 14.02j } m/s Thus, the magnitude of \ B / A is v B / A = V'( —15) 2 + ( —14.02) 2 = 20.53m/s = 20.5 m/s Ans. And its direction is defined by angle 8 , Fig. a /14 02 \ 8 = tan ^ ) = 43.06° = 43.1°^ Ans. Vb 30 2 Relative Acceleration. Here, (a B ), = 2 m/s 2 and (a B )„ = ^ = 4.50 m/s 2 and their directions are shown in Fig. ft. Then, express a B as a Cartesian vector, a B = (-2 sin 30° - 4.50 cos 30°)i + (2 cos 30° - 4.50 sin 30°)j = { —4.8971i - 0.5179j } m/s 2 Applying the relative acceleration equation with a A = 0 , ir a s — a A + a B / A —4.8971i - 0.5179j = 0 + a B/A &B/A = { ~4.8971i - 0.5179j } m/s 2 Thus, the magnitude of a B / A is a B/A = V(-4.8971) 2 + (-0.5179) 2 = 4.9244 m/s 2 = 4.92 m/s 2 And its direction is defined by angle 8', Fig. c, /o 5179\ 8' = tan ^ 1 -- = 6.038° = 6.04°^ V 4.8971/ Ans: v B / A = 20.5 m/s 8 V = 43.1° IP’ d B / A = 4.92 m/s 2 8 n = 6.04° 5^ 236 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12-228. At the instant shown, cars A and B are traveling at velocities of 40 m/s and 30 m/s, respectively. If A is increasing its speed at 4 m/s 2 , whereas the speed of B is decreasing at 3 m/s 2 , determine the velocity and acceleration of B with respect to A. The radius of curvature at B is p g = 200 m. SOLUTION Relative velocity. Express v A and v B as Cartesian vector, v^ = {40j } m/s \ B = {— 30 sin 30°i + 30cos30°j} m/s Applying the relative velocity equation, {—15i + 15V / 3j} m/s = v A + \ B/A —15i + 15V3j = 40j + y B/A \ B / A — { — 15i — 14.02} m/s Thus the magnitude of \ B / A is v B / A = V( —15) 2 + ( —14.02) 2 = 20.53 m/s = 20.5 m/s Ans. And its direction is defined by angle 6 , Fig a. ( 14 02 \ 0 = tan~ x ( —-J = 43.06° = 43.1°?" Ans. y 2 ^ 3 q 2 Relative Acceleration. Here ( a B ) t = 3 m/s 2 and (a B )„ = -p- = = 4.5 m/s 2 and their directions are shown in Fig. b. Then express a B as a Cartesian vector, a B = (3 sin 30° - 4.50 cos 30°)i + (-3 cos 30° - 4.50 sin 30°)j = {—2.3971i - 4.848ljj m/s 2 Applying the relative acceleration equation with a A = {4j} m/s 2 , a B = a A + a B/A —2.3971i - 4.8481] = 4j + a B/A *b/a = {— 2.3971i - 8.8481j| m/s 2 Thus, the magnitude of a B / A is a B/A = V(-2.3971) 2 + (-8.8481) 2 = 9.167 m/s 2 = 9.17 m/s 2 Ans. And its direction is defined by angle S', Fig. c ( 8 8481 \ O' = tan~M -- = 74.84° = 74.8°?" Ans. V2.3971/ Ans: Vb/a = 20.5 m/s 6 = 43.1° ?" a B /A = 9.17 m/s 2 O' = 74.8°?" 237 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 238 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 230 . A man can swim at 4 ft/s in still water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. If the river flows with a velocity of 2 ft/s, determine the speed of the man and the time needed to make the crossing. Note : While in the water he must not direct himself toward point B to reach this point. Why? SOLUTION Relative Velocity: V m = V,. + V m/r 3 4 p m i + —v m j = 2i + 4 sin Oi + 4 cos 0j Equating the i and j components, we have 3 — v m = 2 + 4 sin 8 4 -v m = 4 cos 0 ( 1 ) ( 2 ) Solving Eqs. (1) and (2) yields 6 = 13.29° v m = 4.866 ft/s = 4.87 ft/s Ans. Thus, the time t required by the boat to travel from points A to B is t sab _ V40 2 + 30 2 v b 4.866 10.3 s Ans. In order for the man to reached point B , the man has to direct himself at an angle 0 = 13.3° with y axis. Ans: v„, = 4.87 ft/s t = 10.3 s 239 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 231 . The ship travels at a constant speed of v s = 20 m/s and the wind is blowing at a speed of v w = 10 m/s, as shown. Determine the magnitude and direction of the horizontal component of velocity of the smoke coming from the smoke stack as it appears to a passenger on the ship. SOLUTION Solution I Vector Analysis: The velocity of the smoke as observed from the ship is equal to the velocity of the wind relative to the ship. Here, the velocity of the ship and wind expressed in Cartesian vector form are v, = [20 cos 45° i + 20 sin 45° j] m/s = [14.14i + 14.14j] m/s and v w = [10 cos 30° i - 10 sin 30° j]= [8.660i - 5j] m/s. Applying the relative velocity equation, = v s + \ w/s 8.660i - 5j = 14.14i + 14.14j + v w/s v w /s = [— 5.482i - 19.14j] m/s Thus, the magnitude of v w/s is given by v w = V(-5.482) 2 + (-19.14) 2 = 19.9 m/s and the direction angle 6 that \ w / s makes with the x axis is 6 = tan 1 19.14 / 5.482/ 74.0° IF* Ans. Ans. Solution II Scalar Analysis: Applying the law of cosines by referring to the velocity diagram shown in Fig. a , v w/s = V20 2 + 10 2 - 2(20)(10) cos 75° = 19.91 m/s = 19.9 m/s Ans. Using the result of v w / s and applying the law of sines, sin /> _ sin 75° 10 ~~ 19.91 29.02° Thus, e = 45° + /> = 74.0° IF Ans. Ans: v w/s = 19.9 m/s 6 = 74.0° F 240 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 12 - 232 . The football player at A throws the ball in the y-z plane at a speed v A = 50 ft/s and an angle 0 A = 60° with the horizontal. At the instant the ball is thrown, the player is at B and is running with constant speed along the line BC in order to catch it. Determine this speed, v B , so that he makes the catch at the same elevation from which the ball was thrown. SOLUTION (i) s = s 0 + v 0 t d ac = 0 + (50 cos 60°) t (+\)v = Vq + a c t -50 sin 60° = 50 sin 60° - 32.2 t t = 2.690 s d ac = 67.24 ft d BC = V(30) 2 + (67.24 - 20) 2 = 55.96 ft _ 55.96 _ 20.8 ft/s Ans. B 2.690 ' Ans: v B = 20.8 ft/s 241 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 242 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 234 . At a given instant the football player at A throws a football C with a velocity of 20 m/s in the direction shown. Determine the constant speed at which the player at B must run so that he can catch the football at the same elevation at which it was thrown. Also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made. Player B is 15 m away from A when A starts to throw the football. SOLUTION Ball: ()s = So + Vo t s c = 0 + 20 cos 60° t (+T) v = v 0 + a c t -20 sin 60° = 20 sin 60° - 9.81 t t = 3.53 s s c = 35.31 m Player B: ( ^ ) S B = S Q + V B t Require, 35.31 = 15 + v B (3.53) v B = 5.75 m/s At the time of the catch ( v c) x = 20 cos 60° = 10 m/s —» ( v c) y = 20 sin 60° = 17.32 m/s I v c = v B + \ C / B lOi - 17.32j = 5.751i + (v C / B ) x \ + (v C /n) y .\ (* ) 10 = 5.75 + (v c/B ) x (+T) —17.32 = (v C / B ) y 0 v C /b)x = 4.25 m/s —> 0 c/B)y = 17-32 m/s i v c/B = V(4.25) 2 + (17.32) 2 = 17.8 m/s * -) - m2 ’ ^ a C = a B + &C/B -9.81 j = 0 + a C / B a C / B — 9.81 m/s 2 1 Ans. Ans. Ans. Ans: v B = 5.75 m/s Ans. u c/fi = 17.8 m/s 6 = 76.2° ^ a C / B = 9.81 m/s 2 J. 243 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12 - 235 . At the instant shown, car A travels along the straight portion of the road with a speed of 25 m/s. At this same instant car B travels along the circular portion of the road with a speed of 15 m/s. Determine the velocity of car B relative to car A. SOLUTION Velocity: Referring to Fig. a, the velocity of cars A and B expressed in Cartesian vector form are Applying the relative velocity equation, = V A + \ b /a 14.49i - 3.882j = 21.65i - 12.5j + \ B/A \ B /a ~ [—7.162i + 8.618j] m/s Thus, the magnitude of y B/A is given by v B/A = V(-7.162) 2 + 8.618 2 = 11.2 m/s The direction angle 0 V of y B/A measured down from the negative x axis, Fig. b is d„ = tan 8.618\ 7.162/ 50.3° ^ &> Ans: v B /A = 11.2 m/s 6 = 50.3° 244 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-1. The 6 -lb particle is subjected to the action of its weight and forces F, = {2i + 6 j - 2tk} lb, F 2 = {t 2 i - 4fj - lk} lb, and F 3 = {—2ri} lb, where t is in seconds. Determine the distance the ball is from the origin 2 s after being released from rest. SOLUTION 2F = nta; (2i + 6j - 2tk) + (ri - 4fj - lk) - 2d - 6k = ( J(a x i + a y j + a A) Equating components: — }a x = t 2 - 2t + 2 (|u v = -4f + 6 (—]a„ = — It - 7 32.2/ x \ 32.2 / y \ 32.2 , Since dv = a dt , integrating from v = 0, t = 0, yields 3 / 6 6 \ t J 2 o - \v x =-r + 2 1 32.2 / x 3 ,v v = — 2 1 2 + 6 1 32.2/ y 6 322 v, = —r — It Since ds = v dt , integrating from s = 0, f = 0 yields r r , — — — — + r 2r ^ 2 s v = —-—I- 3r - _£ _ 32.2 l Sz ~ 3 2 32.2 y~ x 12 3 ' y 32.2 y" y 3 When r = 2 s then, = 14.31 ft, s y = 35.78 ft Sj. = —89.44 ft Thus, x = V(14.31) 2 + (35.78) 2 + (—89.44) 2 = 97.4 ft Ans. •4 ^ 2 ^ Ans: x = 97.4 ft 245 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 2 . The two boxcars A and B have a weight of 20 000 lb and 30 000 lb, respectively. If they are freely coasting down the incline when the brakes are applied to all the wheels of car A, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is ju k = 0.5. The wheels of car B are free to roll. Neglect their mass in the calculation. Suggestion: Solve the problem by representing single resultant normal forces acting on A and B, respectively. SOLUTION Caryl: +\2,F y = 0; N a - 20 000 cos 5° = 0 N A = 19 923.89 lb +/1 I F X = ma x \ 0.5(19 923.89) - T - 20 000 sin 5° 20 000 \ - c 32.2 J Both cars: +/"%F X = ma x ; 0.5(19 923.89) - 50 000 sin 5° 50 000 \ - u 32.2 J Solving, a = 3.61 ft/s 2 T = 5.98 kip ( 1 ) Ans. Ans: T = 5.98 kip 246 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 3 . If the coefficient of kinetic friction between the 50-kg crate and the ground is /t , k = 0.3, determine the distance the crate travels and its velocity when t = 3 s. The crate starts from rest, and P = 200 N. P SOLUTION Free-Body Diagram: The kinetic friction Ft = fj. k N is directed to the left to oppose the motion of the crate which is to the right. Fig. a. Equations of Motion: Here, a y = O.Thus, + 12^ = 0; N - 50(9.81) + 200 sin 30° = 0 N = 390.5 N -±> = ma x \ 200 cos 30° - 0.3(390.5) = 50a a = 1.121 m/s 2 Kinematics: Since the acceleration a of the crate is constant, v — v 0 + a c t v — 0 + 1.121(3) = 3.36 m/s Ans. 1 2 s = s 0 + v 0 t + -a c t 5 = 0 + 0 + i(1.121)(3 2 ) = 5.04 m Ans. and ■i* H Ans: v = 3.36 m/s 5 = 5.04 m 247 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 4 . If the 50-kg crate starts from rest and achieves a velocity of v = 4 m/s when it travels a distance of 5 m to the right, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is = 0.3. P SOLUTION Kinematics: The acceleration a of the crate will be determined first since its motion is known. ( ^ ) v 2 = v 0 2 + 2 a c (s - s 0 ) 4 2 = 0 2 + 2a(5 - 0) a = 1.60 m/s 2 —> Free-Body Diagram: Here, the kinetic friction Ff = fj. k N = 0.3/V is required to be directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: + T ^F y = ma y ; N + P sin 30° - 50(9.81) = 50(0) N = 490.5 - 0.5 P Using the results of N and a, ■** ZF X = ma x ; P cos 30° - 0.3(490.5 - 0.5P) = 50(1.60) P = 224 N Ans. Ans: P = 224 N 248 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 5 . If blocks A and B of mass 10 kg and 6 kg, respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are /jl / i = 0.1 and fj. B = 0.3. Neglect the mass of the link. SOLUTION Free-Body Diagram: Here, the kinetic friction (Ff) A = ^ A N A = 0.1 N A and (Ff) B = /jl b N b = 0.3 N b are required to act up the plane to oppose the motion of the blocks which are down the plane. Since the blocks are connected, they have a common acceleration a. Equations of Motion: By referring to Figs, (a) and (b), +S'ZFy = may ; N A - 10(9.81) cos 30° = 10(0) N a = 84.96 N 10(9.81) sin 30° - 0.1(84.96) — F = 10a 40.55 — F = 10a and +/1jFy = may : \+2iv = ma r ': N b - 6(9.81) cos 30° = 6(0) N b = 50.97 N F + 6(9.81) sin 30° - 0.3(50.97) = 6 a F + 14.14 = 6 a Solving Eqs. (1) and (2) yields a = 3.42 m/s 2 F = 6.37 N ( 1 ) ( 2 ) Ans. Ans: F = 6.37 N 249 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 6 . The 10-lb block has a speed of 4 ft/s when the force of u = 4 ft/s F = (8 1 2 ) lb is applied. Determine the velocity of the block when t = 2 s. The coefficient of kinetic friction at the surface F = (St 2 ) lb is /jL k = 0.2. SOLUTION Equations of Motion. Here the friction is Ff = p. /c N = 0.2 N. Referring to the FBD of the block shown in Fig. a. + T£F v = ma y ; N - 10 = — (0) N = 10 lb .+ %F X = ma x ; 8 1 2 - 0.2(10) = a = 3.22(8 1 2 - 2) ft/s z Kinematics. The velocity of the block as a function of t can be determined by integrating dv = a dt using the initial condition v = 4 ft/s at t = 0. nV pt dv 4 ft/s / 3.22 (8t 2 Jo 2 )dt v - 4 = 3.22 ( ^ r' 2( j v = { 8.5867f 3 - 6.44t + 4 } ft/s When t = 2 s, v = 8.5867(2 3 ) - 6.44(2) + 4 = 59.81 ft/s = 59.8 ft/s Ans. Ans: v = 59.8 ft/s 250 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 7 . The 10-lb block has a speed of 4 ft/s when the force of w = 4 ft/s F = (8 1 2 ) lb is applied. Determine the velocity of the block when it moves s = 30 ft. The coefficient of kinetic friction at F= (8^) lb the surface is /jl s = 0.2. SOLUTION Equations of Motion. Here the friction is Ft — /r k N = 0.2 N. Referring to the FBD of the block shown in Fig. a, + T= ma y \ N - 10 = -^-(0) A = 101b 32.2 r=6t ± XF t = 8 1 1 ~ 0.2(10) = 32.2 a = 3.22(8 1 2 - 2) ft/s 2 Kinematics. The velocity of the block as a function of t can be determined by integrating dv = adt using the initial condition v = 4 ft/s at t = 0. f dv = [ 3.22(8 1 2 - 2)dt J 4 ft/s J 0 v - 4 = 3.22 0 1 3 - 21 v = {8.5867f 3 - 6.44t + 4} ft/s The displacement as a function of t can be determined by integrating ds = vdt using the initial condition s = 0 at t = 0 [ ds = [ (8.5867 1 3 - 6.44 1 + 4)dt Jo Jo s = {2.1467r 4 - 3.22 1 2 + 4f} ft At s = 30 ft, 30 = 2.1467f 4 - 3.22 1 2 + 4 1 Solved by numerically, t = 2.0089 s O) Thus, at x = 30 ft, v = 8.5867(2.0089 3 ) - 6.44(2.0089) + 4 = 60.67 ft/s = 60.7 ft/s Ans. Ans: v = 60.7 ft/s 251 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-8. The speed of the 3500-lb sports car is plotted over the 30-s time period. Plot the variation of the traction force F needed to cause the motion. SOLUTION Kinematics: For 0 < t < 10 s. n = — t = (6/j ft/s. Applying equation a = we have a dv dt 6 ft/s 2 For 10 < t < 30 s, dv a = —— , we have dt v — 60 t - 10 a Equation of Motion: For 0 < t < 10 s For 10 < t < 30 s 80 - 60 30 - 10’ v = [t + 50} ft/s. Applying equation dv dt = 1 ft/s 2 (6) = 652 lb ns) = max ; F 109 lb Ans. F(lh > 652 - + 10 -H-/ (S) 30 Ans: c + XF x = ma x ; F = 652 lb XF x = ma x , F = 109 lb 252 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 9 . The conveyor belt is moving at 4 m/s. If the coefficient of static friction between the conveyor and the 10-kg package B is /jl s = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt SOLUTION = ma x \ 0.2(98.1) = 10 a a = 1.962 m/s 2 = v 0 + a c t 4 = 0 + 1.962 t t = 2.04 s Ans: t = 2.04 s <184 ii - >a “p’F-aam.O Ans. B <+1-0 (u 1 CO CO CO 253 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-10. The conveyor belt is designed to transport packages of various weights. Each 10-kg package has a coefficient of kinetic friction /jL k = 0.15. If the speed of the conveyor is 5 m/s, and then it suddenly stops, determine the distance the package will slide on the belt before coming to rest. SOLUTION + XF x = ma x ; 0.15 «7(9.81) = ma a = 1.4715 m/s 2 (iO v 2 = vl + 2 a c (s - s 0 ) 0 = (5) 2 + 2(—1.4715)(s - 0) s = 8.49 m Ans: s = 8.49 m < 18.1 ii T Ans. c 254 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-11. Determine the time needed to pull the cord at B down 4 ft starting from rest when a force of 10 lb is applied to the cord. Block A weighs 20 lb. Neglect the mass of the pulleys and cords. SOLUTION +1 tF y = ma y ; 40 — 20 = - a A 32.2 A a A = 32.2 ft/s 2 s B + 2 s c = /; a B = ~2a c 2 s a — sc = l'\ 2a A = Oq a B = ~^ a A a B = 128.8 ft/s 2 (+1) 5 = s 0 + v 0 t + - a c t 2 1 4 = 0 + 0 + -(128.8) t 2 t = 0.249 s Ans: t = 0.249 s 255 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-12. Cylinder B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the block’s vertical position y so that when F is applied the block rises with a constant acceleration a /; . Neglect the mass of the cord and pulleys. SOLUTION + | XF y = ma y ; 2Fcos 0 — mg = ma B where cos d = 2F (wrw) ~ " m “ m(a B + g)V4v 2 + d 2 F= 4 y Ans. Ans: F = m(a B + g)y/ 4y 2 + d 2 4y 256 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 13 . Block A has a weight of 8 lb and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic friction is fj. k = 0.2. If the spring has a stiffness of k = 20 lb/ft, and it is compressed 0.2 ft, determine the acceleration of each block just after they are released. A B SOLUTION Block A: + XF x = ma x ; g 4 — 1.6 = - a A 32.2 A Block B: a A = 9.66 ft/s 2 <— +j 1,F X = ma x ; 6 4 - 12 = a B 32.2 " a B = 15.0 ft/s 2 —» Ans. Ans. Ans: a A = 9.66 ft/s 2 <— a B = 15.0 ft/s 2 —» 257 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-14. The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for a distance of 10 m before coming to rest. Determine the constant horizontal force developed in the coupling C, and the frictional force developed between the tires of the truck and the road during this time. The total mass of the boat and trailer is 1 Mg. SOLUTION Kinematics: Since the motion of the truck and trailer is known, their common acceleration a will be determined first. p 2 = v 0 2 + 2a c (s - s 0 ) 0 = 15 2 + 2a(10 - 0) a = —11.25 m/s 2 = 11.25 m/s 2 <— Free-Body Diagram :The free-body diagram of the truck and trailer are shown in Figs, (a) and (b), respectively. Here, F representes the frictional force developed when the truck skids, while the force developed in coupling C is represented by T. Equations of MotiomUsing the result of a and referrning to Fig. (a), YF X = ma x ; -T = 1000(-11.25) T = 11 250 N = 11.25 kN Ans. Using the results of a and T and referring to Fig. (b), +12 F x = ma x ; 11 250 - F = 2000(-11.25) F = 33 750 N = 33.75 kN Ans. Oy Ans: T = 11.25 kN F = 33.75 kN 258 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-15. The motor lifts the 50-kg crate with an acceleration of 6 m/s 2 . Determine the components of force reaction and the couple moment at the fixed support A. SOLUTION Equation of Motion. Referring to the FBD of the crate shown in Fig. a , + t 2F V = ma y \ T - 50(9.81) = 50(6) T = 790.5 N y Equations of Equilibrium. Since the pulley is smooth, the tension is constant throughout entire cable. Referring to the FBD of the pulley shown in Fig. b, +> XF x = 0; 790.5 cos 30° - B x = 0 B x = 684.59 N + T lF y = 0; B y - 790.5 - 790.5 sin 30° = 0 B y = 1185.75 N Consider the FBD of the cantilever beam shown in Fig. c, A XF x = 0; 684.59 - A x = 0 A = 684.59 N = 685 N Ans. + T2F V = 0; A y - 1185.75 = 0 A y = 1185.75 N = 1.19 kN Ans. C +XM a = 0 ; M a - 1185.75(4) = 0 M a = 4743 N • m = 4.74 kN • m Ans. Ans: A x = 685 N A y = 1.19 kN M a = 4.74 kN • m 259 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 16 . The 75-kg man pushes on the 150-kg crate with a horizontal force F. If the coefficients of static and kinetic friction between the crate and the surface are /u s = 0.3 and /x k = 0.2, and the coefficient of static friction between the man’s shoes and the surface is = 0.8, show that the man is able to move the crate. What is the greatest acceleration the man can give the crate? SOLUTION Equation of Equilibrium. Assuming that the crate is on the verge of sliding (F f ) c = Us Nc = 0.3 N C - Referring to the FBD of the crate shown in Fig. a , + | = 0; N c ~ 150(9.81) = 0 N c = 1471.5 N %F X = 0; 0.3(1471.5) - F = 0 F = 441.45 N Referring to the FBD of the man, Fig. b. + |SF V = 0; N m - 75(9.81) = 0 N B = 735.75 N -4 %F X = 0; 441.45 - (F f ) m = 0 (F f ) m = 441.45 N Since (7y)„, < /J.' s N m = 0.8(735.75) = 588.6 N, the man is able to move the crate. Equation of Motion. The greatest acceleration of the crate can be produced when the man is on the verge of slipping.Thus, (iy)„, = = 0.8(735.75) = 588.6 N. ±,%F X = 0; F — 588.6 = 0 F = 588.6 N Since the crate slides, (Ff) c = = 0.2(1471.5) = 294.3 N. Thus, XF x = ma x ; 588.6 - 294.3 = 150 a a = 1.962 m/s 2 = 1.96 m/s 2 Ans. Ans: a = 1.96 m/s 2 260 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-17. Determine the acceleration of the blocks when the system is released. The coefficient of kinetic friction is ix k , and the mass of each block is m. Neglect the mass of the pulleys and cord. SOLUTION Free Body Diagram. Since the pulley is smooth, the tension is constant throughout the entire cord. Since block B is required to slide, = /%1V. Also, blocks A and B are attached together with inextensible cord, so a A = a B = a. The FBDs of blocks A and B are shown in Figs, a and b, respectively. Equations of Motion. For block A, Fig. a, + |2Fy = ma y ; T — mg = m(—a) (1) For block B, Fig. b , +1 XFy = ma y \ N — mg = m( 0) N = mg = ma x \ T — g. k mg = ma ( 2 ) Solving Eqs. (1) and (2) a = |(1 - Hk) g Ans. ^ = -(1 + fi k ) mg T / ,/! V p % 1 (A) Ans: a = « ( x _ Bk) g 261 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 18 . A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C? SOLUTION 40 + \ = m a x ; 40 sin 30° = - a x x 3 2.2 a = 16.1 ft/s 2 = Vq + 2 a c (s - s 0 ); vj, = 0 + 2(16.1)(20) v B = 25.38 ft/s (+\) v = v 0 + a c t ; 25.38 = 0 + 16.1 t AB t AB = 1.576 s (^ )s x = fo)o + (v x ) 0 1 R = 0 + 25.38 cos 30 °(t BC ) (+!)«, = (Sy) 0 + (Vy) 0 t + a c t 2 4 = 0 + 25.38 sin 30° t BC + ^(32.2 ){t BC ) 2 t BC = 0.2413 s R = 5.30 ft Total time = t AB + t BC = 1.82 s folk N I Ans. Ans. Ans: R = 5.30 ft Tic = 1-82 s 262 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 263 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 20 . The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is v A = 2.5 m/s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is /x*. = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Take d = 30°. SOLUTION /+%F y = ma y ; N c ~ 12(9.81) cos 30° = 0 N c = 101.95 N F\1F X = ma x ; 12(9.81) sin 30° - 0.3(101.95) = 12 a c a c = 2.356 m/s 2 (+\) v 2 b = v\ + 2 a c (s B ~ s A ) v\ = (2.5) 2 + 2(2.356)(3 - 0) v B = 4.5152 = 4.52 m/s v A = 2.5 m/s Ans. Ans: v B = 4.52 m/s 264 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 21 . The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is v A = 2.5 m/s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is gi k = 0.3, determine the smallest incline 0 of the ramp so that the crates will slide off and fall into the cart. SOLUTION (+\) v\= v\ + 2 a c (s B - s A ) 0 = (2.5) 2 + 2(« c )(3 - 0) a c = 1.0417 v A = 2.5 m/s /+%Fy = ma y ; = ma x . Solving, 6 = 22 . 6 ° N c ~ 12(9.81) cos d = 0 N c = 117.72 cos 0 12(9.81) sin 0 - 0.3(N C ) = 12 (1.0417) 117.72 sin 6 - 35.316 cos e - 12.5 = 0 Ans. Ans: 6 = 22 . 6 ° 265 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 22 . The 50-kg block A is released from rest. Determine the velocity of the 15-kg block B in 2 s. SOLUTION Kinematics. As shown in Fig. a, the position of block B and point A are specified by s B and s A respectively. Flere the pulley system has only one cable which gives + s B + 2(s B - a) = l s A + 3 s B = I + 2a (1) Taking the time derivative of Eq. (1) twice, a A + 3 a B = 0 ( 2 ) Equations of Motion. The FBD of blocks B and A are shown in Fig. b and c. To be consistent to those in Eq. (2), a A and a B are assumed to be directed towards the positive sense of their respective position coordinates s A and s B . For block B, + T SF V = mOy\ 3 T - 15(9.81) = 15 (~a B ) (3) For block A, + tSF v = ma y \ T - 50(9.81) = 50(-a A ) (4) Solving Eqs. (2), (3) and (4), a B = -2.848 m/s 2 = 2.848 m/s 2 f a A = 8.554 m/s 2 T = 63.29 N The negative sign indicates that a B acts in the sense opposite to that shown in FBD. The velocity of block B can be determined using +1 v B = (v A )o + a B t; v B = 0 + 2.848(2) v B = 5.696 m/s = 5.70 m/s \ Ans. T A J jK 50 ( 9 - 60 /I (C) (a) (h Ans: v B = 5.70 m/s | 266 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 23 . If the supplied force F = 150 N, determine the velocity of the 50-kg block A when it has risen 3 m, starting from rest. SOLUTION Equations of Motion. Since the pulleys are smooth, the tension is constant throughout each entire cable. Referring to the FBD of pulley C, Fig. a. of which its mass is negligible. + T tF y = 0; 150 + 150 - T = 0 T = 300 N Subsequently, considered the FBD of block A shown in Fig. b , + t2F v = ma y \ 300 + 300 - 50(9.81) = 50a a = 2.19 m/s 2 f Kinematics. Using the result of a, ( + T) v 2 = Vq + 2 a c s; v 2 = 0 2 + 2(2.19)(3) v = 3.6249 m/s = 3.62 m/s Ans. I50hI 150/J 4 - A T (a.) Ans: v = 3.62 m/sf 267 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 24 . Kinematics. From A to C, the suitcase moves along the inclined plane (straight line). (+/) v 2 = vl + 2a c s\ v 2 = 0 2 + 2(4.905)(5) v = 7.0036 m/s (+/) s = s 0 + v 0 t + | a c t 2 ; 5 = 0 + 0 + | (4.905)1 \ c t AC = 1.4278 s From C to B , the suitcase undergoes projectile motion. Referring to x—y coordinate system with origin at C, Fig. b, the vertical motion gives (+J-) Sy = (so)y + Vyt + y Cl y t 2 ; 2.5 = 0 + 7.0036 sin 30° t CB + | (9.81 )t 2 CB 4.905 t 2 CB + 3.5018 t CB - 2.5 = 0 Solve for positive root, t CB = 0.4412 s Then, the horizontal motion gives («t) s* = (^o)x + v x t; R = 0 + 7.0036 cos 30° (0.4412) = 2.676 m = 2.68 m Ans. The time taken from A to B is t AB = t AC + t CB = 1.4278 + 0.4412 = 1.869 s = 1.87 s Ans. 6omi)rl ft O) 1 Ans: ( c + ) s x = 2.68 m t AB = 1-87 s 268 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 25 . \ + Sly = ma y \ N - 60(9.81) cos 30° = 60(0) N = 509.74 N +,/ 2iy = ma x .\ 60(9.81) sin 30° - 0.2(509.74) = 60 a a = 3.2059 m/sV Kinematics. From A to C, the suitcase moves along the inclined plane (straight line). (+/) v 2 = Vq + 2 a c s; v 2 = 2 2 + 2(3.2059)(5) v = 6.0049 m/s i/ {+/) s = j 0 + v 0 t + | a c t 2 ; 5 = 0 + 2t AC + y (3.2059)t^ c 1.6029 t 2 AC + 2 t AC — 5 = 0 Solve for positive root, t AC = 1.2492 s From C to B, the suitcase undergoes projectile motion. Referring to x—y coordinate system with origin at C, Fig. b , the vertical motion gives ( + 1) s y = (s 0 ) y + v y t + | Oyt 2 ; boCWOd 'Fsr-o-zfJ 2.5 = 0 + 6.0049 sin 30° t CB + y(9.81 )t 2 CB 4.905 t 2 CB + 3.0024 t CB - 2.5 = 0 Solve for positive root, t CB = 0.4707 s Then, the horizontal motion gives («t) s x = (s Q ) x + v x t\ R = 0 + 6.0049 cos 30° (0.4707) = 2.448 m = 2.45 m The time taken from A to B is t AB = t AC + t CB = 1.2492 + 0.4707 = 1.7199 s = 1.72 s Ans: R = 2.45 m tAB = 1-72 s 269 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 26 . The 1.5 Mg sports car has a tractive force of F = 4.5 kN. If it produces the velocity described by v-t graph shown, plot the air resistance R versus t for this time period. SOLUTION Kinematic. For the v-t graph, the acceleration of the car as a function of t is a = — = {—O.lr + 3}m/s 2 at Equation of Motion. Referring to the FBD of the car shown in Fig. a, (<t)XF x = ma x ; 4500 - R = 1500(—O.lr + 3) R = {150f}N The plot of R vs t is shown in Fig. b v (m/s) t( s) Ans: R = {1501} N 270 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 271 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 28 . At the instant shown the 100-lb block A is moving down the plane at 5 ft/s while being attached to the 50-lb block B. If the coefficient of kinetic friction between the block and the incline is = 0.2, determine the acceleration of A and the distance A slides before it stops. Neglect the mass of the pulleys and cables. SOLUTION Block A: 2F X = ma x ; -T A - 0.2N A + 1000) = +\ 2F y = ma y \ N a - 1000) = 0 Thus, T a - 44 = -3.1056fl^ Block B: + t 2F y = nitty, T b - 50 = Tjj — 50 = 1.553« B Pulleys at C and D: + ]2F y = 0- 2T a — 2T b = 0 Ta = T b Kinematics: + 2 S C = l Sd + - s B ) = r s c + d + s D = d' Thus, ii A ~ 2 ciq 2a D = a B a c = ~a D ’ so that ci A — u B Solving Eqs. (1)—(4): a A = a B = —1.288 ft/s 2 T a = T b = 48.0 lb Thus, a A = 1.29 ft/s 2 (+/) v 2 = vl + 2 a c 0 - 5 0 ) 0 = (5) 2 + 2(-1.288)0 - 0) s = 9.70 ft T ff T 6 Ans. Ans: a A = 1.29 ft/s 2 s = 9.70 ft 272 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 29 . The force exerted by the motor on the cable is shown in the graph. Determine the velocity of the 200-lb crate when f = 2.5 s. SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force F must overcome the weight of the crate. Thus, the time required to move the crate is given by +1 ZF y = 0; lOOt - 200 = 0 t = 2 s F( lb) t{ s) Equation of Motion: For 2 s < t < 2.5 s, F 250 2.5 ‘ (lOOt) lb. By referring to Fig. a, + ]'LF y = ma y ; lOOt - 200 =- a 32.2 a = (16.lt - 32.2) ft/s 2 Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 2 s < / < 2.5 s, v = 0 at ( = 2 s will be used as the lower integration limit. Thus, (+T) When f = 2.5 s. 32.2 )dt v = (8.05t 2 = (8.05t 2 32.2 f) 32.21 + 32.2)ft/s v = 8.05(2.5 2 ) - 32.2(2.5) + 32.2 = 2.01 ft/s Ans. 1 I (a.) Ans: v = 2.01 ft/s 273 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 30 . The force of the motor M on the cable is shown in the graph. F (N) Determine the velocity of the 400-kg crate A when t = 2 s. 2500 / / -F — 625 t 1 J SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force 2F must overcome its weight. Thus, the time required to move the crate is given by + T2F y = 0; 2(625t 2 ) - 400(9.81) = 0 t = 1.772 s (s) Equations of Motion: F = (625t 2 ) N. By referring to Fig. a, + T ~ZF y = ma y \ 2(625t 2 ) - 400(9.81) = 400a a = (3.125t 2 - 9.81) m/s 2 Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 1.772 sS(<2s,» = 0aU = 1.772 s will be used as the lower integration limit. Thus, (+t) 9.81 )dt v = (l.0417t 3 9.81t) t 1.772 s = (l.0417t 3 - 9.81f + 11.587) m/s When t = 2 s, v = 1.0417(2 3 ) - 9.81(2) + 11.587 = 0.301 m/s Ans. Y Ans: v = 0.301 m/s 274 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 31 . The tractor is used to lift the 150-kg load B with the 24-m- long rope, boom, and pulley system. If the tractor travels to the right at a constant speed of 4 m/s, determine the tension in the rope when s A — 5 m. When = 0, s B = 0. SOLUTION 12 — s B + V4 + (12) 2 = 24 ~Sb + (4 + 144 )^ (s a s a J = 0 -s B - (4 + 144) 3 (s A s^J + (4 + 144) 3 ^4) + (4 + 144) 3 ^.44^ = 0 S B ~ 2 -2 SaSa s A + S A S A (4 + 144)3 (4 + 144)1 a B — (5) 2 (4) 2 (4) 2 + 0 (( 5)2 + 144)5 (( 5)2 + 144)5 = 1.0487 m/s 2 + t 24 = ma y ; T - 150(9.81) = 150(1.0487) T = 1.63 kN Ans. Ans: T = 1.63 kN 275 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 32 . The tractor is used to lift the 150-kg load B with the 24-m- long rope, boom, and pulley system. If the tractor travels to the right with an acceleration of 3 m/s 2 and has a velocity of 4 m/s at the instant = 5 m, determine the tension in the rope at this instant. When s A = 0, s B = 0. SOLUTION 12 = s B + Vs 2 a + (12) 2 = 24 144)-’ (s A s A J = 0 S B ~ 2 -2 Sa s a {sa + 144)1 ■Ui 4- s A s A (s 2 a + 144 )* a B = (5)W (( 5) 2 + 144 )* (4) 2 + (5)(3) ((5) 2 + 144)* 2.2025 m/s 2 + T ^F y = ma y ; T - 150(9.81) = 150(2.2025) T = 1.80 kN Ans. Ans: T = 1.80 kN 276 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-33. Block A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that it will not slide on A. Also, what is the corresponding acceleration? The coefficient of static friction between A and B is /jl s . Neglect any friction between A and the horizontal surface. SOLUTION Equations of Motion. Since block B is required to be on the verge to slide on A, Ff = h s N b . Referring to the FBD of block B shown in Fig. a , +1 ~%F y = ma y \ N b cos 8 — /Ji s N B sin 8 — mg = m( 0) N„ = mg cos 8 — i*l s sin 8 c + SF r = ma x ; P — N B sin 8 — /jl s N b cos 8 = ma P — N b (sin 8 T /jl s cos 8) = ma ( 1 ) ( 2 ) Substitute Eq. (1) into (2), P - sin 8 + gL s cos 8 vcos 8 — /UjSin 8 Referring to the FBD of blocks A and B shown in Fig. b .+ 1,F X = ma x ; P = 2 ma Solving Eqs. (2) into (3), /sin 8 + /jl s cos 8 mg = ma P = 2 mg I - \cos 8 — ^sin 8 /sin 8 + /jl s cos 8\ \cos 8 — gi s sin 8)^ (3) (4) Ans. Ans. m Co-) ? Ans: P = a = / sin 8 + gi s cos 8 2 mg \ -;- \cos 8 — /jl s sin 8 /sin 8 + /x s cos 8\ \cos 8 — fj. s sin 6/ 277 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-34. The 4-kg smooth cylinder is supported by the spring having a stiffness of k AB = 120 N/m. Determine the velocity of the cylinder when it moves downward s = 0.2 m from its equilibrium position, which is caused by the application of the force F = 60 N. SOLUTION Equation of Motion. At the equilibrium position, realizing that F sp = kxa = 12(k 0 the compression of the spring can be determined from + t £F V = 0; 12(kb - 4(9.81) = 0 Xq = 0.327 m Thus, when 60 N force is applied, the compression of the spring is x = v + *o = ‘ s + 0.327. Thus, F sp = kx = 120(s + 0.327). Then, referring to the FBD of the collar shown in Fig. a, + t £F V = ma y \ 120(s + 0.327) - 60 - 4(9.81) = 4(-a) a = { 15 — 30 v} m/s 2 Kinematics. Using the result of a and integrate f vdv = ads with the initial condition v = 0 at s = 0, f vdv = f (15 — 30 s)ds Jo Jo v 2 — = 15 x — 15 s 2 2 v = { V30(s - s 2 )} m/s At s = 0.2 m, v = V30(0.2 - 0.2 2 ) = 2.191 m/s = 2.19 m/s Ans. Ans: v = 2.19 m/s 278 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 35 . The coefficient of static friction between the 200-kg crate and the flat bed of the truck is /jl s = 0.3. Determine the shortest time for the truck to reach a speed of 60 km/h, starting from rest with constant acceleration, so that the crate does not slip. SOLUTION Free-Body Diagram: When the crate accelerates with the truck, the frictional force Ff develops. Since the crate is required to be on the verge of slipping, F f = n s N = 0.31V. Equations of Motion: Here, a y = 0. By referring to Fig. a , + T 'ZF y = may, N - 200(9.81) = 200(0) N = 1962 N -±> ~EF X = ma x , -0.3(1962) = 200(-a) a = 2.943 m/s 2 <— Kinematics: The final velocity of the truck is v — I 60 kmY1000 mV lh h J\ 1km /\3600s 16.67 m/s. Since the acceleration of the truck is constant, ( V ) v = Vq + a c t 16.67 = 0 + 2.943r t = 5.66 s Ans. % 200(%OH Ans: t = 5.66 s 279 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 36 . The 2-lb collar C fits loosely on the smooth shaft. If the spring is unstretched when s = 0 and the collar is given a velocity of 15 ft/s, determine the velocity of the collar when s = 1 ft. SOLUTION F s = kx; F s = 4(Vl + s 2 - l) -[2x 2 - 4VlT7£ = 3^(« 2 - 15 2 ) v = 14.6 ft/s Ans. Ans: v = 14.6 ft/s 280 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 37 . The 10-kg block A rests on the 50-kg plate B in the position shown. Neglecting the mass of the rope and pulley, and using the coefficients of kinetic friction indicated, determine the time needed for block A to slide 0.5 m on the plate when the system is released from rest. SOLUTION Block A: + \ 2-fy = ma y \ +/ 1,F X = ma x ; Block B\ + \ 2T v = ma y ; +/ 1,F X = ma x \ N a - 10(9.81) cos 30° =0 N a = 84.96 N -T + 0.2(84.96) + 10(9.81) sin 30° = 10 a A T - 66.04 = -10a„ N b - 84.96 - 50(9.81) cos 30° = 0 N b = 509.7 N -0.2(84.96) - 0.1(509.7) — T + 50(9.81 sin 30°) 177.28 — T = 50 a B S A + S B — l As a = -As b a A = ~a B Solving Eqs. (1) - (3): a B = 1.854 m/s 2 a A = -1.854 m/s 2 T = 84.58 N In order to slide 0.5 m along the plate the block must move 0.25 m. Thus, (+i/) s B = + S B / A — + 0.5 A^ = —0.25 m , , 1 , (+/) s .4 = % + v 0 t + - a A t l -0.25 = 0 + 0 + — (—1.854)t 2 t = 0.519 s Ans. Ans: t = 0.519 s 281 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 38 . The 300-kg bar B , originally at rest, is being towed over a series of small rollers. Determine the force in the cable when t = 5 s, if the motor M is drawing in the cable for a short time at a rate of v = (0.4 1 1 ) m/s, where t is in seconds (0 < t < 6 s). How far does the bar move in 5 s? Neglect the mass of the cable, pulley, and the rollers. SOLUTION -±» = ma- T = 300a v = 0.4 1 2 dv a = —— = 0.8f dt When t = 5 s, a = 4 m/s 2 T = 300(4) = 1200 N = 1.20 kN ds = v dt f ds = I 0.4f 2 ds Jo Jo s = | J(5) 3 = 16.7 m Ans. f T Ans. Ans: s = 16.7 m 282 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 39 . An electron of mass m is discharged with an initial horizontal velocity of v Q . If it is subjected to two fields of force for which F x = F 0 and F y = 0.3 F 0 , where F 0 is constant, determine the equation of the path, and the speed of the electron at any time t. SOLUTION 2F x = ma x ; + t 2F y = ma y ; Thus, F o = ma x 0.3 F o = ma y r dVx = f—dt Jv„ jo m F ° t + V x = - t + V 0 m r> f'o.3F 0 / dVy = f dt Jo Jo m 0.3 F n t + Vq I + = — \/l.09 Fgt 2 + 2F 0 tmv 0 + m 2 v jj jf dx = + v 0 )dt 2m [ y f ‘ 0.3T 0 dy= - -tdt Jo Jo m y = t = 0.3 F 0 t 2 2 m 2m 0.3F 0 F 0 ( 2m X = ^n\V3¥ 0 ]y + V ° x = {3 + v \i^k) yl 2m 0.3 F, y o.^F 0 Ans. Ans: v = — Vl.09F5r + 2 F 0 tmv 0 + m 2 v jj 283 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 40 . The 400-lb cylinder at A is hoisted using the motor and the pulley system shown. If the speed of point B on the cable is increased at a constant rate from zero to v B = 10 ft/s in t = 5 s, determine the tension in the cable at B to cause the motion. SOLUTION 2sA + S B = 1 2 a a = — a B ( X ) v = v Q + a B t 10 = 0 + a B (5) a B = 2 ft /s 2 a A = —1 ft/s 2 + J2/y = may, 400 - 27 = Thus, T = 206 lb Ans: T = 206 lb 284 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 41 . Block A has a mass m A and is attached to a spring having a stiffness k and unstretched length l 0 . If another block B, having a mass m B , is pressed against A so that the spring deforms a distance d, determine the distance both blocks slide on the smooth surface before they begin to separate. What is their velocity at this instant? SOLUTION Block A: -L Si 7 ,. = ma x ; — k(x — d) — N = m A a A Block B: -L 'ZF X = ma x ; N = m B a B Since a A = a B = a, — k(x — d) — m B a = m A a k(d — x) km B (d — x) a = 7-r N = - -- (m A + m B ) (m A + m B ) N = 0 when d — x = 0, or x = d v dv = a dx f v f d k(d - x) / v dv = / - - dx Jo Jo ( m A + m B ) 1 2 k 2 V (m A + m B ) kd 2 (d)x - -x 2 (m A + m B ) 1 kd o 2{m A + m B ) Ans. Ans. Ans: x = d v kd 2 m A + m B 285 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 42 . Block A has a mass m A and is attached to a spring having a stiffness k and unstretched length l 0 . If another block B, having a mass m B , is pressed against A so that the spring deforms a distance d , show that for separation to occur it is necessary that d>2/jLi c g(m A + m B )/k, where is the coefficient of kinetic friction between the blocks and the ground. Also, what is the distance the blocks slide on the surface before they separate? SOLUTION Block A: -L E F x = ma x ; — k(x — d) — N — /JL k m A g = m A a A Block B: ^F x = ma x ; N - n k m B g = m B a B Since a A = a B = a , k(d — x) — fjik g(m A + m B ) kid — x) (m A + m B ) (m A + m B ) ^ k 8 km R (d — x) N =-—-- (m A + m B ) aJ 7 N = 0, then x = d for separation. At the moment of separation: Ans. v dv = a dx / v dv = Jo Jo 1 , k 2 V k(d - x) {m A + m B ) d-kg dx (m A + m B ) 1 (d)x - -x 2 - fx k g x kd~ — 2fi k g(m A + m B )d {m A + m B ) Require v > 0, so that kd 2 — 2 Hk g(m A + m B )d > 0 Thus, kd > 2fx k g(m A + m B ) 2 gt g d > ——— (m A + m B ) Q.E.D. Ans: x = d for separation. 286 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 287 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-44. If the motor draws in the cable with an acceleration of 3 m/s 2 , determine the reactions at the supports A and B.The beam has a uniform mass of 30 kg/m, and the crate has a mass of 200 kg. Neglect the mass of the motor and pulleys. SOLUTION Sc + (S c - S p ) 2v c = v p 2a c Up 2 a c = 3 m/s 2 a c = 1.5 m/s 2 + T 2 F y = ma y C +ZM a = 0; TtSf/, = 0; ± YF X = 0; 2 T - 1962 = 200(1.5) T = 1,131 N B y (6) - (1765.8 + 1,131)3 - (1,131)(2.5) = 0 B y = 1,919.65 N = 1.92 kN Ay - 1765.8 - (2)(1,131) + 1919.65 = 0 A y = 2108.15 N = 2.11 kN A x = 0 0.5 m Ans. Ans. Ans. * so(fcK5.n ')=■/?( r-Fti 3 >v»v b / 3 A/ 15/aJ 288 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-45. If the force exerted on cable AB by the motor is F = (100t 3 / 2 ) N, where t is in seconds, determine the 50-kg crate’s velocity when f = 5 s. The coefficients of static and kinetic friction between the crate and the ground are = 0.4 and fj. k = 0.3, respectively. Initially the crate is at rest. SOLUTION Free-Body Diagram: The frictional force F ; is required to act to the left to oppose the motion of the crate which is to the right. Equations of Motion: Here, a v = O.Thus, + T 'ZF y = ma y \ N - 50(9.81) = 50(0) N = 490.5 N Realizing that F f = /jL k N = 0.3(490.5) = 147.15 N, + T 2F X = ma x \ 100f 3 / 2 - 147.15 = 50a a = (2t 3 / 2 - 2.943) m/s Equilibrium: For the crate to move, force F must overcome the static friction of Ff = fji s N = 0.4(490.5) = 196.2 N. Thus, the time required to cause the crate to be on the verge of moving can be obtained from. -±> ~ZF X = 0; 100t 3 / 2 - 196.2 = 0 somo* J N t = 1.567 s Kinematics: Using the result of a and integrating the kinematic equation dv — a dt with the initial condition v = 0 at t = 1.567 as the lower integration limit. ( ) J dv = J adt [ dv = [ (2t 3 / 2 - 2.943W Jo .71.567 s t v = (o.8f 5 / 2 - 2.943f) 1.567 s v = (o.8t 5 / 2 - 2.943 1 + 2.152) m/s When t — 5 s, 0.8(5) 5 / 2 - 2.943(5) + 2.152 = 32.16 ft/s = 32.2 ft/s Ans. Ans: v = 32.2 ft/s 289 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-46. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth. SOLUTION Require Cl a = (I ff = Ci Block A: + t2f 7 > , = 0; N cos6 — mg = 0 ^~Y.F X = ma x ; N sin 9 = ma a = g tan# Block B: 'S,F X = ma x ; P — N sin 0 = ma P — mg tan 9 = mg tan 6 P = 2 mg tan 9 Ans. Ans: P = 2 mg tan 9 290 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-47. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip on B. The coefficient of static friction between A and B is jx s . Neglect any friction between B and C. SOLUTION Require Cl a — CL b — Cl Block A: + UF y = 0 ; P- if F x = ma x ; N cos 8 — /jl s N sin 8 — mg = 0 N sin 8 + /jl s N cost? = ma mg N = cos 6 — /ji s sin d a = g /sin 6 + /jl s cos 6 V cos 6 — n- s sin 6 Block B: = ma- P — /jl s N cos d — N sin 8 = ma ( sin 6 + cos d P — mg\ -;- V cos 6 — /X, sin 0 ( sin 6 + ix. cos 8 P = 2 mg -—- \cos 8 — /x s sin 8 / sin 8 + /jl s cos 8 mg\ -;- \ cos 8 - /x s sin 8 Ans: P = 2m, sin 8 + /x* cos 8 cos 8 - /x s sin 8 291 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-48. The smooth block B of negligible size has a mass m and rests on the horizontal plane. If the board AC pushes on the block at an angle 6 with a constant acceleration a 0 , determine the velocity of the block along the board and the distance s the block moves along the board as a function of time t. The block starts from rest when s = 0, t = 0. SOLUTION / 1 +'ZF X = m a x ; 0 = m a B sin 4> a B ~ a AC + a B/AC a B = a 0 + a B/AC F /"+ Thus, a B sin <f> = — a 0 sin 0 + a B uc 0 = m(—a 0 sin 0 + ab/tc) a B/AC ~ a o s * n 0 r v B/AC pt / dv B / AC — / a 0 sin 0 dt Jo Jo v b /ac = 0o sin 6 t s b/ac ~ s ~ a o sin 9 t dt Jo I , n s = — a 0 sin 6 t Ans. Ans. Ans: v B /AC = «o sin e t 1 • 9 j = — a n sin d t 2 292 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-49. If a horizontal force P = 12 lb is applied to block A determine the acceleration of block B. Neglect friction. SOLUTION Block A: -±> 'ZF X = ma x ; 12 - N B sin 15° = Block B: +12 Fy = ma y ; IVg cos 15° — 15 = s B = Uttan 15° a B = ci a tan 15° Solving Eqs. (l)-(3) a A = 28.3 ft/s 2 N b = 19.2 lb a B = 7.59 ft/s 2 ( 1 ) ( 2 ) (3) Ans. '.ink * A, Nt t a a Ans: a B = 7.59 ft/s 2 293 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-50. A freight elevator, including its load, has a mass of 1 Mg. It is prevented from rotating due to the track and wheels mounted along its sides. If the motor M develops a constant tension T = 4 kN in its attached cable, determine the velocity of the elevator when it has moved upward 6 m starting from rest. Neglect the mass of the pulleys and cables. SOLUTION Equation of Motion. Referring to the FBD of the freight elevator shown in Fig. a, + t £F V = ma y ; 3(4000) - 1000(9.81) = 1000a a = 2.19 m/s 2 ! Kinematics. Using the result of a, (+t) v 2 = vl + 2 as; v 2 = 0 2 + 2(2.19)(6) v = 5.126 m/s = 5.13 m/s Ans. M, Ans: v = 5.13 m/s JtiOON 294 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-51. The block A has a mass m A and rests on the pan B , which has a mass m B . Both are supported by a spring having a stiffness k that is attached to the bottom of the pan and to the ground. Determine the distance d the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched. SOLUTION For Equilibrium +12F V = ma y ; F s = (m A + m B )g _ F s _ (m A + m B )g yeq k k Block: + t21 7 j, = ma y \ —m A g + N = m A a Block and pan + t ZF y = ma y ; ~(m A + m B )g + k{y eq + y) = (m A + m B )a Thus, ~{m A + m B )g + k Require y = d, N = 0 m A + m B \ k ) g + y = (m A + m B ) -m A g + N m A kd = ~(m A + m B )g Since d is downward, d = (m A + m B )g Ans. X Ans: d = (m A + m B )g k 295 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r — 5 m from the platform’s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is /jl = 0.2. SOLUTION Equation of Motion: Since the girl is on the verge of slipping, Ff = q, S N = 0.2 N. Applying Eq. 13-8, we have ZF b = 0; N - 15(9.81) = 0 N = 147.15 N = ma n ; 0.2(147.15) = 15 v = 3.13 m/s Ans. z b Ans: v = 3.13 m/s 296 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-53. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block is given a speed of v = 10 m/s, determine the radius r of the circular path along which it travels. SOLUTION Free-Body Diagram :The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A, i.e., T = 15(9.81) N = 147.15 N. Here, a n must be directed towards the center of the circular path (positive n axis). Equations of Motion: Realizing that a„ P 10 2 and referring to Fig. (a), 'ZF n = ma n ; 147.15 = 2 r = 1.36 m Ans. b Ans: r = 1.36 m 297 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-54. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block travels along a circular path of radius r = 1.5 m, determine the speed of the block. SOLUTION Free-Body Diagram :The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A , i.e., T = 15(9.81) N = 147.15 N. Here, a (1 must be directed towards the center of the circular path (positive n axis). 2 2 V V Equations of Motion: Realizing that a n = — = — and referring to Fig. (a), = ma„ 147.15 = 2' v = 10.5 m/s Ans. Ans: v = 10.5 m/s 298 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-55. Determine the maximum constant speed at which the pilot can travel around the vertical curve having a radius of curvature p = 800 m, so that he experiences a maximum acceleration a n = 8g = 78.5 m/s 2 . If he has a mass of 70 kg, determine the normal force he exerts on the seat of the airplane when the plane is traveling at this speed and is at its lowest point. SOLUTION 78.5 = 800 v = 251 m/s + T = ma n - N - 70(9.81) = 70(78.5) N = 6.18 kN Ans. Ans. Ans: N = 6.18 kN 299 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 56 . Cartons having a mass of 5 kg are required to move along the assembly line at a constant speed of 8 m/s. Determine the smallest radius of curvature, p, for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are p, s = 0.7 and p, k = 0.5, respectively. SOLUTION + T ZF b = ma b \ N-W = 0 N = W F x = 0.7W , W 8 2 ^2 F n = ma n ; OJW = — (-) 9.81 p p = 9.32 m Ans. Ans: p = 9.32 m 300 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 301 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 58 . The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is = 0.2. If the spool is located 0.25 m from A, determine the minimum constant speed the spool can have so that it does not slip down the rod. SOLUTION p = 0.25^ = 0.2 m = m fl " ; ^'(5) “ °- 2 ^( 5 ) = 2 (ol) + T = m a„- + 0.2A&Q - 2(9.81) = 0 N s = 21.3 N v = 0.969 m/s z Ans. b I Ans: v = 0.969 m/s 302 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 59 . The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is fi s = 0.2. If the spool is located 0.25 m from A, determine the maximum constant speed the spool can have so that it does not slip up the rod. SOLUTION p = 0.25(|) = 0.2 m ^2 F n = ma n ; N s ( |) + 0.2A S (|) = 2(^) + UF b = m a b ; N^) - 0.21V S (|) - 2(9.81) = 0 N s = 28.85 N v = 1.48 m/s z Ans. Ans: v = 1.48 m/s 303 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 60 . At the instant 0 = 60°, the boy’s center of mass G has a downward speed v G = 15 ft/s. Determine the rate of increase in his speed and the tension in each of the two supporting cords of the swing at this instant. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords. SOLUTION +\'ZF t = ma t ; 60 cos 60° = ^~^ a t a t ~ 16.1 ft/s 2 /+^F n = ma n ; 27-60 sin 60° = — f—) T = 46.9 lb " 32.2\ 10 / Ans. Ans. Ans: a, = 16.1 ft/s 2 T = 46.9 lb 304 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 61 . At the instant 0 = 60°, the boy’s center of mass G is momentarily at rest. Determine his speed and the tension in each of the two supporting cords of the swing when 6 = 90°. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords. SOLUTION 60 +\2, = ma t \ 60 cost! = a, a, = 32.2 cos(9 /+Y,F n = ma ■ 2T — 60 sin 9 = ~^—(— " " 32.2 V 10 v dv = a ds however ds = IGdd r 90° v dv — 322 cos 6 d6 J 60° v = 9.289 ft/s From Eq. (1) 2 T - 60 sin 90° = 60 / 9.289 2 32.2 V 10 T = 38.0 lb ( 1 ) Ans. Ans. Ans: v = 9.29 ft/s T = 38.0 lb 305 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 306 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 63 . The pendulum bob B has a weight of 5 lb and is released from rest in the position shown, 6 = 0°. Determine the tension in string BC just after the bob is released, 8 = 0°, and also at the instant the bob reaches point D, 8 = 45°. Taker = 3 ft. SOLUTION Equation of Motion: Since the bob is just being release, v = FBD(a), we have 0. Applying Eq. 13-8 to = ma n ; T = (r 32.2 V~3 = 0 Ans. Applying Eq. 13-8 to FBD(b), we have 1,F t = ma t ; 5 cos 8 = ^ ^ a ‘ a < ~ ^2-2 cos 6 2 F n = ma n ; T — 5 sin 8 = 32.2 V 3 [ 1 ] Kinematics'. The speed of the bob at the instant when 8 = 45° can be determined using vdv = a t ds. However, ds = 3 d8, then vdv = 3 a t dO. / vdv = 3(32.2) / cos OdO Jo Jo v 2 = 136.61 ft 2 /s 2 Substitute 6 = 45° and v 2 = 136.61 ft 2 /s 2 into Eq.  yields T - 5 sin 45° = 32.2 T = 10.6 lb 136.61 3 Ans. n - y t Cci) SI (b ) Ans: T = 0 T = 10.6 lb 307 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 64 . The pendulum bob B has a mass m and is released from rest when 8 = 0°. Determine the tension in string BC immediately afterwards, and also at the instant the bob reaches the arbitrary position 8 . SOLUTION Equation of Motion: Since the bob is just being release, v = 0. Applying Eq. 13-8 to FBD(a), we have = ma n ; T = m = 0 Ans. Applying Eq. 13-8 to FBD(b), we have = ma t ; mg cos 8 = ma, a t = g cos 8 = ma n ; T — mg sin 8 = m [11 Kinematics: The speed of the bob at the arbitrary position 8 can be detemined using vdv = a, ds. However, ds = rdO. then vdv = a, rd8. nV / vdv — gr cos 8 dd Jo Jo v 2 = 2 gr sin 8 Substitute v 2 = 2 gr sin 6 into Eq.  yields T — mg sin 8 = m T = 3mg sin 8 2 gr sin 8 r Ans. G> t (CL) ■>- T (b ) Ans: T = 0 T = 3mg sin 8 308 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 65 . Determine the constant speed of the passengers on the amusement-park ride if it is observed that the supporting cables are directed at 0 = 30° from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what are the components of force in the n, t , and b directions which the chair exerts on a 50-kg passenger during the motion? SOLUTION Si 7 ,, = m a n ; + T5 lF b = 0; T sin 30° = 80( 4 + 6 sin 30° T cos 30° - 80(9.81) = 0 T = 906.2 N E F n = m a n ; E F t = m a t ; EFj = m cif ,; v = 6.30 m/s (6.30) 2 Fn = 50( — ) = 283 N F t = 0 F b - 490.5 = 0 F b = 490 N Ans. Ans. Ans. Ans. ?o(9.?1) +E tSui3o° Fa Ans: v = 6.30 m/s F n = 283 N F, = 0 F b = 490 N 309 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 66 . A motorcyclist in a circus rides his motorcycle within the confines of the hollow sphere. If the coefficient of static friction between the wheels of the motorcycle and the sphere is p, s = 0.4, determine the minimum speed at which he must travel if he is to ride along the wall when 6 = 90°. The mass of the motorcycle and rider is 250 kg, and the radius of curvature to the center of gravity is p = 20 ft. Neglect the size of the motorcycle for the calculation. SOLUTION + 2 F„ = ma n ; + ! ~ZF b = ma b ; Solving, N = 250 20 0.4 N - 250(9.81) = 0 v = 22.1 m/s Ans. «-4f_ 250(9- St)M Ans: v = 22.1 m/s 310 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 67 . The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle is traveling at a constant speed of 80 km/h along a circular curved road of radius 100 m, determine the tilt angle 6 of the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver. SOLUTION Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a). Here, a n must be directed towards the center of the circular path (positive n axis). kmV1000 mV lh Equations of Motion: The speed of the passenger is v — ^ 80 ^ j ^ ^ ^ / V 3600 = 22.22 m/s. Thus, the normal component of the passenger’s acceleration is given by u 2 22.22 z a n = ~ = ^ = 4.938 m/s 2 . By referring to Fig. (a), +12F b = 0; N cos 8 - m( 9.81) = 0 N = = ma„ N cos 6 — m( 9.81) = 0 9.81m cos 6 sin 0 = m(4.938) 9.81m cos 6 6 = 26.7° Ans. b Ans: 0 = 26.7° 311 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 68 . The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m/s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car. SOLUTION dy d 2 y Geometry: Here, — = — 0.00625.x and —, = —0.00625. The slope angle 9 at point dx dx A is given by tan 9 = dx = -0.00625(80) 9 = -26.57° and the radius of curvature at point A is [1 + (dy/dxff/ 2 [1 + (—0.00625.x) 2 ] 3 / 2 | d 2 y/dx 2 |—0.00625| = 223.61 m Equations of Motion: Here, a t = 0. Applying Eq. 13-8 with 9 = 26.57° and p = 223.61 m, we have 1.F, = ma t ; 800(9.81) sin 26.57° - F f = 800(0) F f = 3509.73 N = 3.51 kN = ma n ; 800(9.81) cos 26.57° - N = 800' N = 6729.67 N = 6.73 kN 223.61 Ans. Ans. 600(9-80/J n Ans: F f = 3.51 kN N = 6.73 kN 312 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 69 . The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A , it is traveling at 9 m/s and increasing its speed at 3 m/s 2 . Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car. SOLUTION d y d 2 y Geometry: Here, — = —0.00625x and —, = —0.00625. The slope angle 9 at point dx dx A is given by tan 9 = dy dx = -0.00625(80) 9 = -26.57° and the radius of curvature at point A is [l + (dy/dxff 11 [l + (—0.00625x) 2 ] 3 / 2 = 223.61 m H \d 2 y/dx 2 \ |-0.00625| Equation of Motion: Applying Eq. 13-8 with 9 = 26.57° and p = 223.61 m, we have = ma t \ 800(9.81) sin 26.57° - F f = 800(3) F f = 1109.73 N = 1.11 kN = m«„; 800(9.81) cos 26.57° — N = 800 N = 6729.67 N = 6.73 kN 223.61 Ans. Ans. 8oo(9&0 a / n Ans: F f = 1.11 kN N = 6.73 kN 313 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 70 . The package has a weight of 5 lb and slides down the chute. When it reaches the curved portion AB, it is traveling at 8 ft/s (6 = 0°). If the chute is smooth, determine the speed of the package when it reaches the intermediate point C (6 = 30°) and when it reaches the horizontal plane (6 = 45°). Also, find the normal force on the package at C. SOLUTION +</ 2T, = ma t ; 5 cos$ =

a t = 32.2 cos

5 v 2,

+\SF„ = ma n ; N — 5 sin 0 = (—)

v dv = a t ds

pV p(j)

v dv = / 32.2 cos 0 (20 drf>)
Jg J 45°

^v 2 - y(8) 2 = 644 (sin 0 - sin 45°)

At 4> = 45° + 30° = 75°,

v c = 19.933 ft/s = 19.9 ft/s
N c = 7.91 lb

At 0 = 45° + 45° = 90°

v B — 21.0 ft/s

Ans.

Ans.

Ans.

Ans:

Vq = 19.9 ft/s
N c = 7.91 lb
v B = 21.0 ft/s

314

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13 - 71 .

The 150-lb man lies against the cushion for which the
coefficient of static friction is /jl s = 0.5. Determine the
resultant normal and frictional forces the cushion exerts on
him if, due to rotation about the z axis, he has a constant
speed-u = 20 ft/s. Neglect the size of the man.Take 8 = 60°.

SOLUTION

= m ( a n)y ; N - 150 cos 60° = sin 60 °

N = 277 lb

+ i/ T'iy = m(a n ) x ; — F + 150 sin 60° = cos

v n,x 32.2V 8 )

F = 13.4 lb

Note: No slipping occurs
Since n s N = 138.4 lb > 13.4 lb

Ans:

N = 277 lb
F = 13.4 lb

315

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* 13 - 72 .

The 150-lb man lies against the cushion for which the
coefficient of static friction is /jl s = 0.5. If he rotates about
the z axis with a constant speed v = 30 ft/s, determine the
smallest angle d of the cushion at which he will begin to
slip off.

SOLUTION

ST,, = ma n ; 0.51V cos 0 + IV sin 6 =

" n 32.2

+1 ST ft = 0; - 150 + IVcosfl - 0.5 N sin0 = 0

M =-^°—

cos 0 — 0.5 sin 6

(0.5 cos 0 + sin 0)150 150 / (30) 2 \

(cosO — 0.5 sin 6 ) 32.2 \ 8 /

0.5 cost! + sin 6 = 3.49378 cos 6 - 1.74689 sin 6

6 = 47.5°

Ans.

z

150 lb

Ans:

0 = 47.5°

316

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317

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13 - 74 .

Determine the maximum constant speed at which the 2-Mg
car can travel over the crest of the hill at A without leaving
the surface of the road. Neglect the size of the car in the
calculation.

y

SOLUTION

Geometry. The radius of curvature of the road at A must be determined first. Here

dy ( 2x \

— = 20 -= —0.004.x

dx V 10000 J

jX)OO(9-80fil

d 2 y

dx 2

= -0.004

At point A,x = O.Thus,

= O.Then

x=0

3/2

dry

dx 1

(1 + 0 2 P

0.004

= 250 m

Equation of Motion. Since the car is required to be on the verge to leave the road
surface, A = 0.

Ans.

= ma n - 2000(9.81) = 2000 (

v = 49.52 m/s = 49.5 m/s

Ans:

v = 49.5 m/s

318

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13 - 75 .

The box has a mass m and slides down the smooth chute
having the shape of a parabola. If it has an initial velocity of
v 0 at the origin, determine its velocity as a function of x.
Also, what is the normal force on the box, and the
tangential acceleration as a function of x?

SOLUTION

x = — X
2

<h_ _

dx x
dry

, = -1
dx

dy^ 2
1 + [ Tx

db

dx 2

[1 + x 2 ]i

l-ll

= (1 + x 2 )i

+i/2 F n = ma n ; mg

+\YF t = ma t ; mg

VT

+ x-

— N = m

(1 + x 2 )*

Vl + x 2

v dv = a t ds = g

Vl + x 2

ds

ds =

1 +

a, = g

Vl + X 2

dx = 1 + x 2 2 dx

V = Vvo + gx 2

From Eq. (1):

N =

m

Vl + x 2

g

(yg + g* 2 r

(1 + X 2 ) .

( 1 )

Ans.

Ans.

Ans.

Ans:

a t = g

Vl + x 2

V

N =

= vVPVV

Vl + x 2 -

vl + gx 2

1 + x 2

319

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* 13 - 76 .

Prove that if the block is released from rest at point f? of a
smooth path of arbitrary shape, the speed it attains when it
reaches point A is equal to the speed i t atta ins when it falls
freely through a distance h\ i.e., v = V2 ~gh.

SOLUTION

+\EF ( = may, mg sin 0 = ma t
v dv = a t ds = g sin 8 ds

v dv = / g dy

ir

2

gh

V2 gh

a, - g sin 6

However dy = ds sin 8

Q.E.D.

Ans:

v = V2 gh

320

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12 - 77 .

The cylindrical plug has a weight of 2 lb and it is free to
move within the confines of the smooth pipe. The spring
has a stiffness k = 14 lb/ft and when no motion occurs
the distance d = 0.5 ft. Determine the force of the spring
on the plug when the plug is at rest with respect to the pipe.
The plug is traveling with a constant speed of 15 ft/s, which
is caused by the rotation of the pipe about the vertical axis.

SOLUTION

<L XF n = rna n \

F s = ks ;

Thus,

2 r (15) 2 -

32.2 [ 3 - d _

F s = 14(0.5 - d)

14(0.5 - d)
(0.5 - d ){3

2 r (is) 2 ■

32.21.3 - d_
d) = 0.9982

1.5 - 3.5 d + d 2 = 0.9982

d 2 - 3.5 d + 0.5018 = 0

Choosing the root < 0.5 ft

d = 0.1498 ft

F s = 14(0.5 - 0.1498) = 4.901b

3 ft-

I-- d

Ans.

Ans:

F s = 4.90 lb

321

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13 - 78 .

When crossing an intersection, a motorcyclist encounters
the slight bump or crown caused by the intersecting road.
If the crest of the bump has a radius of curvature p = 50 ft,
determine the maximum constant speed at which he can
travel without leaving the surface of the road. Neglect the
size of the motorcycle and rider in the calculation. The rider
and his motorcycle have a total weight of 450 lb.

SOLUTION

, 450 f v 2 \

+i ^F n = ma n - 450 - 0 = —^-]

v = 40.1 ft/s

On’

Nr'O

Ans:

v = 40.1 ft/s

322

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13 - 79 .

The airplane, traveling at a constant speed of 50 m/s, is
executing a horizontal turn. If the plane is banked at
6 — 15°, when the pilot experiences only a normal force on
the seat of the plane, determine the radius of curvature p of
the turn. Also, what is the normal force of the seat on the
pilot if he has a mass of 70 kg.

SOLUTION

+T 2 F b = ma b ; N P sin 15° - 70(9.81) = 0

N P = 2.65 kN

/50 2 \

'^ J F n = ma n ; N P cos 15° = 70(^- j

p = 68.3 m

Ans.

70(9.81) N

4

Ans.

Ans:

N P = 2.65 kN
p = 68.3 m

323

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* 13 - 80 .

The 2-kg pendulum bob moves in the vertical plane with
a velocity of 8 m/s when 8 = 0°. Determine the initial
tension in the cord and also at the instant the bob reaches
8 = 30°. Neglect the size of the bob.

SOLUTION

Equations of Motion. Referring to the FBD of the bob at position 0 = 0°, Fig. a,

XF n = ma„; T = 2[ y ] = 64.0 N
For the bob at an arbitrary position 8, the FBD is shown in Fig. b.
= ma t ; —2(9.81) cos 8 = 2a t
a t = —9.81 cos 8

„2s

Ans.

"%F n = ma n ; T + 2(9.81) sin 8 = 2

T = v 2 - 19.62 sin 8 (1)

Kinematics. The velocity of the bob at the position 8 = 30° can be determined by
integrating vdv = a t ds. However, ds = rd8 = 2 dd.

Then,

n

r> 30 °

' 8 m/s

vdv = / —9.81 cos 8 (zd8)

= -19.62 sin 8

8 m/s

30 °

0 °

V 1 o

— - — = — 19.62(sin 30° - 0)

v 2 = 44.38 m 2 /s 2

Substitute this result and 8 = 30° into Eq. (1),

T = 44.38 - 19.62 sin 30°

= 34.57 N = 34.6 N Ans.

t

Ot)

Ans:

T = 64.0 N
T = 34.6 N

324

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13 - 81 .

The 2-kg pendulum bob moves in the vertical plane
with a velocity of 6 m/s when 0 = 0°. Determine the angle 0
where the tension in the cord becomes zero.

SOLUTION

Equation of Motion. The FBD of the bob at an arbitrary position 0 is shown in
Fig. a. Here, it is required that T = 0.

%F t = ma t \ —2(9.81) cos 0 = 2a,
a, = —9.81 cos 0

%F n = ma n \

2(9.81) sin 0 = 2

v 2 = 19.62 sin 0 (1)

Kinematics. The velocity of the bob at an arbitrary position 0 can be determined by
integrating vdv = a t ds. However, ds = rd0 = 2d0.

Then

9.81 cos 0(2d0)

= -19.62 sin 0

6 m/s

v 2 = 36 - 39.24 sin 0
Equating Eqs. (1) and (2)

19.62 sin 0 = 36
58.86 sin 0 = 36

39.24 sin 0

0 = 37.71°

37.7°

( 2 )

Ans.

t

(&)

Ans:

0 = 37.7

o

325

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13 - 82 .

The 8-kg sack slides down the smooth ramp. If it has a speed
of 1.5 m/s when y = 0.2 m, determine the normal reaction
the ramp exerts on the sack and the rate of increase in the
speed of the sack at this instant.

SOLUTION

y = 0.2 x = 0
y = 0.2e*
dy

dx

d 2 y

dx 2

= 0.2e x

= 0.2

= 0.2e x

= 0.2

r=0

1 + 1 T

ax

d 2 y

dx 1

[l + (0.2) z ]2
|0-2|

= 5.303

d = tan^ 1 (0.2) = 11.31°

+\2F„ = ma n ; N B - 8(9.81) cos 11.31° = 8

( 1 - 5) 2

5.303

+/"ZF t = ma t ;

N B = 80.4 N
8(9.81) sin 11.31° = 8 a,
a t = 1.92 m/s 2

y

Ans.

Ans.

Ans:

N B = 80.4 N
a, = 1.92 m/s 2

326

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13 - 83 .

The ball has a mass m and is attached to the cord of length /.
The cord is tied at the top to a swivel and the ball is given a
velocity v 0 . Show that the angle 9 which the cord makes with
the vertical as the ball travels around the circular path
must satisfy the equation tan 9 sin 9 = vl/gl. Neglect air
resistance and the size of the ball.

SOLUTION

, / ^
-L ~ZF„ = ma„\ T sin 9 = ml

+ t 2F b = 0; T cos 9 — mg = 0

mv §

Since r = / sin 6 T =

I sin 2 9
f mv / cos 9

1 , M ■ 2 n I - m S

V / Asin z e/

tan 9 sin 9 =

4

gt

Q.E.D.

b

Ans:

Vo

tan 9 sin 9 = —
gl

327

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* 13 - 84 .

The 2-lb block is released from rest at A and slides down
along the smooth cylindrical surface. If the attached spring
has a stiffness k = 2 lb/ft, determine its unstretched length
so that it does not allow the block to leave the surface until
6 = 60°.

A

SOLUTION

+ /^F n = ma„\ F s + 2 cos 0 = Tfyfy

+ \2.F r = ma t ; 2 sin d = yy,

a, = 32.2 sin 0

v dv = a, ds\ v dv = / 32.2(sin 8)2d6

v 2 = 64.4(—cos 0 + 1)

When 0 = 60°
v 2 = 64.4

From Eq. (1)

F s + 2 cos 60°

F s = 1 lb

F s = ks; 1 = 2 5 ; s = 0.5 ft

l 0 = 1 - s = 2 — 0.5 = 1.5 ft

( 1 )

Ans.

Ans:

/<)-

1.5 ft

328

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13 - 85 .

The spring-held follower AB has a weight of 0.75 lb and
moves back and forth as its end rolls on the contoured
surface of the cam, where r = 0.2 ft and z = (0.1 sin d) ft. If
the cam is rotating at a constant rate of 6 rad/s, determine
the force at the end A of the follower when d = 90°. In this
position the spring is compressed 0.4 ft. Neglect friction at
the bearing C.

SOLUTION

z = 0.1 sin 2d
z = 0.2 cos 266

z = —0.4 sin 2dd 2 + 0.2 cos 266
6 = 0

z = -14.4 sin 26

~^F Z = ma z ; F A — 12(z + 0.3) = mi
F a - 12(0.1 sin 2d + 0.3) = |^(“ 14 - 4 sin 2d)

For d = 45°,

F a - 12(0.4) = ~~|( 14.4-)

F a = 4.46 lb Ans.

Ans:

F a = 4.46 lb

329

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13 - 86 .

Determine the magnitude of the resultant force acting on a
5-kg particle at the instant t = 2 s, if the particle is moving
along a horizontal path defined by the equations
r = (2t + 10) m and 6 = (1.5t 2 - 6 1) rad, where t is in
seconds.

SOLUTION

r = It + 10| r=2s = 14

r = 2
r = 0

e = 1.5 1 2 - 6 1
6 = 3t - 6|( =2j = 0
6 = 3

a r = r - r6 2 = 0 - 0 = 0
ag = r'e + 2 rd = 14(3) + 0 = 42

Hence,

= ma r ; F r = 5(0) = 0
= inag, Fg = 5(42) = 210 N

F = V(F r ) 2 + (Fg ) 2 = 210 N

Ans:

F = 210 N

330

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13 - 87 .

The path of motion of a 5-lb particle in the horizontal plane
is described in terms of polar coordinates as r = (2 1 + 1) ft
and 0 = (0.5t 2 — t) rad, where t is in seconds. Determine
the magnitude of the unbalanced force acting on the particle
when t = 2 s.

SOLUTION

r = 2t + l| r = 2 s = 5 ft r — 2 ft/s r = 0

6 = 0.5f 2 — t\ t = 2 s — 0 rad 9 = t — 1| (=2 , = 1 rad/s 0 = 1 rad/s 2
a r = r - rd 2 = 0 - 5(1) 2 = -5 ft/s 2
a e = rb + 2rd = 5(1) + 2(2)(1) = 9 ft/s 2

= ma r \ F r = ^ (-5) = -0.7764 lb
1,F e = ma e \ F

f = VfJ =

Ans:

F = 1.60 lb

•-3l2 (9) - L3981b

V(-0.7764) 2 + (1.398) 2 = 1.60 lb Ans.

331

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* 13 - 88 .

Rod OA rotates counterclockwise with a constant angular
velocity of 6 = 5 rad/s. The double collar B is pin-
connected together such that one collar slides over the
rotating rod and the other slides over the horizontal curved
rod, of which the shape is described by the equation
r = 1.5(2 - cos 8) ft. If both collars weigh 0.75 lb,
determine the normal force which the curved rod exerts on
one collar at the instant 8 = 120°. Neglect friction.

SOLUTION

Kinematic: Here, 6 = 5 rad/s and 6 = 0. Taking the required time derivatives at
8 = 120°, we have

r = 1.5(2 - cos 0)| fl =i2o° = 3.75 ft

r = 1.5 sin 00| e =i2o° = 6.495 ft/s

r = 1.5(sin + cos 88 2 )|e=i 20 ° = -18.75 ft/s 2

Applying Eqs. 12-29, we have

a r = r - rd 2 = -18.75 - 3.75(5 2 ) = -112.5 ft/s 2

a e = rd + 2 r 'd = 3.75(0) + 2(6.495)(5) = 64.952 ft/s 2

Equation of Motion: The angle ip must be obtained first.
1.5(2 — cos 8)

tan ip =

dr/dd

1.5 sin 8

= 2.8867

>P = 70.89°

Applying Eq. 13-9, we have
2 F r = ma r ; -N cos 19.11° = ^ (-112.5)

N = 2.773 lb = 2.77 lb

= ma » : F oa + 2.773 sin 19.11° = (64.952)

F oa = 0.605 lb

Ans.

Ans:

N = 2.77 lb

332

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333

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13 - 90 .

The 40-kg boy is sliding down the smooth spiral slide such
thatz = —2 m/s and his speed is 2 m/s. Determine the r,d,z
components of force the slide exerts on him at this instant.
Neglect the size of the boy.

SOLUTION

r = 1.5 m
r = 0
r = 0

v e = 2 cos 11.98° = 1.9564 m/s
v z = -2 sin 11.98° = -0.41517 m/s
v e = rd\ 1.9564 = 1.5 0

Sly = ma r \ -F r = 40(0 - 1.5(l.3043) 2 )

F r = 102 N

Sly = ma e ; N b sin 11.98° = 40 (a e )

XF Z = ma z ; -N b cos 11.98° + 40(9.81) = 40a-

Require tan 11.98° = —, a e = 4.7123a z

a e

Thus,

a z = 0.423 m/s 2
a e = 1.99 m/s 2
N b = 383.85 N

N z = 383.85 cos 11.98° = 375 N
N e = 383.85 sin 11.98° = 79.7 N

Ans.

2-TrO>5)

Ans.

Ans.

Ans:

F r = 102 N
F z = 375 N
F e = 79.7 N

334

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13 - 91 .

Using a forked rod, a 0.5-kg smooth peg P is forced to move
along the vertical slotted path r = (0.5 6) m, where 8 is in
radians. If the angular position of the arm is 8 = (ft 2 ) rad,
where t is in seconds, determine the force of the rod on the
peg and the normal force of the slot on the peg at the instant
t = 2 s. The peg is in contact with only one edge of the rod
and slot at any instant.

SOLUTION

dr

Equation of Motion. Here, r = 0.5 8. Then — = 0.5. The angle ip between the

dO

extended radial line and the tangent can be determined from

tan ip =

r _ 0.5 8
dr/dd 0.5

= e

At the instant t = 25, 8 = T-(2 2 ) = -^-rad
8 2

tan ip = y ip = 57.52°

sense of 8 (counter clockwise) to the tangent. Then the FBD of the peg shown in
Fig. a can be drawn.

%F r = ma r ; N sin 57.52° - 0.5(9.81) = 0.5 a r (1)

%F e = ma e ; F - N cos 57.52° = 0.5a fl (2)

Kinematics. Using the chain rule, the first and second derivatives of r and 8 with
respect to t are

r . « = o--e

osmod

7T

r = —t

77

8 =

When t = 2 s,

fl - f (2=) - f „ d

r= 8 (2) = 4 m/S

8 = j(2) = y rad/s

Thus,

77 / 2
= -m/s

7T , z <■>

4

/ \ 2

77 77 I 77

a r = r - rd 2 = = -1.5452 m/s 2

a e = rd + 2rd = = 3 - 0843 m /s 2

Substitute these results in Eqs. (1) and (2)

N = 4.8987 N = 4.90 N
F = 4.173 N = 4.17 N

Ans.

Ans.

Ans:

N = 4.90 N
F = 4.17 N

335

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* 13 - 92 .

The arm is rotating at a rate of = 4rad/s when
0 = 3 rad/s 2 and 6 = 180°. Determine the force it must
exert on the 0.5-kg smooth cylinder if it is confined to move
along the slotted path. Motion occurs in the horizontal plane.

SOLUTION

Equation of Motion. Here, r = —. Then

0 dd

extended radial line and the tangent can be determined from

= — 3^. The angle i fj between the

tan i/f

r

dr/dd

i/e

-2/e 2 -

At 0 = 180° = t r rad,

tan if) = —77 ip = —72.34°

negative sense of 6 (clockwise) to the tangent. Then, the FBD of the peg shown in
Fig. a can be drawn.

XF r = ma r \ —N sin 72.34° = 0.5a r (1)

£F e = ma e \ F - N cos 72.34° = 0.5a fl (2)

Kinematics. Using the chain rule, the first and second time derivatives of r are

r = 20- 1

r - -2,-H . -(!>

r = -2(-26 r 3 e 2 + e~ 2 e) = ^ ( 20 2 - 00 )

When 6 = 180° = tt rad, 6 = 4 rad/s and 6 = 3 rad/s 2 . Thus

r =

— m = 0.6366 m

77

r =

4t 2 ( 42 )

77

-0.8106 m/s
77 ( 3 )] = 1.4562 m/s 2

Thus,

a r = r - r0 2 = 1.4562 - 0.6366(4 2 ) = -8.7297 m/s 2
a g = r’0 + 2 rd = 0.6366(3) + 2(-0.8106)(4) = -4.5747 m/s 2
Substitute these result into Eqs. (1) and (2),

N = 4.5807 N

F = -0.8980 N = -0.898 N Ans.

The negative sign indicates that F acts in the sense opposite to that shown in
the FBD.

Ans:

F = -0.898 N

336

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13 - 93 .

If arm OA rotates with a constant clockwise angular
velocity of 8 = 1.5 rad/s. determine the force arm OA
exerts on the smooth 4-lb cylinder B when 8 = 45°.

SOLUTION

Kinematics: Since the motion of cylinder B is known, a r and a, ( will be determined
4

first. Here, — = cos 8 or r = 4 sec 6 ft. The value of r and its time derivatives at the
r

instant 6 = 45° are

r = 4secd| 9=45 = = 4 sec 45° = 5.657 ft
r = 4sec0(tan0)0| e=45 ° = 4 sec 45° tan45°(1.5) = 8.485 ft/s
r = 4 [sec0(tan 8)8 + f)(sec 8 sec 2 88 + tan 0 seed tan 0$)] = 4[sec0(tan0)0 + sec 3 00 2 + sec0tan 2 00 ? = 4[sec45°tan45°(0) + sec 3 45°(1.5) 2 + sec45°tan 2 45°(1.5) 2 ] = 38.18 ft/s 2 Using the above time derivatives, a r = r - r8 2 = 38.18 - 5.657(l.5 2 ) = 25.46 ft/s 2 a e = r'8 - 2r8 = 5.657(0) + 2(8.485)(1.5) = 25.46 ft/s 2 Equations of Motion: By referring to the free-body diagram of the cylinder shown in Fig. a, 'ZF. = ma,\ N cos 45° — 4 cos 45° = ^^(25.46) r r 32.2 v N = 8.472 lb ZF e = ma e ; F OA - 8.472 sin 45° - 4 sin 45° = -^(25.46) Fqa ~ 12-0 lb Ans. Ans: F oa = 12.0 lb 337 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 94 . Determine the normal and frictional driving forces that the partial spiral track exerts on the 200-kg motorcycle at the instant 6 = |irrad, 6 = 0.4 rad/s, and 6 = 0.8 rad/s 2 . Neglect the size of the motorcycle. SOLUTION e = I -771 = 300° 6 = 0.4 6 = 0.8 r = 56> = 5( - it ) = 26.18 r = 56 = 5(0.4) = 2 r = 56 = 5(0.8) = 4 a r = r - r6 2 = 4 - 26.18(0.4) 2 = -0.1888 a e = r6 + 2 r'd = 26.18(0.8) + 2(2)(0.4) = 22.54 tan i ft = r dr/dd = 5.236 ifi = 79.19° +\2F r = ma r ; +S1,Fh = ma „; F sin 10.81° - N cos 10.81° + 200(9.81) cos 30° = 200(-0.1888) F cos 10.81° - 200(9.81) sin 30° + N sin 10.81° = 200(22.54) F = 5.07 kN Ans. N = 2.74 kN Ans. Ans: F = 5.07 kN N = 2.74 kN 338 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 339 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 96 . The spring-held follower AB has a mass of 0.5 kg and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.15 m and z = (0.02 cos 20) m. If the cam is rotating at a constant rate of 30 rad/s, determine the force component F „ at the end A of the follower when 6 = 30°. The spring is uncompressed when 0 = 90°. Neglect friction at the bearing C. SOLUTION Kinematics. Using the chain rule, the first and second time derivatives of z are z = (0.02 cos 26) m z = 0.02[—sin 20(2f?)] = [—0.04(sin 26)6] m/s ^ z = —0.04[cos 26(26)6 + (sin 20)0] = [—0.04(2 cos 26(d) 1 + sin 20(0))] m/s 2 Here, 0 = 30 rad/s and 0 = O.Then z = —0.04[2 cos 20(3O 2 ) + sin 20(0)] = (—72 cos 20) m/s 2 Equation of Motion. When 0 = 30°, the spring compresses x = 0.02 + 0.02 cos 2(30°) = 0.03 m. Thus, F sp = kx = 1000(0.03) = 30 N. Also, at this position a z = z = —72 cos 2(30°) = —36.0 m/s 2 . Referring to the FBD of the follower. Fig. a, %F Z = ma z ; N - 30 = 0.5(-36.0) N = 12.0 N Ans. w Ans: N = 12.0 N 340 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 97 . The spring-held follower AB has a mass of 0.5 kg and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.15 m and z = (0.02 cos 26) m. If the cam is rotating at a constant rate of 30 rad/ s, determine the maximum and minimum force components F_ the follower exerts on the cam if the spring is uncompressed when 0 = 90°. SOLUTION Kinematics. Using the chain rule, the first and second time derivatives of z are z = (0.02 cos 26) m f\j z. = 0.02[—sin 26(26)] = (—0.04 sin 266) m/s z = —0.04[cos 26(26)6 + sin 200] = [—0.04(2 cos 26(d) 2 + sin 26(6))] m/s 2 Here 6 = 30 rad/s and 6 = O.Then, z = —0.04[2 cos 2#(30 2 ) + sin 26(0)] = (—72 cos 26) m/s 2 Equation of Motion. At any arbitrary 6 , the spring compresses x = 0.02(1 + cos 26). Thus, F sp = kx = 1000[0.02(1 + cos 26)] = 20 (1 + cos 26). Referring to the FBD of the follower, Fig. a , 2,F Z = ma z ; N - 20(1 + cos 26) = 0.5(-72 cos 26) N = (20 - 16 cos 26) N N is maximum when cos 26 = — 1. Then (A)max = 36.0 N Ans. N is minimum when cos 26 = 1. Then (AOmin = 4.00 N Ans. w £ Ans: (AO max = 36.0 N (AOmin = 4 - 00N 341 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 98 . The particle has a mass of 0.5 kg and is confined to move along the smooth vertical slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when 8 = 30°. The rod is rotating with a constant angular velocity 8=2 rad/s. Assume the particle contacts only one side of the slot at any instant. SOLUTION 0.5 r = -= 0.5 sec 0 , r = 0.5 sec 8 tan 08 cos 8 r = 0.5 sec 8 tan 88 + 0.5 sec 3 88 2 + 0.5 sec 8 tan 2 88 2 At 8 = 30°. 8 = 2 rad/s 8 = 0 r = 0.5774 m r = 0.6667 m/s r = 3.8490 m/s 2 a r = r - rd 2 = 3.8490 - 0.5774(2) 2 = 1.5396 m/s 2 a g = rd + 2r0 = 0 + 2(0.6667)(2) = 2.667 m/s 2 + /2/y = ma r ; N P cos 30° - 0.5(9.81)sin 30° = 0.5(1.5396) N P = 3.7208 = 3.72 N = ma e ; F - 3.7208 sin 30° - 0.5(9.81) cos 30° = 0.5(2.667) F = 7.44 N 0.5 m Ans. Ans. Ans: N s = 3.72 N F r = 7.44 N 342 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 99 . A car of a roller coaster travels along a track which for a short distance is defined by a conical spiral, r = |z, 0 = —1.5z, where r and z are in meters and 0 in radians. If the angular motion 8=1 rad/s is always maintained, determine the r, 8, z components of reaction exerted on the car by the track at the instant z — 6 m. The car and passengers have a total mass of 200 kg. SOLUTION r = 0.75z r = 0.75z r = 0.75'z 8 = —1.5z 8 = — 1.5z 8 = — 1.5'z 0=1 = — 1.5z z = —0.6667 m/s z = 0 At z = 6 m, r = 0.75(6) = 4.5 m r = 0.75(-0.6667) = -0.5 m/s r = 0.75(0) = 0 0 = 0 a, = r - rd 2 = 0 - 4.5(1) 2 = -4.5 m/s 2 a 0 = r8 + 2 rd a z = z = 0 = 4.5(0) + 2(—0.5)(1) = -1 m/s 2 1,F t = ma r ; F r = 200(—4.5) 7y = -900 N Ans. HFg = ma # ; = 200(—1) = -200N Ans. S /%. = F z - 200(9.81) = 0 F z = 1962 N = 1.96 kN Ans. l Ans: F r = -900 N F 0 = -200 N F z = 1.96 kN 343 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 100 . The 0.5-lb ball is guided along the vertical circular path r = 2 r c cos 8 using the arm OA. If the arm has an angular velocity 8 = 0.4 rad/s and an angular acceleration 8 = 0.8 rad/s 2 at the instant 8 = 30°, determine the force of the arm on the ball. Neglect friction and the size of the ball. Set r c = 0.4 ft. SOLUTION r = 2(0.4) cos 8 = 0.8 cos 8 r = -0.8 sin 88 r = -0.8 cos 88 2 - 0.8 sin 88 At 8 = 30°, 8 = 0.4 rad/s, and 8 = 0.8 rad/s 2 r = 0.8 cos 30° = 0.6928 ft r = —0.8 sin 30°(0.4) = -0.16 ft/s r = —0.8 cos 30°(0.4) 2 - 0.8 sin 30°(0.8) = -0.4309 ft/s 2 a r = r - r8 2 = -0.4309 - 0.6928(0.4) 2 = -0.5417 ft/s 2 a e = r8 + 2r8 = 0.6928(0.8) + 2(-0.16)(0.4) = 0.4263 ft/s 2 +/’ZF r = ma r \ N cos 30° - 0.5 sin 30° = ^ (-0.5417) N = 0.2790 lb \+"ZF e = ma e ; F OA + 0.2790 sin 30° - 0.5 cos 30° = -^(0.4263) F oa = 0.300 lb Ans. Ans: F oa = 0.300 lb 344 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 101 . The ball of mass m is guided along the vertical circular path r = 2 r c cos 8 using the arm OA. If the arm has a constant angular velocity 8 0 , determine the angle 8 < 45° at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball. SOLUTION r = 2 r c cos 8 r = -2 r c sin 88 r = —2r c cos 88 2 — 2 r c sin 88 Since 8 is constant, 8 = 0. a r = r — rd 2 = —2r c cos 88 § — 2 r c cos 88 q = — 4r c cos 88q +/'ZF r = ma r \ tan 8 mg sin 8 = 4r c flo S m(— 4r c cos 88q) 8 = tan 4f c 0q ^ Ans. Ans: 8 = tan 345 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 102 . Using a forked rod, a smooth cylinder P, having a mass of 0.4 kg, is forced to move along the vertical slotted path r = (0.60) m, where 0 is in radians. If the cylinder has a constant speed of v c = 2 m/s, determine the force of the rod and the normal force of the slot on the cylinder at the instant 0 = tt rad. Assume the cylinder is in contact with only one edge of the rod and slot at any instant. Hint: To obtain the time derivatives necessary to compute the cylinder’s acceleration components a r and a 8 , take the first and second time derivatives of r = 0.60. Then, for further information, use Eq. 12-26 to determine 0. Also, take the time derivative of Eq. 12-26, noting that v c = 0, to determine 9. SOLUTION r = 0.60 r = 0.60 r = 0.60 v r = r = 0.60 v e = r9 = 0.600 v 2 = r 2 + ( r9 9 = 2 = 0.60 + 0.600 0.6\/l + 0 2 0 = 0.7200 + 0.361 200 3 + 20 2 00 0 = - ee 1 At 0 = tt rad, 0 = 0.6 VT + TT l + 0 2 = 1.011 rad/s (tt)(1.011) 2 , 0 = - -V = -0.2954 rad/s 2 1 + T7 2 ' r = 0.6 (tt) = 0.6 tt m r = 0.6(1.011) = 0.6066 m/s r = 0.6(—0.2954) = -0.1772 m/s 2 a r = r - rd 2 = -0.1772 - 0.6tt(1.0U) 2 = -2.104 m/s 2 a B = rd + 270 = 0.6tt(- 0.2954) + 2(0.6066)(1.011) = 0.6698 m/s 2 tan tfj r _ 0.60 dr/d9 0.6 ifj = 72.34° ^ 2F r = ma r ; -N cos 17.66° = 0.4(-2.104) N = 0.883 N + |2E S = ma e ; -F + 0.4(9.81) + 0.883 sin 17.66° = 0.4(0.6698) F = 3.92 N Ans. Ans. Ans: N = 0.883 N F = 3.92 N 346 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 103 . The pilot of the airplane executes a vertical loop which in part follows the path of a cardioid, r = 200(1 + cosd) m, where 6 is in radians. If his speed at A is a constant v p = 85 m/s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at A. He has a mass of 80 kg. Hint: To determine the time derivatives necessary to calculate the acceleration components a r and a g , take the first and second time derivatives of r = 200(1 + cos 0). Then, for further information, use Eq. 12-26 to determine 6. SOLUTION Kinematic. Using the chain rule, the first and second time derivatives of r are r = 200(1 + cos 6) r = 200(—sin 6){6 ) = —200(sin 6)6 r = -200[(cos 6)(6) 2 + (sin 0)(0)] When 6 = 0°, r = 200(1 + cos 0°) = 400 m r = —200(sin 0°) 6 = 0 r = -2OO[(cosO°)(0) 2 + (sin O°)(0) ] = -200 6 2 Using Eq. 12-26 v = Vr 2 + ( rd) 2 v 2 = r 2 + ( rO) 2 85 2 = 0 2 + (4OO0) 2 6 = 0.2125 rad/s Thus, a r = r — rd 2 = -200(0.2125 2 ) - 400(0.2125 2 ) = -27.09 m/s 2 Equation of Motion. Referring to the FBD of the pilot, Fig. a, = ma r ; 80(9.81) - N = 80(-27.09) N = 2952.3 N = 2.95 kN Ans. 80 ( 9 - 81)6 Ca) Ans: N = 2.95 N 347 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 104 . The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = (e s ) m, where 8 is in radians. Determine the tangential force F and the normal force N acting on the collar when 8 = 45°, if the force F maintains a constant angular motion 8 = 2 rad/s. SOLUTION r = e e r = e e 8 r = e\8) 2 + e e 8 At 8 = 45° 8 = 2 rad/s 8 = 0 r = 2.1933 r = 4.38656 r = 8.7731 a, = r — r(8) 2 = 8.7731 - 2.1933(2) 2 = 0 a e = r8 + 2r8 = 0 + 2(4.38656)(2) = 17.5462 m/s 2 N = 24.8 N Ans. F = 24.8 N Ans. Ans: N = 24.8 N F = 24.8 N 348 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 105 . The particle has a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when 6 = 30°. The rod is rotating with a constant angular velocity 0=2 rad/s. Assume the particle contacts only one side of the slot at any instant. SOLUTION 0.5 r = -= 0.5 sec 6 cos 6 r = 0.5 sec 6 tan 06 r = 0.5 { [ (sec 0 tan 00 )tan 0 + sec 0( sec 2 00) ] 0 + sec 0 tan 06 } = 0.5 [sec 0 tan 2 00 2 + sec 3 00 2 + sec 0 tan 00] When 0 = 30°, 6 = 2 rad/s and 6 = 0 r = 0.5 sec 30° = 0.5774 m r = 0.5 sec 30° tan 30°(2) = 0.6667 m/s r = 0.5 [ sec 30° tan 2 30°(2) 2 + sec 3 30°(2) 2 + sec 30° tan 30°(0) ] = 3.849 m/s 2 a T = r — rO 2 = 3.849 - 0.5774(2) 2 = 1.540 m/s 2 a e = rO + 2rd = 0.5774(0) + 2(0.6667)(2) = 2.667 m/s 2 /+tF r = ma r ; N cos 30° - 0.5(9.81) cos 30° = 0.5(1.540) N = 5.79 N +\2F fl = ma e ; F + 0.5(9.81)sin 30° - 5.79 sin 30° = 0.5(2.667) F = 1.78 N 0.5 m Ans. Ans. Ans: F r = 1.78 N N , = 5.79 N 349 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 350 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 107 . The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a lima 5 on, r = (2 + cos 6) ft. If 9 = (0.5 t 2 ) rad, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = Is. The fork and path contact the particle on only one side. SOLUTION r = 2 + cos 9 r = —sin 99 r = -cos 99 2 - sin 88 8 = 0.5f 2 8 = t 9 = 1 rad/s 2 At t = 1 s, 8 = 0.5 rad, 8=1 rad/s, and 9 = 1 rad/s 2 r = 2 + cos 0.5 = 2.8776 ft r = -sin 0.5(1) = -0.4974 ft/s 2 r = —cos0.5(l) 2 — sin 0.5(1) = —1.357 ft/s 2 a r = r - rd 2 = -1.375 - 2.8776(1) 2 = -4.2346 ft/s 2 a„ = r8 + 2 rd = 2.8776(1) + 2(-0.4794)(l) = 1.9187 ft/s 2 tan = 2 + cos 8 dr/dd —sin 8 = -6.002 i)i = -80.54° -I-/SF r = ma r ; +\'ZFg = ma g ; -N cos 9.46° = -(-4.2346) N = 0.2666 lb 32.2 v ’ 2 F - 0.2666 sin 9.46° = — (1.9187) F = 0.163 lb 6 F Ans: F = 0.163 lb 351 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 108 . The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a parabola r = 4/(1 — cos 8), where 0 is in radians and r is in feet. If the collar’s angular rate is constant and equals (9 = 4 rad/s, determine the tangential retarding force P needed to cause the motion and the normal force that the collar exerts on the rod at the instant 8 = 90°. SOLUTION 4 1 — cos 8 -4 sin 8 8 (1 - cos (9) 2 —4 sin 6 8 - 4 cos 8(d) 2 8 sin 2 8 8 2 (1 - cos 8) 2 (1 - cos 8) 2 (1 - cos 8) 3 At 8 = 90°, 8 = 4, 8 = 0 r = 4 r = -16 r = 128 a, = r — r(8) 2 = 128 - 4(4) 2 = 64 a e = rO + 2 r8 = 0 + 2(-16)(4) = -128 4 1 — cos 8 dr —4 sin 8 dd (1 — cos 8) 2 r 4 1 - cos 6) -4 sin d (1 - COS0) 2 0=90° (/, = - 45° = 135° 4 —4 = - 1 + T 2F r — m a r ; 4^ = ma g ; 3 P sin 45° - N cos 45° = -(64) 32.2 v ’ - P cos 45° - N sin 45° = — (-128) 32.2 v ’ Solving, P = 12.6 lb Ans. N = 4.22 lb Ans. Ans: P = 12.6 lb N = 4.22 lb 352 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 109 . Rod OA rotates counterclockwise at a constant angular rate 8 = 4rad/s. The double collar B is pin-connected together such that one collar slides over the rotating rod and the other collar slides over the circular rod described by the equation r = (1.6 cos 6) m. If both collars have a mass of 0.5 kg, determine the force which the circular rod exerts on one of the collars and the force that OA exerts on the other collar at the instant 8 = 45°. Motion is in the horizontal plane. SOLUTION r = 1.6 cos 8 r = —1.6 sin 88 r = —1.6 cos 88 2 - 1.6 sin 88 At 8 = 45°, 8 = 4 rad/s and 8 = 0 r = 1.6 cos 45° = 1.1314 m r = —1.6 sin 45°(4) = -4.5255 m/s r = —1.6 cos 45°(4) 2 - 1.6sin45°(0) = -18.1019 m/s 2 a r = r - rd 2 = -18.1019 - 1.1314(4) 2 = -36.20 m/s 2 a B = rd + 2 i-8 = 1.1314(0) + 2(-4.5255)(4) = -36.20 m/s 2 /+XF r = ma r ; -N c cos 45° = 0.5(-36.20) A C = 25.6N +\XF g = ma„; F OA - 25.6 sin 45° = 0.5(-36.20) F OA = 0 O Ans. Ans. Ans: F r = 25.6 N Fqa = 0 353 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 354 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-111. A 0.2-kg spool slides down along a smooth rod. If the rod has a constant angular rate of rotation 9 = 2 rad/s in the vertical plane, show that the equations of motion for the spool are r — 4 r — 9.81 sin 9 = 0 and 0.8 r + N s — 1.962 cos 9 = 0 , where N s is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is r = Cie~ 2t + C 2 e 2 ' - (9.81/8) sin 2t. If r, r, and 9 are zero when t = 0, evaluate the constants Ci and C 2 to determine r at the instant 9 = tt/4 rad. SOLUTION Kinematic: Here, 9. = 2 rad/s and 9 = 0. Applying Eqs. 12-29, we have a r = r - r9 2 = r - r(2 2 ) = r - 4r a e = rd + 2 r6 = r( 0) + 2r(2) = 4r Equation of Motion: Applying Eq. 13-9, we have = ma r \ 1.962 sin 9 = 0.2(r — 4 r) r — 4r — 9.81 sin 9 = 0 (Q.E.D.) (1) = ma e \ 1.962 cos 9 — N s = 0.2(4 r) 0.8 r + N s - 1.962 cos 9 = 0 (: QE.D .) (2) / 2 dt, 9 = 2t. The solution of the differential 0 equation (Eq.(l)) is given by Since 9. = 2 rad/s, then / 9 Jo 02(9-81)=/962 tJ ,, 9.81 r = C 1 e 2 ‘ + C? e 2t -sin 2 1 (3) Thus, r = —2 C\e 2t + 2 C 2 e 2t — cos 2 1 (4) At t = 0, r = 0. From Eq.(3) 0 = C x (1) + C 2 (1) - 0 9.81 At t = 0, r = 0. From Eq.(4) 0 = -2 C, (1) + 2C 2 (1)-— (5) ( 6 ) Solving Eqs. (5) and (6) yields Ci = 9.81 16 C 2 = 9.81 16 ^ Thus, 9.81 „ 9.81 9.81 . „ r = -— e 2t + —— e 2 -sin 2 1 16 16 8 9.81 (—e~ 2t + e 2 ‘ 9.81 At 9 = 2t = — sin 2 1 (sin h 2r — sin 2 1) 9.81 4’ 8 sin h-sin—3 ) = 0.198 m 4 4 Ans. Ans: r = 0.198 m 355 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-112. The pilot of an airplane executes a vertical loop which in part follows the path of a “four-leaved rose,” r = (—600 cos 20) ft, where 0 is in radians. If his speed at A is a constant v P = 80 ft/s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at A. He weighs 130 lb. Hint: To determine the time derivatives necessary to compute the acceleration components a r and a e , take the first and second time derivatives of r = 400(1 + cos 0). Then, for further information, use Eq. 12-26 to determine 0. Also, take the time derivative of Eq. 12-26, noting that v c = 0* to determine 0. SOLUTION r = —600 cos 20 r — 1200 sin 2 00 r = 1200(2 cos 200 1 + sin 200} At 0 = 90° r = -600 cos 180° = 600 ft r = 1200 sin 18O°0 = 0 r = 1200(2 cos 180°e 2 + sin 180°<9) = -2400<9 2 v r = r = 0 vg = rO = 6000 2 2 2 v p = v r + v e 80 2 = 0 2 + (600 e) 2 0 = 0.1333 rad/s r = —2400(0.1333) 2 = -42.67 ft/s 2 a, = r ™ rO 2 = -42.67 - 600(0.1333) 2 = -53.33 ft/s 2 130 + T2E r = ma r \ -N - 130 = ^(~ 53 - 33 ) N = 85 - 3 lb Ans: N = 85.3 lb a! Ans. A r = —600 cos 26 356 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-113. The earth has an orbit with eccentricity e — 0.0167 around the sun. Knowing that the earth’s minimum distance from the sun is 146(10 6 ) km, find the speed at which the earth travels when it is at this distance. Determine the equation in polar coordinates which describes the earth’s orbit about the sun. SOLUTION Ch 2 GM S 1 where C = — I — 1 - v 0 = GM s r 0 [GMs (e + 1) GM S ) r 0 vl. r o ( r 0 v 0 ) 2 GMs rovl and h = r 0 v 0 ( rgvl \GM S - 1 rovo GMs = e + 1 r o 66.73(10~ 12 )(1.99)(10 30 )(0.0167 + 1) 146(10 9 ) GM S = 30409 m/s = 30.4 km/s Ans. 1 - 1 GM S r 0 v 0 cos 0 + 2 2 rm 1 - 66.73(10~ 12 )(1.99)(10 30 ) 146(10 9 ) V" 151.3(10 9 )(30409) 2 0.348(10 -12 ) cos 6 + 6.74(10 -12 ) cos 8 + 66.73(10~ 12 )(1.99)(10 3 °) [ 146(10 9 )] 2 (30409) 2 Ans. Ans: v a = 30.4 km/s - = 0.348 (10- 12 ) cos 8 + 6.74 (l0~ 12 ) r 357 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-114. A communications satellite is in a circular orbit above the earth such that it always remains directly over a point on the earth’s surface. As a result, the period of the satellite must equal the rotation of the earth, which is approximately 24 hours. Determine the satellite’s altitude h above the earth’s surface and its orbital speed. SOLUTION The period of the satellite around the circular orbit of radius r 0 = h + r e = \h + 6.378(10 6 )] m is given by 2irr 0 T = -- V s 2v\h + 6.378(10 6 )] 24(3600) = Vs 2ir[h + 6.378(10 6 ) V- = -o- 86.4(10 3 ) ( 1 ) The velocity of the satellite orbiting around the circular orbit of radius r 0 = h + r e = \ji + 6.378(10 6 )] m is given by v s = 66.73(10~ 12 )(5.976)(10 24 ) h + 6.378(10 6 ) ( 2 ) Solving Eqs.(l) and (2), h = 35.87(10 6 ) m = 35.9 Mm v s — 3072.32 m/s = 3.07 km/s Ans. Ans: h = 35.9 mm v s = 3.07 km/s 358 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-115. The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13-25. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit 800 km from the earth’s surface. SOLUTION For a 800-km orbit v 0 = 66.73(10~ 12 )(5.976)(10 24 ) (800 + 6378)(10 3 ) = 7453.6 m/s = 7.45 km/s Ans. Ans: Vq = 7.45 km/s 359 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 360 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-117. Prove Kepler’s third law of motion. Hint: Use Eqs. 13-19, 13-28,13-29, and 13-31. SOLUTION From Eq. 13-19, 1 GM S — — C cos 0 H- r h For 0 = 0° and 6 = 180°, 1 GM S — = C H-- u h 2 1 GM S - = -C + — r, h Eliminating C, from Eqs. 13-28 and 13-29, 2 a _ 2 GM S b 2 ~ h 2 From Eq. 13-31, T = l ( 2 a)(b) Thus, b 2 T 2 h 2 4tt 2 <j 2 4tt 2 o 3 GM s T 2 h 2 ~ h 2 Q.E.D. Ans: 361 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-118. The satellite is moving in an elliptical orbit with an eccentricity e = 0.25. Determine its speed when it is at its maximum distance A and minimum distance B from the earth. SOLUTION Ch 2 GM e where C = r o 1 - GM e r 0 Vo and h = r 0 Vp. 1 GM e r 0 r o rpVo GM P = e + 1 1 - r 0 Vp GM e v 0 = GM, Vp (ro - 1 GM e (e + 1) r 0 where r 0 = r p = 2(l0 6 ) + 6378(l0 3 ) = 8.378(l0 6 ) m. /66.73(10^ 12 )(5.976)(10 24 )(0.25 + 1) v B = v Q = _ rp_ 2 GM e r 0 v 0 8.378(10 6 ) 8.378(10 6 ) - 1 - 1 v A = — v B = , r a 13.96(10 6 ) 2(66.73)(10 _lz )(5.976)(10 24 ) 8.378(10 6 )(7713) 2 8.378(10 6 ) = 7713 m/s = 7.71 km/s = 13.96(l0 6 ) m (7713) = 4628 m/s = 4.63 km/s Ans. Ans. Ans: v B = 7.71 km/s v A = 4.63 km/s 362 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-119. The rocket is traveling in free flight along the elliptical orbit. The planet has no atmosphere, and its mass is 0.60 times that of the earth. If the rocket has the orbit shown, determine the rocket’s speed when it is at A and at B. SOLUTION Applying Eq. 13-27, (2 GMIr p vl) - 1 2 GM rptf 2 GM r p v 2 p - 1 = r p + r a 2GM r a r P (fp + r a ) The elliptical orbit has r p = 7.60(10 6 ) m, r a 18.3(10 6 ) m and v p = v A . Then v A = 66.73(l0~ 12 ) ] [0.6(5.976)(lO 24 ) ] [ 18.3(l0 6 ) 7.60( 10 s ) [ 7.60( 10 6 ) + 18.3(l0 6 )] = 6669.99 m/s = 6.67(l0 3 ) m/s In this case, h = r p v A = r a v B 7.60( 10 s ) (6669.99) = 18.3(lO s )v B v B = 2770.05 m/s = 2.77(l0 3 ) m/s Ans. Ans. Ans: v A = 6.67(l0 3 ) m/s v B = 2.77(10 3 ) m/s 363 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-120. Determine the constant speed of satellite S so that it circles the earth with an orbit of radius r = 15 Mm. Hint: Use Eq. 13-1. SOLUTION m , m, /u?\ F = G —j— Also F = m s ( — \ Hence m S = G yj 66.73(1(T 12 ) ( 5 ^|q 6 ) ^) = 5156 m/s = 5.16 km/s Ans. Ans: v = 5.16 km/s 364 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-121. The rocket is in free flight along an elliptical trajectory A' A. The planet has no atmosphere, and its mass is 0.70 times that of the earth. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A. SOLUTION Central-Force Motion: Use r a = -----.with r n = r„ = 6( 10 6 ) m and (2 GM/r 0 i>o) —1 M = 0.70 M e , we have 9(l0 6 ) 6 ( 10) 6 /2(66.73) (10~ 12 ) (0.7) [5.976(10 24 )] V 6(10 6 )u 2 v A = 7471.89 m/s = 7.47 km/s 1 Ans. Ans: v A = 7.47 km/s 365 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-122. The Viking Explorer approaches the planet Mars on a parabolic trajectory as shown. When it reaches point A its velocity is 10 Mm/h. Determine r 0 and the required velocity at A so that it can then maintain a circular orbit as shown. The mass of Mars is 0.1074 times the mass of the earth. SOLUTION When the Viking explorer approaches point A on a parabolic trajectory, its velocity at point A is given by v A 2 GM m r 0 10 ( 10 6 ) m 1 h 3600 s 2(66.73)(10 12 ) 0.1074(5.976)(10 24 ) 'o r 0 = 11.101(10 6 ) m = 11.1 Mm Ans. When the explorer travels along a circular orbit of r 0 = 11.101(10 6 ) m, its velocity is VA' GM r ro 66.73(10 12 ) 0.1074(5.976)(10 44 ) ,24x 11 . 101 ( 10 6 ) = 1964.19 m/s Thus, the required sudden decrease in the explorer’s velocity is = v A ~ va - 10 < mt >(ss)" 1964 19 = 814 m/s Ans. Ans: r 0 = 11.1 Mm Av a = 814 m/s 366 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-123. The rocket is initially in free-flight circular orbit around the earth. Determine the speed of the rocket at A. What change in the speed at A is required so that it can move in an elliptical orbit to reach point A'l SOLUTION The required speed to remain in circular orbit containing point A of which r 0 = 8(l0 6 ) + 6378(l0 3 ) = 14.378(l0 6 ) m can be determined from Me = GM e fo 66.73(10 12 )] [5.976(l0 24 ) 14.378(l0 6 ) = 5266.43 m/s = 5.27(l0 3 ) m/s Ans. To more from A to A', the rocket has to follow the elliptical orbit with r p = 8(l0 6 ) + 6378(l0 3 ) = 14.378(l0 6 ) m and r a = 19(l0 6 ) + 6378(l0 3 ) = 25.378(l0 6 ) m. The required speed at A to do so can be determined using Eq. 13-27 (2 GMJ r P°p) ~ 1 2 GM e r - 1 = - r p v p r a 2 GM e _ r p + r a r p v 2 p r a r P ( r P + r a) Here, v p = (v A ) e . Then = /2[66.73(l0^ 12 )][5.976(l0 24 )][25.378(l0 6 )] ' 14.378(l0 6 )[l4.378(l0 6 ) + 25.378(l0 6 )] = 5950.58 m/s Thus, the required change in speed is An = (v A ) e — (v A ) c = 5950.58 — 5266.43 = 684.14 m/s = 684 m/s Ans. Ans: {va)c = 5.27(l0 3 ) m/s Av = 684 m/s 367 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 124 . The rocket is in free-flight circular orbit around the earth. Determine the time needed for the rocket to travel from the inner orbit at A to the outer orbit at A'. SOLUTION To move from A to A’, the rocket has to follow the elliptical orbit with r p = 8(l0 6 ) + 6378(l0 3 ) = 14.378(l0 6 ) m and r a = 19(l0 6 ) + 6378(l0 3 ) = 25.378(10 6 ) m.The required speed at A to do so can be determined using Eq. 13-27 “ (2 GMJr P v 2 p ) - 1 2 GM P r p v l - 1 = 2GM, r p + r a r p v p r p( r p + r a) Here, v p = r^.Then /2[66.73(10~ 12 )][5.976(10 24 )][25.378(10 6 )] v a = \l . , r. . .. = 5950.58m/s 14.378(l0 6 )[l4.378(l0 6 ) + 25.378(l0 6 )] Then h = v A r p = 5950.58[14.378(10 6 )] = 85.5573(l0 9 ) m 2 /s The period of this elliptical orbit can be determined using Eq. 13-31. T = + r a )V^ a [14.378(10 6 ) + 25.378( 10 6 )] V [l4.378( 10 6 )] [25.378( 10 6 )] 85.5573(10 9 ) = 27.885(l0 3 ) s Thus, the time required to travel from A to A' is T 27.885(l0 3 ) r 2 = 13.94(10 3 ) s = 3.87 h Ans. Ans: t = 3.87 h 368 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-125. A satellite is launched with an initial velocity v 0 = 2500 mi/h parallel to the surface of the earth. Determine the required altitude (or range of altitudes) above the earth’s surface for launching if the free-flight trajectory is to be (a) circular, (b) parabolic, (c) elliptical, and (d) hyperbolic. Take G = 34.4(10~ 9 )(lb-ft 2 )/slug 2 , M e = 409(10 21 ) slug, the earth’s radius r e = 3960 mi, and 1 mi = 5280 ft. SOLUTION v 0 = 2500 mi/h = 3.67(10 3 ) ft/s (a) C-h e = -= 0 or C = 0 GM e = r 0 v o GM e = 34.4(10 9 )(409)(10 21 ) = 14.07(10 15 ) ro GM e 14.07(10 15 ) ~ [3.67(10 13 )] 2 1.047(10 9 ) = 1.046(10 9 ) ft (b) 5280 C 2 h , e =-= 1 GM e TTTU (7oPo)f ~ GM e \r 0 - 3960 = 194(10 ) mi 1 - GM, = 1 rova 2GM e _ 2(14.07)(10 15 ) ~ [3.67(10 3 )] 2 r = 396(10 3 ) - 3960 = 392(10 3 ) mi ro = = 2.09(10 9 ) ft = 396(10 3 ) mi Ans. Ans. (c) e < 1 194(10 3 ) mi < r < 392(10 3 ) mi Ans. (d) e > 1 r > 392( 10 3 ) mi Ans. Ans: (a) r = 194 (10 3 ) mi (b) r = 392 (10 3 ) mi (c) 194 (10 3 ) mi < r (d) r > 392 (10 3 ) mi < 392 (10 3 ) mi 369 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-126. The rocket is traveling around the earth in free flight along the elliptical orbit. If the rocket has the orbit shown, determine the speed of the rocket when it is at A and at B. SOLUTION Here r p Here v p 20(l0 6 ) m and r a = 30(l0 6 ) m. Applying Eq. 13-27, (2 GM e /r p v 2 p ) - 2 GM e r P r P v \ - 1 = r a 2 GM e r p + r a r p v p r a v 1 ' 2GM e r a Vp y r p( r p + r a) « 4 .Then v A = 2[66.73(l(T 12 )] [5.976( 10 24 )] [30(l0 6 )] 20(l0 6 )[20(l0 6 ) + 30(l0 6 )] = 4891.49 m/s = 4.89(l0 3 ) m/s For the same orbit h is constant. Thus, ^ ^ eft Cl [20(l0 6 )] (4891.49) = [30(l0 6 )]v B v B = 3261.00 m/s = 3.26(l0 3 ) m/s Ans. Ans. Ans: v A = 4.89(10 3 ) m/s v B = 3.26(l0 3 ) m/s 370 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-127. An elliptical path of a satellite has an eccentricity e = 0.130. If it has a speed of 15 Mm/h when it is at perigee, P, determine its speed when it arrives at apogee, A. Also, how far is it from the earth’s surface when it is at A? P SOLUTION e = 0.130 v p = v 0 = 15 Mm/h = 4.167 km/s = Ctf = jV _ GM e $$r 2 0 v 2 0 \ 6 GM e ^ r 0 vl )\GM e ) _ (e + 1 )GM e vl 1.130(66.73) (10~ 12 ) (5.976) (10 24 ) [4.167(10 3 )] 2 = 25.96 Mm GM e _ 1 r 0 vl e + 1 ro r 0 r 0 (e + 1) 'a = i - 1 — e 25.96(10 6 )(1.130) 0.870 = 33.71 (10 6 )m = 33.7 Mm v o r o 'A = - r A 15(25.96)(10 6 ) 33.71(10 6 ) = 11.5 Mm/h Ans. d = 33.71(10 6 ) - 6.378(10 6 ) = 27.3 Mm Ans. Ans: v A = 11.5 Mm/h d = 27.3 Mm 371 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *13-128. A rocket is in free-flight elliptical orbit around the planet Venus. Knowing that the periapsis and apoapsis of the orbit are 8 Mm and 26 Mm, respectively, determine (a) the speed of the rocket at point A', (b) the required speed it must attain at A just after braking so that it undergoes an 8-Mm free-flight circular orbit around Venus, and (c) the periods of both the circular and elliptical orbits. The mass of Venus is 0.816 times the mass of the earth. SOLUTION a) M„ = 0.816(5.976(10 24 )) = 4.876(10 24 ) (2GMv^ _ \ \OAv i ) 26(10) 6 = 8(10 6 / 2(66.73) (10~ 12 )4.876(10 24 ) 8 ( 10 6 )^ - 1 b) 81.35(10 6 ) -,-= 1.307 va v A = 7887.3 m/s = 7.89 km/s OA v A _ 8(10 6 )(7887.3) OA' 26(10 6 ) 2426.9 m/s = 2.43 m/s Ans. c) VA" 66.73(10~ lz )4.876(10 24 ) 8 ( 10 6 ) v A " = 6377.7 m/s = 6.38 km/s Circular orbit: T = x r 2-nOA v A - 2tt8(10 6 ) -= 7881.41 s = 2.19 h 6377.7 Ans. Ans. Elliptic orbit: T e = —^—(OA + OA')V(OA)(OA') =--( 8 + 26)(l0 6 )(V(8)(26))(l0 6 OAv A y 8(10 6 )(7886.8) V A A T e = 24414.2 s = 6.78 h Ans. Ans: v A = 2.43 m/s v A " = 6.38 km/s T c = 2.19 h T e = 6.78 h 372 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 129 . The rocket is traveling in a free flight along an elliptical trajectory A'A. The planet has no atmosphere, and its mass is 0.60 times that of the earth. If the rocket has the orbit shown, determine the rocket’s velocity when it is at point A. SOLUTION Applying Eq. 13-27, _ _ r _p _ “ (2 GM/r p v 2 p ) - 1 2 GM rptf 2 GM _ r P + r a r p vl r a 2 GMr a r p {r p + r a ) The rocket is traveling around the elliptical orbit with r p = 70( 10 6 ) m, r a = 100(l0 6 ) m and v p = u^.Then v A = [66.73(l0~ 12 )][0.6(5.976)(l0 24 )][l00(l0 6 )] 70(10 6 )[70(10 6 ) + 100(l0 6 )] = 2005.32 m/s = 2.0l(l0 3 ) m/s Ans. Ans: v A = 2.0l(l0 3 ) m/s 373 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13-130. If the rocket is to land on the surface of the planet, determine the required free-flight speed it must have at A' so that the landing occurs at B. How long does it take for the rocket to land, going from A' to B? The planet has no atmosphere, and its mass is 0.6 times that of the earth. SOLUTION Applying Eq. 13-27, ( 2 GM/r p vl) - 1 r„v„ - 1 = - • pvp 'a 2 GM _ r P + r a r p Vp r a 2GMr a r p {r p + r a ) To land on B, the rocket has to follow the elliptical orbit A’B with r p = 6(l0 6 ), r a = 100(l0 6 ) m and v p = v B . v B = In this case 2[66.73( 10- 12 )][0.6(5.976)(10 24 )] [l00( 10 6 )] 6(l0 6 )[6(l0 6 ) + 100(l0 6 )] = 8674.17 m/s h = r p v B = i- a v A , 6(l0 6 ) (8674.17) = 100(l0 6 )^. v A ' = 520.45 m/s = 521 m/s The period of the elliptical orbit can be determined using Eq. 13-31. T = — (r + r '\\/y r h Up ^ r a ) V r p r a = 6(10 6 )(8674 17) [6(1 ° 6) + 100 ( 106 )] V [ 6 ( 106 )][ 10 °( 106 )] = 156.73(l0 3 ) s Thus, the time required to travel from A' to B is Ans. t = — = 78.365(10 3 ) s = 21.8 h Ans. Ans: v A ’ = 521 m/s t = 21.8 h 374 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 - 131 . The rocket is traveling around the earth in free flight along an elliptical orbit AC. If the rocket has the orbit shown, determine the rocket’s velocity when it is at point A. SOLUTION For orbit AC, r p = 10(l0 6 ) m and r a = 16(l0 6 ) m. Applying Eq. 13-27 Here v p = t^.Then “ (2 GM e /r p v 2 p ) - 1 2GM„ 2 r V P P ~ 1 = 2 GM e r p + r a r p v 2 p 2GM e r a r p( r p + r a ) Va = 2[66.73(l0^ 12 )] [5.976(l0 24 )] [l6(l0 6 )] 10(l0 6 )[l0(l0 6 ) + 16(l0 6 )] = 7005.74 m/s = 7.0l(l0 3 ) m/s Ans. Ans: v A = 7.01 (10 3 ) m/s 375 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 13 - 132 . The rocket is traveling around the earth in free flight along the elliptical orbit AC. Determine its change in speed when it reaches A so that it travels along the elliptical orbit AB. SOLUTION Applying Eq. 13-27, r _ _ r _P_ _ r “ ~ (2GM e /r pV p) - 1 2GM e r —r- 1 = - r p ir p r a 2 GM e _r p + r a r a v 2 p G 2 GM e r a rJr + r a ) For orbit AC, r p = 10(l0 6 ) m ,r a = 16(l0 6 ) m and v p = (u y4 ) > i C .Then 12 [66.73( 1(T 12 )] [5.976( 10 24 )] [l6( 10 6 )] ( Va)ac = 10(l0 6 )[l0(l0 6 ) + 16(l0 6 )] For orbit AB. r p = 8(l0 6 ) m, r a = 10(l0 6 ) m and tij, = ■Ug.Then = 7005.74 m/s v B = 2[66.73( 10 12 )] [5.976(l0 24 )] [l0(l0 6 )] 8(l0 6 )[8(l0 6 ) + 10(l0 6 )] = 7442.17 m/s Since h is constant at any position of the orbit, h f'p'Vp ?cfta 8(10 6 ) (7442.17) = 10(10 6 )(« A ) AB (va)ab = 5953.74 m/s Thus, the required change in speed is = (v A ) AH - (v A ) AC = 5953.74 - 7005.74 = —1052.01 m/s = —1.05 km/s Ans. The negative sign indicates that the speed must be decreased. Ans: Aw = -1.05 km/s 376 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 1 . The 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N. When s = 15 m, the crate is moving to the right with a speed of 8 m/s. Determine its speed when s = 25 m. The coefficient of kinetic friction between the crate and the ground is p, k = 0.25. SOLUTION Equation of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = /x k N = 0.251V. Applying Eq. 13-7, we have + T 2 F v = ma y ; N + 100 sin 30° - 20(9.81) = 20(0) N = 146.2 N 2(9.81) N Principle of Work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do no work. Applying Eq.14-7, we have T 1 + 2 ^ 1-2 = T 2 ^ p25 m yr (20)(8 2 ) + / 100 cos 30° ds 2 J 15 m I* 25 m ^ - / 36.55 ds = ' (20) v 2 J 15 m 2 v = 10.7 m/s Ans. Ans: v = 10.7 m/s 377 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 2 . For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is F = (90(10 3 )x 1 ^ 2 ) lb, where x is in ft, determine the car’s maximum penetration in the barrier. The car has a weight of 4000 lb and it is traveling with a speed of 75 ft/s just before it hits the barrier. F( lb) /F= 90(10) 3 X 1/2 - -* (h) SOLUTION Principle of Work and Energy: The speed of the car just before it crashes into the barrier is = 75 ft/s. The maximum penetration occurs when the car is brought to a stop, i.e., i> 2 = 0. Referring to the free-body diagram of the car, Fig. «, W and N do no work; however, F 6 does negative work. T\ + 2tfr_ 2 = T 2 1/4000 2 V 32.2 ( 75 z ) + 90(10 3 )x 1 '' 2 dx = 0 Ans. w= 4-000 lb Ans: -t max 3.24 ft 378 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 3 . The crate, which has a mass of 100 kg, is subjected to the action of the two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m/s. The coefficient of kinetic friction between the crate and the surface is p k = 0.2. 1000 N SOLUTION Equations of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = p k N = 0.2 N. Applying Eq. 13-7, we have +12E y = ma y -, N + 1000^0 - 800 sin 30° - 100(9.81) = 100(0) IV = 781 N Principle of Work and Energy: The horizontal components of force 800 N and 1000 N which act in the direction of displacement do positive work, whereas the friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of 800 N and 1000 N force and the weight of the crate do not displace, hence they do no work. Since the crate is originally at rest, T | = 0. Applying Eq. 14-7, we have I0OW&OF 30 - Ss v l Fj-OZhi IOON a A/ F + 2^1-2 = F 0 + 800 cos 30 o (s) + lOOoQjs - 156.2.V = |(100)(6 2 ) s = 1.35m Ans. Ans: s = 1.35 m 379 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 4 . The 100-kg crate is subjected to the forces shown. If it is originally at rest, determine the distance it slides in order to attain a speed of v = 8 m/s. The coefficient of kinetic friction between the crate and the surface is = 0.2. SOLUTION Work. Consider the force equilibrium along the y axis by referring to the FBD of the crate, Fig. a, TtSFy = 0; N + 500 sin 45° - 100(9.81) - 400 sin 30° = 0 N = 827.45 N Thus, the friction is fy = p k N = 0.2(827.45) = 165.49 N. Here, F 1 and F 2 do positive work whereas 7y does negative work. W and N do no work U Fl = 400 cos 30° s = 346.41 s U Fl = 500 cos 45° s = 353.55 s U Ff = -165.49 s Principle of Work And Energy. Applying Eq. 14-7, 7j + Xtfr-2 = T 2 0 + 346.41 s + 353.55 s + (-165.49 s) = ^(100)(8 2 ) s = 5.987 m = 5.99 m Ans. I* (A; Ans: s = 5.99 m 380 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 5 . Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill A it will reach a speed of 100 km/h when it comes to the bottom B. Also, what should be the minimum radius of curvature p for the track at B so that the passengers do not experience a normal force greater than 4mg = (39.24«;) N? Neglect the size of the car and passenger. SOLUTION ioo(io 3 ) 100 km/h = -'-- = 27.778 m/s 1 3600 ' T\ + Xt/i-2 = T 2 0 + m(9.81)h = i/«(27.778) 2 h = 39.3 m ,* Vr , .... ((21.118f\ + | Zr n = ma n \ 39.24 m — mg = - J p = 26.2 m Ans: h = 39.3 m p = 26.2 m A 381 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 6 . When the driver applies the brakes of a light truck traveling 40 km/h, it skids 3 m before stopping. How far will the truck skid if it is traveling 80 km/h when the brakes are applied? SOLUTION 40(l0 3 ) 40 km/h = = 11.11 m/s 80 km/h = 22.22 m/s 3600 7/ + 21/i- 2 = T 2 i/w( ll.ll) 2 - fi k mg( 3) = 0 kg = 20.576 71 + 2t/r- 2 = T 2 |/w(: 22.22) 1 - (20.576)m(d) = 0 d = 12 m Ans: d = 12 m ^Cr Ans. 382 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 7 . As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block if it has an initial speed of 5 m/s and travels 10 m, both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x' axis and moving at a constant velocity of 2 m/s relative to A. Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy. SOLUTION A • - x B • - x' 2 m/s Observer A: 7/ + = T 2 |(10)(5) 2 + 6(10) = |(10)wi v 2 — 6.08 m/s Observer B: F = ma 6 = 10a a = 0.6 m/s 2 () s = s 0 + v 0 t + -a c t 2 10 = 0 + 5t + i(0 tyt 2 t 2 + 16.67r - 33.33 = 0 t = 1.805 s At v = 2 m/s, s' = 2(1.805) = 3.609 m Block moves 10 - 3.609 = 6.391 m Thus 7\ + = T 2 ^(10)(3) 2 + 6(6.391) = i(10)v§ = 4.08 m/s Note that this result is 2 m/s less than that observed by A. Ans. Ans. Ans: Observer A: v 2 = 6.08 m/s Observer B\v 2 = 4.08 m/s 383 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 8 . A force of F = 250 N is applied to the end at B. Determine the speed of the 10-kg block when it has moved 1.5 m, starting from rest. SOLUTION Work, with reference to the datum set in Fig. a. T 2 Sp = 1 8S W + 2 8s F = 0 (1) Assuming that the block moves upward 1.5 m, then 8S W = —1.5 m since it is directed in the negative sense of S w . Substituted this value into Eq. (1), — 1.5 + 2 8s F = 0 8s F = 0.75 m Thus, U F = F8S f = 250(0.75) = 187.5 J U w = -W8S w = —10(9.81)(1.5) = -147.15 J Principle of Work And Energy. Applying Eq. 14-7, T\ + t/1-2 = T 2 0 + 187.5 + (-147.15) = |(10)u 2 v = 2.841 m/s = 2.84 m/s Ans. (a; Ans: v = 2.84 m/s 384 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 9 . The “air spring” A is used to protect the support B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D. The force developed by the air spring as a function of its deflection is shown by the graph. If the block has a mass of 20 kg and is suspended a height d = 0.4 m above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails. Neglect the mass of the pulley and belt. I D j f B SOLUTION Work. Referring to the FBD of the tensioning weight, Fig. a, W does positive work whereas force F does negative work. Flere the weight displaces downward S w = 0.4 + Xmax where ^ is the maximum compression of the air spring. Thus U w = 20(9.81)(0.4 + 4, ax ) = 196.2(0.4 + x^x) The work of F is equal to the area under the F-S graph shown shaded in Fig. b. Flere F 1500 - = -; F = 7500x max . Thus ■*max 0.2 Up = -2(7500 Xmax)(-*max) = -3750xjL,x Principle of Work And Energy. Since the block is at rest initially and is required to stop momentarily when the spring is compressed to the maximum, 7i = T 2 = 0. Applying Eq. 14-7, 71 + St/r-2 = T 2 0 + 196.2(0.4 + x max ) + (-3750Xm a x) = 0 3750 xLx — 196.2x max — 78.48 = 0 Xmax = 0.1732 m = 0.173 m < 0.2 m (O.K!) Ans. Ans: Xmax = 0.173 m 385 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 10 . The force F, acting in a constant direction on the 20-kg block, has a magnitude which varies with the position x of the block. Determine how far the block must slide before its velocity becomes 15 m/s. When x = 0 the block is moving to the right at v = 6 m/s. The coefficient of kinetic friction between the block and surface is p k = 0.3. SOLUTION Work. Consider the force equilibrium along y axis, by referring to the FBD of the block, Fig. a, +\ tF y = 0; N - 20(9.81) = 0 N = 196.2 N Thus, the friction is fy = fi k N = 0.3(196.2) = 58.86 N. Here, force F does positive work whereas friction Fj does negative work. The weight W and normal reaction N do no work. U F = 1 100 3 50x2 ds = -x 2 3 U Ff = -58.86 x Principle of Work And Energy. Applying Eq. 14-7, 71 + St/!- 2 = T 2 |(20)(6 2 ) + fxi + (-58.86x) = i(20)(15 2 ) 100 3 —X 2 - 58.86x - 1890 = 0 F(N) F=50s 1 P x(m) Solving numerically, x = 20.52 m = 20.5 m Ans. Ans: x = 20.5 m 386 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 11 . The force of F = 50 N is applied to the cord when s = 2m. If the 6-kg collar is orginally at rest, determine its velocity at s = 0. Neglect friction. SOLUTION Work. Referring to the FBD of the collar, Fig. a , we notice that force F does positive work but W and N do no work. Here, the displacement of F is s = V2 2 + 1.5 2 - 1.5 = 1.00 m U F = 50(1.00) = 50.0 J Principle of Work And Energy. Applying Eq. 14-7, 71 + 2£/i- 2 = T 2 0 + 50 = i(6)v 2 v = 4.082 m/s = 4.08 m/s Ans. 1.5 Ans: v = 4.08 m/s 387 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 12 . Design considerations for the bumper B on the 5-Mg train car require use of a nonlinear spring having the load- deflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to 0.2 m when the car, traveling at 4 m/s, strikes the rigid stop. Neglect the mass of the car wheels. E(N) F=ks 2 -i- (m) SOLUTION 1 r 02 —(5000)(4) 2 — / ks 2 ds = 0 2 Jo (0.2) 3 40 000 - k -—- = 0 3 k = 15.0 MN/m 2 Ans. 5(10 3 )(9.81)N t N C Ans: k = 15.0 MN/m 2 388 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 13 . The 2-lb brick slides down a smooth roof, such that when it is at A it has a velocity of 5 ft/s. Determine the speed of the brick just before it leaves the surface at B , the distance d from the wall to where it strikes the ground, and the speed at which it hits the ground. SOLUTION t a + ZU A - B = t b K3i2) (5)2 + 2(15) = l(M)* B v B = 31.48 ft/s = 31.5 ft/s ( -** J S = S 0 + Vot d = 0 + 31 . 48^1 ( + 1) s = s 0 + v 0 t - -a c t 2 30 = 0 + 31.480jf + i (32.2 )t 2 16.1 1 2 + 18.888t - 30 = 0 Solving for the positive root, t = 0.89916 s d = 31.48^(0.89916) = 22.6 ft T a + c = T c K^) (s)2+2(45 > = v c = 54.1 ft/s Ans. Ans. Ans. Ans: Vg = 31.5 ft/s d = 22.6 ft v c = 54.1 ft/s 389 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 14 . Block A has a weight of 60 lb and block B has a weight of 10 lb. Determine the speed of block A after it moves 5 ft down the plane, starting from rest. Neglect friction and the mass of the cord and pulleys. SOLUTION 2 s A + s B = l 2 A s A + As B = 0 2v a + v B = 0 7} + XUi- 2 = T 2 0 + 60 10 ( 10 ) v A = 7.18 ft/s Ans: v A = 7.18 ft/s 390 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 15 . The two blocks A and B have weights W A = 60 lb and W B = 10 lb. If the kinetic coefficient of friction between the incline and block A is /x k = 0.2, determine the speed of A after it moves 3 ft down the plane starting from rest. Neglect the mass of the cord and pulleys. SOLUTION Kinematics: The speed of the block A and B can be related by using position coordinate equation. ■^,4 T O'M a /;) ^ 2.s' 7 i s b I 2\s a — A s B = 0 A s B = 2\s A = 2(3) = 6 ft 2v A - Vg = 0 (1) Equation of Motion: Applying Eq. 13-7, we have +2Fy = may ; N - 600 j ^ (0) N = 48.0 lb Principle of Work and Energy: By considering the whole system, W A which acts in the direction of the displacement does positive work. W B and the friction force Ff — HkN = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite direction to that of displacement Here, W A is being displaced vertically (downward) 3 —As^ and W B is being displaced vertically (upward) A s B . Since blocks A and B are at rest initially, 7\ = 0. Applying Eq. 14-7, we have Ti + 2 ^ 1-2 = T 2 (3 \ 11 0 + Wa { 5 ASA j ” F f^ SA ~ W B As B = 2 m A V A + f m B Vg - 9.60(3) - 10(6) = | 1236.48 = 60«i + 10v 2 b (2) Eqs. (1) and (2) yields v A = 3.52 ft/s Ans. v B = 7.033 ft/s 60 (3) 60 322 v\ + 1 10 2 \ 32.2 vl Ans: v A = 3.52 ft/s 391 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 16 . A small box of mass m is given a speed of v = Vjgr at the top of the smooth half cylinder. Determine the angle 8 at which the box leaves the cylinder. SOLUTION Principle of Work and Energy: By referring to the free-body diagram of the block, Fig. a, notice that N does no work, while W does positive work since it displaces downward though a distance of h = r — r cos 8. T i + St/r-2 = T 2 1 (1 2 \4 — ml — gr I + mg(r — r cos 8) = — mv 2 v- = gr I — — 2 cos 8 ( 1 ) v 2 gr[^-2 cost? Equations of Motion: Here, a n — — = referring to Fig. a, ? g[--2cos8 ). By = ma„ ; mg cos 8 2 cos 8 W= N = mg\ 3 cos 8 ~ It is required that the block leave the track. Thus, N = 0. 0 = mg\ 3 cos 8 — — Since mg ^ 0, 9 3 cos 8 -=0 4 8 = 41.41° = 41.4° Ans. Ans: 8 = 41.4° 392 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 17 . If the cord is subjected to a constant force of F = 30 lb and the smooth 10-lb collar starts from rest at A , determine its speed when it passes point B. Neglect the size of pulley C. F = 30 lb x SOLUTION Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B , W displaces upward through a distance h = 4.5 ft, while force F displaces a distance of s = AC - BC = vVi 2 + 4.5 2 -2 - 5.5 ft. The work of F is positive, whereas W does negative work. T A + 'ZU A - B — T b (a) j~~30 lb v B = 27.8 ft/s Ans. Ans: v B = 27.8 ft/s 393 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 18 . When the 12-lb block A is released from rest it lifts the two 15- lb weights B and C. Determine the maximum distance A will fall before its motion is momentarily stopped. Neglect the weight of the cord and the size of the pulleys. SOLUTION Consider the entire system: t = Vy 2 + 4 2 T + %U X - 2 = r 2 (0 + 0 + 0) + 12y - 2(15)(Vy 2 + 4 2 - 4) = (0 + 0 + 0) 0.4y = Vy 2 + 16 - 4 (0.4y + 4) 2 = y 2 + 16 —0.84y 2 + 3.20y + 16 = 16 —0.84y + 3.20 = 0 y = 3.81 ft Ans: y = 3.81 ft 394 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 19 . If the cord is subjected to a constant force of F = 300 N and the 15-kg smooth collar starts from rest at A, determine the velocity of the collar when it reaches point B. Neglect the size of the pulley. SOLUTION Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a. Principle of Work and Energy: Referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B, W displaces vertically upward a distance h — (0.3 + 0.2) m = 0.5 m, while force F displaces a distance of s = AC — BC = \/0.7 2 + 0.4 2 — Vo.2 2 + 0.2 2 = 0.5234 m. Here, the work of F is positive, whereas W does negative work. T a + 21 / a-b ~ T b 0 + 300(0.5234) + [—15(9.81)(0.5)] = ^(15)v B 2 v B = 3.335 m/s = 3.34 m/s Ans. Ans: v B = 3.34m/s 395 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 20 . The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having a weight of 4000 lb will penetrate the barrier if it is originally traveling at 55 ft/s when it strikes the first barrel. SOLUTION T\ + ST/,, = T 2 1 2 /4000 \ V 32.2 ) (55) 2 Area = 0 Area = 187.89 kip • ft 2(9) + (5 - 2)(18) + x(27) = 187.89 x = 4.29 ft < (15 - 5) ft Thus 5 = 5 ft + 4.29 ft = 9.29 ft (O.K!) Ans. w ■fooovb Ans: s = 9.29 ft 396 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 397 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 22 . The 25-lb block has an initial speed of v 0 = 10 ft/s when it is midway between springs A and B. After striking spring B, it rebounds and slides across the horizontal plane toward spring A, etc. If the coefficient of kinetic friction between the plane and the block is /x k = 0.4, determine the total distance traveled by the block before it comes to rest. SOLUTION Principle of Work and Energy: Here, the friction force /■/ = /x /c N = 0.4(25) = 10.0 lb. Since the friction force is always opposite the motion.it does negative work. When the block strikes spring B and stops momentarily, the spring force does negative work since it acts in the opposite direction to that of displacement. Applying Eq. 14—7, we have T, + 2^1-2 = T 2 K^) (io)2 - io(i+sJ 4 (6o)s; - 0 ,v, = 0.8275 ft Assume the block bounces back and stops without striking spring A. The spring force does positive work since it acts in the direction of displacement. Applying Eq. 14-7, we have T 2 + 2 ^ 2-3 = *3 0 + | (60)(0.8275 2 ) - 10(0.8275 + s 2 ) = 0 s 2 = 1.227 ft Since s 2 = 1.227 ft < 2 ft, the block stops before it strikes spring A. Therefore, the above assumption was correct. Thus, the total distance traveled by the block before it stops is s Xot = 2s 1 + s 2 + 1 = 2(0.8275) + 1.227 + 1 = 3.88 ft Ans. Ans: s Tot = 3.88 ft 398 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 23 . The 8-kg block is moving with an initial speed of 5 m/s. If the coefficient of kinetic friction between the block and plane is = 0.25, determine the compression in the spring when the block momentarily stops. SOLUTION Work. Consider the force equilibrium along y axis by referring to the FBD of the block, Fig. a + = 0; N - 8(9.81) = 0 N = 78.48 N Thus, the friction is iy = p, k N = 0.25(78.48) = 19.62 N and F sp = kx = 200 x. Here, the spring force F sp and iy both do negative work. The weight W and normal reaction N do no work. U Fsp = — I 200 x dx = —100 x 2 Jo U Ff = — 19.62(x + 2) Principle of Work And Energy. It is required that the block stopped momentarily, T 2 = 0. Applying Eq. 14-7 7) + 2 Ut - 2 = T 2 |(8)(5 2 ) + (-100x 2 ) + [—19.62(jt + 2)] = 0 100x 2 + 19.62x - 60.76 = 0 Solved for positive root, x = 0.6875 m = 0.688 m Ans. Ans: x = 0.688 m N (a) 5 m/s - - 2 m -- k A = 200 N/m B mmm A 399 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 24 . At a given instant the 10-lb block A is moving downward with a speed of 6 ft/s. Determine its speed 2 s later. Block B has a weight of 4 lb, and the coefficient of kinetic friction between it and the horizontal plane is /j. k = 0.2. Neglect the mass of the cord and pulleys. SOLUTION Kinematics: The speed of the block A and B can be related by using the position coordinate equation. S A + ( S A S B ) — l 2 S A Sg — l 2k.S A — A Sg = 0 A Sg = 2\s A v b = 2 V A [11 [ 2 ] Equation of Motion: +ZFy = may ■ Ng - 4 = 3^ (0) N B = 4.00 lb Principle of Work and Energy: By considering the whole system, W A , which acts in the direction of the displacement, does positive work. The friction force Ff = p k N B = 0.2(4.00) = 0.800 lb does negative work since it acts in the opposite direction to that of displacement. Here, W A is being displaced vertically (downward) As^. Applying Eq. 14-7, we have =4- It 4 T i + 1-2 = T 2 1 1 -m A ( v 2 a ) 0 + -m B (u|) 0 + W A \s A - F f \s B 1 , 1 , = 2 m A V A + ^ni B Vg From Eq. , (v B ) 0 = 2(u^) 0 = 2(6) = 12 ft/s. Also, As^ = (Ta)o + v A [ 3 ] ( 2 ) = (u^)o + v A — 6 + v A and A s B = 2As^ = 12 + 2v A (Eq. ). Substituting these values into Eq.  yields 1 / 10 2 \32 + \ +io ( 6 + y *) - °- 8o °( i2 + 2v ^ 1 t 10 2\322 2 1 A 2 V32.2 v A = 26.8 ft/s Ans. Ans: v A = 26.8 ft/s 400 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-25. The 5-lb cylinder is falling from A with a speed v A = 10 ft/s onto the platform. Determine the maximum displacement of the platform, caused by the collision. The spring has an unstretched length of 1.75 ft and is originally kept in compression by the 1-ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision. v A = 10 ft/s 1 SOLUTION Ti + 2 t/r-2 = T 2 1 , -(400)(0.75 + sY - -(400)(0.75) = 0 200 s 2 + 295 s - 22.76 = 0 i = 0.0735 ft < 1 ft (O.K!) x = 0.0735 ft Ans. 3 ft k = 400 lb/ft Ans: s = 0.0735 ft 401 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 26 . The catapulting mechanism is used to propel the 10-kg slider A to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the left by means of a piston P. If the piston applies a constant force F = 20 kN to rod BC such that it moves it 0.2 m, determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod BC. SOLUTION 2 s c + s A = l 2 A sc + A S 4 = 0 2(0.2) = — A -0.4 = A s A Ti + = T 2 0 + (10 000)(0.4) = |(10)(uJ 2 v A = 28.3 m/s Ans: v A = 28.3 m/s Ans. - 4 - T, io(1o 3 ;n 402 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 27 . The “flying car” is a ride at an amusement park which consists of a car having wheels that roll along a track mounted inside a rotating drum. By design the car cannot fall off the track, however motion of the car is developed by applying the car’s brake, thereby gripping the car to the track and allowing it to move with a constant speed of the track, v t = 3 m/s. If the rider applies the brake when going from B to A and then releases it at the top of the drum. A, so that the car coasts freely down along the track to B {8 = i t rad), determine the speed of the car at B and the normal reaction which the drum exerts on the car at B. Neglect friction during the motion from A to B. The rider and car have a total mass of 250 kg and the center of mass of the car and rider moves along a circular path having a radius of 8 m. SOLUTION t a + XU A B = t b i(250)(3) 2 + 250(9.81)(16) = ^(250 ){v B f v B = 17.97 = 18.0 m/s + T = ma n N B - 250(9.81) = 250 N b = 12.5 kN ^ (17.97) 2 ~j A Ans. Ans. S - Ans: v B = 18.0 m/s N b = 12.5 kN 403 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 28 . The 10-lb box falls off the conveyor belt at 5-ft/s. If the coefficient of kinetic friction along AB is p k = 0.2, determine the distance x when the box falls into the cart. SOLUTION Work. Consider the force equilibrium along the y axis by referring to Fig. a, "4 + TSly = 0; 2V — 10 ( — I = 0 N = 8.00 lb Thus, Ff = p k N = 0.2(8.00) = 1.60 lb. To reach B, W displaces vertically downward 15 ft and the box slides 25 ft down the inclined plane. U w = 10(15) = 150 ft • lb U Ff = -1.60(25) = -40 ft-lb Principle of Work And Energy. Applying Eq. 14-7 T a + 2 U A R = T b v B = 27.08 ft/s 2 V32.2 Kinematics. Consider the vertical motion with reference to the x-y coordinate system, ( + t ) (Sc)y = (S B )y + (v B ) y t + - a y t 2 ; 5 = 30 - 27.08 (jjr + |(-32.2)t 2 16.lt 2 + 16.25? - 25 = 0 Solve for positive root, t = 0.8398 s Then, the horizontal motion gives -L (S c )x = ( S B ) X + (v B ) x t ; ' 4 ... Ans. X = 0 + 27.08( - )(0.8398) = 18.19 ft = 18.2 ft fr-o-zti A / (A) Ans: x = 18.2 ft 404 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 29 . The collar has a mass of 20 kg and slides along the smooth rod. Two springs are attached to it and the ends of the rod as shown. If each spring has an uncompressed length of 1 m and the collar has a speed of 2 m/s when s = 0, determine the maximum compression of each spring due to the back- and-forth (oscillating) motion of the collar. SOLUTION Ti + St/r-2 = t 2 i(20)(2) 2 - i(50)(,) 2 - |(100)(,) 2 = 0 s = 0.730 m Ans. X Ans: j = 0.730 m 405 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 30 . The 30-lb box A is released from rest and slides down along the smooth ramp and onto the surface of a cart. If the cart is prevented from moving determine the distance s from the end of the cart to where the box stops. The coefficient of kinetic friction between the cart and the box is p, k = 0.6. SOLUTION Principle of Work and Energy: W A which acts in the direction of the vertical displacement does positive work when the block displaces 4 ft vertically. The friction force Ff = p< k N = 0.6(30) = 18.0 lb does negative work since it acts in the opposite direction to that of displacement Since the block is at rest initially and is required to stop, T A — T c = 0. Applying Eq. 14—7, we have T A + 'ZUa-c = T c 0 + 30(4) - 18.0s' = 0 s' = 6.667 ft Thus, s = 10 — s' = 3.33 ft Ans. Ans: s = 3.33 ft 406 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 31 . Marbles having a mass of 5 g are dropped from rest at A through the smooth glass tube and accumulate in the can at C. Determine the placement R of the can from the end of the tube and the speed at which the marbles fall into the can. Neglect the size of the can. SOLUTION T A + 2 U A-B = T B 0 + [0.005(9.81)(3 - 2)] = | (0.005)v| Vb — 4.429 m/s (+J) s = s 0 + vqI + -a c t 2 • A 2 = 0 + 0 = | (9.81)f 2 t = 0.6386 s s = s 0 + v 0 t R = 0 + 4.429(0.6386) = 2.83 m Ans. Ta + 2 U A -c = T ] 0 + [0.005(9.81)(3) = ' 2 (0.005)4 v c = 7.67 m/s Ans. Ans: R = 2.83 m v c = 7.67 m/s 407 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 32 . The block has a mass of 0.8 kg and moves within the smooth vertical slot. If it starts from rest when the attached spring is in the unstretched position at A, determine the constant vertical force F which must be applied to the cord so that the block attains a speed v B = 2.5 m/s when it reaches B\ s B = 0.15 m. Neglect the size and mass of the pulley. Hint: The work of F can be determined by finding the difference A/ in cord lengths AC and BC and using U F = F A/. SOLUTION l AC = V(0.3) 2 + (0.4) 2 = 0.5 m l BC = V(0.4 - 0.15) 2 + (0.3) 2 = 0.3905 m Ta+^U A h = T b 0 + F(0.5 - 0.3905)-|(100)(0.15) 2 - (0.8)(9.81)(0.15) = ^(0.8)(2.5) 2 F = 43.9 N Ans. Ans: F = 43.9 N 408 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 33 . The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at A. If the plane is smooth, B determine the distance d, measured from the wall, to where 3 ft the block strikes the ground. Neglect the size of the block. A/\ k = 100 lb/ftX/ 1 — 4 ft-- -- SOLUTION t a + zu a _ b = t b 0 + 2 (100)(2) 2 - (10)0) = =2(322)^ v B = 33.09 ft/s ( ^ ) S = S 0 + V 0 t d = 0 + 33.09 0j t ( + T) s = So + v 0 t + ^a c f -3 = 0 + (33.09)0 + i(-32.2)r 2 16.lt 2 - 19.853f -3 = 0 Solving for the positive root, t = 1.369 s d = 33.090^(1.369) = 36.2 ft folk Ans. Ans: d = 36.2 ft 409 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 34 . The spring bumper is used to arrest the motion of the 4-lb block, which is sliding toward it at v = 9 ft/s. As shown, the spring is confined by the plate P and wall using cables so that its length is 1.5 ft. If the stiffness of the spring is k = 50 lb/ft, determine the required unstretched length of the spring so that the plate is not displaced more than 0.2 ft after the block collides into it. Neglect friction, the mass of the plate and spring, and the energy loss between the plate and block during the collision. SOLUTION t x + St/i-j = t 2 1 2 (9 ) 2 |(50 )(s - 1.3) 2 - |(50)(s 0.20124 = s 2 - 2.60 s + 1.69 - (s 2 - 3.0 s + 2.25) 0.20124 = 0.4 j - 0.560 j = 1.90 ft Ans. Ans: s = 1.90 ft 410 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 35 . When the 150-lb skier is at point A he has a speed of 5 ft/s. Determine his speed when he reaches point B on the smooth slope. For this distance the slope follows the cosine curve shown. Also, what is the normal force on his skis at B and his rate of increase in speed? Neglect friction and air resistance. SOLUTION ( TT \ y — 50 cos - x 7 Viooy = 22.70 ft t=35 ^=tan0=—Sof^W^lx dx \woj Viooy \ = -ff)sm(i^)x = -1.3996 e = -54.45° cry dx 2 200 ) C ° S ( 10o) X = -0.02240 K0 d y [l + (-1.3996) 2 ]i |-0.02240| = 227.179 dx 2 Ta+^U a r = T b kb # + - -°> - v B = 42.227 ft/s = 42.2 ft/s +/'2F„ = ma,—N + 150 cos 54.45° = c = ( 150^ (42.227)' V 32.2 A 227.179 N = 50.6 lb +\'2F t = ma ,; 150 sin 54.45° = 150 \ 32 JZP Ans. Ans. a, = 26.2 ft/s 2 Ans. y Ans: v B = 42.2 ft/s N = 50.6 lb a, = 26.2 ft/s 2 411 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 412 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 37 . If the track is to be designed so that the passengers of the roller coaster do not experience a normal force equal to zero or more than 4 times their weight, determine the limiting heights h A and h c so that this does not occur. The roller coaster starts from rest at position A. Neglect friction. SOLUTION Free-Body Diagram :The free-body diagram of the passenger at positions B and C are shown in Figs, a and h, respectively. ip' Equations of Motion: Here, a n = —. The requirement at position B is that P N b = 4 mg. By referring to Fig. a, + T ~SjF n = ma n \ Amg — mg = m v B 2 = 45g At position C, N c is required to be zero. By referring to Fig. b, + l2i 7 „ = ma n \ mg — 0 = m A % 2 = 20g Principle of Work and Energy: The normal reaction N does no work since it always acts perpendicular to the motion. When the rollercoaster moves from position A to B, W displaces vertically downward h = h A and does positive work. We have T a + 21/ A _ B — T b 0 + mgh A = i/n(45g) h A = 22.5 m Ans. When the rollercoaster moves from position A to C, W displaces vertically downward h = h A — h c = (22.5 — h c ) m. T A + "2U A - B = T b 0 + mg( 22.5 - h c ) = | m(20g) h c = 12.5 m Ans. Ans: h A = 22.5 m h c = 12.5 m 413 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 38 . If the 60-kg skier passes point A with a speed of 5 m/s, determine his speed when he reaches point B. Also find the normal force exerted on him by the slope at this point. Neglect friction. SOLUTION Free-Body Diagram: The free-body diagram of the skier at an arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, we notice that N does no work since it always acts perpendicular to the motion. When the skier slides down the track from A to B, W displaces vertically downward h = y A ~ y B = 15 — [o.025(o 2 ) + 5] = 10 m and does positive work. T A + 21/ A -B = T b i(60)(5 2 ) + [60(9.81)(10)] = |(60)v B 2 vg = 14.87 m/s = 14.9 m/s Ans. dy/dx = 0.05x d 2 y/dx 2 = 0.05 [1 + 0] 3 / 2 +1= ma„ ; 20 m N - 60(9.81) (14.87) 2 \ 20 J N = 1.25 kN Ans. Ans: v B = 14.9 m/s N = 1.25 kN 414 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 39 . If the 75-kg crate starts from rest at A, determine its speed when it reaches point B. The cable is subjected to a constant force of F = 300 N. Neglect friction and the size of the pulley. SOLUTION Free-Body Diagram: The free-body diagram of the crate and arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, notice that N, W, and R do no work. When the crate moves from A to B, force F displaces through a distance of s = AC - BC = V8 2 + 6 2 - V2 2 + 6 2 = 3.675 m. Here, the work of F is positive. T 1 + 2 = T 2 0 + 300(3.675) = | (75)yg 2 F=3oo a| vg = 5.42 m/s Ans. Ans: v B = 5.42 m/s 415 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 40 . If the 75-kg crate starts from rest at A, and its speed is 6m/s when it passes point B, determine the constant force F exerted on the cable. Neglect friction and the size of the pulley. SOLUTION Free-Body Diagram: The free-body diagram of the crate and cable system at an arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, notice that N, W, and R do no work. When the crate moves from A to B, force F displaces through a distance of s = AC - BC = V8 2 + 6 2 - V2 2 + 6 2 = 3.675 m. Here, the work of F is positive. T\ + St/i-2 = T 2 0 + F(3.675) = | (75)(6 2 ) F = 367 N Ans. Ans: F = 367 N 416 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 41 . A 2-lb block rests on the smooth semicylindrical surface. An elastic cord having a stiffness k = 2 lb/ft is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A (6 = 0°), determine the unstretched length of the cord so that the block begins to leave the semicylinder at the instant 0 = 45°. Neglect the size of the block. SOLUTION Ans: l 0 = 2.77 ft 417 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 42 . The jeep has a weight of 2500 lb and an engine which transmits a power of 100 hp to all the wheels. Assuming the wheels do not slip on the ground, determine the angle 6 of the largest incline the jeep can climb at a constant speed v = 30 ft/s. SOLUTION P = Fjv 100(550) = 2500 sin 0(30) 6 = 47.2° Ans. 2, 500 lb Nj Ans: 6 = 47.2° 418 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 43 . Determine the power input for a motor necessary to lift 300 lb at a constant rate of 5 ft/s. The efficiency of the motor is e = 0.65. SOLUTION Power: The power output can be obtained using Eq. 14-10. P = F v = 300(5) = 1500 ft-lb/s Using Eq. 14-11, the required power input for the motor to provide the above power output is power output power input = - = 1^00 = 2307.7 ft • lb/s = 4.20 hp Ans. 0.65 ' F Ans: Pj = 4.20 hp 419 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 44 . An automobile having a mass of 2 Mg travels up a 7° slope at a constant speed of v = 100 km/h. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the automobile has an efficiency e = 0.65. SOLUTION Equation of Motion: The force F which is required to maintain the car’s constant speed up the slope must be determined first. +SJv = ma x .\ F - 2(10 3 )(9.81) sin T = 2(10 3 )(0) F = 2391.08 N Power: Here, the speed of the car is v = 100(10 3 ) m h lh 3600 s = 27.78 m/s. The power output can be obtained using Eq. 14-10. P = F v = 2391.08(27.78) = 66.418(10 3 ) W = 66.418 kW Using Eq. 14-11, the required power input from the engine to provide the above power output is power output power input =- 2(10 3 )(9.81)N 66.418 0.65 = 102 kW Ans. Ans: power input = 102 kW 420 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 45 . The Milkin Aircraft Co. manufactures a turbojet engine that is placed in a plane having a weight of 13000 lb. If the engine develops a constant thrust of 5200 lb. determine the power output of the plane when it is just ready to take off with a speed of 600 mi/h. SOLUTION At 600 ms/h. Ans. Ans: P = 8.32 (10 3 ) hp 421 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 46 . To dramatize the loss of energy in an automobile, consider a car having a weight of 5000 lb that is traveling at 35 mi/h. If the car is brought to a stop, determine how long a 100-W light bulb must burn to expend the same amount of energy. (1 mi = 5280 ft.) SOLUTION Energy: Here, the speed of the car is u — ^ 51.33 ft/s. Thus, the kinetic energy of the car is 35 mi\ J X / 5280 ft \ / lh \ V 1 mi / \3600 s J 1 , If 5000\/ ^ U = -mv 2 = 2\^)( 5133 ) = 204.59(l0 3 )ft-lb / lhp \ /550 ft-lb/s The power of the bulb is P bu , h = 100 W X (^ 746 W J x - 73.73 ft-lb/s.Thus, U 204.59110 3 ) t = —— = ———“—r-= 2774.98 s = 46.2 min Ans. P bulb 73.73 Ans: t = 46.2 min 422 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 47 . The escalator steps move with a constant speed of 0.6 m/s. If the steps are 125 mm high and 250 mm in length, determine the power of a motor needed to lift an average mass of 150 kg per step. There are 32 steps. SOLUTION Step height: 0.125 m 4 The number of steps: - = 32 Total load: 32(150)(9.81) = 47 088 N If load is placed at the center height, h 4 2 2 m,then U = 47 0881 94.18 kJ = v sin 0 = 0.6 V(32(0.25)) 2 + 4 2 = 0.2683 m/s h_ _ 2 v y 0.2683 U _ 94.18 7 “ 7.454 7.454 s 12.6 kW Also, P = Fv = 47 088(0.2683) = 12.6kW Ans. Ans. Ans: P = 12.6 kW 423 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 48 . The man having the weight of 150 lb is able to run up a 15-ft-high flight of stairs in 4 s. Determine the power generated. How long would a 100-W light bulb have to bum to expend the same amount of energy? Conclusion: Please turn off the lights when they are not in use! SOLUTION Power: The work done by the man is U ~ Wh — 150(15) = 2250 ft-lb Thus, the power generated by the man is given by U 2250 P = — = 1 man ^ = 562.5 ft - lb/s = 1.02 hp The power of the bulb is P bu i b = 100 W X = 73.73 ft-lb/s. Thus, ( Ihp \746 W 550 ft-lb/s 1 hp t = U 2250 Pbuib 73.73 = 30.5 s Ans. Ans. Ans: Pmcm = 1-02 hp t = 30.5 s 424 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 425 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 50 . Determine the power output of the draw-works motor M necessary to lift the 600-lb drill pipe upward with a constant speed of 4 ft/s. The cable is tied to the top of the oil rig, wraps around the lower pulley, then around the top pulley, and then to the motor. SOLUTION 2s P + Sm = I 2v P = ~v M 2 (— 4 ) = —v M v M = 8 ft/s P a = Fv = = 2400 ft-lb/s = 4.36 hp Ans: P a = 4.36 hp 426 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 51 . The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, determine the power that must be supplied to the motor at the instant the load has been hoisted s = 15 ft starting from rest. The motor has an efficiency of e = 0.65. SOLUTION Equation of Motion. Referring to the FBD of the elevator. Fig. a, +12F V = may, 3(500) - 1000 = 1000 32.2 a = 16.1 ft/s 2 When S = 15 ft, + t v 2 = Vq + 2a c (S - Sq); v 2 = 0 2 + 2(16.1)(15) v = 21.98 ft/s Power. Applying Eq. 14-9, the power output is P out = F-V = 3(500)(21.98) = 32.97(l0 3 ) lb-ft/s The power input can be determined using Eq. 14-9 P,„ r <out „„ 32.97(10 3 ) S = —: 0.65 = - P in = [50.72(10 3 ) lb-ft/s] 1 hp 550 lb • ft/s = 92.21 hp = 92.2 hp Ans. 600\\> 500lk Sot)Ik Ans: P = 92.2 hp 427 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 52 . The 50-lb crate is given a speed of 10 ft/s in t = 4 s starting from rest. If the acceleration is constant, determine the power that must be supplied to the motor when t = 2 s. The motor has an efficiency s = 0.65. Neglect the mass of the pulley and cable. SOLUTION +1 SF y = ma y \ 2T - 50 = -^a ( + T) v = v 0 + a c t 10 = 0 + o(4) a = 2.5 ft/s 2 T = 26.94 lb Int = 2 s ( + T) v = v 0 + a c l v = 0 + 2.5(2) = 5 ft/s s c + (*c - Sp) = 1 2v c = v P 2(5) = v P = 10 ft/s P 0 = 26.94(10) = 269.4 269 4 Pl = W = 4145ft ' lb/S Pi = 0.754 hp zr ha i Ans. Ans: Py = 0.754 hp 428 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 429 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 430 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 55 . The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of e = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of v E — 4 m/s. SOLUTION Elevator: Since a = 0, + T ZF y = 0; 60(9.81) + 3 T - 400(9.81) = 0 T = 1111.8 N 2s E + (s E - Sp) = l 3v e = v P Since v E — —4 m/s, v P _ F-v P _ (1111.8)(12) ' e 0.6 — 12 m/s 22.2 kW Ans. m ^ ^ datum -M p i Se 60(9.81)N (400X9.81) N Ans: P t = 22.2 kW 431 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 56 . The 10-lb collar starts from rest at A and is lifted by applying a constant vertical force of F = 25 lb to the cord. If the rod is smooth, determine the power developed by the force at the instant 0 = 60°. SOLUTION Work of F U Y 2 = 25(5 - 3.464) = 38.40 lb • ft T\ + St / 12 = T 2 s 0 + 38.40 - 10(4 - 1.732) = v = 10.06 ft/s P = Fv = 25 cos 60°(10.06) = 125.76 ft • lb/s P = 0.229 hp TuA f Si 3ft HH F W ,4=5fr 3fJM—HF -7- F=J5tl Ans: P = 0.229 hp 432 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 57 . The 10-lb collar starts from rest at A and is lifted with a constant speed of 2 ft/s along the smooth rod. Determine the power developed by the force F at the instant shown. SOLUTION +1 2F y = ma y ; F( - | - 10 = 0 F = 12.5 lb P = Fv = 12.5^- )(2) = 20 lb-ft/s = 0.0364 hp Ans. N io It. Ans: P = 0.0364 hp 433 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 58 . The 50-lb block rests on the rough surface for which the coefficient of kinetic friction is ix k = 0.2. A force F = (40 + s 2 ) lb, where s is in ft, acts on the block in the direction shown. If the spring is originally unstretched (s = 0) and the block is at rest, determine the power developed by the force the instant the block has moved s = 1.5 ft. SOLUTION +1 lF y = 0; N b - (40 + s 2 ) sin 30° - 50 = 0 N b = 70 + 0.5s 2 71 + XC/j-z = T 2 r LD i 0 + / (40 + s 2 ) cos 30° ds - (20)(1.5) 2 - 0.2 / (70 + 0.5 s 2 )ds = Jo 2 J 0 0 + 52.936 - 22.5 - 21.1125 = 0.7764n! v 2 = 3.465 ft/s When s = 1.5 ft, F = 40 + (1.5) 2 = 42.25 lb P = F - v = (42.25 cos 30°)(3.465) P = 126.79 ft- lb/s = 0.231 hp k = 20 lb/ft I sou, " 0.2 NJq Ne 50 2 V3Z2 V2 Ans. Ans: P = 0.231 hp 434 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 59 . The escalator steps move with a constant speed of 0.6 m/s. If the steps are 125 mm high and 250 mm in length, determine the power of a motor needed to lift an average mass of 150 kg per step. There are 32 steps. SOLUTION Step height: 0.125 m 4 The number of steps: = 32 Total load: 32(150)(9.81) = 47 088 N If load is placed at the center height, h 4 2 2 m, then U = 47 088| - 94.18 kJ t P vsin d — 0.61 - 4 - ) V ( 32 ( 0 . 25)) 2 + 4 2 ) h v y U t 2 0.2683 94.18 7.454 " = 7.454 s 12.6 kW 0.2683 m/s Also, P = F v = 47 088(0.2683) = 12.6 kW v Ans. Ans. Ans: P = 12.6 kW 435 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 60 . If the escalator in Prob. 14-47 is not moving, determine the constant speed at which a man having a mass of 80 kg must walk up the steps to generate 100 W of power—the same amount that is needed to power a standard light bulb. SOLUTION p = U 1-2 _ (80)(9.81)(4) t t = 100 t = 31.4 s v = - = V(32(0.25)) 2 + 4 2 31.4 = 0.285 m/s v Ans. Ans: v = 0.285 m/s 436 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 61 . If the jet on the dragster supplies a constant thrust of T = 20 kN, determine the power generated by the jet as a function of time. Neglect drag and rolling resistance, and the loss of fuel. The dragster has a mass of 1 Mg and starts from rest. SOLUTION Equations of Motion: By referring to the free-body diagram of the dragster shown in Fig. a, ^>'ZF X = ma x ; 20(10 3 ) = 1000(a) a = 20 m/s 2 Kinematics: The velocity of the dragster can be determined from v = v Q + a c t v = 0 + 20f = (20r) m/s Power: P = Fv = 20(10 3 )(20f) = [400(10 3 )t] W Ans. Ans: P = | 400(10 3 )t i W /ooocfdO a / 437 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 62 . An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the power applied as a function of time and the work done in t = 0.3 s. SOLUTION For 0 < 1 < 0.2 F = 800 N 20 V= M t= 66 ' 67r P = F v = 53.31 kW For 0.2 < 1 < 0.3 F = 2400 - 80001 v = 66.671 P = F-v = (160f - 5331 2 )kW a0.3 U= Pdt Jo n0.2 ^0.3 U= / 53.31 dt + / (1601 - 533l 2 ) dt Jo J 0.2 = + 1 f[(0.3) 2 - (0.2) 2 ] - ^[(O.S) 3 - (0.2) 3 ] = 1.69 kJ F (N) 800- H-N-i(s) 0.2 0.3 v (m/s) 20 - Ans. Ans. Ans. Ans: P = kW U = 1.69 kJ 438 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 63 . An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the maximum power developed during the 0.3-second time period. SOLUTION See solution to Prob. 14-62. P = 160 1 - 533 f 2 dP — = 160 - 1066.6 t = 0 dt t = 0.15 s < 0.2 s Thus maximum occurs at t = 0.2 s P max = 53.3(0.2) = 10.7 kW E(N) 800-- H--f (s) 0.2 0.3 v (m/s) 20 f(s) Ans. ? Ans: P max = 10.7 kW 439 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 64 . The block has a weight of 80 lb and rests on the floor for which /ju k = 0.4. If the motor draws in the cable at a constant rate of 6 ft/s, determine the output of the motor at the instant 8 = 30°. Neglect the mass of the cable and pulleys. SOLUTION 2 (Vs 2 B + 3 2 j + S P = 1 Time derivative of Eq. (1) yields: 2 SgS B Vs 2 b + 0 2s B v B , _ n V4 + 9 Vs 2 b + 9 + s P = 0 Where s B = v B and s P = v P + v P = 0 v B = 2s R Vp Time derivative of Eq. (2) yields: / 2 ^ 9 ) 3/2 + 9 ) s b - 2 s 2 b s 2 b + 2s b (s 2 b + 9 ) 52 ,] + s B = 0 where s p = a P = 0 and s B = a B 2(s 2 b + 9)v 2 b - 2s 2 B v B2 + 2 s b {s 2 b + 9 )a B = 0 s b v 2 b ~ v B (sl) + 9 v B = Sb(s 2 b + 9) At 8 = 30°, s B = -- = 5.196 ft B tan 30° V5 196 2 + 9 From Eq. (3) v B = - 9(f . 1%) —(6) = -3.464 ft/s From Eq. (4) a B = 5.196 2 (-3.464) 2 - (-3.464 2 )(5.196 2 + 9) 5.196 ( 5.196 2 + 9) 80 XF x = ma ; p - 0.4(80) = — (-0.5773) P = 30.57 lb F 0 = 8- v = 30.57(3.464) = 105.9 ft-lb/s = 0.193 hp Also, tF x = 0 —F + 2T cos 30° = 0 ^ 30.57 _, rll T = -= 17.65 lb 2 cos 30 F 0 = T-v = 17.65(6) = 105.9 ft • lb/s = 0.193 hp Ans. Ans. Ans: F 0 = 0.193 hp F„ = 0.193 hp 440 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-65. The block has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are /jl s = 0.5 and [x k = 0.4, respectively. If a force F = (60t 2 ) N, where t is in seconds, is applied to the cable, determine the power developed by the force when t = 5 s. Hint: First determine the time needed for the force to cause motion. SOLUTION 2 F x = 0; 2 F - 0.5(150)(9.81) = 0 F = 367.875 = 60; 2 t = 2.476 s ^2 F x = m a x \ 2(60t 2 )-0.4(150)(9.81) = 150a p a p = O.St 2 - 3.924 dv = a dt [ dv = / (0.8/ 2 - 3.924) dt Jo J2.476 V /q o\ ^ v = ( — y - 3.924f = 19.38 m/s V 3 / 2.476 Sp + ( S P ~ Sp) = / ISO(1.81) N I ^ h - 0.5N f +-?— 150(1.81) N 2 v P = v F v F = 2(19.38) = 38.76 m/s F = 60(5) 2 = 1500 N P = F- v = 1500(38.76) = 58.1 kW Ans. Ans: P = 58.1 kW 441 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 66 . The girl has a mass of 40 kg and center of mass at G. If she is swinging to a maximum height defined by 9 = 60°, determine the force developed along each of the four supporting posts such as AB at the instant 9 = 0°. The swing is centrally located between the posts. SOLUTION The maximum tension in the cable occurs when 9 = 0°. t 1 + v 1 = t 2 + v 2 1 0 + 40(9.81)(—2 cos 60°) = -(40)?/ + 40(9.81)(-2) v = 4.429 m/s f 4 4?9 2 \ + UF n = ma n ; T - 40(9.81) = C 40 )^^ J + T 2F y = 0; 2(2F) cos 30° - 784.8 = 0 T = 784.8 N F = 227 N Ans. it 4«Cf8lJ id ZF ?\ZF T?7fa3rJ Ans: F = 227 N 442 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 67 . The 30-lb block A is placed on top of two nested springs B and C and then pushed down to the position shown. If it is then released, determine the maximum height h to which it will rise. SOLUTION Conservation of Energy: T 1 + V 1 = T 2 + v 2 1 ; mv\ + V, 1 (^)i = 2 mv 2 + + [v e ) 2 1 0 + 0 + - (200)(4) 2 + - (100)(6) = 0 + h{ 30) + 0 h = 113 in. Ans. Da-hum 0 &) Ans: h = 133 in. 443 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 68 . The 5-kg collar has a velocity of 5 m/s to the right when it is at A. It then travels down along the smooth guide. Determine the speed of the collar when it reaches point B, which is located just before the end of the curved portion of the rod. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod. SOLUTION Potential Energy. With reference to the datum set through B the gravitational potential energies of the collar at A and B are {V g ) A = mgh A = 5(9.81)(0.2) = 9.81 J (V g h = 0 At A and B, the spring stretches x A = Vo.2 2 + 0.2 2 - 0.1 = 0.1828 m and x B = 0.4 — 0.1 = 0.3 m respectively. Thus, the elastic potential energies in the spring at A and B are (V e ) A = \kx\ = ~ (50)(0.1828 2 ) = 0.83581 (VQ fl = \kx 2 B = y (50)(0.3 2 ) = 2.251 Conservation of Energy. T A + v a = t b + v b i(5)(5 2 ) + 9.81 + 0.8358 = | (5)u| + 0 + 2.25 v B = 5.325 m/s = 5.33 m/s Ans. Equation of Motion. At B , F sp = kx B = 50(0.3) = 15 N. Referring to the FBD of the collar, Fig. a , / 5 325 2 \ = ma n ; * + 15 = 5(—) N = 693.95 N = 694 N Ans. 5(W0A ri a) Ans: v B = 5.33 m/s N = 694 N 444 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 69 . The 5-kg collar has a velocity of 5 m/s to the right when it is at A. It then travels along the smooth guide. Determine its speed when its center reaches point B and the normal force it exerts on the rod at this point. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod. SOLUTION Potential Energy. With reference to the datum set through B the gravitational potential energies of the collar at A and B are (V g ) A = mgh A = 5(9.81)(0.2) = 9.81 J (V g ) B = 0 At A and B , the spring stretches x A = \/o.2 2 + 0.2 2 — 0.1 = 0.1828 m and x B = 0.4 — 0.1 = 0.3 m respectively. Thus, the elastic potential energies in the spring at A and B are (V e ) A = \kx\ = i(50)(0.1828 2 ) = 0.8358 J (V e ) B = \kx | = |(50)(0.3 2 ) = 2.25 J Conservation of Energy. T a + V a = T b + V b 1 1 — (5)(5 2 ) + 9.81 + 0.8358 = - (5)v 2 B + 0 + 2.25 v B = 5.325 m/s = 5.33 m/s Ans. Equation of Motion. At B. F sp = kx B = 50(0.3) = 15 N. Referring to the FBD of the collar, Fig. a , „ / 5.325 2 \ XF„ = ma n - ^ +15 = 5 (^^J N = 693.95 N = 694 N Ans. 5(9S0tl (a > Ans: N = 694 N 445 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 70 . The ball has a weight of 15 lb and is fixed to a rod having a negligible mass. If it is released from rest when 8 = 0°, determine the angle 6 at which the compressive force in the rod becomes zero. SOLUTION T 1 + V 1 = T 2 + U 2 0 + 0 = \( J ^ ) v2 ~ 15 ( 3 ) (1 “ cos v 2 = 193.2(1 - cos 8) + i/£F„ = ma n 15 cos 8 = 15 322 193.2(1 - cos 8) cos 8 = 2 — 2 cos 8 F 8 = 48.2° Ans. Ans: 8 = 48.2° 446 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 71 . The car C and its contents have a weight of 600 lb, whereas block B has a weight of 200 lb. If the car is released from rest, determine its speed when it travels 30 ft down the 20° incline. Suggestion: To measure the gravitational potential energy, establish separate datums at the initial elevations of B and C. SOLUTION 2 s B + s c = / 2A s B = — Asc A S B = -y = -15 ft 2v b = -v c Establish two datums at the initial elevations of the car and the block, respectively. T l + V 1 = T 2 + V 2 ° + ° - KS) (vd!+ KHX^) 2+2 “ <i5) - 600 ” 2o " (3o) v c = 17.7 ft/s Ans. Ans: v c = 17.7 ft/s 447 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 72 . The roller coaster car has a mass of 700 kg, including its passenger. If it starts from the top of the hill A with a speed v A = 3 m/s, determine the minimum height h of the hill crest so that the car travels around the inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take p B = 7.5 m and pc = 5 m. A SOLUTION Equation of Motion. Referring to the FBD of the roller-coaster car shown in Fig. a, XF n = ma n ; N + 700(9.81) = 700 (yj (1) When the roller-coaster car is about to leave the loop at B and C, N = 0. At B and C, p B = 7.5 m and p c = 5 m. Then Eq. (1) gives 0 + 700(9.81) = 700 v 2 b = 73.575 m 2 /s 2 and 0 + 700(9.81) = 700 j v 2 c = 49.05 m 2 /s 2 Judging from the above results, the coster car will not leave the loop at C if it safely passes through B. Thus N b = 0 Ans. Conservation of Energy. The datum will be set at the ground level. With v B = 73.575 m 2 /s 2 , T a + V a = T b + V b N 1 1 — (700)(3 2 ) + 700(9.81)6 = - (700)(73.575) + 700(9.81)(15) h = 18.29 m = 18.3 m And from B to C, T b + V B = T c + V c | (700)(73.575) + 700(9.81)(15) = | (700)w 2 + 700(9.81)(10) v 2 c = 171.675 m 2 /s 2 > 49.05 m 2 /s 2 Substitute this result into Eq. 1 with p c = 5 m. N c + 700(9.81) N c 700 171.675 17.17(10 3 ) N = 17.2 kN Ans. (O.K!) Ans. Ans: N b = 0 h = 18.3 m N c = 17.2 kN 448 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 73 . The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, determine the minimum height h of the hill crest so that the car travels around both inside the loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take p B = 7.5 m and Pc = 5 m. A SOLUTION Equation of Motion. Referring to the FBD of the roller-coaster car shown in Fig. a, 1F n = ma n ; N + 700(9.81) = 700 j (1) When the roller-coaster car is about to leave the loop at B and C, N = 0. At B and C, p B = 7.5 m and p c = 5 m. Then Eq. (1) gives 0 + 700(9.81) = 700^||^) v\ = 73.575 m 2 /s 2 and 0 + 700(9.81) = 700 (y) = 49.05 m 2 /s 2 Judging from the above result the coaster car will not leave the loop at C provided it passes through B safely. Thus N b = 0 Ans. N Conservation of Energy. The datum will be set at the ground level. Applying Eq. 14- from A to B with v B = 73.575 m 2 /s 2 , T a + V a = T b + V b 1 0 + 700(9.81 )h = - (700)(73.575) + 700(9.81)(15) h = 18.75 m Ans. And from B to C, 7fl + V B = T c + Vc | (700)(73.575) + 700(9.81)(15) = y(700)?£ + 700(9.81)(10) v 2 c = 171.675 m 2 /s z > 49.05 m 2 /s 2 (O.K!) Substitute this result into Eq. 1 with p c = 5 m, N c + 700(9.81) = 700 ( - j -J N c = 17.17(10 3 )N = 17.2 kN Ans. Ans: N b = 0 h = 18.75 m N c = 17.2 kN 449 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 74 . The assembly consists of two blocks A and B which have a mass of 20 kg and 30 kg, respectively. Determine the speed of each block when B descends 1.5 m. The blocks are released from rest. Neglect the mass of the pulleys and cords. SOLUTION 3s A + s B = 1 3As,4 = —A Sg 3v a = -v b 71 + y lS t 2 + u 2 (0 + 0) + (0 + 0) = \m{v A ) 2 + i(30)(-3^) 2 + 20(9.81)^) - 30(9.81)( 1.5) v A = 1.54 m/s Ans. v B = 4.62 m/s Ans. Ans: v A = 1.54 m/s v B = 4.62 m/s 450 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 75 . The assembly consists of two blocks A and B , which have a mass of 20 kg and 30 kg, respectively. Determine the distance B must descend in order for A to achieve a speed of 3 m/s starting from rest. SOLUTION 3 ^ + s B = l 3\s A = —A s B 3v a = ~v B v B = -9 m/s 7/ + Vl = t 2 + V 2 (0 + 0) + (0 + 0) = !-(20)(3) 2 + |(30)(-9) 2 + 20(9.81)^) - 30(9.81)( % ) s B = 5.70 m Ans. Ans: s B = 5.70 m 451 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 76 . The spring has a stiffness k = 50 N/m and an unstretched length of 0.3 m. If it is attached to the 2-kg smooth collar and the collar is released from rest at A (8 = 0°), determine the speed of the collar when 0 = 60°. The motion occurs in the horizontal plane. Neglect the size of the collar. SOLUTION Potential Energy. Since the motion occurs in the horizontal plane, there will be no change in gravitational potential energy when 6 = 0°, the spring stretches x 1 = 4 — 0.3 = 3.7 m. Referring to the geometry shown in Fig. a, the spring stretches = 4 cos 60° — 0.3 = 1.7 m. Thus, the elastic potential energies in the spring when 0 = 0° and 60° are z (K), ’ kx] : i(50)(3.7 2 ) = 342.25 J (Veh = \kx\ = ^ (50)(1.7 2 ) = 72.25 J Conservation of Energy. Since the collar is released from rest when 9 = 0°, 7', = 0. 7i + V 1 = T 2 + V 2 \ 0 + 342.25 = - (2)v 2 + 72.25 v = 16.43 m/s = 16.4 m/s Ans. Ans: v = 16.4 m/s 452 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 77 . The roller coaster car having a mass m is released from rest at point A. If the track is to be designed so that the car does not leave it at B, determine the required height h. Also, find the speed of the car when it reaches point C. Neglect friction. SOLUTION A Equation of Motion: Since it is required that the roller coaster car is about to leave 2 2 v B v B the track at B, N B = 0. Here, a n =-=-. By referring to the free-body Pb 7.5 diagram of the roller coaster car shown in Fig. a, V = 73W/s* Potential Energy: With reference to the datum set in Fig. b. the gravitational potential energy of the rollercoaster car at positions A, B, and C are {V g ) A ~ m gh A ~ m(9.81)/z = 9.81 mh, = mgh B = m(9.81)(20) = 196.2 m, and (v g )c = mgh c = m(9.81)(0) = 0. Conservation of Energy: Using the result of v B 2 and considering the motion of the car from position A to B, t a + v a = t b + v b (a) 1 2 mv A 2 + (V g ) A = 1 2 mv B 2 + [V g ) B 0 + 9.81 mh = - m(73.575) + 196.2m h = 23.75 m Ans. Also, considering the motion of the car from position B to C, T b + V b = T c + Vc 2 ™s 2 + (Vg) B = \ rnv c 2 + (v g ) c 1 1 — m(73.575) + 196.2m = —mv c 2 + 0 Vc — 21.6 m/s Ans. A C bj Ans: h = 23.75 m v c = 21.6 m/s 453 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 78 . The spring has a stiffness k = 200 N/m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and the collar is released from rest at A , determine the speed of the collar when it reaches B. Neglect the size of the collar. SOLUTION Potential Energy. With reference to the datum set through B, the gravitational potential energies of the collar at A and B are (V g ) A = mgh A = 3(9.81)(2) = 58.86 J (V g ) B = 0 At A and B. the spring stretches x A = Vl.S 2 + 2 2 — 0.5 = 2.00 m and x B = 1.5 — 0.5 = 1.00 m. Thus, the elastic potential energies in the spring when the collar is at A and B are (V e ) A = \kx\ = |(200)(2.00 2 ) = 400 J (K) b = l kxl = i(200)(l.00 2 ) = 100 J Conservation of Energy. Since the collar is released from rest at A, T A = 0. T a +V a = T b + V b \ 0 + 58.86 + 400 = - (3)v 2 B + 0 + 100 v B = 15.47 m/s = 15.5 m/s Ans. Ans: v B = 15.5 m/s 454 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 79 . A 2-lb block rests on the smooth semicylindrical surface at A. An elastic cord having a stiffness of k = 2 lb/ft is attached to the block at B and to the base of the semicylinder at C. If the block is released from rest at A , determine the longest unstretched length of the cord so the block begins to leave the cylinder at the instant 8 = 45°. Neglect the size of the block. SOLUTION Equation of Motion: It is required that N = 0. Applying Eq. 13-8, we have SE, = ma ■ 2 cos 45° = —— (—— ^ v 2 = 34.15 m 2 /s 2 Potential Energy: Datum is set at the base of cylinder. When the block moves to a position 1.5 sin 45° = 1.061 ft above the datum, its gravitational potential energy at this position is 2(1.061) = 2.121 ft • lb. The initial and final elastic potential energy are ^(2) \tt (1.5) — /] 2 and^(2) [0.75 tt( 1.5) — /] 2 , respectively. Conservation of Energy: 0 + |(2) [7r(1.5) - if l = 2.77 ft Ans. X7i + %V X = 27) + XC 2 = K# 3415) + Zm + ^t 0 - 75 ^ 1 ’ 5 ) " ^ Ans: / = 2.77 ft 455 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 80 . When 5 = 0, the spring on the firing mechanism is unstretched. If the arm is pulled back such that 5 = 100 mm and released, determine the speed of the 0.3-kg ball and the normal reaction of the circular track on the ball when 0 = 60°. Assume all surfaces of contact to be smooth. Neglect the mass of the spring and the size of the ball. SOLUTION Potential Energy. With reference to the datum set through the center of the circular track, the gravitational potential energies of the ball when 0 = 0° and 8 = 60° are (V g ) i = -mgh x = —0.3(9.81)(1.5) = -4.4145 J (V g ) 2 = -mgh 2 = — 0.3(9.81)(1.5 cos 60°) = -2.20725 J When 0 = 0°, the spring compress x 1 = 0.1 m and is unstretched when 6 = 60°. Thus, the elastic potential energies in the spring when 6 = 0° and 60° are (Veh = \kx] = i(1500)(0.1 2 ) = 7.50 J (vy z = 0 Conservation of Energy. Since the ball starts from rest, 1] = 0. t 1 + v 1 = t 2 + y 2 o + (-4.4145) + 7.50 = ^ (0.3)w 2 + (-2.20725) + 0 v 2 = 35.285 m 2 /s 2 v = 5.94 m/s Equation of Motion. Referring to the FBD of the ball, Fig. a. 2 35.285' XF n = ma n ; N - 0.3(9.81) cos 60° = 0.3 1.5 Ans. N = 8.5285 N = 8.53 N Ans. Ans: v = 5.94 m/s N = 8.53 N 456 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 81 . When s = 0, the spring on the firing mechanism is unstretched. If the arm is pulled back such that s = 100 mm and released, determine the maximum angle 8 the ball will travel without leaving the circular track. Assume all surfaces of contact to be smooth. Neglect the mass of the spring and the size of the ball. SOLUTION Equation of Motion. It is required that the ball leaves the track, and this will occur provided 8 > 90°. When this happens, A = 0. Referring to the FBD of the ball, Fig. a = ma n ; 0.3(9.81) sin (8 - 90°) = v 2 = 14.715 sin (8 - 90°) (1) Potential Energy. With reference to the datum set through the center of the circular track Fig. b, the gravitational potential Energies of the ball when 8 = 0 ° and 8 are (V g \ = -mgh x = —0.3(9.81)(1.5) = -4.4145 J (V g ) 2 = mgh 2 = 0.3(9.81)[1.5 sin (8 - 90°)] = 4.4145 sin (8 - 90°) When 8 = 0°, the spring compresses x j = 0.1 m and is unstretched when the ball is at 8 for max height. Thus, the elastic potential energies in the spring when 8 = 0° and 8 are (K)i = \kx\ = i(1500)(0.1 2 ) = 7.501 (v e ) 2 = 0 Conservation of Energy. Since the ball starts from rest, 7) = 0. t 1 + v 1 = t 2 + v 2 1 o + (-4.4145) + 7.50 = - (0.3)v 2 + 4.4145 sin (8 - 90°) + 0 v 2 = 20.57 - 29.43 sin (8 - 90°) (2) Equating Eqs. (1) and (2), 14.715 sin (8 - 90°) = 20.57 - 29.43 sin (8 - 90°) sin (8 - 90°) = 0.4660 8 - 90° = 21.IT 8 = 111.IT = 118° Ans. t Ans: 8 = 118° 457 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 82 . If the mass of the earth is M e , show that the gravitational potential energy of a body of mass m located a distance r from the center of the earth is V g = -GM e m/r. Recall that the gravitational force acting between the earth and the body is F= G(M e m/r 2 ), Eq. 13-1. For the calculation, locate the datum at r —» oo. Also, prove that F is a conservative force. SOLUTION The work is computed by moving F from position r 1 to a farther position r 2 . = —G M e m = - G M e m _ 1 _ As r 1 —» oo , let r 2 = r u F 2 = F 2 , then V a -G m To be conservative, require F = -V y g = d ( GM e m \ dr \ r ) -G M e m Q.E.D. \ Ans: F = — G M e m 458 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 83 . A rocket of mass m is fired vertically from the surface of the earth, i.e., at r = r } . Assuming no mass is lost as it travels upward, determine the work it must do against gravity to reach a distance r 2 . The force of gravity is F = GM e m/r 2 (Eq. 13-1), where M e is the mass of the earth and r the distance between the rocket and the center of the earth. SOLUTION r = g- M P m f f ' 2 d r F i _ 2 = j F dr = GM e m J 2 = GM„m 1 _ 1 G r 2 Ans. Ans: F = GM e m 459 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 84 . The 4-kg smooth collar has a speed of 3 m/s when it is at 5 = 0. Determine the maximum distance s it travels before it stops momentarily. The spring has an unstretched length of 1 m. SOLUTION Potential Energy. With reference to the datum set through A the gravitational potential energies of the collar at A and B are (V g ) A = 0 (V g ) B = —mgh B = -4(9.81) S max = -39.24 S max At A and B, the spring stretches x A = 1.5 — 1 = 0.5 m and x B = V S^ ax + 1.5 2 — 1. Thus, the elastic potential Energies in the spring when the collar is at A and B are (V e ) A = \kx\ = i (100)(0.5 2 ) = 12.5 J (V e ) B = ^ kx 2 H = |(100)(VSL* + 1-5 2 - l) 2 = 50 (S 2 max - 2Vs 2 max + 1.5 2 + 3.25) Conservation of Energy. Since the collar is required to stop momentarily at B , 7 /( = 0 . T a + V a =T b + v b i(4)(3 2 ) + 0 + 12.5 = 0 + (—39.24 S max ) + 50 (S 2 max - 2 Vs 2 max + 1.5 2 + 3.25) 50 S 2 max - lOOVsLx + 1-5 2 - 39.24 S max + 132 = 0 Solving numerically, Smax = 1.9554 m = 1.96 m Ans. 1.5 m- k = 100 N/i — I—3 m/s ifr Ans: ^max 1.96 m 460 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 85 . A 60-kg satellite travels in free flight along an elliptical orbit such that at A, where r A = 20 Mm, it has a speed v A = 40 Mm/h. What is the speed of the satellite when it reaches point B, where r B = 80 Mm? Hint: See Prob. 14-82, where M e = 5.976(10 24 ) kg and G = 66.73(10~ 12 ) m 3 /(kg • s 2 ). SOLUTION v A = 40 Mm/h = 11 111.1 m/s Since V = GM e m r t 1 + v 1 = t 2 + v 2 66.73(10)~ 12 (5.976)(10) 23 (60) 1 i(60)(ll 111.1 r - 2 V 20(10) 6 v B = 9672 m/s = 34.8 Mm/h = ^(60 )v 2 b 66.73(10)~ 12 (5.976)(10) 24 (60) 80(10) 6 Ans. Ans: v B = 34.8 Mm/h 461 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-86. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed v B when he reaches B. Also, compute the distance v to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass of 70 kg. SOLUTION Ta+ V A = T b + V b 0 + 70(9.81) (46) = | (70)i> 2 + 0 v = 30.04 m/s = 30.0 m/s i 1 9 (+1) S y = (5y)o + (VoV + 2 a <f 4 + x sin 30° = 0 + 0 + j (9.81 )t 2 (<±)S X = v x t s cos 30° = 30.04f s = 130 m t = 3.75 s Ans: s = 130 m A Ans. ( 1 ) ( 2 ) Ans. 462 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14 - 87 . The block has a mass of 20 kg and is released from rest when 5 = 0.5 m. If the mass of the bumpers A and B can be neglected, determine the maximum deformation of each spring due to the collision. SOLUTION Datum at initial position: T 1 + V 1 = T 2 + V 2 0 + 0 = 0 + ^(500)4 + ^(800)4 + 20(9.81)[-(s a + s B ) - 0.5] Also, = 5005 A = 800 j b s a = 1.6 s B Solving Eqs. (1) and (2) yields: s B = 0.638 m s A = 1.02 m ( 1 ) ( 2 ) Ans. Ans. Ans: S B = ■sa = = 500 N/m = 800 N/m 0.638 m 1.02 m 463 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 14 - 88 . The 2-lb collar has a speed of 5 ft/s at A. The attached spring has an unstretched length of 2 ft and a stiffness of k = 10 lb/ft. If the collar moves over the smooth rod, determine its speed when it reaches point B, the normal force of the rod on the collar, and the rate of decrease in its speed. SOLUTION Datum at B: t a + v a = t b + v b + i <10) < 4 - 5 - 2)2 + 2(4 - 5) ■ Kiu )'-"' 2 + i (10)<3 - 2)2 + 0 v B = 34.060 ft/s = 34.1 ft/s Ans. y I A [ -3 ft- \ y dy — = tan# = — x dx 0 = -71.57° = —3 x=3 d*y dx 2 2 4 ' IT = -i [i + (-3) 2 r> d y dx 2 Ml = 31.623 ft +AYF n = ma n ; -N + 10 cos 18.43° + 2 cos 71.57° / 2 W (34.060) 2 \ \32.2 / \ 31.623 / N = 7.84 lb +\~ZF, = ma t ; 2 sin 71.57° - 10 sin 18.43° = a t = —20.4 ft/s 2 Ans. Ans. 2 lb Ans: v B = 34.1 ft/s N = 7.84 lb a, = -20.4 ft/s 2 464 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-89. When the 6-kg box reaches point A it has a speed of v A = 2 m/s. Determine the angle d at which it leaves the smooth circular ramp and the distance s to where it falls into the cart. Neglect friction. SOLUTION At point B: +/~ZF n — ma n \ 6(9.81) cos <t> = 6 Datum at bottom of curve: t a + v a = t b + v b ^(6)(2) 2 + 6(9.81)(1.2 cos 20°) = |(6)(v B ) 2 + 6(9.81)(1.2 cos </>) 13.062 = 0.5v| + 11.772 cos </> Substitute Eq. (1) into Eq. (2), and solving for v B , v B = 2.951 m/s .( (2.951) 2 \ ThUS ’ * = C ° S (l^ 81 ) j = 4Z29 ° 0 = (j) - 20° = 22.3° ( + T) s = s 0 + v 0 t + \a c t 2 1 o -1.2 cos 42.29° = 0 - 2.951(sin 42.29°)t + -(-9.81)f 2 4.905f 2 + 1.9857f - 0.8877 = 0 Solving for the positive root: t = 0.2687 s ( j s = 5 0 + v 0 t s = 0 + (2.951 cos 42.29°)(0.2687) s = 0.587 m U1.71) M Ans. Ans: 0 = 22.3° i = 0.587 m 465 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-90. When the 5-kg box reaches point A it has a speed v A = 10 m/s. Determine the normal force the box exerts on the surface when it reaches point B. Neglect friction and the size of the box. SOLUTION Conservation of Energy. At point B.y = x 1 1 XP- + Xi = 3 9 x = — m 4 y Then y = — m. With reference to the datum set to coincide with the x axis, the 4 gravitational potential energies of the box at points A and B are <y.) A = o {V„) B = mgh B = 5(9.81) - = 110.3625 J Applying the energy equation, T a +V a = T b + V b f (5)(l0 2 ) + 0 = ^ (5)«s + 110.3625 v B = 55.855 m 2 /s 2 Equation of Motion. Here, y = ( 3 — x 2 ) 2 . Then, ^ = 2(3 - x*)y-^x 2 xi - 3 „ 3 , d 2 y 3 _3 3 A . „ 9 _ = --— =1-- and —r = — x 2 = —At point B.x = — m. Ihus, Xi Xi d x ^ 2xi ^ tan d B = d 2 y dx 2 dy dx 9 ^ / 9\1 x=?m (ij 2 = -1 e B = -45° = 45° 2(5)* = 0.4444 The radius of curvature at B is [1 + (dy/dx) 2 fi [1 + (-l) 2 ]f B \d 2 y/dx 2 \ Referring to the FBD of the box, Fig. a 0.4444 = 6.3640 m lF n = ma n ; N - 5(9.81) cos 45° = 5| N = 78.57 N = 78.6 N 55.855 \ 6.3640/ Ans. Ans: N = 78.6 N 466 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-91. When the 5-kg box reaches point A it has a speed v A — 10 m/s. Determine how high the box reaches up the surface before it comes to a stop. Also, what is the resultant normal force on the surface at this point and the acceleration? Neglect friction and the size of the box. SOLUTION Conservation of Energy. With reference to the datum set coincide with x axis, the gravitational potential energy of the box at A and C (at maximum height) are y (V g ) A = 0 (V g )c = mgh c = 5(9.81)0) = 49.05y It is required that the box stop at C. Thus, T c = 0 T a + V a =T c + V c j(5)(10 2 ) + 0 = 0 + 49.05y y = 5.0968 m = 5.10 m Ans. Then, + 5.09682 = 3 x = 0.5511 m Equation of Motion. I Icrc.y = (3 — x 2 ) 2 .Then,— = 2(3 — x 2 ) (— —x 2 dx X2“ - 3 = 1 - dy X 2 — and—r = —x 2 = —- At point C,x = 0.5511m. x* dx 2 2 x 2 Thus tan 9 r = dy dx = 1 - *= 0.5511 m 0.5511 2 T = -3.0410 d c = -71.80° = 71.80° Referring to the FBD of the box, Fig. a. 1F n = ma n ; N - 5(9.81) cos 71.80° = 5 N = 15.32 N = 15.3 N = ma, ■ -5(9.81) sin 71.80° = 5 a, a, = -9.3191 m/s 2 = 9.32 m/s 2 \ Since a n = 0, Then a = a t = 9.32 m/s 2 \ Ans. Ans. Ans: y = 5.10 m N = 15.3 N a = 9.32 m/s 2 \ 467 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *14-92. The roller-coaster car has a speed of 15 ft/s when it is at the crest of a vertical parabolic track. Determine the car’s velocity and the normal force it exerts on the track when it reaches point B. Neglect friction and the mass of the wheels. The total weight of the car and the passengers is 350 lb. SOLUTION l , y =-(40 000 - x 2 ) J onn v ’ dy dx dx 1 Datum at A: = -2, 6 = tan _1 (—2) = -63.43° i=200 1 Too l - X 100 Ta + V A = T b + if 350 \ is? + o= 3 f 350 w - 2 V 32.2 2 V 32.2 {v b Y ~ 350(200) v B = 114.48 = 114 ft/s . + 1 f' ax [1 + (-2) z p d 2 y dx 2 1 'Too = 1118.0 ft +/^F n = ma n \ 350 cos 63.43° — N B = N b = 29.1 lb _ / 350 \(H4.48) 2 “ \322J 1118.0 y v A = 15 ft/s Ans. Ans. Ans: v B = 114 ft/s N b = 29.1 lb 468 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-93. The 10-kg sphere C is released from rest when 0 = 0° and the tension in the spring is 100 N. Determine the speed of the sphere at the instant 0 = 90°. Neglect the mass of rod AB and the size of the sphere. SOLUTION Potential Energy: With reference to the datum set in Fig. a , the gravitational potential energy of the sphere at positions (1) and (2) are (UgL = mghi = 10(9.81)(0.45) = 44.145 J and (y g ) 2 = mgh 2 = 10(9.81)(0) = 0. When the sphere is at position (1), the spring stretches sj = = 0.2 m. Thus, the unstretched length of the spring is l 0 = \/o.3 2 + 0.4 2 — 0.2 = 0.3 m, and the elastic potential energy of the spring is ( Vg)i = ^ksi = ^(500)(0.2 2 ) = 10 J. When the sphere is at position (2), the spring stretches s 2 = 0.7 — 0.3 = 0.4 m, and the elastic potential energy of the spring is {V e ) 2 = \ks 2 1 = |(500)(0.4 2 ) = 40 J. Conservation of Energy: t 1 + v 1 = t 2 + v 2 2 m *\ v s)i + K)i + {Ve) i = 2 m A v s)i + [V g )i + ( v e ) 2 0 + (44.145 + 10) = -(10)(u s ) 2 2 + (0 + 40) (v s ) 2 = 1-68 m/s Ans. 469 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 470 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-95. The cylinder has a mass of 20 kg and is released from rest when h = 0. Determine its speed when h = 3 m. Each spring has a stiffness k = 40 N/m and an unstretched length of 2 m. SOLUTION t 1 + v 1 = t 2 + v 2 '1 0 + 0 = 0 + 2 v = 6.97 m/s (40) (V3 2 + 2 2 - 2 2 ) - 20(9.81)(3) + -(20)v 2 Ans. Ans: v = 6.97 m/s 471 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *14-96. If the 20-kg cylinder is released from rest at h = 0, determine the required stiffness k of each spring so that its motion is arrested or stops when h = 0.5 m. Each spring has an unstretched length of 1 m. SOLUTION T\ + V\ = T 2 + V 2 0 + 2 k(2 - l) 2 = 0 - 20(9.81)(0.5) + 2 ;k(V(2) 2 + (0.5) 2 - l) 2 k = -98.1 + 1.12689 k k = 773 N/m Ans. Ans: k = 773 N/m 472 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 14-97. A pan of negligible mass is attached to two identical springs of stiffness k = 250 N/m. If a 10-kg box is dropped from a height of 0.5 m above the pan, determine the maximum vertical displacement d. Initially each spring has a tension of 50 N. SOLUTION Potential Energy: With reference to the datum set in Fig. a, the gravitational potential energy of the box at positions (1) and (2) are (v g )i = mghy = 10(9.81)(0) = 0 and (V^2 ~ m Shi — 10(9.81)[ — (o.5 + rZ)] = —98.l(o.5 + rf). Initially, the spring 50 stretches ,?! = = 0.2 m. Thus, the unstretched length of the spring is l Q = 1 — 0.2 = 0.8 m and the initial elastic potential of each spring is (y e )j = (2)— ks\ 2 = 2(250 / 2)(0.2 2 ) = 10 J. When the box is at position (2), the spring stretches s 2 = ^VrZ 2 + l 2 — 0.8^) m. The elastic potential energy of the springs when the box is at this position is (v%) 2 =(2)|W = 2(250/2) Conservation of Energy: T l + V l + T 2 + V 2 Vd 2 + 1 - 0.8 250 (d 2 - 1.6 Vd 2 + 1 + 1.64^. 1 2 -mvf + y g ) { + Kh = - mv 2 + V, {v e ) 2 0 + (o + 10) = 0 + —98.l(o.5 + d) + 250^rZ 2 - 1.6 V d 2 + 1 + 1.64 250rZ 2 - 98.1rZ - 400Vd 2 + 1 + 350.95 = 0 Solving the above equation by trial and error, d = 1.34 m Ans. L=(0.5idjnn Ca; Ans: d = 1.34 m 473 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-1. A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a distance of 12 m away. Determine the impulse of his foot on the ball at A. Neglect the impulse caused by the ball’s weight while it’s being kicked. SOLUTION Kinematics. Consider the vertical motion of the ball where (%)_,, = s y = 0, (r» 0 )j, = v sin 60° | and a y = 9.81 m/s 2 i , (+t) s y = (s 0 ) y + (v 0 ) y t + ^ a y t 2 ; 0 = 0 + v sin 60°f + y(-9.81)f 2 t(v sin 60° - 4.905f) = 0 Since t A 0, then v sin 60° — 4.905f = 0 t = 0.1766 v Then, consider the horizontal motion where (v 0 ) x = v cos 60°, and (%)* ( ^ ) = (■%)* + ( v 0 ) x t ■ 12 = 0 + v cos 60°f 24 t = — v Equating Eqs. (1) and (2) 0.1766 v = 24 v = 11.66 m/s Principle of Impulse and Momentum. (+/*) mv i + % / Fdt = mv 2 0 + I = 0.15 (11.66) 7 = 1.749 N-s = 1.75 N-s ( 1 ) 0 , ( 2 ) Ans. \/ Ans: v = 1.75 N-s 474 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 2 . A 20-lb block slides down a 30° inclined plane with an initial velocity of 2 ft/s. Determine the velocity of the block in 3 s if the coefficient of kinetic friction between the block and the plane is = 0.25. SOLUTION (+$$ m{v v ') + 2 f Fydt = m(Vy ') 2 • It 1 0 + N( 3) - 20 cos 30°(3) = 0 N = 17.32 lb [h V/) + 2 / F x dt = m(v y) 2 Jt 1 20 20 —(2) + 20 sin 30°(3) - 0.25(17.32)(3) = — v v = 29.4 ft/s Ans. Ans: v = 29.4 ft/s 475 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 476 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 4 . Each of the cables can sustain a maximum tension of 5000 lb. If the uniform beam has a weight of 5000 lb, determine the shortest time possible to lift the beam with a speed of 10 ft/s starting from rest. SOLUTION + T SFy = 0: Pmax ~ 2^(5000) = 0 Pmax = 8000 lb (+ \ ) mv | + X iF dt = mv 2 0 + 8000(r) - 5000(0 = ^(10) t = 0.518 s Ans. Ans: t = 0.518 s 477 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 5 . A hockey puck is traveling to the left with a velocity of Vi = 10 m/s when it is struck by a hockey stick and given a velocity of t) 2 = 20 m/s as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of 0.2 kg. SOLUTION Vi = {—10i} m/s v 2 = {20 cos 40°i + 20 sin 40°jj m/s I = m\v = (0. 2) {[20 cos 40° - (— 10)]i + 20 sin 40°]j = {5.0642i + 2.5712j) kg • m/s I = V(5.0642) 2 + (2.5712) 2 = 5.6795 = 5.68 kg-m/s Ans. Ans: I = 5.68 N • s 478 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 6 . A train consists of a 50-Mg engine and three cars, each having a mass of 30 Mg. If it takes 80 s for the train to increase its speed uniformly to 40 km/h, starting from rest, determine the force T developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resultant frictional tractive force F which gives the train forward motion, whereas the car wheels roll freely. Also, determine F acting on the engine wheels. SOLUTION ( v x ) 2 = 40 km/h = 11.11 m/s Entire train: ■J* m(v x ) i + 2 0 + F(80) = [50 + 3(30)]( 10 3 )(11.11) F = 19.4 kN Ans. Three cars: 3 ^ m{v x )\ + 2 J F x dt = m(v x ) 2 0 + 7X80) = 3(30)(l0 3 )(ll.ll) T = 12.5 kN Ans. Ans: F = 19.4 kN T = 12.5 kN 479 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 7 . Crates A and B weigh 100 lb and 50 lb, respectively. If they start from rest, determine their speed when r = 5 s. Also, find the force exerted by crate A on crate B during the motion. The coefficient of kinetic friction between the crates and the ground is p k = 0.25. SOLUTION Free-Body Diagram: The free-body diagram of crates A and B are shown in Figs, a and b , respectively. The frictional force acting on each crate is (Ff) A ~ = 0.25 Af* and ( Ff) B = p k N B = 025N B . Principle of Impulse and Momentum: Referring to Fig. a, (+t) ?(«i)v + 2 / F v dt = m(v 2 ) y H ( o) + ^ (5) - 100(5) = ll (0) N a = 100 lb ( _+ ) m(v l ) x + 2 / F x dt = m(v 2 ) Jh 100 322 v = 40.25 - 1.61 F^ By considering Fig. b , — (0) + 50(5) - 0.25(100)(5) - F AB ( 5) = ^v (+t) i(Vi) y + 2 / F y dt = m(v 2 ) y ( 1 ) (i) ii«» + N ‘< 5 > - 50 < 5 > - 5U (0) N r = 50 lb m(v i) x + 2 / F x dt = m(v 2 ) x ^(0) + F ab { 5) - 0.25(50)(5) = ^ v v = 322F AB - 40.25 ( 2 ) Solving Eqs. (1) and (2) yields F AB = 16.67 lb = 16.7 lb v = 13.42 ft/s = 13.4 ft/s Ans. Ans: F ab = 16.71b v = 13.4 ft/s 480 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 8 . The automobile has a weight of 2700 lb and is traveling forward at 4 ft/s when it crashes into the wall. If the impact occurs in 0.06 s, determine the average impulsive force acting on the car. Assume the brakes are not applied. If the coefficient of kinetic friction between the wheels and the pavement is pL k = 0.3, calculate the impulsive force on the wall if the brakes were applied during the crash.The brakes are applied to all four wheels so that all the wheels slip. SOLUTION Impulse is area under curve for hole cavity. I = j Fdt = 4(3) + | (8 + 4)(6 - 3) + |(8)(10 - 6) = 46 lb • s For starred cavity: I = Fdt = 6(8) + ^ (6)(10 - 8) = 54 lb • s Ans. Ans. ran Ans: I = 46 lb • s I = 54 lb • s 481 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 9 . The 200-kg crate rests on the ground for which the coefficients of static and kinetic friction are p s = 0.5 and p k = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the crate when 1 = 4 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the crate. SOLUTION Equilibrium. The time required to move the crate can be determined by considering the equilibrium of the crate. Since the crate is required to be on the verge of sliding, Ff = p, s N = 0.5 N. Referring to the FBD of the crate, Fig. a, + | lF y = 0; N - 200(9.81) = 0 N = 1962 N XF x = 0; 2(400ri) - 0.5(1962) = 0 t = 1.5037 s Principle of Impulse and Momentum. Since the crate is sliding, Ff = p, k N = 0.4(1962) = 784.8 N. Referring to the FBD of the crate, Fig. a / 'h F x dt = m(v x ) 2 h rA s 0 + 2 / 400 fidt - 784.8(4 - 1.5037) = 200w J 1.5037 s T(N) ZOotf-BO^ v = 6.621 m/s = 6.62 m/s Ans. O) Ans: v = 6.62 m/s 482 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 10 . The 50-kg crate is pulled by the constant force P. If the crate starts from rest and achieves a speed of 10 m/s in 5 s, deter¬ mine the magnitude of P. The coefficient of kinetic friction between the crate and the ground is pu k = 0.2. SOLUTION Impulse and Momentum Diagram: The frictional force acting on the crate is F f = p k N = 0.2N. Principle of Impulse and Momentum: P*2 (+1) rn(vi) y + 2 / F y dt = m(v 2 ) y Jh 0 + N( 5) + P{ 5) sin 30° - 50(9.81)(5) = 0 N = 490.5 - 0.5P rh ( X ) m M x + 2 / F x dt = m(v 2 ) x Jt i 50(0) + P{ 5) cos 30° - 0.2W(5) = 50(10) 4.3301P — N = 500 Solving Eqs. (1) and (2), yields N = 387.97 N P = 205 N ( 1 ) ( 2 ) Ans. 50 ( 0 ) + HiP) Ans: P = 205 N 483 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 11 . During operation the jack hammer strikes the concrete surface with a force which is indicated in the graph. To achieve this the 2-kg spike S is fired into the surface at 90 m/s. Determine the speed of the spike just after rebounding. SOLUTION Principle of Impulse and Momentum. The impulse of the force F is equal to the area under the F-t graph. Referring to the FBD of the spike, Fig. a F (kN) 0 0.1 0.4 t (ms) rh ( + t) >n(v y ) 1 + 2 / Fy dt = m(v y ) 2 ■' h 2(—90) + — [0.4(10 3 ) ] [1500(10 3 ) ] = 2v v = 60.0 m/s | Ans. v / V F Ans: v = 60.0 m/s 484 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 12 . For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is F D = (600f 2 ) N, where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t — 5 s. SOLUTION Principle of Impulse and Momentum: The initial speed of the van is = 20(10 3 ) lh 3600 s (i) = 5.556 m/s. Referring to the free-body diagram of the van shown in Fig. a , m(v 0* + 2 / F x dt = m(v 2 ) x ■it i p5s 2500(5.556) + / 600t dt = 2500 v 2 Jo v 2 = 15.6 m/s Ans. V 2?oo(W/J -x | fc6oot A/ W Ans: v 2 = 15.6 m/s 485 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 13 . The 2.5-Mg van is traveling with a speed of 100 km/h when the brakes are applied and all four wheels lock. If the speed decreases to 40 km/h in 5 s, determine the coefficient of kinetic friction between the tires and the road. SOLUTION Free-Body Diagram: The free-body diagram of the van is shown in Fig. a. The frictional force is Ff = p k N since all the wheels of the van are locked and will cause the van to slide. Principle of Impulse and Momentum: The initial and final speeds of the van are «i = 100 ( 10 3 ) — 1 h 3600 s = 27.78 m/s and v 2 = 40(10 3 ) — 1 h 3600 s = 11.11 m/s. Referring to Fig. a, (+t) m(vi) y + 2 / F y dt = m(v 2 ) y ■it, 2500(0) + N( 5) - 2500(9.81)(5) = 2500(0) N = 24 525 N rh (<+ ) m{v r ) x + 2 / F x dt = m(v 2 ) x Jti 2500(27.78) + [-^(24525) (5)] = 2500(11.1) p k = 0.340 Ans. Ans: p, k = 0.340 486 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 14 . A tankcar has a mass of 20 Mg and is freely rolling to the right with a speed of 0.75 m/s. If it strikes the barrier, determine the horizontal impulse needed to stop the car if the spring in the bumper B has a stiffness (a) k —» oo (bumper is rigid), and (b) k = 15 kN/m. v = 0.75 m/s SOLUTION a) b) ( -L ) mv l + X F dt = mv 2 20(l0 3 )(0.75) Fdt = 15 kN • s Ans. The impulse is the same for both cases. For the spring having a stiffness k = 15 kN/m, the impulse is applied over a longer period of time than for k —» °°. Ans: I = 15 kN • s in both cases. 487 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 15 . The motor, M, pulls on the cable with a force F = (lOr 2 + 300) N, where t is in seconds. If the 100 kg crate is originally at rest at t = 0, determine its speed when t = 4 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. SOLUTION Principle of Impulse and Momentum. The crate will only move when 3(l0f 2 + 300) = 100(9.81). Thus, this instant is t = 1.6432 s. Referring to the FBD of the crate, Fig. a. rh ( + T) rn(y y )i + X / F y dt = m(v y ) 2 J f, n4 s 0 + / 3(l0t 2 + 300) dt - 100(9.81)(4 - 1.6432) = 100u J 1.6432 s 4 s - 2312.05 = 100u v = 4.047 m/s = 4.05 m/s | Ans. F Ff (A) Ans: v = 4.05 m/s 488 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 16 . The choice of a seating material for moving vehicles depends upon its ability to resist shock and vibration. From the data shown in the graphs, determine the impulses created by a falling weight onto a sample of urethane foam and CONFOR foam. SOLUTION CONFOR foam: l c = J Fdt = = 6.55 N • ms |(2)(0.5) + ^(0-5 + 0.8)(7 - 2) + |(0.8)(14 - 7) (io- 3 ) Ans. F( N) t (ms) Urethane foam: I„ = F dt = |(4)(0.3) + |(1.2 + 0.3)(7 - 4) + |(1.2 + 0.4)(10 - 7) + Ans: I c = 6.55 N • ms I v = 6.05 N • ms 489 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 17 . The towing force acting on the 400-kg safe varies as shown on the graph. Determine its speed, starting from rest, when t = 8 s. How far has it traveled during this time? SOLUTION Principle of Impulse and Momentum. The FBD of the safe is shown in Fig. a. For 0<f<5s,F = = 120f. Ph ( ^ ) m(v x ) i + X F x dt = m(y x ) 2 A F( N) 750 — 600 ■t( s) 0 + / 120 t dt = 400?; Jo n/ At t = 5 s, v = { 0.15r 2 } m/s v = 0.15(5 2 ) = 3.75 m/s F - 600 750 - 600 For 5 s < t < 8 s,-= -, F = 50f + 350. Here, t - 5 8-5 ( v x )i = 3.75 m/s and f — 5 s. rh ( ) m{v x \ + X I F x dt = m(v x ) 2 J t. 400(3.75) + / (50f + 350) dt = 400w J 5s v = {0.0625 1 2 + 0.875? - 2.1875} m/s At t = 8 s, (a) = 0.0625(8 2 ) + 0.875(8) - 2.1875 = 8.8125 m/s = 8.81 m/s Ans. 490 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-17. Continued Kinematics. The displacement of the safe can be determined by integrating ds = v dt. For 0 < t < 5 s, the initial condition is s = 0 at t = 0. f ds = f 0.15f 2 dt Jo Jo s= { 0.05? 3 } m At t = 5 s, j = 0.05(5 3 ) = 6.25 m For 5 s < t < 8 s, the initial condition is s = 6.25 m at t = 5 s. [ ds = [ ( 0.0625? 2 + 0.875 t - 2.1875) dt J 6.25 m J 5 s 5 - 6.25 = (0.02083r 3 + 0.4375r 2 - 2.1875f) s = { 0.02083f 3 + 0.4375f 2 - 2.1875? + 3.6458} m i = 0.02083(8 3 ) + 0.4375(8 2 ) - 2.1875(8) + 3.6458 = 24.8125 m = 24.8 m Ans. At t = 8 s, Ans: v = 8.81 m/s s = 24.8 m 491 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 18 . The motor exerts a force F on the 40-kg crate as shown in the graph. Determine the speed of the crate when t = 3 s and when t = 6s.Whent = 0, the crate is moving downward at 10 m/s. SOLUTION Principle of Impulse and Momentum. The impulse of force F is equal to the area F - 150 450 - 150 under the F-t graph. At t = 3 s, FBD of the crate, Fig. a 3-0 6-0 F = 300 N. Referring to the rh ( + T ) m(v y ) l + 2 / F v dt = m(v y ) 2 ■J h 40(—10) + 2 -(150 + 300)(3) - 40(9.81)(3) = 4Qv = —5.68 m/s = 5.68 m/s J. Ans. At f = 6 s, rh ( + T) m(Vy)i + 2 / F y dt = m(v y ) 2 40(—10) + 2 ^(450 + 150)(6) 40(9.81)(6) = 40?; v = 21.14 m/s = 21.1m/s \ Ans. Qx) Ans: p|i= 3 s = 5.68 m/s I Hr= 6s = 21.1 m/s t 492 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 19 . The 30-kg slider block is moving to the left with a speed of 5 m/s when it is acted upon by the forces F 1 and F 2 . If these loadings vary in the manner shown on the graph, determine the speed of the block at t = 6 s. Neglect friction and the mass of the pulleys and cords. SOLUTION Principle of Impulse and Momentum. The impulses produced by F 1 and F, are equal to the area under the respective F-t graph. Referring to the FBD of the block Fig. a, ( ) m(v x )i + S / F x dx = m(v x ) 2 J h -30(5) + 4 10(2) + -(10 + 30)(4 - 2) + 30(6 - 4) -40(4) - -(10 + 40)(6 - 4) v = 4.00 m/s = 30u Ans. F(N) 30($3/)Al 1 /

F,

F

F

f)

Ans:

v = 4.00 m/s

493

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* 15 - 20 .

The 200-lb cabinet is subjected to the force F = 20(f + 1) lb
where t is in seconds. If the cabinet is initially moving to
the left with a velocity of 20 ft/s, determine its speed when
t = 5 s. Neglect the size of the rollers.

SOLUTION

Principle of Impulse and Momentum. Referring to the FBD of the cabinet, Fig. a

rh

( ) m(v x )i + 2 / F x dt = m(v x ) 2

A

200 I 5 s 200

—(- 20 ) + 20 cos 30 °./, = ^

v = 28.80 ft/s = 28.8 ft/s —* Ans.

to.)

Ans:

v = 28.8 ft/s —»

494

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15 - 21 .

If it takes 35 s for the 50-Mg tugboat to increase its speed
uniformly to 25 km/h, starting from rest, determine the
force of the rope on the tugboat.The propeller provides the
propulsion force F which gives the tugboat forward motion,
whereas the barge moves freely. Also, determine F acting on
the tugboat. The barge has a mass of 75 Mg.

SOLUTION

2s O = 6 - 944m/s

System:

( -L ) mv j + S jF dt = mv 2

[0 + 0] + F( 35) = (50 + 75)(10 3 )(6.944)
F = 24.8 kN

Barge:

( -L ) mVl +v J F dt = mv 2

0 + T(35) = (75)(10 3 )(6.944)

T = 14.881 = 14.9 kN

Also, using this result for T,

Tugboat:

( -L ) mVl +2 J F dt = mv 2

0 + F(35) - (14.881)(35) = (50)(10 3 )(6.944)
F = 24.8 kN

Ans.

Ans.

Ans.

Ans:

T = 14.9 kN
F = 24.8 kN

495

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15 - 22 .

The thrust on the 4-Mg rocket sled is shown in the graph.
Determine the sleds maximum velocity and the distance the
sled travels when t = 35 s. Neglect friction.

SOLUTION

Principle of Impulse And Momentum. The FBD of the rocket sled is shown in
Fig. a. For 0 < r < 25 s,

nh

(±>) m(v x )i + 1 F x dt = m(v x ) 2
J t\

0 + [ 4(l0 3 )f2~^ = 4(l0 3 )

Jo

4(l0 3 )(%i) ' = 4(10 3 )

v = i - v- f m/s

At t = 25 s.

v = -(25)s = 83.33 m/s

T - 0 20(l0 3 ) - 0 . ,,

For 25 s < t < 35 s,^—— = ^ _ 3g or T = 2(l0 3 )(35 - t).

Here, ( v x )i = 83.33 m/s and tj = 25 s.

(A)

(*)

rh

m(v x ) i + 11 F x dt

m(v x ) 2

4(10 3 )(83.33) + / 2( 10 3 )(35 - t)dt = 4(l0 3 ) v

J 25 s

v = {-0.25 1 2 + 17.5/ - 197.9167} m/s
The maximum velocity occurs at t = 35 s.Thus,

«max = -0.25(35 2 ) + 17.5(35) - 197.9167
= 108.33 m/s = 108 m/s

Ans.

Ans:

«max = 108 m/s

496

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15-22. Continued

Kinematics. The

ds = vdt. For 0 <

displacement of the sled can be determined by integrating
t < 25 s, the initial condition is s = 0 at t = 0.

At t = 25 s,

5 = ^(25)t = 833.33 m

For 25 < t < 35 s, the initial condition is s = 833.33 at t = 25 s.
r s r l

,2

ds = / (-0.25r + 17.5/ - 197.9167) dt

f 833.33 m

S

S

833.33 m

'25 s

= (—0.08333f 3 + 8.75 1 2 - 197.9167 1)

25 s

{ -0.08333r 3 + 8.75 1 2 - 197.9167t + 1614.58} m

At t = 35 s,

5 = -0.08333(35 3 ) + 8.75(35 2 ) - 197.9167(35) + 1614.58
= 1833.33 m = 1833 m Ans.

Ans:

j = 1.83 km

497

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15-23.

The motor pulls on the cable at A with a force
F = (30 + f 2 ) lb, where t is in seconds. If the 34-lb crate is
originally on the ground at t = 0, determine its speed in
t = 4 s. Neglect the mass of the cable and pulleys. Hint:
First find the time needed to begin lifting the crate.

SOLUTION

30 + t 2 = 34

t = 2 s for crate to start moving

( + |) mv\ + X J Fdt = mv 2

0 + J (30 + t 2 )dt - 34(4 - 2)

30r + - 1 3
3

34

— 68 = - Vi

2 32.2 2

v 2 = 10.1 ft/s

$Ans. Ans: v = 10.1 ft/s 498 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *15-24. The motor pulls on the cable at A with a force F = (e 2r ) lb, where t is in seconds. If the 34-lb crate is originally at rest on the ground at t = 0, determine the crate’s velocity when t = 2 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. SOLUTION F = e 2 ‘ = 34 t = 1.7632 s for crate to start moving ( + T) mvi + 2 J Fdt = mv 2 34 0 ■/ e 2, dt - 34(2 - 1.7632) = —« 2 ' 1.7632 (•2 8.0519 = 1.0559 v 2 1 1.7632 v 2 = 2.13 m/s Ans. Ans: v 2 = 2.13 m/s 499 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 25 . The balloon has a total mass of 400 kg including the passengers and ballast. The balloon is rising at a constant velocity of 18 km/h when h = 10 m. If the man drops the 40-kg sand bag, determine the velocity of the balloon when the bag strikes the ground. Neglect air resistance. v A = 18 km/h SOLUTION Kinematic. When the sand bag is dropped, it will have an upward velocity of ( km\/1000m\/ lh \ . Vn — I 18 —— - - = 5 m/s T. When the sand bag strikes the \ h / \ 1 km J \ 3600 s J ground s = 10 m I. The time taken for the sand bag to strike the ground can be determined from ( + T) s = s 0 + v 0 t + -a c t 2 ; -10 = 0 + 5t + ^(-9.81t 2 ) 4.905t 2 - 5t - 10 = 0 Solve for the positive root, t = 2.0258 s Principle of Impulse and Momentum. The FBD of the ballon when the ballon is rising with the constant velocity of 5 m/s is shown in Fig. a ( + t) m(v y )i + 2 f F y dt = m{v y ) 2 A 400(5) + T(t) - 400(9.81)f = 400(5) T = 3924 N When the sand bag is dropped, the thrust T = 3924 N is still maintained as shown in the FBD, Fig. b. rh ( + t) m(v y )i + 2 / Fydt = m{v y ) 2 J u 360(5) + 3924(2.0258) - 360(9.81)(2.0258) = 360v v = 7.208 m/s = 7.21 m/s I J Ans. Ans: v = 7.21 m/s | 500 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 26 . A •- x B *- x' 2 m/s 5 m/s 6 N-► SOLUTION Observer A: As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which slides along the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block in 4 s if it has an initial speed of 5 m/s measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x' axis that moves at a constant velocity of 2 m/s relative to A. ( ) m V\ + 2 J F dt — m V 2 10(5) + 6(4) = lOv v = 7.40 m/s Observer B: 5 m/s Ans. ( ) m V\ + 2 J F dt = m V 2 10(3) + 6(4) = lOn v = 5.40 m/s Ans. Ans: Observer A:v = 7.40 m/s Observer B:v = 5.40 m/s 501 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-27. The 20-kg crate is lifted by a force of F = (100 + 5t 2 ) N, where t is in seconds. Determine the speed of the crate when t = 3 s, starting from rest. SOLUTION Principle of Impulse and Momentum. At i = 0, F = 100 N. Since at this instant, 2 F = 200 N > W = 20(9.81) = 196.2 N, the crate will move the instant force F is applied. Referring to the FBD of the crate, Fig. a, (+T) rh m(Vy) ] + 2 / F y dt = m(Vy) 2 r 3s 0+2 (100 + 5 t 2 )dt - 20(9.81)(3) = 20v Jo 3 s - 588.6 = 20v o v = 5.07 m/s Ans. 2^100? + |f 3 j 20CW) (a.) Ans: v = 5.07 m/s 502 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 28 . The 20-kg crate is lifted by a force of F = (100 + 5t 2 ) N, where t is in seconds. Determine how high the crate has moved upward when t = 3 s, starting from rest. SOLUTION Principle of Impulse and Momentum. At t = 0, F = 100 N. Since at this instant, 2 F = 200 N > W = 20(9.81) = 196.2 N, the crate will move the instant force F is applied. Referring to the FBD of the crate, Fig. a rh ( + T) m(v y )! + X / F y dt = m(v y ) 2 Jt 1 0 + 2 [ (100 + 5 t 2 )dt - 20(9.81)f = 2Qv Jo 2 ( lOOt + -r 3 - 196.2f = 20v v = {0.16671 3 + 0.19?} m/s Kinematics. The displacement of the crate can be determined by integrating ds = v dt with the initial condition s = 0 at t = 0. At f = 3 s, [ ds = I (0.1667t 3 + 0.19f) dt Jo Jo s = { 0.04167f 4 + 0.095f 2 } m s = 0.04167(3 4 ) + 0.095(3 2 ) = 4.23 m Ans. P-OCW) (a.) Ans: s = 4.23 m 503 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 29 . In case of emergency, the gas actuator is used to move a 75-kg block B by exploding a charge C near a pressurized cylinder of negligible mass. As a result of the explosion, the cylinder fractures and the released gas forces the front part of the cylinder, A, to move B forward, giving it a speed of 200 mm/s in 0.4 s. If the coefficient of kinetic friction between B and the floor is /j, k = 0.5, determine the impulse that the actuator imparts to B. SOLUTION Principle of Linear Impulse and Momentum: In order for the package to rest on top of the belt, it has to travel at the same speed as the belt. Applying Eq. 15-4, we have + 2 F y dt = my Vy ) 2 v B = 200 mm/s B (+1) 6(0) + Nt - 6(9.81) t = 6(0) N = 58.86 N m(v x ) i + rh 2 / F x dt = m(v x ) 2 6(3) + [—0.2(58.86)f] = 6(1) t = 1.02 s vr^jyw r ^,o.s(lO9.T0 Ans. () m (v*)! + 2 J F x dt = m{v x ) 2 0 + F dt - (0.5)(9.81)(75)(0.4) = 75(0.2) F dt = 162 N • s Ans. Ans: t = 1.02 s I = 162 N • s 504 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 30 . A jet plane having a mass of 7 Mg takes off from an aircraft carrier such that the engine thrust varies as shown by the graph. If the carrier is traveling forward with a speed of 40 km/h, determine the plane’s airspeed after 5 s. SOLUTION The impulse exerted on the plane is equal to the area under the graph. v\ = 40 km/h = 11.11 m/s () >n(v x )i +2 F x dt = m(v x ) 2 F (kN) 40 km/h -► (7)( 10 3 )(11.11) - ^(2)(5)(10 3 ) + |(15 + 5)(5 - 2)(10 3 ) = 7(Hf\ 2 v 2 = 16.1 m/s Ans. Ans: v = 16.1 m/s 505 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 31 . Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity (v B ) 1 = 3 ft/s at t = 0, determine the velocity of A when f = Is. Assume that the horizontal plane is smooth. Neglect the mass of the pulleys and cords. SOLUTION S A + 2s B = / v A = -2v b ( ) mvi +2 JF dt = mv 2 - tu (2)(3) - r(1) - (+1) mvi +2 J F dt = mv 2 5n< 3 > + 3(,) - 2T(1) ‘ - 32.2 T - 1Q(v a ) 2 = 60 - 64.4T + 1.5(v a ) 2 = -105.6 T = 1.40 lb (v A )i ~ —10.5 ft/s = 10.5 ft/s —> 1016 [±>T T T Ans. 9 3f6 Ans: (v A ) 2 = 10.5 ft/s —» 506 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 32 . Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity (v B )i = 3 ft/s at t = 0, determine the velocity of A when t = 1 s. The coefficient of kinetic friction between the horizontal plane and block A is /Ji A = 0.15. SOLUTION S4 + 2 S B — l v A = -2v b () mv 1 + 2 JF dt = mv 2 - ^(2)0) - T(l) + 0.15(10) = ^(v A ) 2 (+ 1 ) mvi + 2 F dt — mv 2 3 32.2 ( 3 ) 3(1) 27(0 = 3 /K) 2 \ 32.2 V 2 ) - 32.2 T - W(v A ) 2 = 11.70 - 64.47 + l.5(v A ) 2 = -105.6 7 = 1.501b (v A )2 = —6.00 ft/s = 6.00 ft/s —» Ans. Ans: ( v A ) 2 = 6.00 ft/s —> 507 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 33 . The log has a mass of 500 kg and rests on the ground for which the coefficients of static and kinetic friction are fjL s = 0.5 and = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the log when t = 5 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log. SOLUTION ^ ZF X = 0; Thus, (*) T (N) 1800 F - 0.5(500)(9.81) = 0 F = 2452.5 N 2T = F 2(200t 2 ) = 2452.5 t = 2.476 s to start log moving m Vi + 2 J F dt = mv 2 0 + 2 f 200f 2 dt + 2(1800)(5 - 3) - 0.4(500)(9.81)(5 - 2.476) = 500w 2 + 2247.91 = 500 v 2 2.476 3 400(—) J 2.476 v 2 = 7.65 m/s Ans. A T -mi at Soofl.ai) M r4-i-»F 0.5Nl 0.4/4.« Tm^ 500(1.81)^ 500(1.81) M I 1 I — 1m.* 500(1.8!) ^ Ans: v = 7.65 m/s 508 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 34 . The 0.15-kg baseball has a speed of v = 30 m/s just before it is struck by the bat. It then travels along the trajectory shown before the outfielder catches it. Determine the magnitude of the average impulsive force imparted to the ball if it is in contact with the bat for 0.75 ms. 100 m SOLUTION ( ) m A (v A )i + m B (v B )! = (m A + m B )v 2 4500 3000 /7500 32.2 ^ 32.2 ^ ~~ 32.2 Vl v 2 = -0.600 ft/s = 0.600 ft/s ^ Ans. Ans: v = 0.6 ft/s «— 509 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 35 . The 5-Mg bus B is traveling to the right at 20 m/s. Meanwhile a 2-Mg car A is traveling at 15 m/s to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision. SOLUTION Conservation of Linear Momentum. ( ) m A v A + m B v B = (m A + m B )v [ 5( 10 3 ) ] (20) + [2(10 3 ) ] (15) = [ 5( 10 3 ) + 2(l0 3 )]v v = 18.57 m/s = 18.6 m/s —* Ans. Ans: v = 18.6 m/s —> 510 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 36 . The 50-kg boy jumps on the 5-kg skateboard with a hori¬ zontal velocity of 5 m/s. Determine the distance .v the boy reaches up the inclined plane before momentarily coming to rest. Neglect the skateboard’s rolling resistance. SOLUTION Free-Body Diagram: The free-body diagram of the boy and skateboard system is shown in Fig. a. Here, W*,W S 6, and N are nonimpulsive forces. The pair of impulsive forces F resulting from the impact during landing cancel each other out since they are internal to the system. Conservation of Linear Momentum: Since the resultant of the impulsive force along the x axis is zero, the linear momentum of the system is conserved along the x axis. (<t) m b (v b \ + m sb (v sb \ = ( m b + m sb )v 50(5) + 5(0) = (50 + 5)v v = 4.545 m/s Conservation of Energy: With reference to the datum set in Fig. b, the gravitational potential energy of the boy and skateboard at positions A and B are (’ Vg) A = (™ b + m sb )gh A = 0 and [V g ) B = (m h + m sb )gh B = (50 + 5)(9.81)(s sin 30°) = 269.775x. T a + V a = T b + V b s = 2.11 m Ans. YJb=so(W0 d Lb) Ans: s = 2.11 m 511 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 37 . The 2.5-Mg pickup truck is towing the 1.5-Mg car using a cable as shown. If the car is initially at rest and the truck is coasting with a velocity of 30 km/h when the cable is slack, determine the common velocity of the truck and the car just after the cable becomes taut. Also, find the loss of energy. 30 km/h SOLUTION Free-Body Diagram: The free-body diagram of the truck and car system is shown in Fig. a. Here, W„W C , N„ and N c are nonimpulsive forces. The pair of impulsive forces F generated at the instant the cable becomes taut are internal to the system and thus cancel each other out. Conservation of Linear Momentum: Since the resultant of the impulsive force is zero, the linear momentum of the system is conserved along the x axis. The initial speed of the truck is (= 30(10 3 ) — 1 h 3600 s = 8.333 m/s. (<L) rn,{v t )i + m c (v c ) i = [m t + m c )v 2 2500(8.333) + 0 = (2500 + 1500)n 2 v 2 = 5.208 m/s = 5.21 m/s <— Ans. Kinetic Energy: The initial and final kinetic energy of the system is 1 ,1 , T\ = + -m c (v c ) i 1 i = - (2500)(8.333 2 ) + 0 = 86 805.56 J and T 2 = (m, + m c )v 2 = i (2500 + 1500)(5.208 2 ) = 54 253.47 Thus, the loss of energy during the impact is Ar = Tj - r 2 = 86 805.56 - 54 253.47 = 32.55(10 3 )J = 32.6 kj Ans. Ans: v = 5.21 m/s <— A T = -32.6 kJ 512 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 38 . A railroad car having a mass of 15 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m/s in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred, and explain qualitatively what happened to this energy. SOLUTION (iL) 'Amv 1 = 1,mv 2 15 000(1.5) - 12 000(0.75) = 27 000(?; 2 ) v 2 = 0.5 m/s Ans. 7) = ^(15 000)(1.5) 2 + y(12 000)(0.75) 2 = 20.25 kj T 2 = ^(27 000)(0.5) 2 = 3.375 kj AT = T 2 - 7j = 3.375 - 20.25 = -16.9kJ Ans. This energy is dissipated as noise, shock, and heat during the coupling. Ans: v = 0.5 m/s AT = -16.9 kJ 513 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 39 . A ballistic pendulum consists of a 4-kg wooden block originally at rest, 0 = 0°. When a 2-g bullet strikes and becomes embedded in it, it is observed that the block swings upward to a maximum angle of 0 = 6°. Estimate the speed of the bullet. SOLUTION Just after impact: Datum at lowest point. T 2 + V 2 = T 3 + y 3 |(4 + 0.002) (v B )i + 0 = 0 + (4 + 0.002)(9.81)(1.25)(1 - cos 6°) (■ v b ) 2 = 0.3665 m/s For the system of bullet and block: ( % ) Zmvx = tmv 2 0.002(n B ) 1 = (4 + 0.002)(0.3665) (%)i = 733 m/s l 0 1.25 m \ — 0 \ 1.25 i \ \ V Ans. Ans: v = 733 m/s 514 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 40 . The boy jumps off the flat car at A with a velocity of v = 4 ft/s relative to the car as shown. If he lands on the second flat car B , determine the final speed of both cars after the motion. Each car has a weight of 80 lb. The boy’s weight is 60 lb. Both cars are originally at rest. Neglect the mass of the car’s wheels. SOLUTION (^) Sm(t) i) = Sm(r 2 ) „ „ 80 60 + ~~ 322° A + 32.2 * v A = 0.75 (v b ) x v b = v A + v b/A ( ^ ) (Vb) x s ~v A + 4 ( (v b ) x = 2.110 ft/s v A = 1.58 ft/s —* ( ) Xm(v-s) = S/ n(v 2 ) 60 / 80 60 , 32.2 2 ' _ \32.2 + 32.2 J V v = 0.904 ft/s v = 4 ft/s Ans. Ans. Ans: v A = 1.58 ft/s —> v = 0.904 ft/s 515 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 41 . A 0.03-lb bullet traveling at 1300 ft/s strikes the 10-lb wooden block and exits the other side at 50 ft/s as shown. Determine the speed of the block just after the bullet exits the block, and also determine how far the block slides before it stops. The coefficient of kinetic friction between the block and the surface is /x k = 0.5. SOLUTION ( -L j = S«j 2 r'2 MW£) + o- 13 10 \ /0.03 4 32.2 J B + ( 32.2 ) ( 5 v B = 3.48 ft/s T\ + 2 = T 2 Kiu) (3 - 4s)J -■ ° d = 0.376 ft Ans. Ans. 20/4 ^ o.s(/$)=siy

loll,

Ans:

v B = 3.48 ft/s
d = 0.376 ft

516

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15 - 42 .

A 0.03-lb bullet traveling at 1300 ft/s strikes the 10-lb
wooden block and exits the other side at 50 ft/s as shown.
Determine the speed of the block just after the bullet exits
the block. Also, determine the average normal force on the
block if the bullet passes through it in 1 ms, and the time the
block slides before it stops. The coefficient of kinetic
friction between the block and the surface is /x k = 0.5.

SOLUTION

(■±>) Smit4 = 'Zm 2 V2

-t)

' J*'

0.03 V 12 \ „ / 10 \ / 0.03 V„V 4

32.2 ) 1 ° \13 ) + ~~ ( 32.2/ 5 + \32.2j^ 5 °\5

v B = 3.48 ft/s Ans.

mvi + 2 JF dt = mv 2

-(if) <1300| (il) - 1®CD(10-’) + «(D(10- 3 ) - d§)<a»(f)

N = 504 lb
mvY+'Z F dt = mv 2

Ans.

P A8) ~ 5{t) = °

t = 0.216 s

Ans.

'Oli>

| r//>

N-/o/b

Ans:

v B = 3.48 ft/s
Vvg = 504 lb
t =‘‘0.216 s

517

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15 - 43 .

The 20-g bullet is traveling at 400 m/s when it becomes 400 m /s

embedded in the 2-kg stationary block. Determine the , n

distance the block will slide before it stops. The coefficient
of kinetic friction between the block and the plane is
Pk = 0 . 2 .

SOLUTION

Conservation of Momentum.

( ) m b v b + m B v B = (m b + m B )v

0.02(400) + 0 = (0.02 + 2)v
v = 3.9604 m/s

Principle of Impulse and Momentum. Here, friction ly = p. k N = 0.2 N. Referring
to the FBD of the blocks, Fig. a,

rh

(+1) m(Vy)i + 2 / F v dt = m(Vy) 2

J h

0 + N(t) - 2.02(9.81 )(t) = 0

N = 19.8162 N
rh

( ^ ) m(v x )\ + 2 / F x dt = m(v x ) 2

2.02(3.9604) + [-0.2(19.8162)1] = 2.02v
v = (3.9604 - 1.962 f} m/s
Thus, the stopping time can be determined from
0 = 3.9604 - 1.962r
t = 2.0186 s

Kinematics. The displacement of the block can be determined by integrating
ds = vdt with the initial condition s — 0 at t = 0.

Z0Z(i$t)tJ n) N O; ds= (3.9604 - 1.962f) dt Jo Jo s = (3.9604 1 - 0.981I 2 } m The block stopped at t = 2.0186 s. Thus s = 3.9604(2.0186) - 0.98l(2.0186 2 ) = 3.9971 m = 4.00 m Ans. Ans: x = 4.00 m 518 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 44 . A toboggan having a mass of 10 kg starts from rest at A and carries a girl and boy having a mass of 40 kg and 45 kg, respectively. When the toboggan reaches the bottom of the slope at B, the boy is pushed off from the back with a horizontal velocity of \ b u = 2 m/s, measured relative to the toboggan. Determine the velocity of the toboggan afterwards. Neglect friction in the calculation. SOLUTION A Conservation of Energy: The datum is set at the lowest point B. When the toboggan and its rider is at A , their position is 3 m above the datum and their gravitational potential energy is (10 + 40 + 45)(9.81)(3) = 2795.85 N • m. Applying Eq. 14-21, we have T 1 + V t = T 2 + V 2 0 + 2795.85 = - (10 + 40 + 45) v\ + 0 v B = 7.672 m/s Relative Velocity: The relative velocity of the falling boy with respect to the toboggan is v b/t = 2 m/s. Thus, the velocity of the boy falling off the toboggan is ✓ - XL V b = V t + v blt (^ ) v b = v, - 2 [11 Conservation of Linear Momentum: If we consider the tobbogan and the riders as a system, then the impulsive force caused by the push is internal to the system. Therefore, it will cancel out. As the result, the linear momentum is conserved along the x axis. m T v B = m b v b + ( m t + m g ) v, () (10 + 40 + 45)(7.672) = 45u 6 + (10 + 40) v t [21 Solving Eqs.  and  yields v t = 8.62 m/s Ans. v b = 6.619 m/s Ans: v t = 8.62 m/s 519 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 520 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 46 . The two blocks A and B each have a mass of 5 kg and are suspended from parallel cords. A spring, having a stiffness of k = 60 N/m, is attached to B and is compressed 0.3 m against A and B as shown. Determine the maximum angles 8 and <fi of the cords when the blocks are released from rest and the spring becomes unstretched. SOLUTION (iL ) Xmv i = S»w2 0 + 0 = — 5v a + 5v b Va = Vb = v Just before the blocks begin to rise: Ty + V 1 = T 2 + V 2 (0 + 0) + i(60)(0.3) 2 = i(5)(t;) 2 + i(5)( V ) 2 + 0 v = 0.7348 m/s For A or B: Datum at lowest point. t 1 + v 1 = t 2 + u 2 i (5)(0.7348) 2 + 0 = 0 + 5(9.81)(2)(1 - cos 8) 8 = 4) = 9.52° Ans. Ans: 0 - <b - 9.52° 521 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 47 . The 30-Mg freight car A and 15-Mg freight car B are moving towards each other with the velocities shown. Determine the maximum compression of the spring mounted on car A. Neglect rolling resistance. 20 km/h 10 km/h 3 MN/m £ inn SOLUTION Conservation of Linear Momentum: Referring to the free-body diagram of the freight cars A and B shown in Fig. a , notice that the linear momentum of the system is con¬ served along the x axis. The initial speed of freight cars A and B are (pa)i = - m 20 ( 10 3 ) — lh = 5.556 m/s and (^ 5)1 = , m 10 ( 10 3 ) — 1 h 3600 s ,3600 s = 2.778 m/s. At this instant, the spring is compressed to its maximum, and no relative motion occurs between freight cars A and B and they move with a common speed. (i) m A (v A )i + »IjWi = ( m A + m B )v 2 30(10 3 )(5.556) + v 2 = 2.778 m/s — —15(10 3 )(2.778) 30(10 3 ) + 15(10 3 ) v 2 L J ^_ n - t 1 H A L (&) Conservation of Energy: The initial and final elastic potential energy of the spring is (V e f = | ks ! 2 = 0 and (V e ) 2 = | ks 2 1 = |(3)(10 6 )s max 2 = 1.5(10 6 ).s max 2 . 2T! + 2 Vi = 2T 2 + SU 2 2 m A (v A ) r 2 + - m B {v B ) i 2 + (Ye)\ = 2 + m B )v 2 + (V e ) 2 i (30)(l0 3 )(5.556 2 ) + i (15)(l0 3 )(2.778 2 ) + 0 = 30(l0 3 ) + 15(l0 3 )j(2.778 2 ) + 1.5(l0 6 )x max 2 Ans. Ans: ^max = 481 mm 522 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 48 . Blocks A and B have masses of 40 kg and 60 kg, respectively. They are placed on a smooth surface and the spring connected between them is stretched 2 m. If they are released from rest, determine the speeds of both blocks the instant the spring becomes unstretched. k = 180 N/m SOLUTION ( -L ) E/771'1 = 2/771/2 0 + 0 — 40 v A — 60 //g Tl + Vl = T 2 + v 2 0 + ^(180)(2) 2 = $$40)M 2 + v A = 3.29 m/s v B = 2.19 m/s 5 = 2m OHmvMZ] ^-QO-n 5=0 Ans. Ans. Ans: v A = 3.29 m/s v B = 2.19 m/s 523 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 49 . A boy A having a weight of 80 lb and a girl B having a weight of 65 lb stand motionless at the ends of the toboggan, which has a weight of 20 lb. If they exchange positions, A going to B and then B going to A’s original position, determine the final position of the toboggan just after the motion. Neglect friction between the toboggan and the snow. A B |*—4 ft— A, SOLUTION A goes to B , ( ^> ) Xmv i = hmv 2 0 = m A v A - (m, + m B )v B 0 = m A s A - (m, + m B )s B Assume B moves x to the left, then A moves (4 — x ) to the right 0 = m A (4 — x) — (m, + m B )x 4 m A x = - m A + m B + m, 4(80) = --—--= 1 939 ft «- 80 + 65 + 20 B goes to other end. ( ) Xmv i = trim 2 0 = ~m B v B + (m, + m A )v A 0 = —m B s B + ( m t + m A )s A Assume B moves x' to the right, then A moves (4 — x') to the left 0 = —m B ( 4 — x') + (m t + m A )x' 4 m B X = - m A + m B + m t 4(65) = 80 + 65 + 20 = L576ft - Net movement of sled is x = 1.939 - 1.576 = 0.364 ft <- Ans. -fiuvt- c. SI “" ;mi-1 oifl) Vfl Ans: x = 0.364 ft «- 524 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 50 . A boy A having a weight of 80 lb and a girl B having a weight of 65 lb stand motionless at the ends of the toboggan, which has a weight of 20 lb. If A walks to B and stops, and both walk back together to the original position of A, determine the final position of the toboggan just after the motion stops. Neglect friction between the toboggan and the snow. 4 ft H SOLUTION A goes to B, ( -^ ) Xmvi = Xmv 2 0 = m A v A - (m, + m B )v B 0 = m A s A - (pi, + m B )s B Assume B moves x to the left, then A moves (4 0 = m A (4 — x) — (in, + m B )x 4 m A x = - m A + m B + m, 4(80) 80 + 65 + 20 - = 1.939 ft x) to the right A and B go to other end. ( ) Xmvx = Xmv 2 0 = —m B v — m A v + m,v, 0 = ~in B s — m A s + m,s, Assume the toboggan moves x' to the right, then A and B move (4 0 = —m B ( 4 — x') — m A ( 4 — x') + ni,x' 4 (m B + m A ) x = - m A + m B + m, 4(65 + 80) 80 + 65 + 20 = 3.515 ft x') to the left Net movement of sled is ( ±> ) x = 3.515 - 1.939 = 1.58 ft -» Ans. Ans: x = 1.58 ft 525 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 51 . The 10-Mg barge B supports a 2-Mg automobile A. If someone drives the automobile to the other side of the barge, determine how far the barge moves. Neglect the resistance of the water. SOLUTION Conservation of Momentum. Assuming that V B is to the left, ( ) m A v A + m B v B = 0 2(l0 3 )v A + 10(l0 3 )va = 0 2 v A + 10 v B = 0 Integrate this equation, 2 + 10 s B = 0 (1) Kinematics. Here, s A / B = 40 m using the relative displacement equation by assuming that s B is to the left, ( ^ ) s A = s B + s A/B *4 = S B + 40 (2) Solving Eq. (1) and (2), s B = —6.6667 m = 6.67 m —> Ans. s A = 33.33 m <— The negative sign indicates that s B is directed to the right which is opposite to that of the assumed. Ans: s B = 6.67 m —> 526 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 52 . The free-rolling ramp has a mass of 40 kg. A 10-kg crate is released from rest at A and slides down 3.5 m to point B. If the surface of the ramp is smooth, determine the ramp’s speed when the crate reaches B. Also, what is the velocity of the crate? SOLUTION TTTTTT Conservation of Energy: The datum is set at lowest point B. When the crate is at point A, it is 3.5 sin 30 = 1.75 m above the datum. Its gravitational potential energy is 10(9.81)(1.75) = 171.675 N-m. Applying Eq. 14-21, we have 7i + Vi = T 2 + V 2 0 + 171.675 = | ( 10)4 + \ ( 40)4 171.675 = 5 v z c + 20 v 2 r (1) Relative Velocity: The velocity of the crate is given by Vc = Vfl + Vc/fi = -Pfii + (vc/r cos 30°i - Vc/r sin 30°j) = (0.8660 v c/R - v R )i - 0.5n c/R j (2) The magnitude of v c is v c = V(0.8660 vq R - v R ) 2 + (-0.5v c/R ) 2 = Vv 2 C / r + v 2 r - U32v r v C /r ( 3 ) Conservation of Linear Momentum: If we consider the crate and the ramp as a system, from the FBD, one realizes that the normal reaction N c (impulsive force) is internal to the system and will cancel each other. As the result, the linear momentum is conserved along the x axis. 0 = m c (v c ) x + m R v R () 0 = 10(0.8660 v C /r ~ v R ) + 40 (~v R ) 0 = 8 . 660 n C / i ? — 50 v R ( 4 ) Solving Eqs. (1), (3), and (4) yields v R = 1.101 m/s = 1.10 m/s v c = 5.43 m/s Ans. Vqr — 6.356 m/s From Eq. (2) v c = [0.8660(6.356) - 1.101 ]i - 0.5(6.356)j = {4.403i - 3.178j} m/s Thus, the directional angle f of Vq is 3 178 d> = tan -1 '-= 35.8° 'Ti d> Ans. 4.403 r O) tie % Ans: Vc/r — 6.356 m/s f = 35.8° ^5 527 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 53 . Block A has a mass of 5 kg and is placed on the smooth triangular block B having a mass of 30 kg. If the system is released from rest, determine the distance B moves from point O when A reaches the bottom. Neglect the size of block A. SOLUTION ( ^ ) tmvi = Xmv-i 0 = 30n B - 5(v a ) x (v A ) x = 6 v B v B = v A + v B/A ( ) v B = ~(v A ) x + ( v B/A ) x v B = -6v b + (v B/A ) x (v B / A ) x = Tvs Integrate { s B /a)x = 7 s B (s B/A ) x = 0.5 m 0.5 m Thus, 0.5 s B = — = 0.0714 m = 71.4 mm —* Ans. Ans: s B = 71.4 mm —* 528 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 54 . Solve Prob. 15-53 if the coefficient of kinetic friction between A and B is fi k = 0.3. Neglect friction between block B and the horizontal plane. SOLUTION +\XF y = 0; N a — 5(9.81) cos 30° = 0 N a = 42.4785 N /+XF x = 0; F a - 5(9.81) sin 30° = 0 F a = 24.525 N F max = fiN A = 0.3(42.4785) = 12.74 N < 24.525 N Block indeed slides. Solution is the same as in Prob. 15-53. Since F A is internal to the system s B = 71.4 mm —» Ans: s B = 71.4 mm —» SW$$N t i ft Ans. 529 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 530 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 56 . Two boxes A and B, each having a weight of 160 lb, sit on the 500-lb conveyor which is free to roll on the ground. If the belt starts from rest and begins to run with a speed of 3 ft/s, determine the final speed of the conveyor if (a) the boxes are not stacked and A falls off then B falls off, and (b) A is stacked on top of B and both fall off together. SOLUTION a) Let v b be the velocity of A and B. 'Znivi = 2 „ / 320 \ , / 500 \ , ~~ \3Z2V3^2/ v b = v c + v b/c v b = ~v c + 3 Thus, v b = 1.83 ft/s - v c = 1.17 ft/s * When a box falls off, it exerts no impulse on the conveyor, and so does not alter the momentum of the conveyor. Thus, a) v c = 1.17 ft/s *— Ans. b) v c = 1.17 ft/s <— Ans. Ans: a) v c = 1.17 ft/s <— b) v c = 1.17 ft/s <— 531 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 57 . The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the 10-g bullet is traveling at 300 m/s when it becomes embedded in the 10-kg block, determine the distance the block will slide up along the plane before momentarily stopping. SOLUTION Conservation of Linear Momentum : If we consider the block and the bullet as a system, then from the FBD, the impulsive force F caused by the impact is internal to the system. Therefore, it will cancel out. Also, the weight of the bullet and the block are nonimpulsive forces. As the result, linear momentum is conserved along the x' axis. mb(vb)x' = (m b + m B ) v x , 0.01(300 cos 30°) = (0.01 + 10) v v = 0.2595 m/s Conservation of Energy: The datum is set at the blocks initial position. When the block and the embedded bullet is at their highest point they are h above the datum. Their gravitational potential energy is (10 + 0.01)(9.81)/; = 98.1981 h. Applying Eq. 14-21, we have tomo bi X' t 1 + v 1 = t 2 + v 2 o + |(10 + 0.01)(0.2595 2 ) = 0 + 98.1981 /j h = 0.003433 m = 3.43 mm d = 3.43 / sin 30° = 6.87 mm Ans. Ans: d = 6.87 mm 532 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 58 . Disk A has a mass of 250 g and is sliding on a smooth horizontal surface with an initial velocity (v A )i = 2 m/s. It makes a direct collision with disk B , which has a mass of 175 g and is originally at rest. If both disks are of the same size and the collision is perfectly elastic (e = 1), determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same. SOLUTION ( ±> ) (0.250)(2) + 0 = (0.250)0^)2 + (0.175)(w fl ) 2 e = 1 = (v b ) 2 - (y A h 2-0 Solving (v A ) 2 = 0.353 m/s ( v B ) 2 = 2.35 m/s Ans. Ans. 7j = j; (0.25)(2) 2 = 0.5 J T 2 = | (0.25)(0.353) 2 + | (0.175)(2.35) 2 = 0.5 J 0 = T 2 QED Ans: ( v A ) 2 = 0.353 m/s (v B ) 2 = 2.35 m/s 533 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 59 . The 5-Mg truck and 2-Mg car are traveling with the free- rolling velocities shown just before they collide. After the collision, the car moves with a velocity of 15 km/h to the right relative to the truck. Determine the coefficient of restitution between the truck and car and the loss of energy due to the collision. 30 km/h 10 km/h SOLUTION Conservation of Linear Momentum: The linear momentum of the system is conserved along the x axis (line of impact). The initial speeds of the truck and car are (v,)± = 30(l0 3 lh 3600 s = 8.333 m/s and (w c )i = 10(l0 3 ) “ lh 3600 s = 2.778 m/s. By referring to Fig. a, ' m t{v t)i + m c (v c )i = m t (v,) 2 + m c (v c ) 2 5000(8.333) + 2000(2.778) = 5000(v t ) 2 + 2000(u c ) 2 5(n ( ) 2 + 2(v c ) 2 = 47.22 ( lh ( 1 ) Coefficient of Restitution: Here, (v c i t ) = Applying the relative velocity equation, (v c )2 = (vr) 2 + (v c /t) 2 ( ^ ) (Vch = (v,h + 4.167 (V c ) 2 - (v,) 2 = 4.167 (2) Applying the coefficient of restitution equation, 15f 10 3 ) — h V 3600: = 4.167 m/s ' V («c)2 ~ (.Vf)2 (V,)l ~ ( V c)\ (Vc)l ~ (Vt)l 8.333 - 2.778 (3) Juirt after' Tmpatt 534 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-59. Continued Substituting Eq. (2) into Eq. (3), 4.167 8.333 - 2.778 0.75 Ans. Solving Eqs. (1) and (2) yields ( v t ) 2 = 5.556 m/s (u c ) 2 = 9.722 m/s Kinetic Energy: The kinetic energy of the system just before and just after the collision are 1 \ T\ = -m t (y t ) i 2 + -m c (v ^ 2 1 , 1 , = —(5000)(8.333 2 ) + — (2000)(2.778 2 ) = 181.33(l0 3 ) J Ti = \m, (v t ) 2 2 + \rn c (v c ) 2 2 = —(5000)(5.556 2 ) + — (2000)(9.722 2 ) = 171.68(l0 3 )j Thus, AT = T 1 — T 2 = 181.33(l0 3 ) - 171.68(l0 3 ) = 9.645(l0 3 )j = 9.65 kj Ans. Ans: e = 0.75 A T = -9.65 kJ 535 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 60 . Disk A has a mass of 2 kg and is sliding forward on the smooth surface with a velocity (?Ui)i = 5 m/s when it strikes the 4-kg disk B , which is sliding towards A at ( v B ) 1 = 2 m/s, with direct central impact. If the coefficient of restitution between the disks is e = 0.4, compute the velocities of A and B just after collision. (v A )i = 5 m/s C> A (Vfl)i = 2 m/s -CD B SOLUTION Conservation of Momentum : m A («a)i + niB (v B )i = m A (v A ) 2 + m B (v B )z () 2(5) + 4(—2) = 2(v a ) 2 + 4(v b ) 2 Coefficient of Restitution: (v B ) 2 ■ ' J*' 0.4 = (v A ) i (v B )2 5 - - («a)2 - (V B )l - (Va )2 (- 2 ) Solving Eqs. (1) and (2) yields (v A ) 2 — —1.53 m/s = 1.53 m/s (v B ) 2 — 1.27 m/s —> ( 1 ) ( 2 ) Ans. Ans: (v a ) 2 = 1-53 m/s <— (v B ) 2 = 1-27 m/s —» 536 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 61 . The 15-kg block A slides on the surface for which m k = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-kg block B. If the unstretched spring has a stiffness k = 1000 N/m, determine the maximum compression of the spring due to the collision. Take e = 0.6. SOLUTION Principle of Work and Energy. Referring to the FBD of block A, Fig. a, motion along the y axis gives N A = 15(9.81) = 147.15 N. Thus the friction is F f = p k N A = 0.3(147.15) = 44.145 N. 71 + St/r-z = T 2 |(15)(10 2 ) + (-44.145) (4) = $$15)(v A )l (v A )i = 8.7439 m/s <— Conservation of Momentum. ( ) m A (v A )i + m B (v B ) 1 = m A (v A ) 2 + m B (v B ) 2 15(8.7439) + 0 = 15(« a ) 2 + 10(v B ) 2 3(v a ) 2 + 2(v b ) 2 = 26.2317 (1) Coefficient of Restitution. f + | ^ ( Ml ~ Ml _ (Vb) 1 ~ Ml 1 e Mi~Mi ■ 8.7439 -0 Mi ~ Mi = 5.2463 (2) Solving Eqs. (1) and (2) (v B ) 2 = 8.3942 m/s <— (v A ) 2 = 3.1478 m/s <— Conservation of Energy. When block B stops momentarily, the compression of the spring is maximum. Thus, T 2 = 0. 71 + Vx = T 2 + V 2 |(10)(8.3942 2 ) + 0 = 0 + y (lOOO)jt 2 Xmax = 0.8394 m = 0.839 m Ans. 15 (W M 5 |po- 3 MA B Ha mmummmm bejor-e Impact 6 A Pffcr Impact (CL) Ans: *max = 0-839 m 537 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-62. The four smooth balls each have the same mass m. If A and B are rolling forward with velocity v and strike C, explain why after collision C and D each move off with velocity v. Why doesn’t D move off with velocity 2v? The collision is elastic, e = 1. Neglect the size of each ball. an A B OQ C D SOLUTION Collision will occur in the following sequence; B strikes C ( ) mv = —mv B + mvc v = -v B + v c (±>) C strikes D (±0 e = 1 = v c + v B v c = v, v B = 0 (±0 mv = —mvc + mv D v D + v c e = 1 = v c = 0 , V V D = V A strikes B (±0 (±0 mv = —mv A + mv B , v B + v A e = 1 =- v B = v, v A = 0 Finally, B strikes C ( ) mv = —mv B + mv c (±0 e = 1 = v c = v, Vc + Vi t v v B = 0 Ans. Ans. Note: If D rolled off with twice the velocity, its kinetic energy would be / 1 , 1 , energy available from the original two A and B: I —mv H— mv A Ans. twice the Ans: v c = 0,v D = v v B = v, v A = 0 v c = v,v B = 0 538 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-63. The four balls each have the same mass m. If A and B are rolling forward with velocity v and strike C, determine the velocity of each ball after the first three collisions. Take e = 0.5 between each ball. v v QQ C D SOLUTION Collision will occur in the following sequence; B strikes C (±>) mv = mv B + mv c v = v B + v c (±>) o c v c~ v B e = 0.5 =- V v c = 0.75v—>, v B = 0.25v^> C strikes D (±0 m( 0.75?>) = mv c + mv D (±0 v D ~ v c e = 0.5 = 0.75?; Vc = 0.1875?) —» Ans. v D = 0.5625?) —> Ans. A strikes B (±0 mv + m(0.25v) = mv A + mv B (±>) or v b~ v A e = 0.5 = --- (?) - 0.25?)) v B = 0.8125?) —* , v A = 0.4375?) — * Ans. Ans: Vq = 0.1875?) —> v D = 0.5625?) —» v B = 0.8125?; —> v A = 0.4375?) —> 539 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 64 . Ball A has a mass of 3 kg and is moving with a velocity of 8 m/s when it makes a direct collision with ball B , which has a mass of 2 kg and is moving with a velocity of 4 m/s. If e = 0.7, determine the velocity of each ball just after the collision. Neglect the size of the balls. 8 m/s 4 m/s SOLUTION Conservation of Momentum. The velocity of balls A and B before and after impact are shown in Fig. a ( ^ ) m A {v A ) x + m B (v B )i = m A (v A ) 2 + m B (v B ) 2 3(8) + 2(—4) = 3v A + 2v b 3v a + 2v b = 16 (1) Coefficient of Restitution. /+N _ (vb)i - (va)2 Vb - v A e Ou), - (*>«), ’ ' 8-(-4) V B - V A = 8.4 Solving Eqs. (1) and (2), v B = 8.24 m/s —> v A = —0.16 m/s = 0.160 m/s <— ( 2 ) Ans. Ans. ( 7 ) Before Impact M'S After I/»j>act (a-) Ans: v B = 8.24 m/s —» v A = 0.160 m/s <— 540 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 65 . A 1-lb ball A is traveling horizontally at 20 ft/s when it strikes a 10-lb block B that is at rest. If the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic friction between the plane and the block is ix k = 0.4, determine the time for the block B to stop sliding. SOLUTION Smj V\ = v 2 1 — ](20) + 0 = ( — )(v A ) 2 + ( — )K) 2 32.2 10 32.2 (v A ) 2 + 10(n B ) 2 = 20 (v b ) 2 ~ (v A ) 2 0.6 = (^)i - (v B )i (y B ) 2 - (v A h 20-0 (v B ) 2 ~ (v A ) 2 = 12 Thus, (v B ) 2 = 2.909 ft/s -> (v A ) 2 = -9.091 ft/s = 9.091 ft/s «- Block B: 4 ^ mvi+'Z v 2 (^)< 2 .909) - 4,-0 t = 0.226 s W */oil9 N=/0fb Ans. Ans: t = 0.226 s 541 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 66 . Block A, having a mass m, is released from rest, falls a distance h and strikes the plate B having a mass 2m. If the coefficient of restitution between A and B is e, determine the velocity of the plate just after collision. The spring has a stiffness k. SOLUTION Just before impact, the velocity of A is A h t 1 + v 1 = t 2 + v 2 1 , 0 + 0 = —mv A — mgh v a = V2gh |, ( v b)i ~ (va) 2 V2 gh eV2gh = (y B ) 2 ~ (v A ) 2 (1) (+1) 'Zmvi = Srav 2 m(v A ) + 0 = m(v A ) 2 + 2 m(v B ) 2 (2) Solving Eqs. (1) and (2) for (y B ) 2 yields; (v B ) 2 = ^V2gh(l + e) Ans. Ans: (v b ) 2 = | Vlgh(l + e) 542 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 67 . The three balls each weigh 0.5 lb and have a coefficient of restitution of e = 0.85. If ball A is released from rest and strikes ball B and then ball B strikes ball C, determine the velocity of each ball after the second collision has occurred. The balls slide without friction. SOLUTION Ball A: Datum at lowest point. Tl+ Vl= T 2+ y 2 0 + (0.5)(3) = f (||)(V)? + 0 (t>a)i = 13.90 ft/s Balls A and B: 1,mvi = 'Zniv 2 (HK13.90) + 0 - (^(v,), + (f/)(vA _ (Vb) 2 ~ Ml) (.V A )l - (v B ) 1 0.85 = {vb)i ~ Mi 13.90 - 0 Solving: (■ v A ) 2 = 1.04 ft/s Ans. (v B ) 2 = 12.86 ft/s Balls B and C: Sow 2 = 3 (HX12.86) + 0 - + (||)(%)3 e 0.85 i v c )3 ( v b) 3 {Vb)2 ~ M2 (%)3 _ ( v b )3 12.86 - 0 Solving: {v B ) 3 = 0.964 ft/s (i>c) 3 = 11.9 ft/s Ans. Ans. Ans: (va )2 = 1.04 ft/s (v B ) 3 = 0.964 ft/s (Pc) 3 = 11-9 ft/s 543 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 68 . A pitching machine throws the 0.5-kg ball toward the wall with an initial velocity v A = 10 m/s as shown. Determine (a) the velocity at which it strikes the wall at B , (b) the velocity at which it rebounds from the wall if e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C. SOLUTION (a) (v B ) x i = 10 cos 30° = 8.660 m/s—» ( ^ ) s = s 0 + v 0 t 3 = 0 + 10 cos 30°? t = 0.3464 s (+t) v = v 0 + a c t (v B ) yt = 10 sin 30° - 9.81(0.3464) = 1.602 m/s f (+T) s = so + v Q t + -a c t 2 h = 1.5 + 10 sin 30°(0.3464) - i(9.81)(0.3464) 2 = 2.643 m (v B )t = V(1.602) 2 + (8.660) 2 = 8.81 m/s 6= tan^f 1 ^ 2 ) = 10.5° Z_ 1 1.8.660/ (b) ,+ x (Vb) 2 - (Va)2 (VBxh - 0 6 (v A ), - (v B y ■ 0 - (8.660) (v B x )2 = 4.330 m/s «- (v B y )2 = (v By )! = 1.602 m/s t (v B ) 2 = \/(4.330) 2 + (1.602) 2 = 4.62 m/s "1.602'' 0? = tan 4.330 = 20.3° A (c) (+t) s = s 0 + v Bt +-a c t 2 -2.643 = 0 + 1.602(f) “ -(9.81)(0 2 t = 0.9153 s ( ^ ) s = s 0 + v 0 t s = 0 + 4.330(0.9153) = 3.96 m m S,6ho -mi/j Ans. A~ Ans. "Wo™ ! £ Ans. Ans. / 2 ■ 643 ^ Ans: Ans. ( v b)i = 8.81 m/s 6»i = 10.5° (v B ) 2 = 4.62 m/s 0 2 = 20.3° A s = 3.96 m 544 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 69 . A 300-g ball is kicked with a velocity of v A = 25 m/s at point A as shown. If the coefficient of restitution between the ball and the field is e = 0.4, determine the magnitude and direction 8 of the velocity of the rebounding ball at B. B A SOLUTION Kinematics: The parabolic trajectory of the football is shown in Fig. a. Due to the symmetrical properties of the trajectory, v B = v A = 25 m/s and f = 30°. Conservation of Linear Momentum: Since no impulsive force acts on the football along the x axis, the linear momentum of the football is conserved along the x axis. m(v B ) x = m(v' B ) x 0.3(25 cos 30°) = 0.3 (»g)j (pb)x = 21.65 m/s <— a/siuA % Coefficient of Restitution: Since the ground does not move during the impact, the coefficient of restitution can be written as (+T) ^-{y' B ) y (• V B )y - 0 -My -25 sin 30° (v' B ) y = 5 m/s t Thus, the magnitude of is v' B - V(«b)j- + (v' B ) y = V21.65 2 + 5 2 = 22.2 m/s and the angle of is 8 = tan 13.0° Ans. Ans. Ans: v' B = 22.2 m/s 8 = 13.0° 545 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 546 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 71 . It was observed that a tennis ball when served horizontally 7.5 ft above the ground strikes the smooth ground at B 20 ft away. Determine the initial velocity V 4 of the ball and the velocity \ B (and 0) of the ball just after it strikes the court at B. Take e = 0.7. SOLUTION () S = Sq + v 0 t 20 = 0 + v A t (+ 1 ) s = s 0 + v 0 t + -a c t 2 7.5 = 0 + 0 + y(32.2 )t 2 t = 0.682524 v A = 29.303 = 29.3 ft/s v Bx\ ~ 29.303 ft/s (+!) v = Vq + a c t v Byl = 0 + 32.2(0.68252) = 21.977 ft/s ( -L ) mv i = mv 2 v B 2 x = v B ix = 29.303 ft/s -» A v x Ans. v By2 e =- v By 1 v By2 . °' 7 = 21 977’ VBy2 = 15 ' 384 ft/s T v b2 = V(29.303) 2 + (15.384) 2 = 33.1 ft/s Ans. , 15.384 6 = tan -1 -= 27.7° Ans. Ans: v A = 29.3 ft/s v B 2 = 33.1 ft/s 6 = 21. T 547 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 548 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 73 . Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine their final velocities just after collision. The coefficient of restitution is e = 0.75. SOLUTION ( -L ) 2mwi = 2tm>2 0.5(4)(|) - 0.5(6) = 0.5(ub) 2jc + 0.5(i^) 2 * , + , ( v a )2 ~~ ( v b )2 \?Bh ~ ( v a )i (v/d2x - (vb)2x ■ " 4 ( 1 ) - (- 6 ) (Va) 2 x = 1-35 m/s -» (y B )ix = 4.95 m/s <- (+T) mvi = mv 2 y (v A )i = 6 m / s 5 J4 °/ («b)i = 4 m /s O— x 0-5(|)(4) = Q.5(v B ) 2y («B) 2 y = 3.20 m/s t v A = 1.35 m/s —> Ans. v B = V(4.59) 2 + (3.20) 2 = 5.89 m/s Ans. , 3.20 6 = tan -1 —- = 32.9 5^ Ans. 4.95 Ans: v A = 1.35 m/s —» v B = 5.89 m/s 6 = 32.9° 5^ 549 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 74 . Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision B travels along a line, 30° counterclockwise from the y axis. SOLUTION Hmvi = ~S,mv 2 ( *) 0.5(4)(|) - 0.5(6) = -0.5K) 2x + 0.5(v a ) 2x -3.60 = ~(v B ) 2x + (v A ) 2x (+t) 0.5(4)(|) = 0.5(v B ) 2y ( v B ) ly = 3.20 m/s t (v B ) 2x = 3.20 tan 30° = 1.8475 m/s <— (*) ( v a) 2 x ~ —1.752 m/s = 1.752 m/s ( v a)2 ~ ( v b )2 Ob)i “ ( v a )i —1.752—(—1.8475) 4(|)—(—6) = 0.0113 (v A )i = 6 m/s 5J4 Bj/* •/ K)i = 4 m/s (Mi )±. 1.10 ">/ s io° Ans. Ans: e = 0.0113 550 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 75 . The 0.5-kg ball is fired from the tube at A with a velocity of v = 6 m/s. If the coefficient of restitution between the ball and the surface is e = 0.8, determine the height h after it bounces off the surface. t = 6m/s ^t30° 2 m C h B SOLUTION Kinematics. Consider the vertical motion from A to B. ( + t) (v B )y = (v A ) 2 y + 2 Oy[(s B )y - (s A ) v ]; {v B f y = (6 sin 30°) 2 + 2(-9.81)(-2 - 0) (v B ) y = 6.9455 m/si Coefficient of Restitution. The y-component of the rebounding velocity at B is (y' B ) y and the ground does not move. Then (v' B ) y = 5.5564 m/s ] Kinematics. When the ball reach the maximum height h at C, (v c ) y = 0. (+T) (v c ) 2 y = (v' B )] + 2 a c [(s c ) y - (s B ) v ] ; 0 2 = 5.55 64 2 + 2(-9.81 )(h - 0) h = 1.574 m = 1.57 m Ans. Ans: h = 1.57 m 551 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 76 . A ball of mass m is dropped vertically from a height h 0 above the ground. If it rebounds to a height of h\ , determine the coefficient of restitution between the ball and the ground. SOLUTION Conservation of Energy: First, consider the ball’s fall from position A to position B. Referring to Fig. a , T A + V A = T B + V b 1 2 1 2 2 mv A + (V g ) A = - mv B + (V g ) B 0 + mg(ho) = jm(v B )i 2 + 0 Subsequently, the ball’s return from position B to position C will be considered. T B + V B = T c + V c | mv B 2 + (V g ) B = | mv c 2 + ( V g ) c 1 2 - m(v B ) 2 +0 = 0 + mgh x (v B ) 2 = V2ghi t Coefficient of Restitution: Since the ground does not move, (Vb) 2 (+T) (Vb)i Vlghi -V2gh 0 Ans. Ans: 552 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 77 . The cue ball A is given an initial velocity (pa)i = 5 m/s. If it makes a direct collision with ball B (e = 0.8), determine the velocity of B and the angle 8 just after it rebounds from the cushion at C (e' = 0.6). Each ball has a mass of 0.4 kg. Neglect their size. SOLUTION Conservation of Momentum: When ball A strikes ball B, we have m A (v A ) i + m H (v B )\ = m A (v A ) 2 + m B (v B ) 2 0.4(5) + 0 = 0A(v a ) 2 + 0.4(O 2 Coefficient of Restitution: («t) (Vb)2 - ( v a)2 (Va)i - («u)i (V B )2 ~ ( V A )l ( 1 ) ( 2 ) Solving Eqs. (1) and (2) yields (v a ) 2 = 0.500 m/s (v B ) 2 = 4.50 m/s Conservation of“y” Momentum: When ball B strikes the cushion at C, we have >n B (v By ) 2 = m B (v By ) 3 (+1) 0.4(4.50 sin 30°) = 0.4(n B ) 3 sin 8 (v B ) 3 sin 8 = 2.25 Coefficient of Restitution (x): _ (Vch ~ («fl,)3 (v Bx h - (v c ) 1 . 0 - [-(«fl)3 COS 61] v ’ 4.50 cos 30° - 0 Solving Eqs. (1) and (2) yields (v B ) 3 = 3.24 m/s 8 = 43.9° ( 3 ) ( 4 ) Ans. Ans: («s) 3 = 3.24 m/s 8 = 43.9° 553 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 78 . Using a slingshot, the boy fires the 0.2-lb marble at the concrete wall, striking it at B. If the coefficient of restitution between the marble and the wall is e = 0.5, determine the speed of the marble after it rebounds from the wall. SOLUTION Kinematics: By considering the x and y motion of the marble from A to B, Fig. a, ( and (*b)x = (ui)* + [v A )xt 100 = 0 + 75 cos 45° t t = 1.886 s (‘S'fi))' {^A)y t 2 t (, s B ) y = 0 + 75 sin 45°(1.886) + ~ (-32.2)(1.886 2 ) and +T = 42.76 ft ( v B ) y = (v A ) y + a y t (■ v B ) y = 75 sin 45° + (-32.2)(1.886) = -7.684 ft/s = 7.684 ft/s i Since (u B ) x = (v A ) x ~ 75 cos 45° = 53.03 ft/s, the magnitude of \ B is v B = V(v B ) x 2 + (v B ) y 2 = V53.03 2 + 7.684 2 = 53.59 ft/s and the direction angle of \ B is My 6 = tan {v B )x = tan^ 1 1 | = 8.244° 53.03 Conservation of Linear Momentum: Since no impulsive force acts on the marble along the inclined surface of the concrete wall {x' axis) during the impact, the linear momentum of the marble is conserved along the x' axis. Referring to Fig. b, V/) m B (v' B ) x ’ = m B (v' B ) X ' — (53.59 sin 21.756°) 32.2 1 ’ v' B cos 4> = 19.862 0.2 32.2 [V B COS ( m Lb (i) 554 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-78. Continued Coefficient of Restitution: Since the concrete wall does not move during the impact, the coefficient of restitution can be written as o -{v'b)/ {v' B )y' ~ 0 — v' B sin <f> “ -53.59 cos 21.756° v' B sin f = 24.885 Solving Eqs. (1) and (2) yields v' B = 31.8 ft/s ( 2 ) Ans. Ans: v'b = 31.8 ft/s 555 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-79. The two disks A and B have a mass of 3 kg and 5 kg, respectively. If they collide with the initial velocities shown, determine their velocities just after impact. The coefficient of restitution is e = 0.65. SOLUTION (v Ax ) = 6 m/s (v Ay )^ = 0 (' v Bx)l ~ -7 cos 60° = - -3.5 m/s { v By)l ~ ~7 cos 60° = (*) m A {v Ax )i + m B( v Bx)\ = m A (v Ax ) 2 + m b( v Bx) 2 3(6)—5(3.5) = 3(v a ) x2 + 5(v b ) x2 ( v Bx)2 ~ (v A x )2 0 65 (VBx)l “ (vAxh V ) (v Ax )! ~ ( v Bx)l °' 65 6—( -3.5) 0 v Bx)l - (vax) 2 = 6.175 Solving, iyAx) 2 = —3.80 m/s ( v Bx) 2 = 2.378 m/s (+T) + m A (v Ay ) 2 O II (N (+T) m B{ v By) 1 + m B{ v A y) 2 ( v B ^j 2 = —6.062 m/s (v A ) 2 = V(3.80) 2 + (0) 2 = = 3.80 m/s *- (v B ) 2 = V(2.378) 2 + (-6.062) 2 = 6.51 m/s ( ^ )2 = tan_1 (l?l) = 68 - 60 6.062 m/s Ans. Ans. Ans. Ans: (v A ) 2 = 3.80 m/s <— (vb)i = 6.51 m/s (Osh = 68.6° 556 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *15-80. A ball of negligible size and mass m is given a velocity of v 0 on the center of the cart which has a mass M and is originally at rest. If the coefficient of restitution between the ball and walls A and B is e, determine the velocity of the ball and the cart just after the ball strikes A. Also, determine the total time needed for the ball to strike A , rebound, then strike B, and rebound and then return to the center of the cart. Neglect friction. SOLUTION After the first collision; ( ) 'tmvx = Xmv2 0 + mv o = mv b + Mv c Vo i v/yvj' - 3 VrU M mv o = mv b H- v c m ev 0 = v c - v b i>o(l + e) = ^1 + ^0 v c no(l + e)m (,m + M) v 0 (l + e)m v b = — ~ ... - ev 0 (m + M) v r = = Vq = V 0 \ m + me — em — eM m + M m — eM^ m + M The relative velocity on the cart after the first collision is v , V 0 «ref = ev 0 Similarly, the relative velocity after the second collision is ev 0 V re f = e 2 V 0 Total time is t d 2d d -1-b ~i — v 0 ev Q e l v 0 Ans. Ans. Ans. Ans: r>o(l + e)m (m + M ) Vb = Po 1 m — m + M t — —| 1 + — Vo 557 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-81. The girl throws the 0.5-kg ball toward the wall with an initial velocity v A = 10 m/s. Determine (a) the velocity at which it strikes the wall at B , (b) the velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C. SOLUTION Kinematics: By considering the horizontal motion of the ball before the impact, we have () s x = {so)* + v x t 3 = 0 + 10 cos 30°t t = 0.3464 s By considering the vertical motion of the ball before the impact, we have (+T) v y = (v 0 ) y + (a c ) y t = 10 sin 30° + (—9.81)(0.3464) = 1.602 m/s The vertical position of point B above the ground is given by (+ t) Sy = (s 0 )y + Oo)y t + 2 K)y t 2 (s B ) y = 1.5 + 10 sin 30°(0.3464) + | (-9.81)(o.3464 2 ) = 2.643 m Thus, the magnitude of the velocity and its directional angle are (Vb)i = \/(10 cos 30°) 2 + 1.602 2 = 8.807 m/s = 8.81 m/s Ans. 6 = tan^ 1 1-602 = 10.48° = 10.5° Ans. 10 cos 30° Conservation of “y” Momentum: When the ball strikes the wall with a speed of (v b )i = 8.807 m/s, it rebounds with a speed of ( v b ) 2 ■ m b (pftji = m b (v by ) 2 ( ^ ) m b (1.602) = m b [(v b ) 2 sin 4>\ (v b ) 2 sin 4 > ~ 1-602 (1) Coefficient of Restitution (x): (v w ) 2 ~ {v bx ) 2 0.5 = [v b )\ ~ (y w \ 0 - [-(ui) 2 cos</)] 10 cos 30° - 0 ( 2 ) 558 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-81. Continued Solving Eqs. (1) and (2) yields 4> = 20.30° = 20.3° (v b ) 2 = 4.617 m/s = 4.62 m/s Ans. Kinematics: By considering the vertical motion of the ball after the impact, we have [ ( + t ) Sy = (S Q ) y + (V Q ) y t + - (a c )y t 2 -2.643 = 0 + 4.617 sin 20.30°t 1 + i(— 9.81 )t\ t\ = 0.9153 s By considering the horizontal motion of the ball after the impact, we have ( ^ ) s* = Oo)* + v x t s = 0 + 4.617 cos 20.30°(0.9153) = 3.96 m Ans. Ans: (a) (v B )i = 8.81 m/s, 8 = 10.5° ■=£ (b) (v B ) 2 = 4.62 m/s, <£ = 20.3° ^ (c) s = 3.96 m 559 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-82. The 20-lb box slides on the surface for which fj. k = 0.3. The box has a velocity v = 15 ft/s when it is 2 ft from the plate. If it strikes the smooth plate, which has a weight of 10 lb and is held in position by an unstretched spring of stiffness k = 400 lb/ft, determine the maximum compression imparted to the spring. Take e = 0.8 between the box and the plate. Assume that the plate slides smoothly. SOLUTION T\ + 2^-2 = T 2 (°-3)( 2 0)(2) - \ (fZ)te)’ v 2 = 13.65 ft/s (*) ^mv i = = 2 mv 2 / 20 > ) (13.65) = ( 20 \ 10 \3Z2y 132 2 VA + 322 Vb v = 15 ft/s I--2 ft 20 lb ! 0.3(20) lb 20 lb (v B ) 2 - (v A h ( pa)i - (v B ) 1 0.8 = Vp ~ Va 13.65 Solving, v P = 16.38 ft/s, v A = 5.46 ft/s 7i + ^ = T 2 + U 2 \ (iu) (1638)2 + 0 = 0 + ^ ( 400)(s) z s = 0.456 ft Ans. Ans: s = 0.456 ft 560 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-83. The 10-lb collar B is at rest, and when it is in the position shown the spring is unstretched. If another 1-lb collar A strikes it so that B slides 4 ft on the smooth rod before momentarily stopping, determine the velocity of A just after impact, and the average force exerted between A and B during the impact if the impact occurs in 0.002 s. The coefficient of restitution between A and B is e = 0.5. SOLUTION Collar B after impact: t 2 + v 2 =t 3 + v 3 + 0 ■ 0 + >>< 5 - 3)2 (v B ) 2 = 16.05 ft/s System: ( -L j HmjVj = Smi v 2 32.2 + ° _ 32.2 (V ^ 2 + 32.2 ^ 16 '° 5 ^ (Ui)t “ M 2 = 160.5 ( fa)? 2 ( v ^) 2 1 ’ ( v A )i ~ (v B )i _ 16.05 - (v A ) 2 (Ui)i “ 0 0.5(u0i + K ) 2 = 16.05 Solving: (v A )! = 117.7 ft/s = 118 ft/s (v A ) 2 = -42.8 ft/s = 42.8 ft/s *- Collar A: mv\ + 2 v 2 (^)(H7.7) - F( 0 . 002 ) - ( 5 ^)—42.8) F = 2492.2 lb = 2.49 kip I lb F Ans. Ans. Ans: (v A ) 2 = 42.8 ft/s <— F = 2.49 kip 561 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 562 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-85. A ball is thrown onto a rough floor at an angle of 8 = 45°. If it rebounds at the same angle 4> = 45°, determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is e = 0.6. Hint: Show that during impact, the average impulses in the x and y directions are related by I x = /il y . Since the time of impact is the same, F x At = /iF y At or F x = /iF y . SOLUTION 0 — [— v 2 sin <f>] v 1 sin 8 — 0 v 2 sin 4> Vi sin 8 m(v x )i + jf F x dx = m{v x ) 2 mv 1 cos 6 — F x At = mv 2 cos <f> mvi cos 8 — mv 2 cos 4> F x = At (+t) f h \ : K)i + 1 F y dx = mlv y Y mvi sin 0 — F y At = -m» 2 sin < mvi sin 6 + mv 2 sin (f> Fy = At Since F x = fxF y , from Eqs. (2) and (3) mvi cos 8 - mv 2 cos (f> ix(mv\ sin 8 + mv 2 sin (f >) At v 2 cos 8 - /jl sin 8 /jl sin 4> + cos 4> Substituting Eq. (4) into (1) yields: sin < f> f cos 8 — /jl sin 8 0.6 = At sin 8 V/rsin 4> + cos </>, sin 45°/cos 45° — /jl sin 45' 0.6 = sin 45° \/jL sin 45° + cos 45' /jl = 0.25 1 F /JL ( 1 ) ( 2 ) (3) (4) Ans. Ans: /jL k = 0.25 563 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-86. Two smooth billiard balls A and B each have a mass of 200 g. If A strikes B with a velocity ( 11 , 4)1 = 1.5 m/s as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.85. Neglect the size of each ball. SOLUTION (v A ) 1 = -1.5 cos 40° = -1.1491 m/s ( v A y )i = —1.5 sin 40° = —0.9642 m/s (±0 m A {v A ) 1 + m B {v B ) 1 = m A (v Ax ) 2 + "‘b(vb x ) 2 -0.2(1.1491) + 0 = 0.2(v A x ) 2 + 0.2 (v B x ) 2 (±>) ( v Ax ) 2 ~ (v Bx ) 2 ( v A ) 2 - (v B ) 2 e = - 7 - 7 -—; 0.85 =- (v Bx )i - (paJi 1.1491 Solving, ( v Ax ) 2 = —0.08618 m/s ( v b x )i = -1-0629 m/s For A: ( + J') ™ A (v Ay )i = m A (v A ) 2 (v A ) 2 = 0.9642 m/s For B: (+T) m B (v B )i = m B (v By ) 2 ( v B y ) 2 = 0 Hence. (v B ) 2 = (v Bx ) 2 = 1.06 m/s «- (v A ) 2 = V(-0.08618) 2 + (0.9642) 2 = 0.968 m/s , , J 0.08618\ A ( d A ) 2 = tan 1 = 5.11 v An \ 0.9642 ) y Ans. Ans. Ans. Ans: (v B ) 2 = 1-06 m/s <— (v A ) 2 = 0.968 m/s ( 0 , 4)2 = 5.11° A 564 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-87. The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has a weight of 47 lb, and the coefficient of restitution between the “stones” is e = 0 . 8 , determine their speeds just after collision. Initially A has a velocity of 8 ft/s and B is at rest. Neglect friction. SOLUTION Line of impact (x-axis): ~%mv\ = ~%mv 2 (+N) 0 + il (8) COS 30 ° = § 2 ^ + § 2 {[Va) * (+$$ e = 0.8 = (« b )2* - (^4)2* 1 cos 30° — 0 Solving: O/O 2 * = 0.6928 ft/s (' v b ) 2 x = 6-235 ft/s Plane of impact (y-axis): Stone A: mv 1 = mx >2 {/+) 3222 (8) Sin 3 °° = 32.2 ^ VA ^ 2y (v A h y = 4 Stone B: mv\ = mv 2 41 (/'+) 0 = ( v B)ly = 0 (>, 4)2 = \/(0.6928) 2 + (4 ) 2 = 4.06 ft/s (v B ) 2 = V(0) 2 + (6.235) 2 = 6.235 = 6.24 ft/s Ans. Ans. Ans: 0 . 4)2 = 4.06 ft/s (v B ) 2 = 6.24 ft/s 565 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *15-88. The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has a weight of 47 lb, and the coefficient of restitution between the “stone” is e = 0.8, determine the time required just after collision for B to slide off the runway. This requires the horizontal component of displacement to be 3 ft. SOLUTION See solution to Prob. 15-87. (■ v B )i = 6.235 ft/s S = S 0 + v (/ 3 = 0 + (6.235 cos 60 °)t t = 0.962 s Ans: t = 0.962 s 566 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-89. Two smooth disks A and B have the initial velocities shown just before they collide. If they have masses m A = 4 kg and m B = 2 kg, determine their speeds just after impact. The coefficient of restitution is e = 0 . 8 . SOLUTION Impact. The line of impact is along the line joining the centers of disks A and B represented by y axis in Fig. a. Thus [(pa)iL ~ 15 (j) = 9 m/s i/ l(v A )i\x = 15 = 12 m/s \ [(pb)i]v = 8 m/s Z 1 [('«fl)iL = 0 Coefficient of Restitution. Along the line of impact (y axis), , , > [(«B) l]y ~ [(«4) 2 ]y „ 0 [K)dy ~ [(v A ) 2 ]y ( / ' ) 6 “ [(^)l]y - [K)l]/ _ -9-8 [Mliy ~ [(« B ) 2 ]y = 13.6 (1) Conservation of ‘y’ Momentum. (+/) m A [(v A )i] y + m B [(v B )i] y = m A [(v A ) 2 ] y + m B [(v B )T\ y 4 (— 9 ) + 2 ( 8 ) = A[(vM y + 2[(v B ) 2 ] y 2[(v A ) 2 ]y + [(v B ) 2 ]y = -10 (2) Solving Eqs. (1) and (2) [( v A ) 2 iy = 1.20 m/s / [('ffl) 2 ]y = —12.4 m/s = 12.4 m/s i/ Conservation of ‘x’ Momentum. Since no impact occurs along the x axis, the component of velocity of each disk remain constant before and after the impact. Thus / [(v A h]x = [0hi)iL = 12 m/s \ [(v B ) 2 ] x = [(v B )y] x = 0 Thus, the magnitude of the velocity of disks A and B just after the impact is { v a)i = '^[( v a)2& + [(PA) 2 ]y = Vl2 2 + 1.20 2 = 12.06 m/s = 12.1 m/s Ans. (v B ) 2 = V[(t ; B ) 2 ] 2 + [(v B ) 2 f y = Vo 2 + 12.4 2 = 12.4 m/s Ans. Ans: (v A ) 2 = 12.1 m/s (v B ) 2 = 12.4 m/s 567 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-90. Before a cranberry can make it to your dinner plate, it must pass a bouncing test which rates its quality. If cranberries having an e > 0.8 are to be accepted, determine the dimensions d and h for the barrier so that when a cranberry falls from rest at A it strikes the incline at B and bounces over the barrier at C. SOLUTION Conservation of Energy: The datum is set at point B. When the cranberry falls from a height of 3.5 ft above the datum, its initial gravitational potential energy is W(3.5) = 3.5 W. Applying Eq. 14-21, we have T 1 + V x = T 2 + U 2 0 + 3W -1 (Is) + 0 (rOi = 15.01 ft/s Conservation of “x' ” Momentum: When the cranberry strikes the plate with a speed of («,.)! = 15.01 ft/s, it rebounds with a speed of (v c ) 2 . ™c (wji = m c KJ 2 (+/) m c (15.01)0 j = m c [(v c ) 2 cos <f>] ( v c )2 cos <f> = 9.008 Coefficient of Restitution (y 1 ): (v P ) 2 ~ (v Cy ,) 2 (\ + ) 0.8 = (v)t “ 0 ~ (v c ) 2 sin f -15.011 -|-0 ( 1 ) ( 2 ) Solving Eqs. (1) and (2) yields 4> = 46.85° (v c ) 2 = 13.17 ft/s Kinematics: By considering the vertical motion of the cranberry after the impact, we have (+T) v y = (v 0 ) y + a c t 0 = 13.17 sin 9.978° + (-32.2) t t = 0.07087 s . 1 , (+ T ) Sy = (so)> + (v 0 )y t + - (a c )y t = 0 + 13.17 sin 9.978° (0.07087) + | (-32.2)(0.07087 2 ) = 0.080864 ft 568 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-90. Continued By considering the horizontal motion of the cranberry after the impact, we have ( ^ ) S x = Oo)* + V x t = 0 + 13.17 cos 9.978° (0.07087) d = 1.149 ft = 1.15 ft Thus, h = s y + ^d = 0.080864 + | (1.149) = 0.770 ft Ans. Ans. Ans: d = 1.15 ft h = 0.770 ft 569 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-91. The 200-g billiard ball is moving with a speed of 2.5 m/s when it strikes the side of the pool table at A. If the coefficient of restitution between the ball and the side of the table is e = 0.6, determine the speed of the ball just after striking the table twice, i.e., at A , then at B. Neglect the size of the ball. SOLUTION At A: (v A ) yl = 2.5(sin45°) = 1.7678 m/s — » ( v A y h ( v A y h £ = (%V °' 6 = 1-7678 (.v Ay ) 2 = 1-061 m/s <h- (va ,)2 = ( v a,)i = 2.5 cos 45° = 1.7678 m/s 1 At B-. (Vfijl _ ( v B x h (y B ) 2 ' 1.7678 (vn x h = 1-061 m/s KJs = (%) 2 = 1-061 m/s Hence, (v B ) 3 = V(1.061) 2 + (1.061) 2 = 1.50 m/s Ans: (v B h = 1-50 m/s 570 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *15-92. The two billiard balls A and B are originally in contact with one another when a third ball C strikes each of them at the same time as shown. If ball C remains at rest after the collision, determine the coefficient of restitution. All the balls have the same mass. Neglect the size of each ball. SOLUTION Conservation of “x” momentum: -L ^ m v = 2 mv' cos 30° v = 2v' cos 30° Coefficient of restitution: v ' v cos 30° Substituting Eq. (1) into Eq. (2) yields: 2v' cos 2 30° 3 Ans: e = 2 3 571 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-93. Disks A and B have a mass of 15 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. The coefficient of restitution between them is e = 0.8. SOLUTION Conservation of Linear Momentum: By referring to the impulse and momentum of the system of disks shown in Fig. a , notice that the linear momentum of the system is conserved along the n axis (line of impact). Thus, +/ m A (v A )„ + m B ( v B ) n = m A (v A ) n + m B ( v B ) n 15(10)^^ — 10(8)^^ = 15u)i cos + 10 ub cos <p B 15^ cos <t> A + 10« B cos 4> b = 42 (1) Also, we notice that the linear momentum of disks A and B are conserved along the t axis (tangent to? plane of impact). Thus, +\ m A (v A ), = m A (v A ), '4 N 15(10) = 15v a sin i v A sin f A = 8 and +\ m B ( v B ), = m B [v B ), '4 ( 2 ) 10(8) V 5 / = 10 Vb sin ^ B v B sin <j) B = 6.4 Coefficient of Restitution: Ill e coefficient of restitution equation written along the n axis (line of impact) gives n m tov' +/ e = 0.8 = ( VB)n - (VA)n (v A ) n ~ ( V B )„ v B cos <p B — V A COS (f> A oJ A 10 v B cos 4> B — v A cos 4> a = 8.64 Solving Eqs. (1), (2), (3), and (4), yeilds v A = 8.19 m/s f A = 102.52° v B = 9.38 m/s d>n = 42.99° Ans: (y A h = 8-19 m/s (v B ) 2 = 9.38 m/s 572 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-94. Determine the angular momentum H 0 of the 6-lb particle about point O. SOLUTION Position and Velocity Vector. The coordinates of points A and B are A(- 8, 8,12) ft and B( 0,18, 0) ft. Then fOB = {18j} ft r OA = {—8i + 8j + 12k} ft z Va = V A \ r AB r AB [0 - (—8)]i + (18 - 8)j + (0 - 12)k V[0 - (—8)] 2 + (18 - 8) 2 + (0 - 12) 2 I 32 . 40 . | V308 1 V308 J ft/s Angular Momentum about Point O. Ho = r OB X m ^A i J k = 0 18 0 6 / 32 \ 6 / 40 \ 6 / 48 \ 32-2 Vv / 3082 32.2VV308/ 32.2V V308/ = { -9.1735i - 6.1156k} slug-ft 2 /s = {-9.17i - 6.12k} slug * ft 2 /s Also, H 0 = r OA X mV A i j = -8 8 6 / 32 \ 6 / 40 \ 32 -2VV / 308/ 32.2VV308/ = { -9.1735i - 6.1156k} slug-ft 2 /s = {— 9.17i — 6.12k} slug-ft 2 /s Ans. Ans. k 12 6 / 48 \ 32-2 \ V308/ Ans: { —9.171 - 6.12k} slug • ft 2 /s 573 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-95. Determine the angular momentum H p of the 6-lb particle about point P. SOLUTION Position and Velocity Vector. The coordinates of points A, B and P are A(-8, 8,12) ft, B( 0,18, 0) ft and P(-8, 0, 0).Then fpB = [0 - (—8)]i + (18 - 0)j] = {8i + 18j} ft z V = [-8 - (—8)]i + (8 - 0)j + (12 - 0)j] = (8j + 12k} ft V A (~8)]i + (18 - 8)j + (0 - 12)k (—8)] 2 + (18 - 8) 2 + (0 - 12) 2 I 32 . 40 . | V308 1 V308 J ft/s Angular Momentum about Point P. H P = r pA X mV A i j k = 0 8 12 6 ( 32 \ 6 ( 40 ) 6 / 48 ) 32-2 \ V308/ 32 -2\V308/ 32 - 2 \ V308/ = { -9.17351 + 4.0771j - 2.7181} slug • ft 2 /s = { —9.171 + 4.08j - 2.72k} slug-ft 2 /s Ans. Also, H P = r pB X mV A i J k = 8 18 0 6 / 32 ) 6 / 40 \ 6 / 48 ) 32 - 2 V V3082 32 - 2 V V308/ 32 -2V V / 3082 = { —9.1735i + 4.0771j - 2.7181k} slug-ft 2 /s = { —9.171 + 4.08j - 2.72k} slug-ft 2 /s Ans. Ans: { —9.171 + 4.08j - 2.72k} slug-ft 2 /s 574 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 96 . Determine the angular momentum H 0 of each of the two particles about point O. SOLUTION Q+(H a ) 0 C +{H b )o (-1.5) 3(8)| (—1)[4(6 sin 30°)] ( 2 ) -57.6 kg • m 2 /s (4)[4 (6 cos 30°)] = -95.14 kg • m 2 /s Thus (H a )o = {-57.6*} kg• m 2 /s (H b ) 0 = {-95.1*} kg• m 2 /s y Ans. Ans. Ans: ( H a ) 0 = {-57.6*} kg• m 2 /s (H b ) q = {-95.1*} kg• m 2 /s 575 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 97 . Determine the angular momentum of each of the two particles about point P. SOLUTION C +(h a ) p C +{ h b ) p (2.5) 3(8)1 (4) [4(6 sin 30°)] ( 7 ) -52.8 kg • m 2 /s 8[4 (6 cos 30°)] = -118.28 kg • m 2 /s Thus, (H A ) p = {-52.8k} kg• m 2 /s ( H B ) P = {-118k} kg• m 2 /s y Ans. Ans. Ans: (H A )p — {-52.8k} kg • m 2 /s ( H b ) p = {-118k} kg• m 2 /s 576 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 98 . Determine the angular momentum H 0 of the 3-kg particle about point O. SOLUTION Position and Velocity Vectors. The coordinates of points A and B A( 2, -1.5, 2) m and B( 3, 3, 0). z y r OB = {3i + 3j} m r OA = {21 - 1.5j + 2k} m V A ( 6 ) (3 - 2)i + [3 V(3 - 2 f + [3 ( —1.5)]j + (0.2)k (—1.5)] 2 + (0 - 2) 2 27 V25.25 V25.25 12 V25.25 m/s Angular Momentum about Point O. Applying Eq. 15 Ho = r OB X m ^A 3 6 V25.257 VV25.25 J 3 27 k 0 3 - 12 V25.25 = { —21.4928i + 21.4928j + 37.6124k} kg-m 2 /s = {— 21.5i + 21.5j + 37.6} kg-m 2 /s Also, H 0 = r OA X mV 4 2 6 V25.25/ VV25.25 J -1.5 27 k 2 3 - 12 V25.25 = { —21.4928i + 21.4928j + 37.6124k} kg-m 2 /s = {—21.5i + 21.5j + 37.6k} kg-m 2 /s Ans. Ans. Ans: {—21.5i + 21.5j + 37.6} kg-m 2 /s 577 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 99 . Determine the angular momentum H P of the 3-kg particle z r PB = [3 - (—l)]i + (3 - 1.5)j + (0 - 2)k = {4i + 1.5j - 2k} m V A (3 - 2)i + [3 - ( — 1.5)]j + (0 - 2)k V(3 - 2) 2 + [3 - (-1.5)] 2 + (0 - 2 f 27 . V25.25 J 12 V25.25 m/s Angular Momentum about Point P. Applying Eq. 15 H p = r pA X mV A 3 6 V25.25 J -3 27 k 0 3 - 12 V25.25 ^V25.25 = {21.4928i + 21.4928j + 59.1052k} kg-m 2 /s = {21.51 + 21.5j + 59.1k} kg-m 2 /s Also, H p = r PB X mV A 4 6 V25.25 J 1.5 27 3 — k -2 12 -■V 25 . 25 J V V25.25 = {21.4928i + 21.4928j + 59.1052k} kg-m 2 /s = {21.51 + 21.5j + 59.1k} kg-m 2 /s Ans. Ans. Ans: {21.51 + 21.5j + 59.1k} kg-m 2 /s 578 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 100 . Each ball has a negligible size and a mass of 10 kg and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque M = (f 2 + 2) N • m, where t is in seconds, determine the speed of each ball when t = 3 s. Each ball has a speed v = 2 m/s when t = 0. SOLUTION Principle of Angular Impulse and Momentum. Referring to the FBD of the assembly, Fig. a (Hzh + - / M z dt = (H z ) 2 A r 3s 2[0.5(10)(2)] + / (f 2 + 2 )dt = 2[0.5(10u)] J o v = 3.50 m/s Ans. Ans: v = 3.50 m/s 579 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 101 . The 800-lb roller-coaster car starts from rest on the track having the shape of a cylindrical helix. If the helix descends 8 ft for every one revolution, determine the speed of the car when t = 4 s. Also, how far has the car descended in this time? Neglect friction and the size of the car. SOLUTION 6 = tan_1 ( TUTT ) = 9.043° VMS)/ %F y = 0; N - 800 cos 9.043° = 0 N = 790.1 lb Ht + 0+ [ 8(790.1 sin 9.043°)df Jo 800 32.2 (8)v, v, = 20.0 ft/s 20.0 cos 9.043° 20.2 ft/s VC?* ?<ntlb Ans. 7/ + £14 - 2 = t 2 1 / '800 0 + 800/z = q 2 \ .321. h = 6.36 ft ( 20 . 2) 2 Ans. Ans: v = 20.2 ft/s h = 6.36 ft 580 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 102 . The 800-lb roller-coaster car starts from rest on the track having the shape of a cylindrical helix. If the helix descends 8 ft for every one revolution, determine the time required for the car to attain a speed of 60 ft/s. Neglect friction and the size of the car. SOLUTION 6 = tan / 8 /o J = 9.043° ZF y = 0; N - 800 cos 9.043° = 0 N = 790.1 lb v t cos 9.043° 60 Vt cos 9.043° v, = 59.254 ft/s H, + Mdt = H- rt 800 0 + / 8(790.1 sin 9.043°)dr = — (8)(59.254) .lr\ jZ.Z t = 11.9 s ut(5 H 1 I 'd'Z' Ans. Ans: t = 11.9 s 581 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 103 . A 4-lb ball B is traveling around in a circle of radius r x = 3 ft with a speed (v B )y = 6 ft/s. If the attached cord is pulled down through the hole with a constant speed v r = 2 ft/s, determine the ball’s speed at the instant r 2 = 2 ft. How much work has to be done to pull down the cord? Neglect friction and the size of the ball. SOLUTION H x = H 2 3 ^ (6) ( 3 ) 4 32.2 v a (2) v e = 9 ft/s v 2 = V9 2 + 2 2 = 9.22 ft/s Ty + 21/i- 2 = T 2 \^ W + ^ 2 = \^ 22) 2 2t/!_ 2 = 3.04 ft-lb Ans. Ans. Ans: v 2 = 9.22 ft/s £14 _ 2 = 3.04 ft-lb 582 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 104 . A 4-lb ball B is traveling around in a circle of radius rq = 3 ft with a speed (v B )i = 6 ft/s. If the attached cord is pulled down through the hole with a constant speed v r = 2 ft/s, determine how much time is required for the ball to reach a speed of 12 ft/s. How far r 2 is the ball from the hole when this occurs? Neglect friction and the size of the ball. SOLUTION V= V(v e ) 2 + (2 f 12 = V(v e ) 2 + ( 2) 2 v„ = 11.832 ft/s Hi = H 2 3*2 (6)(3) = 3i2 (1L832)( ^ r 2 = 1.5213 = 1.52 ft A r = v r t (3 - 1.5213) = It t = 0.739 s Ans: r 2 = 1.52 ft t = 0.739 s Ans. Ans. 583 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 105 . The two blocks A and B each have a mass of 400 g. The blocks are fixed to the horizontal rods, and their initial velocity along the circular path is 2 m/s. If a couple moment of M = (0.6) N • m is applied about CD of the frame, determine the speed of the blocks when t = 3 s. The mass of the frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks. SOLUTION (H„h + 2 j ' 2 M 0 dt = (H 0 ) 2 ■> h 2[0.3(0.4)(2)] + 0.6(3) = 2[0.3(0.4)v] v = 9.50 m/s Ans. Ans: v = 9.50 m/s 584 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 585 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 107 . If the rod of negligible mass is subjected to a couple moment of M = (30 1 1 ) N • m and the engine of the car supplies a traction force of F = (15f) N to the wheels, where t is in seconds, determine the speed of the car at the instant f = 5 s. The car starts from rest. The total mass of the car and rider is 150 kg. Neglect the size of the car. SOLUTION Free-Body Diagram: The free-body diagram of the system is shown in Fig. a. Since the moment reaction IY1 V has no component about the z axis, the force reaction F ? acts through the z axis, and the line of action of W and N are parallel to the z axis, they produce no angular impulse about the z axis. Principle of Angular Impulse and Momentum: r‘i (2/ M z dt = (h 2 ) z J t 2 />5s /*5 s 0+ / 30t 2 dt + / I5t(4)dt = 150v(4) Jo Jo v = 3.33 m/s Ans. (a) N Ans: v = 3.33 m/s 586 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 108 . When the 2-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. If the force F on the cord is increased, the bob rises and then rotates around the horizontal circular path B. Determine the speed of the bob around path B. Also, find the work done by force F. SOLUTION Equations of Motion: By referring to the free-body diagram of the bob shown in Fig. a , + = 0 ; F cos 8 — 2(9.81) = 0 ,,2 = ma F sin 8 = 2 v I sin 8 ( 1 ) ( 2 ) Eliminating F from Eqs. (1) and (2) yields sin 2 6 _ v 2 cos 8 9.81/ 1 - cos 2 0 _ v 2 cos 0 9.81/ When / = 0.6 m, v = V! = 5 m/s. Using Eq. (3), we obtain 1 - cos 2 /?! 1.5 2 cos 8i 9.81(0.6) cos 2 ^ + 0.3823 cos 8 1 — 1 = 0 b Solving for the root < 1, we obtain 0, = 34.21° Conservation of Angular Momentum: By observing the free-body diagram of the system shown in Fig. b, notice that W and F are parallel to the z axis, M s has no z component, and F s acts through the z axis. Thus, they produce no angular impulse about the z axis. As a result, the angular momentum of the system is conserved about the z axis. When 8 = 6 1 = 34.21° and 8 = 0 2 , r = r 1 = 0.6 sin 34.21° = 0.3373 m and r = r 2 = 0.3 sin d 2 ■ Thus, K)i = fah r^mv i = r 2 mv2 0.3373(2)(1.5) = 0.3 sin d 2 (2)v 2 v 2 sin d 2 = 1.6867 ( 4 ) £ iP) 587 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *15-108. Continued Substituting / = 0.3 and 0 = 0 2 v = v 2 into Eq. (3) yields 1 — cos 2 0 2 v 2 cos 6 2 9.81(0.3) 1 — COS 2 62 v ? 2 -- = (5) cos 62 2.943 Eliminating v 2 from Eqs. (4) and (5), sin 3 0 2 tan0 2 — 0.9667 = 0 Solving the above equation by trial and error, we obtain d 2 = 57.866° Substituting the result of d 2 into Eq. (4), we obtain v 2 = 1.992 m/s = 1.99 m/s Ans. Principle of Work and Energy: When 6 changes from 0 l to h 2 , W displaces vertically upward h = 0.6 cos 34.21° — 0.3 cos 57.866° = 0.3366 m. Thus, W does negatives work. Ti + j.Ui-2 = r 2 | (2)(1.5 2 ) + U F - 2(9.81)(0.3366) = | (2)(1.992) 2 U F = 8.32 N • m Ans. Ans: v 2 = 1.99 m/s U F = 8.32 N-m 588 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 109 . The elastic cord has an unstretched length 1 0 = 1.5 ft and a stiffness k = 12 lb/ft. It is attached to a fixed point at A and a block at B , which has a weight of 2 lb. If the block is released from rest from the position shown, determine its speed when it reaches point C after it slides along the smooth guide. After leaving the guide, it is launched onto the smooth horizontal plane. Determine if the cord becomes unstretched. Also, calculate the angular momentum of the block about point A, at any instant after it passes point C. SOLUTION Tr + V H = T c + V c 0 + 1(12)(5 - 1.5)1 . 1 + 1(12X3 - 1.5)1 v c = 43.95 = 44.0 ft/s There is a central force about A, and angular momentum about Ha = 3 Ta( 43 - 9 ~’)( 3 ) = 8-19 slug-ft 2 /s If cord is slack AD = 1.5 ft (H a ) i = (fl A ) 2 8 - 19 = ^MdC 1 - 5 ) (v g ) D = 88 ft/s But Ans. is conserved. Ans. Tc + Vc — T d + Vj } 1{^) (4 3.95)1 + i(12)(3 - ..5)1 - v D = 48.6 ft/s Since v D < (v e ) D cord will not unstretch. C v D f + 0 Ans. Ans: Vc = 44.0 ft/s H a = 8.19 slug -ft 2 /s. The cord will not unstretch. 589 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 590 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 111 . A box having a weight of 8 lb is moving around in a circle of radius r A — 2 ft with a speed of = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of v r = 4 ft/s, determine the speed of the box at the instant r B — 1 ft. How much work is done after pulling in the rope from A to B1 Neglect friction and the size of the box. SOLUTION (H z ) A = (H z ) b - (^) ( 2)(5) = (^)(l)Mtangent = T b -T a (y b )tangent 10 v B = V(10) 2 + (4) 2 = 10.77 = 10.8 ft/s u “ - Ki) (,077) 4(3li) (5)! U AB = 11.3 ft-lb Ans. Ans. Ans: v B = U AB 10.8 ft/s = 11.3 ft-lb 591 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 112 . A toboggan and rider, having a total mass of 150 kg, enter horizontally tangent to a 90° circular curve with a velocity of v A = 70 km/h. If the track is flat and banked at an angle of 60°, determine the speed v B and the angle 0 of “descent," measured from the horizontal in a vertical x-z plane, at which the toboggan exists at B. Neglect friction in the calculation. SOLUTION v A = 70 km/h = 19.44 m/s (H a \ = (H b ) z 150(19.44)(60) = 150(i/ b ) cos 0(57) Datum at B: Ta + V a = T b + V b i ( 150 )( 19 . 44) 2 + 150 ( 9 . 81 )/. = ^( 150)(^) 2 + 0 Since h = ( r A — r B ) tan 60° = (60 — 57) tan 60° = 5.196 Solving Eq. (1) and Eq (2): v B = 21.9 m/s 0 = 20.9 Ans: v B = 21.9 m/s 0 = 20.9 z ( 1 ) h (2) r*-r E Ans. Ans. 592 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 113 . An earth satellite of mass 700 kg is launched into a free- flight trajectory about the earth with an initial speed of v A — 10 km/s when the distance from the center of the earth is r A = 15 Mm. If the launch angle at this position is 4>a — 70°, determine the speed v B of the satellite and its closest distance r B from the center of the earth. The earth has a mass M e = 5.976(10 24 ) kg. Hint: Under these conditions, the satellite is subjected only to the earth’s gravitational force, F = GM e m s /r 2 , Eq. 13-1. For part of the solution, use the conservation of energy. SOLUTION (Ho), = (H 0 ) 2 m s (v A sin <t> A )r A = m s (v B )r B 700[10(10 3 ) sin 70°](15)(10 6 ) = 700 (v B )(r B ) T A + V A = T R + 1 , GM P m, 1 - m s ( v A f -= - m s (v B ) - Z Y a Z 2 GM e m s 1 r A 2 r B 12\/r (700)[10(10 3 )] 2 - 66.73(10 1Z )(5.976)(10 Z4 )(700) 1 = ^(700 ){v B f [15(10 6 )] 66.73(10 12 )(5.976)(10 24 )(700) r B Solving, v B = 10.2 km/s r B = 13.8 Mm ( 1 ) ( 2 ) Ans. Ans. Ans: v B = 10.2 km/s r B = 13.8 Mm 593 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-114. The fire boat discharges two streams of seawater, each at a flow of 0.25 m 3 /s and with a nozzle velocity of 50 m/s. Determine the tension developed in the anchor chain needed to secure the boat. The density of seawater is p sw = 1020 kg/m 3 . SOLUTION Steady Flow Equation: Here, the mass flow rate of the sea water at nozzles A and dm A dm R B are —-— = —-— = psw Q = 1020(0.25) = 225 kg/s. Since the sea water is col- dt dt lected from the larger reservoir (the sea), the velocity of the sea water entering the control volume can be considered zero. By referring to the free-body diagram of the control volume (the boat), i dm a / \ dm ^ ZF r = -v^( V A ) x + dt v ' dt T cos 60° = 225(50 cos 30°) + 225(50 cos 45°) T = 40 114.87 N = 40.1 kN Ans. to.) Ans: T = 40.1 kN 594 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-115. = dt D x + 4968 = 600 (12.0 - 0) D x = 2232N = 2.23 kN + tsr,= ^,v. out v l/ in Vi D v = 600[0 - (-12.0)] D v = 7200 N = 7.20 kN Ans. Ans. The chute is used to divert the flow of water, Q = 0.6 m 3 /s. If the water has a cross-sectional area of 0.05 m 2 , determine the force components at the pin D and roller C necessary 0.12 m for equilibrium. Neglect the weight of the chute and weight of the water on the chute. p w = 1 Mg/nT. SOLUTION 2m Equations of Steady Flow: Here, the flow rate Q = 0.6 m 2 /s. Then, v = — = = 12.0 m/s. Also, — p w Q — 1000 (0.6) = 600 kg/s. Applying Eqs. 15-26 and 15-28, we have dm C + CM A = —- ( d DB v B — d DA v A ); -C x (2) = 600 [0 - 1.38(12.0)] C x = 4968 N = 4.97 kN Ans. dm / (Tb, - v A J; Cx ■iZ-Or./i K \ | - ■ ; ■— 1 / - : dm Ans: C x = 4.97 kN D x = 2.23 kN D y = 7.20 kN 595 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 116 . The 200-kg boat is powered by the fan which develops a slipstream having a diameter of 0.75 m. If the fan ejects air with a speed of 14 m/s, measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest. Assume that air has a constant density of p a = 1.22 kg/nr 1 and that the entering air is essentially at rest. Neglect the drag resistance of the water. 0.75 m SOLUTION Equations of Steady Flow: Initially, the boat is at rest hence v B = v a = 14 m/s. Then, Q = v B A = 14 '(o.75 2 ,, dm = 6.185 m 3 /s and — = p a Q at i/b = 1.22(6.185) = 7.546 kg/s. Applying Eq. 15-26, we have dm = — (v B - v A ); —F = 7.546(—14 - 0) F = 105.64 N at x 1 Equation of Motion: 'ZF X = ma x \ 105.64 = 200n a = 0.528 m/s 2 Ans. D-- 14-mls X- F--/05M-/J 2 Ans: a = 0.528 m/s 2 596 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15-117. The nozzle discharges water at a constant rate of 2 ft 3 /s.The cross-sectional area of the nozzle at A is 4 in 2 , and at B the cross-sectional area is 12 in 2 . If the static gauge pressure due to the water at B is 2 lb/in 2 , determine the magnitude of force which must be applied by the coupling at B to hold the nozzle in place. Neglect the weight of the nozzle and the water within it. y w = 62.4 lb/ft 3 . SOLUTION ^ = pG = (t2) (2) = 3 - 876slUg/S ( ^ ) = £=12^44 = 24 ft/S ( ^ ) = ° ( V Ay) = ^44 = 72 ft/S {VAx) = 0 F B =p B A B = 2(12) = 241b Equations of steady flow: ^F x = dm (v Ax - v Bx )' 24 — F x = 3.876(0 - 24) F x = 117.01 lb at + nF y = d 2(v Ay - v By ); F y = 3 - 876 ( 72 “ °) * 279 - 06 Ib F = Vf 2 x + F y = V117.01 2 + 279.06 2 = 303 lb Ans: F = 303 lb 597 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 118 . The blade divides the jet of water having a diameter of 4 in. If one-half of the water flows to the right while the other half flows to the left, and the total flow is Q = 1.5 ft 3 /s, determine the vertical force exerted on the blade by the jet, = 62.4 lb/ft 3 . SOLUTION dm Equation of Steady Flow. Here ' = p w Q = dt 62.4 322 (1.5) = 2.9068 slug/s. The velocity of the water jet is Vj = — = —= — ft/s. Referring to the FBD of the Ki ) 2 control volume shown in Fig. a, + np y = ~ (tu),]; F = 2.9068 0 - - 54- 7T = 49.96 lb = 50.0 lb Ans. F Ans: F = 50.0 lb 598 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 119 . The blade divides the jet of water having a diameter of 3 in. If one-fourth of the water flows downward while the other three-fourths flows upwards, and the total flow is Q = 0.5 ft 3 /s, determine the horizontal and vertical components of force exerted on the blade by the jet, y w = 62.4 lb/ft 3 . SOLUTION Equations of Steady Flow: Here, the flow rate (9 = 0.5 ft 2 /s. Then, Q 0.5 dm 62.4 v = ^ = 77 YU = 10.19 ft/s. Also, — = Pw Q = — (0.5) = 0.9689 slug/s. 4 (. 12 ,) Applying Eq. 15-25 we have = 2^ (n out — Ujn) ; — F x = 0 — 0.9689 (10.19) F x = 9.87 lb Ans. dt v 1 ’’ ^F y = 2^(u outy - u ir J ; F y = | (0.9689)(10.19) + ^ (0.9689)(-10.19) F y = 4.93 lb Ans. 3 in. Ans: F x = 9.87 lb F y = 4.93 lb 599 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 120 . The gauge pressure of water at A is 150.5 kPa. Water flows through the pipe at A with a velocity of 18 m/s, and out the pipe at B and C with the same velocity v. Determine the horizontal and vertical components of force exerted on the elbow necessary to hold the pipe assembly in equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of 50 mm at A, and at B and C the diameter is 30 mm. p w = 1000 kg/m 3 . SOLUTION Continuity. The flow rate at B and C are the same since the pipe have the same diameter there. The flow rate at A is v B 18 m/s Q a = v a A a = (18)[tt( 0.025 2 )] = 0.0112577 m 3 /s Continuity negatives that Qa = Qb + Qc, 0.01125tt = 2 Q Q = 0.00562577 m 3 /s Thus, Q 0.00562577 v c =v B = - = A t7(0.015 2 ) = 25 m/s Equation of Steady Flow. The force due to the pressure at A is P = Pa A a = (150.5)(l0 3 )[7r(0.025 2 )] = 94.0625 t 7 N. Here, ^ - = p w Q A at = 1000(0.0112577) = 11.2577 kg/s and dmA = dMc = Pw Q = 1000(0.00562577) dt dt = 5.62577 kg/s. dmn dm r dm A F r = (5.625t7)(25) + (5.62577) 25 = 795.22 N = 795 N (11.25t7)(0) Ans. dm i j dmr dm A + T2/y = + —(v C )y - —My-, dt dt 94.0625tt F y = (5.625t7)(0) + (5.62577) (11.25t7)(18) F y = 1196.75 N = 1.20 kN Ans. Ans: F x = 795 N F y = 1.20 kN 600 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 121 . The gauge pressure of water at C is 40 lb/irf. If water flows out of the pipe at A and B with velocities va— 12 ft/s and vb — 25 ft/s, determine the horizontal and vertical components of force exerted on the elbow necessary to hold the pipe assembly in equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of 0.75 in. at C, and at A and B the diameter is 0.5 in. y w = 62.4 lb/ft 3 . SOLUTION dm A 62.4 / 0.25 V ~dT = 32l (12) Hld = 0-03171 slug/s dm B 62.4 f 0.25'. ~df~ = 32l (25)(77) (lT J =°- 06606slU ^ dm ( dt = 0.03171 + 0.06606 = 0.09777 slug/s v c A c = v a A a + v b A b J 0.375 V J 0.25 V , / 0.25 V Wc(7r) ( 12 ) =12(7r) (irJ + 25(77) (-’ 12 J v c = 16.44 ft/s , dm B dm A dmc F r = —r~ v B + . v As -— v c . dt dt dt 40(ir)(0.375) - F x = 0 - 0.03171(12) - - 0.09777(16.44) F y = 19.5 lb Ans. . dm B dm A dmc + T2T V = —;— V B H-:— V A - ;—Vc y dt y dt Ay dt Cy F y = 0.06606(25) + 0.03171 - (12)-0 F v = 1.9559 = 1.961b Ans. Ans: = 25 ft/s 19.5 lb 1.96 lb 601 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 122 . The fountain shoots water in the direction shown. If the water is discharged at 30° from the horizontal, and the cross-sectional area of the water stream is approximately 2 in 2 , determine the force it exerts on the concrete wall at B. y w = 62.4 lb/ft 3 . SOLUTION (-±») s = So + v 0 t 20 = 0 + v A cos 30°t (+T) V = v 0 + a c t — (va sin 30°) = sin 30°) — 32.2 1 Solving, t = 0.8469 s V A = v B = 27.27 ft/s At B: ^ = PVA= (tf) (27 - 27 (^) = °- 7340slu g/ s „ dm + = — (v A - v B ) -F = 0.7340(0 - 27.27) F = 20.0 lb Ans. Ans: F = 20.0 lb 602 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 123 . A plow located on the front of a locomotive scoops up snow at the rate of 10 ft 3 /s and stores it in the train. If the locomotive is traveling at a constant speed of 12 ft/s, determine the resistance to motion caused by the shoveling. The specific weight of snow is y s = 6 lb/ft 3 . SOLUTION F = 22.4 lb Ans. Ans: F = 22.4 lb 603 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 124 . The boat has a mass of 180 kg and is traveling forward on a river with a constant velocity of 70 km/h, measured relative to the river. The river is flowing in the opposite direction at 5 km/h. If a tube is placed in the water, as shown, and it collects 40 kg of water in the boat in 80 s, determine the horizontal thrust T on the tube that is required to overcome the resistance due to the water collection and yet maintain the constant speed of the boat. p w = 1 Mg/nT. SOLUTION dm dt 40 — = 0.5 kg/s 80 s/ v D /t 19.444 m/s 2F. dv dm i m ht + VD ^ T = 0 + 19.444(0.5) = 9.72 N Ans. Ans: T = 9.72 N 604 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 125 . Water is discharged from a nozzle with a velocity of 12 m/s and strikes the blade mounted on the 20-kg cart. Determine the tension developed in the cord, needed to hold the cart stationary, and the normal reaction of the wheels on the cart. The nozzle has a diameter of 50 mm and the density of water is p w = 1000 kg/m 3 . SOLUTION Steady Flow Equation: Here, the mass flow rate at sections A and B of the control 2 ) (12) = 7.5tt kg/s control volume shown in Fig. a, 7.5rr(12cos45° - 12) 82.81 N 7.57r(12sin45° - 0) 199.93 N Equilibrium: Using the results of F c and F v and referring to the free-body diagram of the cart shown in Fig. b , ZF X = 0; 82.81 -7 = 0 T = 82.8 N Ans. + T VFy = 0; N - 20(9.81) - 199.93 = 0 N = 396 N Ans. dm volume is —— = p w Q = p w Av = 1000 dt f(°-05 : Referring to the free-body diagram of the ^ Fx = d ^ ^ Vb)x ~ + UF y = ^[(v B ) y -(v A ) y ]; -F x = F r = Fy = Ans: T = 82.8 N N = 396 N 605 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 126 . A snowblower having a scoop S with a cross-sectional area of A s = 0.12 m 3 is pushed into snow with a speed of v s = 0.5 m/s. The machine discharges the snow through a tube T that has a cross-sectional area of A T = 0.03 m 2 and is directed 60° from the horizontal. If the density of snow is p s = 104 kg/m 3 , determine the horizontal force P required to push the blower forward, and the resultant frictional force F of the wheels on the ground, necessary to prevent the blower from moving sideways. The wheels roll freely. SOLUTION — = pvAs = (104)(0.5)(0.12) _ dmf 1 \ _ ( 6.24 \ Vs ~ ~dt\pA r ) ~ V 104(0.03)/ 6.24 kg/s 2.0 m/s „ dm, = li {VT > ~ Vs J -F = 6.24(—2 cos 60° - 0) F = 6.24 N dm, XF ’ ~ *<•”■, vs 2 ) -P = 6.24(0 - 0.5) P = 3.12 N Ans. Ans. Ans: F = 6.24 N P = 3.12 N 606 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 127 . The fan blows air at 6000 ft 3 /min. If the fan has a weight of 30 lb and a center of gravity at G, determine the smallest diameter d of its base so that it will not tip over. The specific weight of air is y = 0.076 lb/ft 3 . SOLUTION Equations of Steady Flow: Here Q = / 6000 ft 3 \ \ min J 3X (1 min \ X 60s ) = 100 ft 3 /s. Then, u = ^ = ltX) F = 56.59 ft/s. Also, ^ = Pa Q = A f (1-5 2 ) Applying Eq. 15-26 we have dm dt 0.076 32.2 (100) = 0.2360 slug/s. a +2A/(9 — dt (d OB v B - d OA v A \, 30^0.5 + £) = 0.2360 [4(56.59) - 0] d = 2.56 ft Ans. Ans: d = 2.56 ft 607 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 128 . The nozzle has a diameter of 40 mm. If it discharges water uniformly with a downward velocity of 20 m/s against the fixed blade, determine the vertical force exerted by the water on the blade. p w = 1 Mg/m 3 . SOLUTION d ™ = pvA = (1000)(20)(tt)( 0.02) 2 = 25.13 kg/s dm T | X Fy ("^[placeholder] ^Ay ) F = (25.13)(20 sin 45° - (-20)) F = 858 N Ans. Ans: F = 858 N 608 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 129 . The water flow enters below the hydrant at C at the rate of 0.75 m 3 /s. It is then divided equally between the two out¬ lets at A and B. If the gauge pressure at C is 300 kPa, deter¬ mine the horizontal and vertical force reactions and the moment reaction on the fixed support at C. The diameter of the two outlets at A and B is 75 mm, and the diameter of the inlet pipe at C is 150 mm. The density of water is p w = 1000 kg/m 3 . Neglect the mass of the contained water and the hydrant. SOLUTION Free-Body Diagram: The free-body diagram of the control volume is shown in Fig. a. The force exerted on section A due to the water pressure is F c = pcA c = 300(10 3 ) dm A dt 4' dm B dt 0.15 2 Pw = 5301.44 N. The mass flow rate at sections A , B , and C, are dm c (! ) - -(f) = 375 kg/s and dt - PwQ - 1000(0.75) = 750 kg/s. The speed of the water at sections A, B , and C are < 2/2 va = v B = A A 0.75/2 Q -= 84.88 m/s v c = —- — (0.075 2 ) c 4 0.75 = 42.44 m/s. (0.15 2 ) Steady Flow Equation: Writing the force steady flow equations along the x and y axes, Mx, , dm A dmn dmc = —- (v A ) x + , (v B ) x - dt v dt v dt C x = -375(84.88 cos 30°) + 375(84.88) - 0 C x = 4264.54 N = 4.26 kN . dm ,i dm B dmc = dt {Va) > + ~dT {Vb) > “ ~dT {Vc) >' -C y + 5301.44 = 375(84.88 sin 30°) + 0 - 750(42.44) Cy = 21 216.93 N = 2.12 kN Ans. iCt) Ans. Writing the steady flow equation about point C, dm A dm R dm r +2M C =- dv A H- dv g- dv c ; dt dt dt — M c = 375(0.65)(84.88 cos 30°) - 375(0.25)(84.88sin30°) + [-375(0.6)(84.88)] - 0 M c = 5159.28 N-m = 5.16 kN-m Ans. Ans: C x = 4.26 kN C y = 2.12 kN M c = 5.16 kN-m 609 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 130 . Sand drops onto the 2-Mg empty rail car at 50 kg/s from a conveyor belt. If the car is initially coasting at 4 m/s, determine the speed of the car as a function of time. SOLUTION Gains Mass System. Here the sand drops vertically onto the rail car. Thus (v L ) x = 0. Then V D =V i + V D/i (X) V = (Vj) x + (v D/i ) x v = 0 + (v D/i ) x (v D /,) x = V dm; Also, —— = 50 kg/s and m = 2000 + 50f dv dm; 2// = m— + (v D/i ) dt dt , dv 0 = (2000 + 50r)— + v(50) dt dv v 50 dt "2000 + 50r Integrate this equation with initial condition u = 4m/satf=0. r *. -so r ' 4 m/s v dt 0 2000 + 50t In v 4 m/s = -In (2000 + 50t) In V = In 4 v 2000 2000 + 50 t 2000 4 2000 + 50r 8000 v = 2000 + 50r m/s Ans. Ans: f 8000 1 , ^ = \ 2000 + 50r J m/s 610 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 131 . Sand is discharged from the silo at A at a rate of 50 kg/s with a vertical velocity of 10 m/s onto the conveyor belt, which is moving with a constant velocity of 1.5 m/s. If the conveyor system and the sand on it have a total mass of 750 kg and center of mass at point G, determine the horizontal and ver¬ tical components of reaction at the pin support B roller sup¬ port A. Neglect the thickness of the conveyor. SOLUTION Steady Flow Equation: The moment steady flow equation will be written about point B to eliminate B x and B v . Referring to the free-body diagram of the control volume shown in Fig. a, din + 2M„ = — (dv B - dv A ); 750(9.81)(4) - A v (8) = 50[0 - 8(5)1 dt * A y = 4178.5 N = 4.18 kN Ans. Writing the force steady flow equation along the x and y axes, ^ ^ [(v n ) x ~ (v A ) x \- -B x = 50(1.5 cos 30° - 0) B x = 1—64.95 N| = 65.0 N —> Ans. + t2F v = ^- (v A )yl B y + 4178.5 - 750(9.81) = 50[1.5 sin 30° - (-10)] By = 3716.25 N = 3.72 kN T Ans. Ca) Ans: Ay = 4.18 kN B x = 65.0 N By = 3.72 kNt 611 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 612 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 133 . The tractor together with the empty tank has a total mass of 4 Mg. The tank is filled with 2 Mg of water. The water is discharged at a constant rate of 50 kg/s with a constant velocity of 5 m/s, measured relative to the tractor. If the tractor starts from rest, and the rear wheels provide a resultant traction force of 250 N, determine the velocity and acceleration of the tractor at the instant the tank becomes empty. SOLUTION The free-body diagram of the tractor and water jet is shown in Fig. a. The pair of thrust T cancel each other since they are internal to the system. The mass of the tractor and the tank at any instant t is given by m = (4000 + 2000) — 50t = (6000 — 50r)kg. I dv dm„ / \dv ETj = m— - v D/e -^\ 250 = (6000 - 50fj— - 5(50) dv _ 10 dt 120 — t ( 1 ) The time taken to empty the tank is t into Eq. (1), 2000 50 40 s. Substituting the result of t a 10 120-40 0.125m/s 2 Integrating Eq. (1), 10 120 -t dt v = —10 ln(l20 = 4.05 m/s Ans. Ans. Ans: a = 0.125 m/s 2 v = 4.05 m/s 613 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 134 . A rocket has an empty weight of 500 lb and carries 300 lb of fuel. If the fuel is burned at the rate of 15 lb/s and ejected with a relative velocity of 4400 ft/s, determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket. SOLUTION tvc - dv dm ' ' I ^ dt ^[placeholder] ^ dm e At n time t, m = ra 0 — ct , where c = ——. In space the weight of the rocket is zero. at 0 = (mo - ct)— - « [plaC eh old er] [placeholder] ^^[placeholder] tn n — ct \dt V ^[placeholder] ^ j (1) The maximum speed occurs when all the fuel is consumed, that is, when 300 500 + 300 15 f = = 20 s. Here, m 0 = -= 24.8447 slug, c = -= 0.4658 slug/s, 15 u 32.2 6 32.2 ^[placeholder] = 4400 ft/s. Substitute the numerical values into Eq. (1); , ( 24.8447 \ max v 24.8447 - 0.4658(20) / Umax = 2068 ft/s = 2.07(l0 3 ) ft/s Ans. Ans: ^max f\ / \ V 2.07 (10 3 ) ft/s 614 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 135 . A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of 6 m/s through an intake unit A, which has a cross-sectional area of A a = 0.25 m 2 , and then discharging it at the ground, B , where the cross-sectional area is A B = 0.35 m 2 . If air at A is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of 15 kg with center of mass at G. Assume that air has a constant density of p a = 1.22 kg/m 3 . SOLUTION dm dt = pA A v A = 1.22(0.25)(6) = 1.83 kg/s dm + UF y = — ((v B ) y - (v A ) y ) pressure = (0.35) - 15(9.81) = 1.83(0 - (-6)) pressure = 452 Pa tsCljOhi 4 ^ 'f(o.-is) Ans. Ans: 452 Pa 615 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 136 . The rocket car has a mass of 2 Mg (empty) and carries 120 kg of fuel. If the fuel is consumed at a constant rate of 6 kg/s and ejected from the car with a relative velocity of 800 m/s, determine the maximum speed attained by the car starting from rest. The drag resistance due to the atmosphere is F d = (6.8v 2 ) N, where v is the speed in m/s. v SOLUTION dv = m dt ~ P[ P iacehoider] [placeholder] dm r At time ty the mass of the car is m 0 — ct\ where c = -= 6 kg/s dt Set F = kv 2 , then dv ~kv = ( m 0 - ct)— - v D/e c dv dt /o ( cv D / e - kv 2 ) Jo ( m o ~ d) = —ln(»i 0 — ct) c = 1 hr ( m ° — Ct c v rn 0 Maximum speed occurs at the instant the fuel runs out 120 t = = 20 s 6 Thus, ( l ' ^ 2 V( 6 )( 800 )( 6 . 8 ) Solving, In i mm +v \ 6.8 MSOOJ 6.8 7 1 /2120 - 6 ( 20 ) ~6 H 2120 v = 25.0 m/s Ans. Ans: v = 25.0 m/s 616 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 137 . If the chain is lowered at a constant speed v = 4 ft/s, determine the normal reaction exerted on the floor as a function of time. The chain has a weight of 5 lb/ft and a total length of 20 ft. SOLUTION At time t , the weight of the chain on the floor is W = mg(vt ) dv —— = U, m, = m(vt) dnii —— = mv dt VC dv . tF s = m - + VD/i dnij dt R — mg(vt) = 0 + v(mv) R = m(gvt + v 2 ) R = 3 2 2 (32.2(4)0) + (4) 2 ) R = (20 1 + 2.48) lb Ans. Ans: R = {20 1 + 2.48} lb 617 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 138 . The second stage of a two-stage rocket weighs 2000 lb (empty) and is launched from the first stage with a velocity of 3000 mi/h. The fuel in the second stage weighs 1000 lb. If it is consumed at the rate of 50 lb/s and ejected with a relative velocity of 8000 ft/s, determine the acceleration of the second stage just after the engine is fired. What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of gravitation. SOLUTION Initially, a = 133 ft/s 2 Ans. Finally, a = 200 ft/s 2 Ans. Ans: a t = 133 ft/s 2 af = 200 ft / s 2 618 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 139 . The missile weighs 40 000 lb. The constant thrust provided by the turbojet engine is T = 15 000 lb. Additional thrust is provided by two rocket boosters B. The propellant in each booster is burned at a constant rate of 150 lb/s, with a relative exhaust velocity of 3000 ft/s. If the mass of the propellant lost by the turbojet engine can be neglected, determine the velocity of the missile after the 4-s burn time of the boosters. The initial velocity of the missile is 300 mi/h. SOLUTION dv dm e m Tt ~ VD,e lh At a time t,m = m 0 — ct , where c = dv T = (mo - ct)— - v D/e c dv = dt ‘/T + cv D/e o \ m 0 - ct T + cv D/l dt In mp m 0 — ct + v 0 dm e dt B ( 1 ) 40 000 Here, m Q = ^ 2 = 1242.24 slug, c = 2 ) = 9 - 3168 slu s/ s ’ v D/e = 3000 ft/s, 300(5280) f = 4s ’" 0 = ^ 65 ^ = 440ft/s - Substitute the numerical values into Eq. (1): Vmnr = 15 000 + 9.3168(3000) \ 9.3168 M 1242.24 1242.24 - 9.3168(4)/ + 440 V max = 580 ft/s Ans. Ans: «max = 580 ft/s 619 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 620 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 - 141 . The rope has a mass m’ per unit length. If the end length y = h is draped off the edge of the table, and released, determine the velocity of its end A for any position y, as the rope uncoils and begins to fall. SOLUTION , | v j, dv dm, + = m— + v D/i dt dmj m'dy dv At a time t, m = m'y and = ——— = m'v. Here, v^/i — v, — = g. dt dt dt , , dv m gy = m y ——h v(m v) dt dv 7 . dy dy gy = y -b v since v = —, then dt = — dt dt v dv , gy = vy-j- + v L dy Multiply both sides by 2 ydy 2gy 2 dy = 2vy 2 dv + 2yv 2 dy f 2gy 2 dy = jd(v 2 y 2 ) I gy 3 + C = v 2 y 2 v = 0 at y = h \gh 2 + C — 0 C = —| gh 3 lgy 3 ~lgh 3 = v 2 y 2 Ans. Ans: 621 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 622 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 623 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 15 - 144 . A four-engine commercial jumbo jet is cruising at a constant speed of 800 km/h in level flight when all four engines are in operation. Each of the engines is capable of discharging combustion gases with a velocity of 775 m/s relative to the plane. If during a test two of the engines, one on each side of the plane, are shut off, determine the new cruising speed of the jet. Assume that air resistance (drag) is proportional to the square of the speed, that is, F D — cv 2 , where c is a constant to be determined. Neglect the loss of mass due to fuel consumption. SOLUTION Steady Flow Equation: Since the air is collected from a large source (the atmosphere), its entrance speed into the engine is negligible. The exit speed of the air from the engine is d^ v e + v p + v t ■Ip 800(10 3 ) — lh = 222.22 m/s. Thus, 3600 s j v e = -222.22 + 775 = 552.78 m/s -> Referring to the free-body diagram of the airplane shown in Fig. a, * ZF, - ^[(»«), - MJ: C(222.22’) - 4 /j/ (552.78 - 0) Fp- Of When the four engines are in operation, the airplane has a constant speed of dm C = 0.044775 — dt When only two engines are in operation, the exit speed of the air is d^ v e = —v p + 775 Using the result for C, * * F * =$ [ W* - MJ; (0.044775 = 2^1-Vp + 775) - 0]

0.044775u p 2 + 2v p - 1550 = 0

Solving for the positive root,
v p = 165.06 m/s = 594 km/h

Ans.

Ans:

v P = 594 km/h

624

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15 - 145 .

The 10-Mg helicopter carries a bucket containing 500 kg of
water, which is used to fight fires. If it hovers over the land in
a fixed position and then releases 50 kg/s of water at 10 m/s,
measured relative to thehelicopter, determine the initial
upward accelerationthe helicopter experiences as the water
is being released.

SOLUTION

+r

dv dm e

m ~dt~ VD/e ~di~

Initially, the bucket is full of water, hence m = 10(l0 3 ) + 0.5(l0 3 ) = 10.5(l0 3 ) kg
0 = 10.5(l0 3 ) a - (10)(50)

a = 0.0476 m/s 2 Ans.

Ans:

a = 0.0476 m/s 2

625

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15 - 146 .

A rocket has an empty weight of 500 lb and carries 300 lb
of fuel. If the fuel is burned at the rate of 1.5 lb/s and ejected
with a velocity of 4400 ft/s relative to the rocket, determine
the maximum speed attained by the rocket starting from rest.
Neglect the effect of gravitation on the rocket.

SOLUTION

, tvr dv dm e

+np ° = ^ ~ VD/e ^~

dt

At a time t , m = m 0 — ct, where c

dv

0 = ("Jo ~ ct) — - v D/e c

r*. f(^\dt

Jo Jo \ m 0 Ct

m 0

v = v D / e In

m 0 — ct

dm e

dt

. In space the weight of the rocket is zero.

( 1 )

The maximum speed occurs when all the

300
1.5

, 300 ^nn

t = = 200 s.

Here, m 0 = 50Q 3 ^ 2 300 = 24.8447 slug, c

Substitute the numerical into Eq. (1):

24.8447

4400 In
2068 ft/s

1.5

32.2

24.8447 - (0.04658(200))/

fuel is consumed, that is, when
0.04658 slug/s, v D / e = 4400 ft/s.

Ans.

Ans:

v nmx = 2.07(10 3 ) ft/s

626

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15 - 147 .

Determine the magnitude of force F as a function of time,
which must be applied to the end of the cord at A to raise
the hook H with a constant speed v = 0.4 m/s. Initially the
chain is at rest on the ground. Neglect the mass of the cord
and the hook. The chain has a mass of 2 kg/m.

SOLUTION

y = vt

m t = my = mvt

dmj

—— = mv
dt

, tv r ///} i dm j
+ 1% F S = m— + VD/i (—)

dt

F — mgvt = 0 + v(mv)
F = m(gvt + i?)

= 2[9.81(0.4)r + (0.4) 2 ]
F = (7.851 + 0.320) N

Ans.

Ans:

F = (7.85t + 0.320} N

627

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* 15 - 148 .

The truck has a mass of 50 Mg when empty. When it is unload¬
ing 5 m 3 of sand at a constant rate of 0.8 m 3 /s, the sand flows
out the back at a speed of 7 m/s, measured relative to the
truck, in the direction shown. If the truck is free to roll, deter¬
mine its initial acceleration just as the load begins to empty.
Neglect the mass of the wheels and any frictional resistance to
motion. The density of sand is p s = 1520 kg/m 3 .

SOLUTION

A System That Loses Mass: Initially, the
m = 50(10 3 ) + 5(1520)

Applying Eq. 15-29, we have

57.6(10 3 ) kg and

total mass of
dm p

— 1 = 0.8(1520) = 1216 kg/s.
at

, dv dm e

~ Fs = m Yt “ VD ^ ;

dt

0 = 57.6(10 3 )a - (0.8 cos 45°)(1216)
a = 0.104 m/s 2 Ans.

Ans:

a = 0.104 m/s 2

628

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15 - 149 .

The car has a mass m 0 and is used to tow the smooth chain
having a total length l and a mass per unit of length m'. If
the chain is originally piled up, determine the tractive force
F that must be supplied by the rear wheels of the car,
necessary to maintain a constant speed v while the chain is
being drawn out.

SOLUTION

4 - v „ dv dnij

±Y.F s = m d t + v n/l ^

dnij m dx

At a time t,m = m n + ct , where c = —— = —:— = m v.

dt dt

dv

Here, v Dji = v, — = 0.

F = ( m 0 - m'v )(0) + v(m'v) = m'v 2

Ans.

Ans:

F = m'v 2

629

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16 - 1 .

The angular velocity of the disk is defined by
co = (5r 2 + 2) rad/s, where t is in seconds. Determine the
magnitudes of the velocity and acceleration of point A on
the disk when t = 0.5 s.

SOLUTION

co = (5 t~ + 2) rad/s
dco

a = —— =10 t
dt

t = 0.5 s

v A = cor = 3 . 25 ( 0 . 8 ) = 2.60 m/s
a z = or = 5 ( 0 . 8 ) = 4 m/s 2
a n = co 2 r = ( 3 . 25 ) 2 ( 0 . 8 ) = 8.45 m/s 2
a A = V ( 4) 2 + ( 8 . 45) 2 = 9.35 m/s 2

Ans.

Ans.

Ans:

v A = 2.60 m/s
a A = 9.35 m/s 2

630

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 2 .

The angular acceleration of the disk is defined by
a = 'it 1 + 12 rad/s, where t is in seconds. If the disk is
originally rotating at tu 0 = 12 rad/s, determine the
magnitude of the velocity and the n and t components of
acceleration of point A on the disk when t = 2 s.

SOLUTION

Angular Motion. The angular velocity of the disk can be determined by integrating
dco = a dt with the initial condition o> = 12 rad/s at t = 0.

fco o2s

dco =

(3 1 1 + 12 )dt

co - 12 = (t 3 + 12 1)

2 s

0

Motion of Point A. The magnitude of the velocity is

v A = cor A = 44.0(0.5) = 22.0 m/s Ans.

At t = 2 s, a = 3(2 2 ) + 12 = 24 rad/s 2 . Thus, the tangential and normal
components of the acceleration are

( a A ) t = ar A = 24(0.5) = 12.0 m/s 2 Ans.

( a A ) n = = (44.0 2 )(0.5) = 968 m/s 2 Ans.

Ans:

v A = 22.0 m/s
( ci A ) t = 12.0 m/s 2
( a A ) n = 968 m/s 2

631

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 3 .

The disk is originally rotating at cu 0 = 12 rad/s. If it is
subjected to a constant angular acceleration of
a = 20 rad/s 2 , determine the magnitudes of the velocity
and the n and t components of acceleration of point A at the
instant t = 2 s.

SOLUTION

Angular Motion. The angular velocity of the disk can be determined using
w = co 0 + a c t; co = 12 + 20(2) = 52 rad/s

Motion of Point A. The magnitude of the velocity is

v A = cor A = 52(0.5) = 26.0 m/s Ans.

The tangential and normal component of acceleration are

( a A ) t = ar = 20(0.5) = 10.0 m/s 2 Ans.

( a A ) n = to 2 r = (52 2 )(0.5) = 1352 m/s 2 Ans.

Ans:

v A = 26.0 m/s
{a A ) t = 10-0 m/s 2
(a A ) n = 1352 m/s 2

632

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* 16 - 4 .

The disk is originally rotating at co 0 = 12 rad/s. If it
is subjected to a constant angular acceleration of
a = 20 rad/s 2 , determine the magnitudes of the velocity
and the n and t components of acceleration of point B when
the disk undergoes 2 revolutions.

SOLUTION

Angular Motion. The angular velocity of the disk can be determined using
co 2 = col + 2a c (d - 6 0 ); co 2 = 12 2 + 2(20)[2(2t r) - 0]

Motion of Point B. The magnitude of the velocity is

v B = cor B = 25.43(0.4) = 10.17 m/s = 10.2 m/s Ans.

The tangential and normal components of acceleration are

( ci B ) t = ar B = 20(0.4) = 8.00 m/s 2 Ans.

(a B )„ = co 2 r B = (25.43 2 )(0.4) = 258.66 m/s 2 = 259 m/s 2 Ans.

Ans:

v B = 10.2 m/s
(i a B ) t = 8.00 m/s 2
( a B ) n = 259 m/s 2

633

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634

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 6 .

A wheel has an initial clockwise angular velocity of 10 rad/s
and a constant angular acceleration of 3 rad/s 2 . Determine
the number of revolutions it must undergo to acquire a
clockwise angular velocity of 15 rad/s. What time is
required?

SOLUTION

" 2 = "o + 2 a c (6 - d 0 )
(15) 2 = (10) 2 + 2(3)(0-O)

6 = 20.83 rad = 20.83

3.32 rev.

Ans.

w = cog + a c t

15 = 10 + 3 1

t = 1.67 s Ans.

Ans:

6 = 3.32 rev
t = 1.67 s

635

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636

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

* 16 - 8 .

If gear A rotates with an angular velocity of co A =

(6 a +1) rad/s, where d A is the angular displacement of
gear A, measured in radians, determine the angular
acceleration of gear D when 0 A = 3 rad, starting from rest.

Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm,
and 75 mm, respectively.

SOLUTION

Motion of Gear A:

a A dO A = (o A d(o A
oi Add A = (0 A + l)d(d A + 1)
a A dO A = (9a + 1) dd A
oia = {9 A + 1 )

At 6 a = 3 rad,

a A = 3 + 1 = 4 rad/s 2

Motion of Gear D: Gear A is in mesh with gear B. Thus,

= Ol A rA

«B = {~) a A = (^) (4) = L20rad / s2

Since gears C and B share the same shaft a c = a H = 1.20 rad/s 2 . Also, gear D is in
mesh with gear C. Thus,

a D r D ~ a c r c

a c = ^—^(1.20) = 0.4 rad/s 2 Ans.

Ans:

a D = 0.4 rad/s 2

637

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16 - 9 .

At the instant w A = 5 rad/s, pulley A is given an angular
acceleration a = (0.86) rad/s 2 , where 6 is in radians.
Determine the magnitude of acceleration of point B on
pulley C when A rotates 3 revolutions. Pulley C has an inner
hub which is fixed to its outer one and turns with it.

SOLUTION

Angular Motion. The angular velocity of pulley A can be determined by integrating
to dto = a dd with the initial condition co A = 5 rad/s at d A = 0.

r Da

to dto =

o.8 ede

(0 A

5 2

= (0A0 2 )

to 2 A j ,

T - 2 = 0M1

&>/, = ! Vo. 8 d A + 251 rad/s

At 0 A = 3(27t) = 67 t rad,

to A = V 0 . 8 ( 6 tt ) 2 + 25 = 17.585 rad/s
a A = 0.8(6tt) = 4.8tt rad/s 2
Since pulleys A and C are connected by a non-slip belt,

"cTc = ^A r A\ <o c ( 40 ) = 17.585(50)

acre = a A r A ; «c(40) = (4.8-7r)(50)

a c = 6 tt rad/s 2

Motion of Point B. The tangential and normal components of acceleration of
point B can be determined from

( a B)t = a c r B = 6tt( 0.06) = 1.1310 m/s 2

(a B )„ = a> 2 c r B = (21.982 2 )(0.06) = 28.9917 m/s 2

Thus, the magnitude of a B is

a B = V(u B ) 2 + (a B )l = Vl.1310 2 + 28.9917 2

= 29.01 m/s 2 = 29.0 m/s 2 Ans.

Ans:

a B = 29.0 m/s 2

638

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16 - 10 .

At the instant v A = 5 rad/s, pulley A is given a constant
angular acceleration a^ = 6rad/s 2 . Determine the
magnitude of acceleration of point B on pulley C when A
rotates 2 revolutions. Pulley C has an inner hub which is
fixed to its outer one and turns with it.

SOLUTION

Angular Motion. Since the angular acceleration of pulley A is constant, we can
apply

ma = ("/Jo + 2 a A [0 A ~ (0a)o];
o 2 a = 5 2 + 2(6)[2(2tt) - 0]

Since pulleys A and C are connected by a non-slip belt,

"c r c = & A r A ; &> c (40) = 13.2588(50)

a c r c = a A r A ; “c(40) = 6(50)

a c = 7.50 rad/s 2

Motion of Point B. The tangential and normal component of acceleration of
point B can be determined from

(a B ), = a c r B = 7.50(0.06) = 0.450 m/s 2

(a B )„ = (a cm = (l6.5735 2 )(0.06) = 16.4809 m/s 2

Thus, the magnitude of a B is

a B = V(a B )j + (a B ) 2 = \/0.450 2 + 16.4809 2

= 16.4871 m/s 2 = 16.5 m/s 2 Ans.

Ans:

a B = 16.5 m/s 2

639

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16 - 11 .

The cord, which is wrapped around the disk, is given an
acceleration of a = (10f)m/s , where t is in seconds.
Starting from rest, determine the angular displacement,
angular velocity, and angular acceleration of the disk when
t = 3 s.

SOLUTION

Motion of Point P. The tangential component of acceleration of a point on the rim
is equal to the acceleration of the cord. Thus

( a ,) = or; 10 1 = a(0.5)

a = { 20 1} rad/s 2

When t = 3 s,

a = 20(3) = 60 rad/s 2 Ans.

Angular Motion. The angular velocity of the disk can be determined by integrating
dco = a dt with the initial condition go = 0 at t = 0.

ptx) ft

dco = / 20 1 dt

Jo

CO

Jo

When t = 3 s,

co = 10(3 2 ) = 90.0 rad/s

Ans.

The angular displacement of the disk can be determined by integrating dO = co dt
with the initial condition 0 = 0 at t = 0.

When t = 3 s,

0

0 = y(3 3 ) = 90.0 rad

Ans.

a = (10f) m/s 2

Ans:

a = 60 rad / s 2

640

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* 16 - 12 .

The power of a bus engine is transmitted using the belt-and-
pulley arrangement shown. If the engine turns pulley A at
(o A = (20f + 40) rad/s, where t is in seconds, determine the
angular velocities of the generator pulley B and the
air-conditioning pulley C when t = 3 s.

SOLUTION

When t = 3 s

co A = 20(3) + 40 = 100 rad/s
The speed of a point P on the belt wrapped around A is
v P = co A r A = 100(0.075) = 7.5 m/s

100 mm

_ v P _ 7.5
0)8 ~ To “ 0.025

Ans.

The speed of a point P' on the belt wrapped around the outer periphery of B is
v' = co B r B = 300(0.1) = 30 m/s

Hence, co c = -=

v p
r c

Ans.

Ans:

641

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642

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 14 .

The disk starts from rest and is given an angular acceleration
a = (2f 2 ) rad/s 2 , where t is in seconds. Determine the
angular velocity of the disk and its angular displacement
when t = 4 s.

SOLUTION

dco

21 2

When t = 4 s,

2 ,

w = —(4) 3 = 42.7 rad/s

When f = 4 s,

0 = i(4) 4 = 42.7 rad

Ans.

Ans.

Ans:

643

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644

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

* 16 - 16 .

The disk starts at too = 1 rad/s when 0 = 0, and is given an
angular acceleration a = (0.30) rad/s 2 , where 0 is in radians.
Determine the magnitudes of the normal and tangential
components of acceleration of a point P on the rim of the
disk when 0 = 1 rev.

SOLUTION

a = 0.30

030dd

(O

1

O.150 2

— - 0.5 = O.150 2
2

to = VO.30 2 + 1

At 0 = 1 rev = 277 rad
to = V03(2ir)^T7

a t = ar = 0.3(2tt) rad/s 2 (0.4 m) = 0.7540 m/s 2
a n = co 2 r = (3.584 rad/s) 2 (0.4 m) = 5.137 m/s 2
a p = V(0.7540) 2 + (5.137) 2 = 5.19 m/s 2

Ans.

Ans.

Ans:

a, = 0.7540 m/s 2
a n = 5.137 m/s 2

645

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

646

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

647

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 19 .

The vacuum cleaner’s armature shaft S rotates with an
angular acceleration of a = 4&P/4 rad/s 2 , where co is in
rad/s. Determine the brush’s angular velocity when t = 4 s,
starting from w 0 = 1 rad/s, at 0 = 0. The radii of the shaft
and the brush are 0.25 in. and 1 in., respectively. Neglect the
thickness of the drive belt.

SOLUTION

Motion of the Shaft: The angular velocity of the shaft can be determined from

t = <o s lf4 -!
= {t + l) 4

When t = 4 s

(o s = 5 4 = 625 rad/s

Motion of the Beater Brush: Since the brush is connected to the shaft by a non-slip
belt, then

(o B r B — (o s r s

co B —

Ans.

Ans:

648

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* 16 - 20 .

A motor gives gear A an angular acceleration of
a A = (4r 3 ) rad/s 2 , where t is in seconds. If this gear is
initially turning at (w A ) 0 = 20 rad/s, determine the angular
velocity of gear B when t = 2 s.

SOLUTION

a A = 4 t 3
dw = a dt

w A = t A + 20

When t = 2 s,

Wa'a = w Br B

36(0.05) = w B (0.15)

w B = 12 rad/s Ans.

Ans:

649

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650

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 22 .

If the motor turns gear A with an angular acceleration of
a A = 2 rad/s 2 when the angular velocity is w A = 20 rad/s,
determine the angular acceleration and angular velocity of
gear D.

SOLUTION

Angular Motion: The angular velocity and acceleration of gear B must be
determined first. Here, co A r A = cj b r B and a A r A = a B rg.Then,

r A ( 40 \

o>b = -^»a= (^J( 2 °) = 8.00 rad/s
a B ~ ^ a A ~ (^)( 2 ) = 0-800 rad/s 2

Since gear C is attached to gear B , then co c = co B = 8 rad/s and
a c ~ a B ~ 0-8 rad/s 2 . Realizing that w c r c = (o D r D and a c r c = a D r D , then

too = ^toc = (^)( 8 - 00 ) = 4.00 rad/s
old = = (y^)(°- 80 °) = 0.400 rad/s 2

Ans.

Ans.

100 mm

Ans:

a D = 0.400 rad/s 2

651

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652

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

* 16 - 24 .

The gear A on the drive shaft of the outboard motor has a
radius r A = 0.5 in. and the meshed pinion gear B on the
propeller shaft has a radius r B = 1.2 in. Determine the
angular velocity of the propeller in t = 1.5 s, if the drive shaft
rotates with an angular acceleration a = (400f 3 ) rad/s 2 ,
where t is in seconds. The propeller is originally at rest and
the motor frame does not move.

SOLUTION

Angular Motion: The angular velocity of gear A at t — 1.5 s must be determined
first. Applying Eq. 16-2, we have

/» 1.5 s

dco =

400r 3 dt

Jo

co a ~ 100t 4 |J' 5s = 506.25 rad/s

However, co A r A — co B r B where oj /; is the angular velocity of propeller. Then,

rA

<*>b = — W A
r R

05

L2

Ans:

653

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16 - 25 .

For the outboard motor in Prob. 16-24, determine the
magnitude of the velocity and acceleration of point P
located on the tip of the propeller at the instant t = 0.75 s.

SOLUTION

Angular Motion: The angular velocity of gear A at t = 0.75 s must be determined
first. Applying Eq. 16-2, we have

/■0.75 s

d(o =

400r dt

/ o

<o A = lOOtX 75 * = 31.64 rad/s

The angular acceleration of gear A at t = 0.75 s is given by
a A = 400(0.75 3 ) = 168.75 rad/s 2

However, w A r A = w B r B and a A r A = a B r B where co B and a B are the angular
velocity and acceleration of propeller. Then,

r A

u>b = — (o A =
rB

(31.64)

as = r f B aA = (ii) (168 - 75)

Motion of P: The magnitude of the velocity of point P can be determined using
Eq. 16-8.

/ 2.20 \

v P = w B r P = 13.181-) = 2.42 ft/s Ans.

The tangential and normal components of the acceleration of point P can be
determined using Eqs. 16-11 and 16-12, respectively.

/ 2.20 \ ,
a r = a B r p = 70 ' 31 ( ^ ~\2J = 12 ' 89ft / s2

a„ = col r P = (B.18 2 )( 2 f) = 31.86 ft/s 2

The magnitude of the acceleration of point P is

a P = Va 2 r + a 2 n = Vl2.89 2 + 31.86 2 = 34.4 ft/s 2 Ans.

Ans:

v P = 2.42 ft/s
a P = 34.4 ft/s 2

654

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16 - 26 .

The pinion gear A on the motor shaft is given a constant
angular acceleration a = 3 rad/s 2 . If the gears A and B
have the dimensions shown, determine the angular velocity
and angular displacement of the output shaft C, when
t = 2 s starting from rest. The shaft is fixed to B and turns
with it.

SOLUTION

co = coq + a c t

oo A = 0 + 3(2) = 6 rad/s
1 ?

9 — 6 0 + co 0 t + -a c t

0A = 0 + 0 + |(3)(2) 2

<°A r A ~ w B r B

6(35) = tug(125)
co c = co B — 1.68 rad/s

9 a r A = 9 b r B

6(35) = S B (125)

9c ~ 9 B — 1-68 rad

Ans.

Ans.

Ans:

655

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16 - 27 .

The gear A on the drive shaft of the outboard motor has a
radius r A = 0.7 in. and the meshed pinion gear B on the
propeller shaft has a radius r B = 1.4 in. Determine the
angular velocity of the propeller in t = 1.3 s if the drive
shaft rotates with an angular acceleration
a = (300 \A) rad/s 2 , where t is in seconds. The propeller is
originally at rest and the motor frame does not move.

SOLUTION

= ot B r B

(300Vf)(0.7) = a p (1.4)
a P = 150Vf

dco — a dt

pCO ft

dco= 150\/t dt
Jo Jo

co = 100f 3 / 2 |, =13 = 148 rad/s

Ans.

Ans:

656

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* 16 - 28 .

The gear A on the drive shaft of the outboard motor has a
radius r A = 0.7 in. and the meshed pinion gear B on the
propeller shaft has a radius r B = 1.4 in. Determine the
magnitudes of the velocity and acceleration of a point P
located on the tip of the propeller at the instant t = 0.75 s.
the drive shaft rotates with an angular acceleration
a = (300Vt) rad/s 2 , where t is in seconds. The propeller is
originally at rest and the motor frame does not move.

SOLUTION

Angular Motion: The angular velocity of gear A at t = 0.75 s must be determined
first. Applying Eq. 16-2, we have

300 Vt dt

co A = 200 f 3 / 2 \° 0 15s = 129.9 rad/s

The angular acceleration of gear A at t = 0.75 s is given by

a A = 300V0J5 = 259.81 rad/s 2

However, co A r A = co B r B and a A r A = a B r B where a> B and a B are the angular
velocity and acceleration of propeller. Then,

0)3 ~ ~r B W a ~ (l£) (129 - 9)

‘ (u) (25, - sl)

Motion of P: The magnitude of the velocity of point P can be determined using
Eq. 16-8.

/ 2.20 \

v P = co B r P = 64.95 ( —- = 11.9 ft/s

The tangential and normal components of the acceleration of point P can be
determained using Eqs. 16-11 and 16-12, respectively.

a, = a B r P = 129.9^^=^) = 23.82 ft/s 2

a„ = co 2 b r P = (64.95 = 773.44 ft/s 2

The magnitude of the acceleration of point P is

a P = Val + a 2 = \/(23.82) 2 + (773.44) 2 = 774 ft/s 2 Ans.

Ans:

a P = 11A ft/s 2

657

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16 - 29 .

A stamp S, located on the revolving drum, is used to label
canisters. If the canisters are centered 200 mm apart on the
conveyor, determine the radius r A of the driving wheel A
and the radius r B of the conveyor belt drum so that for each
revolution of the stamp it marks the top of a canister. How
many canisters are marked per minute if the drum at B is
rotating at (o B = 0.2 rad/s? Note that the driving belt is
twisted as it passes between the wheels.

SOLUTION

1 = 2ir(r A )

200

r A = -= 31.8 mm Ans.

2rr

For the drum at B:
l = 2 ir(r B )

200

r n =-= 31.8 mm Ans.

2tt

In t = 60 s

6 — 6g + (Ogt

e = 0 + 0.2(60) = 12 rad
/ = dr B = 12(31.8) = 382.0 mm

Hence,

382 0

n = -— = 1.91 canisters marked per minute Ans.

OAA r

Ans:

r A = 31.8 mm
r B = 31.8 mm
1.91 canisters per minute

658

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16 - 30 .

At the instant shown, gear A is rotating with a constant
angular velocity of ai A = 6 rad/s. Determine the largest
angular velocity of gear B and the maximum speed of
point C.

SOLUTION

( B)max A ) mux 50V2 mm

(r B )min = ( r A )min = 50 mm
When r A is max., r B is min.

w b(/b) ~ “>A r A

M max = e( J) = e(^)

V C = i M B)max r C = 8.49(0.05 V2)

v c = 0.6 m/s Ans.

Ans:

(«c)max = 0.6 m/s

659

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660

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

661

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

662

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16 - 34 .

For a short time a motor of the random-orbit sander drives
the gear A with an angular velocity of
oj a = 40(t 3 + 6f) rad/s, where t is in seconds. This gear is
connected to gear B, which is fixed connected to the shaft
CD. The end of this shaft is connected to the eccentric
spindle EF and pad P, which causes the pad to orbit around
shaft CD at a radius of 15 mm. Determine the magnitudes
of the velocity and the tangential and normal components
of acceleration of the spindle EF when t — 2 s after
starting from rest.

SOLUTION

w a r A ~ r B

"a (10) = w b (40)
1

M B — —m a

v E = co B r E = ^(0.015) = i(40)(f 3 + 6f) (0.015)

v E = 3 m/s

a A = d ^ A = ^ [40(t 3 + 6/)] = 120t 2 + 240

a A r A = “b r B

a A (10) = a B (40)
1

a B = ^ a A

(a E ) t = * B r E = j(l20 f 2 + 240)(0.015)

t=2

(a E ), = 2.70 m/s 2

( a E)n ~ m b r E =

( a E)n = 600 m/s 2

i(40)(r 3 + 6t)

(0.015)

40 mm

Ans.

Ans.

Ans.

Ans:

v E = 3 m/s
( a E ) t = 2.70 m/s 2
( ci E ) n = 600 m/s 2

663

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16 - 35 .

If the shaft and plate rotates with a constant angular velocity
of to = 14 rad/s, determine the velocity and acceleration of
point C located on the corner of the plate at the instant
shown. Express the result in Cartesian vector form.

SOLUTION

We will first express the angular velocity to of the plate in Cartesian vector form. The
unit vector that defines the direction of to is

—0.3i + 0.2j + 0.6k 3 2 6

“cm = , , , , = + a j + = k

V(-0.3) 2 + 0.2 2 + 0.6 2 111

Thus,

3. 2. 6,

oo - um OA - 14 --i + - j + -k

[-6i + 4j + 12k] rad/s

Since to is constant

a = 0

For convenience, r c = [—0.3i + 0.4j] m is chosen. The velocity and acceleration of
point C can be determined from

v c = to X r c

= (—6i + 4j + 12k) X (—0.31 + 0.4j)

= [—■4.8i — 3.6j — 1.2k] m/s Ans.

and

a c = a X r c + to X (to X r c )

= 0 + (—6i + 4j + 12k) X [(—6i + 4j + 12k) X (—0.31 + 0.4j)]

= [38.41 - 64.8j + 40.8k]m/s 2 Ans.

z

Ans:

\ c = { —4.8i - 3.6j - 1.2k} m/s
a c = {38.41 - 64.8j + 40.8k} m/s 2

664

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* 16 - 36 .

At the instant shown, the shaft and plate rotates with an
angular velocity of co = 14 rad/s and angular acceleration
of a = 7 rad/s 2 . Determine the velocity and acceleration of
point D located on the corner of the plate at this instant.
Express the result in Cartesian vector form.

SOLUTION

We will first express the angular velocity w of the plate in Cartesian vector form. The
unit vector that defines the direction of m and a is

—0.3i + 0.2j + 0.6k 3 2 6

U ° A = / 7 , W = + T J + 7 k

V(-0.3) 2 + 0.2 2 + 0.6 2 111

Thus,

to = oon OA = 14^-| i + |j + ykj = [— 6i + 4j + 12k] rad/s

a = au 0 A = 7^-|i + |J + = [ _3i + 2 j + 6k] rad/s

For convenience, r D = [—0.3i + 0.4j] m is chosen. The velocity and acceleration of
point D can be determined from

Vd ~ w X r D

= (—6i + 4j + 12k) X (0.31 - 0.4j)

= [4.8i + 3.6j + 1.2k] m/s Ans.

and

a D = a X i D - co 2 i D

= (—3i + 2j + 6k) X (—0.3i + 0.4j) + (-6i + 4j + 12k) X [(-6i + 4j + 12k) X
= [—36.0i + 66.6j — 40.2k] m/s 2 Ans.

z

(—0.3i + 0.4j)]

Ans:

\ D = [4.8i + 3.6j + 1.2k] m/s
a D = [—36.0i + 66.6j — 40.2k] m/s 2

665

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16 - 37 .

The rod assembly is supported by ball-and-socket joints at
A and B. At the instant shown it is rotating about the y axis
with an angular velocity u> = 5 rad/s and has an angular
acceleration a = 8 rad/s 2 . Determine the magnitudes of
the velocity and acceleration of point C at this instant.
Solve the problem using Cartesian vectors and Eqs. 16-9
and 16-13.

SOLUTION

v c = co X r

v c ~ 5j X (— 0.4i + 0.3k) = jl.5i + 2k} m/s
v c = Vl.5 2 + 2 2 = 2.50 m/s
a c = a X r — <u 2 r

= 8j X (— 0.4i + 0.3k)— 5 2 (-0.4i + 0.3k)
= {12.4i - 4.3k} m/s 2
a c = Vl2.4 2 + (—4.3) 2 = 13.1 m/s 2

Ans:

Vc = 2.50 m/s
a c = 13.1 m/s 2

Ans.

Ans.

666

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16 - 38 .

The sphere starts from rest at 8 = 0° and rotates with an
angular acceleration of a = (40 + 1) rad/s 2 , where 0 is in
radians. Determine the magnitudes of the velocity and
acceleration of point P on the sphere at the instant

SOLUTION

i dco = a dd

/ co dco = / (40 + 1) dd
Jo Jo

co = 20

a = 4(6) + 1 = 25 rad/s 2 , co = V / 4(6) 2 + 2(6) = 12.49 rad/s

v = ar' = 12.49(8 cos 30°) = 86.53 in./s

v = 7.21 ft/s

v 2 (86.53) 2
a r = — =

Ans.

r 2 (8 cos 30°)

= 1080.8 in./s 2

= ar 2 = 25(8 cos 30°) = 173.21 in./s 2

a = V(1080.8) 2 + (173.21) 2 = 1094.59 in./s 2
a = 91.2 ft/s 2

Ans.

Ans:

v = 7.21 ft/s
a = 91.2 ft/s 2

667

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16 - 39 .

The end A of the bar is moving downward along the slotted
guide with a constant velocity \ A . Determine the angular
velocity to and angular acceleration a of the bar as a
function of its position y.

SOLUTION

Position coordinate equation:

sinfl = —

y

Time derivatives:

cos 08 =- y however, cos 0 =

y

Vy^V 2

Vy^V 2

y

rv A

and y = — v A , 0 = co

co = ~^v A co =

y yvy

a = (o = rv A [-y 2 y{y 2 - r 2 ) i + (y ^(-^(y 2 - r 2 ) i(2yy)]

rv 2 A (2y 2 - r 2 )
y 2 (y 2 -r 2 )i

Ans.

Ans.

Ans:

rv A

co = - ,

yVy 2 — r 2

rv A (2y 2 - r 2 )
“ = y 2 (y 2 - r 2 ) 3 / 2

668

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* 16 - 40 .

At the instant 8 = 60°, the slotted guide rod is moving
to the left with an acceleration of 2 m/s 2 and a velocity of
5 m/s. Determine the angular acceleration and angular
velocity of link AB at this instant.

SOLUTION

Position Coordinate Equation. The rectilinear motion of the guide rod can be
related to the angular motion of the crank by relating x and 8 using the geometry
shown in Fig. a , which is

x = 0.2 cos 8 m

Time Derivatives. Using the chain rule,

x = —0.2(sin 8)8

(1)

x = — 0.2[(cos 9)8 2 + (sin 0)6]

(2)

Here x = v,x = a, 8 = co and 8 = a when 8 = 60°. Realizing that the velocity
and acceleration of the guide rod are directed toward the negative sense of x,
v = —5 m/s and a = —2 m/s 2 .Then Eq (1) gives

— s = (—0.2(sin 60°)<u

Ans.

Subsequently, Eq. (2) gives

-2 = —0.2[cos 60°(28.87 2 ) + (sin 60°)a]

Ans.

The negative sign indicates that a is directed in the negative sense of 8.

Ca>

Ans:

a = 470 rad/s 2 *)

669

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16 - 41 .

At the instant 8 = 50°, the slotted guide is moving upward
with an acceleration of 3 m/s 2 and a velocity of 2 m/s.
Determine the angular acceleration and angular velocity of
link AB at this instant. Note: The upward motion of the
guide is in the negative y direction.

SOLUTION

y = 0.3 cos 8

y = v y = —0.3 sin 88

y= a y = —O.3(sin0f9 + cos86 2 'j

Here v y = —2 m/s, a y — —3 m/s 2 , and 6 = to, 6 = to, 8 = a, 6 = 50°.
—2 = —0.3 sin 50°(cu) to = 8.70 rad/s

-3 = —0.3[sin 50°(a) + cos 50°(8.70) 2 ] a = -50.5 rad/s 2

Ans:

670

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16 - 42 .

At the instant shown, 6 = 60°, and rod AB is subjected to a
deceleration of 16 m/s 2 when the velocity is 10 m/s.
Determine the angular velocity and angular acceleration of

SOLUTION

x = 2(0.3) cos 6

x = —0.6 sin d(d)

x = -0.6 cos 0(#) 2 - 0.6 sin 0 ( 0 )

Using Eqs. (1) and (2) at 8 = 60°, x, = 10 m/s, x = —16 m/s 2 .

10 = -0.6 sin 60°(co)

co = -19.245 = -19.2 rad/s

-16 = -0.6 cos 60°(—19.245) 2 - 0.6 sin 60°(a)

Ans.

Ans.

Ans:

671

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16 - 43 .

The crank AB is rotating with a constant angular velocity of
4 rad/s. Determine the angular velocity of the connecting
rod CD at the instant 8 = 30°.

SOLUTION

Position Coordinate Equation: From the geometry,

0.3 sin = (0.6 — 0.3 cos (f >) tan 8

[11

Time Derivatives: Taking the time derivative of Eq. , we have

d(f> , dd ( 7 dd d<f>

0.3 cos d)— = O.6sec 2 0--0.3 cos 8 sec 2 0—-tan 0 sin 6 —

dt dt V dt dt

cW_

dt

0.3(cos <fr — tan 6 sin 4>)

O.3sec 2 0(2 — cos <j>)

dtf)

dt

[ 2 ]

d8 d(j>

However, —• = o ) BC , — = co AB = 4 rad/s. At the instant 8 = 30°, from Eq. ,
dt dt

4> = 60.0°. Substitute these values into Eq.  yields

m bc

0.3(cos 60.0° - tan 30° sin 60.0°)
0.3 sec 2 30° (2 - cos 60.0°)

(4) = 0

Ans.

Ans:

W AB = 0

672

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* 16 - 44 .

Determine the velocity and acceleration of the follower
rod CD as a function of 6 when the contact between the cam
and follower is along the straight region AB on the face of
the cam. The cam rotates with a constant counterclockwise
angular velocity co.

SOLUTION

Position Coordinate: From the geometry shown in Fig. a,

Time Derivative: Taking the time derivative,
v cd = x c = r sec 6 tan 66

Flere, 6 = +co since co acts in the positive rotational sense of d.Thus, Eq. (1) gives
v C d ~ rco sec 6 tan 6 —> Ans.

The time derivative of Eq. (1) gives

a CD — *c ~ r{secdtandd + 0[sec$(sec 2 00) + tan0(sec$tan0(i)]}
a CD ~ r [sec 6 tan 60 + (sec 3 6 + sec 6 tan 2 6)6 2 ]

Since 6 = co is constant, 6 = a = 0. Then,

a CD ~ /"[sec 6 tan 0(0) + (sec 3 0 + seed tan 2 0)<u 2 ]
= ra 2 (sec 3 0 + seed tan 2 —»

Ans.

Ans:

v CD = rco sec 6 tan 6 —*

a CD = rco 2 {sec 2 6 + sec 6 tan 2 6) —»

673

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16 - 45 .

Determine the velocity of rod R for any angle 8 of the cam
C if the cam rotates with a constant angular velocity co. The
pin connection at O does not cause an interference with the
motion of A on C.

SOLUTION

Position Coordinate Equation: Using law of cosine.

(A + A) 2 = x 2 + rf ~ 2 r x x cos 8
Time Derivatives: Taking the time derivative of Eq. (1). we have

dx .dd dx

0 = 2x— :- 2r,\ — x sin — + cos 0——

dt \ dt dt

However v = —r and <u = 44 From Eq.(2),
dt dt

0 = xv — r$$v cos 6 — xw sin 6) riXh) sin 8 v =- /q cos 8 — x However, the positive root of Eq.(l) is x — r\ cos 8 + Videos 2 # + r\ + 2rir 2 0} ( 3 ) Substitute into Eq.(3),we have ( rjco sin 28 \ v = — - - + riw sin© Ans. \2V r 2 cos 2 8 + r 2 + 2r jr 2 / Note: Negative sign indicates that v is directed in the opposite direction to that of positive x. Ans: v = rivj sin 28 = + sin 8 cos 2 8 + r■? + 2 r,r 2 674 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 46 . The circular cam rotates about the fixed point O with a constant angular velocity to. Determine the velocity v of the follower rod AB as a function of 6. SOLUTION x = d cos 6 + \/(R + r) 2 — (d sin 0) 2 x = v AB = —d sin 60 — —v = —d sin 6((o) — d 2 sin 20 2V(R + r) 2 - d 2 sin 2 0 d 2 sin 26 6 Where 6 = co and v AB = ~v 2V(R + r) 2 - d 2 sin 2 6 Ans: = tod sin 6 + d sin 26 2 V(f? + r) 2 — d 2 sin 2 6 675 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 47 . Determine the velocity of the rod R for any angle 8 of cam C as the cam rotates with a constant angular velocity at. The pin connection at O does not cause an interference with the motion of plate A on C. SOLUTION x = r + r cos 8 x = -r sin 66 v = —rw sin 8 Ans: v = —no sin 6 OJ Ans. 676 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 48 . Determine the velocity and acceleration of the peg A which is confined between the vertical guide and the rotating slotted rod. SOLUTION Position Coordinate Equation. The rectilinear motion of peg A can be related to the angular motion of the slotted rod by relating y and 8 using the geometry shown in Fig. a , which is y = b tan 8 Time Derivatives. Using the chain rule, y = b(sec 2 8)8 (1) y = b[2 sec 8( sec 8 tan 88)8 + sec 2 88] y = b(2 sec 2 8 tan 88 2 + sec 2 8d) y = b sec 2 0(2 tan 88 2 + d) (2) Flere, y = v,y = a. 8 = co and 8 = cr.Then Eqs. (1) and (2) become v = cob sec 2 8 Ans. a = b sec 2 (2<w 2 tan 8 + a) Ans. Ans: v = cob sec 2 8 a = b sec 2 d(2co 2 tan 8 + cr) 677 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 49 . Bar AB rotates uniformly about the fixed pin A with a constant angular velocity co. Determine the velocity and acceleration of block C, at the instant 0 = 60°. SOLUTION L cos 0 + L cos ep = L cos 0 + cos cp = 1 sin 8 <9] + sirup ep = 0 (1) cos 8(8) 2 + sin 88 + sirup <P + cos <p ( ep) 2 = 0 (2) When 8 = 60°, <p = 60°, thus, 8 = ~(p= or (from Eq. (1)) (9 = 0 <p = — 1.155<w 2 (from Eq.(2)) Also, sc — L sin (p - L sin 8 v c = L cos <p <p — L cos 8 8 a c = — L sin <p (ip) 2 + L cos <p (cp) — L cos 8(6) + L sin 8(d) 2 At 8 = 60°, 4 , = 60° ■sc = 0 v c — L(cos 60°)(— a>) — L cos 60°(co) = —Leo = Lent Ans. a c = — L sin 60°(— w) 2 + L cos 60°(—1.155m 2 ) + 0 + L sin 60°(tu) 2 a c = -0.577 Leo 2 = 0.577 Leo 2 ] Ans. B Ans: v c = Leo] a c = 0.577 Leo 2 ] 678 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 50 . The center of the cylinder is moving to the left with a constant velocity v () . Determine the angular velocity at and angular acceleration a of the bar. Neglect the thickness of the bar. SOLUTION Position Coordinate Equation. The rectilinear motion of the cylinder can be related to the angular motion of the rod by relating x and 9 using the geometry shown in Fig. a, which is = r cot 9/2 tan 9/2 Time Derivatives. Using the chain rule, (-esc 2 9/ 2)( — 9 x = --(esc 2 9/2)9 ( 1 ) x = — 2 2 esc 9/ 2(—esc 9/2 cot 9/ 2)( — 9 j8 + (esc 2 9/2)9 r X ~ 2 (esc 2 0/2cot 9/2)d 2 — (esc 2 9/2)b r esc 2 9/2 (cot 9/2)9 2 - 9 ( 2 ) Here x = — v 0 since v 0 is directed toward the negative sense of x and 9 = w. Then Eq. (1) gives, —Vq = —— (csc 2 9/2)w CO = Ans. 1 679 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 680 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 51 . The pins at A and B are confined to move in the vertical and horizontal tracks. If the slotted arm is causing A to move downward at v^, determine the velocity of B at the instant shown. SOLUTION Position coordinate equation: h d tan (? = — = — x y Time derivatives: Ans. Ans: 681 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 52 . The crank AB has a constant angular velocity to. Determine the velocity and acceleration of the slider at C as a function of 8 . Suggestion: Use the x coordinate to express the motion of C and the <fi coordinate for CB. x = 0 when <fi = 0°. SOLUTION x — I + b — (L cos 4 > + b cos #) l sin <f> = b sin # or sin 4 > = — sin 6 vc = x = / sin 4 >4> + b sin ## cos 4 >4> = — cos 00 Since cos <f> = Vl — sin 2 4> = A /l — ( ) sin 2 0 then. 4 > = cos Oco 1 - | y ) sin 2 # Vc — boo sin 8 cos 6 1 - ( j) sin 2 8 + boo sin 8 From Eq. (1) and (2): a c ~ Pc = ^4> s i n 4> + l<t> cos <p(f> + b cos #( 8 'b —sin (pcfr + cos sin 88 2 <]> = ' 9 . u 1 4 > sin of> — —oo sin 8 COS (f) Substituting Eqs. (1), (2), (3) and (5) into Eq. (4) and simplifying yields a ( = b(o z y^fcos2# + ^y^) sin 4 # 1 - + cos # sin 2 # Ans. ( 3 ) Ans. ( 4 ) ( 5 ) Ans: v c = boo y I sin # cos 6 -V 1 _ (y) 2sin20 . + boo sin # a c = boo“ cos 26 + sin 4 6 + cos # 1 - sin 2 # 682 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 53 . If the wedge moves to the left with a constant velocity v, determine the angular velocity of the rod as a function of 8 . SOLUTION Position Coordinates: Applying the law of sines to the geometry shown in Fig. a, Xa _ L sin(0 - 8 ) sin(l80° - r/j) L sin (<fi — 8) x a = —7-r sin(l80° - <j>) However, sin(180° — </>) = sim/>. Therefore, L sin (cf> — 8) xa — -.—:- sin 4 > Time Derivative: Taking the time derivative, L cos (<f> - 8)(-8 ) sin cf> L cos (4> — 8)8 xa va = x a sin 4 > ( 1 ) Since point A is on the wedge, its velocity is v A = —v. The negative sign indicates that is directed towards the negative sense of x A . Thus, Eq. (1) gives v sin d> 8 = —-—-- Ans. L cos (<p — 8) Ans: 8 = - v sin (f> L cos (4> - 8) 683 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 54 . The crate is transported on a platform which rests on v , rollers, each having a radius r. If the rollers do not slip, determine their angular velocity if the platform moves forward with a velocity v. TT— c-d" & SOLUTION Position coordinate equation: From Example 163, s G = rd. Using similar triangles s A = 2s g = 2 rd Time derivatives: s A = v = 2r d Where d = co A w = v 2r Ans. Ans: aj v 2r 684 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 55 . Arm AB has an angular velocity of to and an angular acceleration of a. If no slipping occurs between the disk D and the fixed curved surface, determine the angular velocity and angular acceleration of the disk. SOLUTION ds = (R + r) dd = r d(f> (R + co' (R + r)w r Ans. (R + r) a a' = - r Ans. to', a' Ans: co cc (R + r)w r (R + r)a r 685 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 56 . At the instant shown, the disk is rotating with an angular velocity of co and has an angular acceleration of a. Determine the velocity and acceleration of cylinder B at this instant. Neglect the size of the pulley at C. SOLUTION 5 = V3 2 + 5 2 - 2(3)(5)cos 9 v B = i = ^ (34 — 30 cos 0)~ J (30 sin 9)6 15 co sin 9 v B =-1 (34 - 30 cos 9y „ _ ■ • . f — — )(15m sin 6 )( 30 sin 99 ) 15 co cos 96 + 15m smS \ 2/ V / a B ~ i =-, =-*-3- V34 - 30 cos 9 (34 _ 30 cos 0 ) s 15 ( co 2 cos 6 + a sin 9) 225 co 2 sin 2 9 (34 - 30 cos 6 Y (34 - 30 cos 0)1 Ans. Ans. Ans: 15 co sin 6 v B = -r (34 - 30 cos 0)2 15 (to 2 cos 9 + a sin 9) 225 m 2 sin 2 9 1 3 (34 — 30 cos 9) 2 (34 — 30 cos 6) 2 686 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 687 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 58 . If the block at C is moving downward at 4 ft/s, determine the angular velocity of bar AB at the instant shown. SOLUTION Kinematic Diagram: Since link AB is rotating about fixed point A , then \ B is always directed perpendicular to link AB and its magnitude is v B = co AB r AB = 2co AB . At the instant shown, v B is directed towards the negative y axis. Also, block C is moving downward vertically due to the constraint of the guide. Then v c is directed toward negative y axis. Velocity Equation: Here, r C / B = {3 cos 30°i + 3sin30°j}ft = {2.598i + 1.50j) ft. Applying Eq. 16-16, we have Vc = y B + W BC X r C/B —4j = -2a) AB ] + (m BC k) x (2.598i + 1.50j) —4j = — 1.50<u BC i + (2.598 <u bc — 2to AB )j Equating i and j components gives 0 = — 1.50ui bc co BC = 0 —4 = 2.598(0) — 2 co AB w AB = 2.00 rad/s Ans. Ans: w AB = 2.00 rad/s 688 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-59. The link AB has an angular velocity of 3 rad/s. Determine the velocity of block C and the angular velocity of link BC at the instant 8 = 45°. Also, sketch the position of link BC when 6 = 60°, 45°, and 30° to show its general plane motion. SOLUTION Rotation About Fixed Axis. For link AB, refer to Fig. a. TB ~ m AB X r AB = (3k) X (0.5 cos 45°i +0.5sin45°j) = {—1.0607i + 1.0607J} m/s General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation. V C — V B + W BC X r C/B -v c i = (—1.0607i + 1.0607j) + (~w BC k ) X (1.5i) —v c i = — 1.0607i + (1.0607 — 1.5tu BC )j Equating i and j components; —Vc = —1.0607 v c = 1.0607m/s = 1.06m/s Ans. 0 = 1.0607 — 1.5w bc co BC = 0.7071 rad/s = 0.707 rad/s Ans. The general plane motion of link BC is described by its orientation when 8 = 30°, 45° and 60° shown in Fig. c. Ans: Vc — 1-06 m/s <— oo B c = 0.707 rad/s/) 689 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 60 . The slider block C moves at 8 m/s down the inclined groove. Determine the angular velocities of links AB and BC , at the instant shown. SOLUTION Rotation About Fixed Axis. For link AB , refer to Fig. a. TS = m AB X r AB v b = X (2i) = -2co AB j General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation. Vs = v c + (Ogc x r B /c -2tu AB j = (8 sin 45°i - 8cos45°j) + (m BC k) X (2j) —2co ab \ = (8 sin 45° — 2tu BC )i — 8 cos 45°j Equating i and j components, 0 = 8 sin 45° — 2co BC co BC = 2.828 rad/s = 2.83 rad/s Ans. -2w ab = —8 cos 45° (o AB = 2.828 rad/s = 2.83 rad/s /) Ans. Ans: w BC = 2.83 rad/s *) w AB = 2.83 rad/s/) 690 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-61. Determine the angular velocity of links AB and BC at the instant 8 = 30°. Also, sketch the position of link BC when 8 = 55°, 45°, and 30° to show its general plane motion. SOLUTION Rotation About Fixed Axis. For link AB, refer to Fig. a. B 6 ft/s V S — °>AB X r AB = ("^.sk) X j = ~ a) AB i Genera! Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, v B = v c + a> BC x r B /c —co AB i = 6j + (<u BC k) X (—3 cos 30°i + 3 sin 30°j) ~ 0) ab* = — 1.5&tgci + (6 — 2.5981 <u BC )j Equating i and j components, 0 = 6 — 2.5981 <ug C ; a> B c = 2.3094 rad/s = 2.31 rad/s *) Ans. ~ w ab = —1.5(2.3094); co AB = 3.4641 rad/s = 3.46 rad/s Ans. The general plane motion of link BC is described by its orientation when 8 = 30°, 45° and 55° shown in Fig. c. Ans: co BC = 2.31 rad/s *) o) AB = 3.46 rad/s 5 691 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 62 . The planetary gear A is pinned at B. Link BC rotates clockwise with an angular velocity of 8 rad/s, while the outer gear rack rotates counterclockwise with an angular velocity of 2 rad/s. Determine the angular velocity of gear A. SOLUTION c Kinematic Diagram : Since link BC is rotating about fixed point C. then v /; is always directed perpendicular to link BC and its magnitude is v B = co BC r BC = 8(15) = 120 in. / s. At the instant shown. v B is directed to the left. Also, at the same instant, point E is moving to the right with a speed of v E = w B r CE = 2(20) = 40 in./s. Velocity Equation : Here, v B / E = co A r B / E = 5 oj a which is directed to the left. Applying Eq. 16-15, we have ' J*' = v £ + \ B/E 120 40 5u>/ - 120 = 40 - 5o a co A = 32.0 rad/s Ans. Ans: w A = 32.0 rad/s 692 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-63. If the angular velocity of link AB is w AB = 3 rad/s, determine the velocity of the block at C and the angular velocity of the connecting link CB at the instant 0 = 45° and 0 = 30°. SOLUTION Vc = + \ C /B _ _ Vc 6 30°^; + m c B (3) 45 (-^» ) — Vc = 6 sin 30° — cu C b (3) cos 45° (+1) 0 = -6 cos 30° + (o CB (3) sin 45° co C b — 2.45 rad/s 5 Ans. Vc — 2.20 ft/s *— Ans. Also, Vc = + (O X r c/B -v c i = (6 sin 30°i — 6 cos 30°j) + (<u CB k) X (3 cos 45°i + 3 sin 45°j) (^) -v c = 3 — 2.12 co C b (+t) 0 = -5.196 + 2.12 (o CB W CB - 2.45 rad/s !) Ans. l'( = 2.20 ft/s <- Ans. Ans: io c b = 2.45 rad/st) v c = 2.20 ft/s <— 693 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-64. The pinion gear A rolls on the fixed gear rack B with an angular velocity co = 4 rad/s. Determine the velocity of the gear rack C. SOLUTION V C = V B + \c/B () Vc = o + 4(0.6) Vc = 2.40 ft/s Also: v c = v B + m X r c/B C Ans. -vci = 0 + (4k) X (0.6j) Vc — 2.40 ft/s Ans. Ans: Vc = 2.40 ft/s v c = 2.40 ft/s 694 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-65. The pinion gear rolls on the gear racks. If B is moving to the right at 8 ft/s and C is moving to the left at 4 ft/s, determine the angular velocity of the pinion gear and the velocity of its center A. SOLUTION c Vc = V B + V C /B (-L ) —4 = 8 — 0.6(to) co = 20 rad/s v A = v B + v A/B (^) v a ~ ~ 20(0.3) v A = 2 ft/s —» Ans. Also: V C = V B + 0 } X r c/B —4i = 8i + (atk) X (0.6j) —4 = 8 — 0.6<u co = 20 rad/s Ans. v A = v B + co X r A/B v A \ = 8i + 20k X (0.3j) v A = 2 ft/s —> Ans. Ans: co = 20 rad/s v A = 2 ft / s —> 695 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 66 . Determine the angular velocity of the gear and the velocity of its center O at the instant shown. SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, Dj = v c + co X r B/c 3i = — 4i + (—mk) X (2.25j) 3i = (2.25m - 4)i Equating the i components yields 3 = 2.25m - 4 (1) m = 3.111 rad/s Ans. (2) For points O and C, V 0 = V C + m X i 0/c = — 4i + (-3.111k) X (l.5j) = [0.6667i] ft/s Thus, v 0 = 0.667 ft/s — * Ans. Ans: m = 3.11 rad/s v 0 = 0.667 ft/s —> 696 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-67. Determine the velocity of point A on the rim of the gear at the instant shown. SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, = v c + w X r B /c 3i = —4i + (-wk) X (2.25j) 3i = (2.25 to - 4)i Equating the i components yields 3 = 2.25 (o - 4 (1) co — 3.111 rad/s (2) For points A and C, \ A = \c + (oX r A/c + (w^)_ v j = — 4i + (—3.111k) X (—1.0611 + 2.561j) (v A ) x i + (v A ) y j = 3.96651 + 3.2998j Equating the i and j components yields (v A ) x = 3.9665 ft/s (v A ) y = 3.2998 ft/s Thus, the magnitude of v A is v A = V(v A ) x 2 + (v A ) y z = V3.9665 2 + 3.2998 2 = 5.16 ft/s and its direction is 0 = tan tan 3.2998 \ 3.9665 J 39.8° Ans. Ans. Ans: v A = 5.16 ft/s 6 = 39.8° 697 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 68 . Knowing that angular velocity of link AB is co AB = 4 rad/s, determine the velocity of the collar at C and the angular velocity of link CB at the instant shown. Link CB is horizontal at this instant. SOLUTION V B = ^AB^B = 4(0.5) = 2 m/s y B = {—2 cos 30°i + 2 sin 30°] ) m/s v c = — v c cos 45°i — n c sin 45°j co = co BC k r c/b = {—0.35i} m v c = + co X i c/B —Vc cos 45°i — Vc sin 45°j = (—2 cos 30°i + 2 sin 30°j ) + (m sc k) x (—0.35i) —Vc cos 45°i — Vc sin 45°j = —2 cos 30°i + (2 sin 30° — 0.35<u BC )j Equating the i and j components yields: —Vc cos 45° = —2 cos 30° Vc = 2.45 m/s Ans. —2.45 sin 45° = 2 sin 30° — 0.35 &j bc <*> B c = 7.81 rad/s Ans. Ans: v c = 2.45 m/s co BC = 7.81 rad/s 698 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-69. Rod AB is rotating with an angular velocity of oo AB = 60 rad/s. Determine the velocity of the slider C at the instant 0 = 60° and 4> — 45°. Also, sketch the position of bar BC when 6 = 30°, 60° and 90° to show its general plane motion. SOLUTION Rotation About Fixed Axis. For link AB. refer to Fig. a. Vb = o) AB X r AB = (60k) X (—0.3 sin 60°i + 0.3 cos 60°j) = {—9i - 9V3j} m/s General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, Vc = + w BC X r C / B — z>cj = (—9i — 9V3j ) + (cu BC k) X (—0.6 sin 45°i — 0.6cos45°j) -®cj = (0.3V2 w BC ~ 9)i + (—0.3 V2<u BC - 9V3)j Equating i components, 0 = 0.3 V2 co BC — 9; co BC = 15 \fl rad/s = 21.2 rad/s 5 Then, equating j components, -v c = (—0.3V2)(15V2) - 9V3; v c = 24.59 m/s = 24.6 m/si Ans. The general plane motion of link BC is described by its orientation when 0 = 30°, 60° and 90° shown in Fig. c. iti hh -(bO rad/i Ans: v c = 24.6 m/s I 699 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 70 . The angular velocity of link AB is w AB = 5 rad/s. Determine the velocity of block C and the angular velocity of link BC at the instant 6 = 45° and 4> = 30°. Also, sketch the position of link CB when d = 45°, 60°, and 75° to show its general plane motion. SOLUTION Rotation About A Fixed Axis. For link AB, refer to Fig. a. TB “ <°AB X r AB = (5k) X (—3 cos 45°i — 3 sin 45°j ) J 15\/2 . 15\/2 . j ( m/s General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, v c = v B + 0) BC X r c/B 15\/2. 15V2.| v c l V C 1 = 2 15V2 ■j + ( w BC k ) X (2 sin 30° i — 2cos30°j) 1 V3 w bc i + w bc ~ 15V2 1 — j Equating j components, O — w BC 15\/2 ’ W BC ~ 15V2 2 2 Then, equating i components, 15V2 rad/s = 10.6rad/s/) vc - —7— + V3 I = 28.98 m/s = 29.0 m/s- Ans. Ans. 700 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-70. Continued The general plane motion of link BC is described by its orientation when 8 60° and 75° shown in Fig. c Cc) Ans: w BC = 10.6 rad/s !) v c = 29.0 m/s —> 701 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-71. The similar links AB and CD rotate about the fixed pins at A and C. If AB has an angular velocity w AB = 8 rad/s, determine the angular velocity of BDP and the velocity of point P. SOLUTION V B = v s + W X I D/B —v D cos 30°i — v D sin 30°j = —2.4 cos 30°i + 2.4 sin 30°j + (rnk) X (0.6i) —v D cos 30° = —2.4 cos 30° — v D sin 30° = 2.4 sin 30° + 0.6m v D = 2.4 m/s w = — 4 rad/s \ P = y B + co X r p/ B \ P = -2.4 cos 30°i + 2.4 sin 30°j + (-4k) X (0.3i - 0.7j) (Vp) x = —4.88 m/s ( Vp) y = 0 v P = 4.88 m/s <— Ans: v P = 4.88 m/s <— 702 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 72 . If the slider block A is moving downward at v A — 4 m/s, determine the velocities of blocks B and C at the instant shown. SOLUTION V B = v A + V B/A ^ = 4i + co AB ( 0.55) (+>) Vfi = {) + oW0.55)(jj ( + T) 0 - —4 + <^(0.55)0) Solving, co AB = 9.091 rad/s v B = 3.00 m/s v D = v A + v D/A \ D = 4 + [(0.3)(9.091) = 2.727] i 4 v c = v D + v c/D Vq = 4 + 2.727 4- ^ce(0-4) ] 03 M 30° 4 ( 4) v c = 0 + 2.727 0) - <u C£ (0.4)(sin 30°) ( + T) 0 = -4 + 2.727 Qj + m C£ (0.4)(cos 30°) to CE = 5.249 rad/s v c = 0.587 m/s Also: Ans. Ans. v B = V A + co AB X r B/A v B i = —4j + (-m AB k) X |0(O.55)i + |(0.55)j| 703 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-72. Continued v b ~ <*> A b{ 0.33) 0 : — —4 + 0.44 co AB ai AB = 9.091 rad/s v B = 3.00 m/s Ans. \ D = V A + <»AB X *B/A v D = — 4j + (-9.091k) X (0.3)1 + | (0.3)j | v D = (1.636i — 1.818j } m/s v c = v d + (Bqe x r C/D V(4 = (1.636i - 1.818j) + (-tu C£ k) X (-0.4cos30°i - 0.4 sin 30°j) Vc — 1.636 — 0.2 (o C e 0 = -1.818 - 0.346(u C£ (o C e = 5.25 rad/s v c = 0.587 m/s Ans. Ans: v B = Vc = V B = v c = 3.00 m/s 0.587 m/s 3.00 m/s 0.587 m/s 704 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 73 . If the slider block A is moving downward at v A = 4 m/s, determine the velocity of point E at the instant shown. SOLUTION See solution to Prob. 16-87. v E = v D + v E/D v*E = 4l + 2.727 + (5.249)(0.3) 4 y\ K 30° ( ) (v E ) x = o + 2.727 (j) + 5.249(0.3)(sin 30°) ( +1) (v E ) y = 4 - 2.727 + 5.249(0.3)(cos 30°) (v E ) x = 2.424 m/s -»■ (v E ) y = 3.182 m/s i y E = V(2.424) 2 + (3.182) 2 3.182\ 6 = tan -1 2.424 = 52.7° 4.00 m/s Also: See solution to Prob. 16-87. Ans. Ans. V E — v n + U>CE X r E/D \ E = (1.6361 - 1.818j) + (-5.25k) X {cos 30°(0.3)i - 0.4 sin 30°(0.3)j) \ E = { 2.424i - 3.182j } m/s v E = V(2.424) 2 + (3.182) 2 = 4.00 m/s Ans. ,/3.182\ ^ = tan~M —— = 52.7° Ans. \2.424 J Ans: v E = 4.00 m/s 0 = 52.7° ^ 705 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 74 . The epicyclic gear train consists of the sun gear A which is in mesh with the planet gear B. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the connecting link DE pinned to B and C is rotating at a ) D e ~ 18 rad/s about the pin at E, determine the angular velocities of the planet and sun gears. SOLUTION Vd = r DE w DE = (0.5)(18) = 9 m/s t The velocity of the contact point P with the ring is zero. V/) = v P + w X r D/P 9j = 0 + (-o> fl k) X (—O.li) io B = 90 rad/s J Let P' be the contact point between A and B. \ P ' = \ P + co X r p'/p vp'i = 0 + (-90k) X (—0.4i) v P ' = 36 m/s T "a = — = = 180 rad/s t> r A u.Z Ans. Ans. Ans: (o B = 90 rad/s /) oo A = 180 rad/s !) 706 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 75 . If link AB is rotating at w AB = 3 rad/s, determine the angular velocity of link CD at the instant shown. SOLUTION y B = W AB X y B/A y C = Wcd X *C/D y C = y B + X T C/B (w CD k) X (—4 cos 45°i + 4sin45°j) = (—3k) X (6i) + (&) BC k) X (—8 sin 30°i — 2.828(o cd — 0 + 6.928co BC — 2.828o) CD — —18 — 4 (jo B q Solving, co BC = —1.65 rad/s (o CD = 4.03 rad/s Ans. Ans: oo C d = 4.03 rad/s 707 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-76. If link CD is rotating at w CD = 5 rad/s, determine the angular velocity of link AB at the instant shown. SOLUTION Vb = W AB X *B/A VC = (BCD X r C/D V B = V C + W BC X r B/C (— co AB k) X (6i) = (5k) X (—4 cos 45°i + 4sin45°j) + (<u BC k) X (8 sin 30°i + 8cos30°j) 0 = -14.142 - 6.9282m BC 6co ab — 14.142 + 4 (o B c Solving, a> AB = 3.72rad/s Ans. to BC = —2.04 rad/s Ans: co AB = 3.72rad/s 708 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-77. The planetary gear system is used in an automatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, w R = 0, and the sun gear S is rotating at co s = 5 rad/s. Determine the angular velocity of each of the planet gears P and shaft A. SOLUTION v A = 5(80) = 400 mm/s <— v B = 0 \ B = \ A + WX r B/A 0 = —400i + (w p k) X (80j) 0 = —400i — 80&jp i cop = —5 rad/s = 5 rad/s v c = + &j X x c/B v c = 0 + (-5k) X (— 40j) = —200i 200 1 A7 A, <°A = Ton = 1 ’ 67 rad / S 40 mm Ans. Ans. Ans: (o P = 5 rad/s (o A = 1.67 rad/s 709 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-78. If the ring gear A rotates clockwise with an angular velocity of (i) A = 30 rad/s, while link BC rotates clockwise with an angular velocity of w BC = 15 rad/s, determine the angular velocity of gear D. SOLUTION Rotation About A Fixed Axis. The magnitudes of the velocity of Point E on the rim and center C of gear D are t>E = M A r A = 30(0.3) = 9 m/s Vc = UBcrBC = 15(0.25) = 3.75 m/s General Plane Motion. Applying the relative velocity equation by referring to Fig. a , v £ = V C + 0) D X r E/c 9i = 3.751 + (-tu^k) X (0.05j) 9i = (3.75 + 0.05m n )i Equating i component, 9 = 3.75 + 0.05 co D co D = 105 rad/s /) Ans. (a.) Ans: o) D = 105 rad/s ^ 710 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-79. The mechanism shown is used in a riveting machine. It consists of a driving piston A, three links, and a riveter which is attached to the slider block D. Determine the velocity of D at the instant shown, when the piston at A is traveling at v A = 20 m/s. SOLUTION Kinematic Diagram: Since link BC is rotating about fixed point B. then v c is always directed perpendicular to link BC. At the instant shown. v c = — Vq cos 30°i + v c sin 30°j = — 0.8660v c i + 0.500u c j. Also, block D is moving towards the negative y axis due to the constraint of the guide. Then. \ D = — v D \. Velocity Equation: Here, \ A = j—20 cos 45°i + 20 sin 45°j)m/s = {—14.14i + 14.14J} m/s and t C / A — {—0.3cos30°i + 0.3sin30°j jm = {—0.2598i + 0.150j} m. Applying Eq. 16-16 to link AC, we have v c = v A + co AC X r c/A —0.8660 Vq'i + 0.500v c j = -14.14i + 14.14j + (oj AC k) X (-0.2598i + 0.150j) —0.8660w c i + 0.500 v c ] = -(14.14 + 0.150&^ c )i + (14.14 - 0.2598&) AC )j Equating i and j components gives —0.8660% = -(14.14 + 0.150w AC )  0.500 v c = 14.14 - 0.2598 at AC  Solving Eqs.  and  yields co AC — 17.25 rad/s % — 19.32 m/s Thus, Vc = {-19.32 cos 30°i + 19.32 sin 30°j) m/s = {—16.73i + 9.659j) m/s and r D /c ~ {-0.15cos 45°i - 0.15sin45°j |m = {—0.1061i - 0.1061]} m. Applying Eq. 16-16 to link CD, we have y D — v c + “co x r n/c -v D ] = —16.73i + 9.659j + (w Cfl k) X (-0.106H - 0.1061j) -v D j = (0.1061 w CD - 16.73) i + (9.659 - 0.1061to CZ) )j Equating i and j components gives 0 = 0.1061<u C n - 16.73 [31 -v D = 9.659 - 0.1061 co CD [41 Solving Eqs.  and  yields wcd — 157.74 rad/s v D = 7.07 m/s Ans. Ans: v D = 7.07 m/s 711 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 80 . The mechanism is used on a machine for the manufacturing of a wire product. Because of the rotational motion of link AB and the sliding of block F, the segmental gear lever DE undergoes general plane motion. If AB is rotating at (o AB = 5 rad/s, determine the velocity of point E at the instant shown. SOLUTION v B = (i)AB r AB = 5(50) = 250 mm/s 45° \ c = y B + v C /b Vq — 250 + co B c( 200) «- 45° ^ 45° IF' ( + t) 0 = 250 sin 45° - m iJC (200) sin 45° (it) v c = 250 cos 45° + m BC (200) cos 45° Solving, v c = 353.6 mm/s; w BC = 1.25 rad/s Vp = Vc + Vc v„ = 353.6 + [(1.25)(20) = 25] - - i VD = Vp + y D / p v D = (353.6 + 25) + 2Qu> de - - i T (it) v D = 353.6 + 0 + 0 (+1) 0 = 0 + (1.25)(20) - <u D£ (20) Solving, v D = 353.6 mm/s; o) DE = 1.25 rad/s v £ = \ D + y E /D v E = 353.6 + 1.25(50) </j^ «- ^45° (it) v E cos 0 = 353.6 - 1.25(50) cos 45° ( + t) v e sin 4> = 0 + 1.25(50) sin 45° Solving, v E = 312 mm/s (j> = 8.13° Ans. Ans. 712 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-80. Continued Also; = “>AB X *B/A = y B + "BC x *c/b —?; c i = ( — 5k) X (—0.05 cos 45°i — 0.05 sin 45°j) + (<n BC k) X (—0.2 cos 45°i + 0.2sin45°j) — v c = —0.1768 — 0.1414u> bc 0 = 0.1768 - 0.1414o>g C to BC = 1.25 rad/s, v c = 0.254 m/s Vp = V C + co BC X r p/c Vd = Vp + (o DE X t D/p Vd = V C + m BC X r p/C + M D E X *D/p v D \ = —0.354i + (1.25k) X (-0.02i) + (co DE k) X (-0.02i) v D = -0.354 0 = -0.025 - co de (0.02) v D = 0.354 m/s, ai DE = 1.25 rad/s Ve = y D + m de x r E/D (v E ) x i + (v E ) y i = —0.354i + (-1.25k) X (-0.05 cos 45°i + 0.05 sin 45°j) (v E ) x = -0.354 + 0.0442 = -0.3098 (v E ) y = 0.0442 v E = V(-0.3098) 2 + (0.0442) 2 = 312 mm/s Ans. Ans. Ans: v E = 312 mm/s <t> = 8.13' v E = 312 mm/s <t> = 8.13' 713 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 81 . In each case show graphically how to locate the instantaneous center of zero velocity of link AB. Assume the geometry is known. SOLUTION a) It, (c) 714 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 82 . Determine the angular velocity of link AB at the instant shown if block C is moving upward at 12 in./s. SOLUTION 4 _ vjc-b _ r ic-c sin 45° ~~ sin 30° ~~ sin 105° r ic-c = 5 -464 in. r IC _ B = 2.828 in. Vc ~ u BC (ri C -c) 12 = <«b C (5.464) to BC = 2.1962 rad/s v B - COBc( r IC-B ) = 2.1962(2.828) = 6.211 in./s V B = W AB r AB 6.211 = C0 AB ( 5) w ab = 1-24 rad/s Ans. Ans: co AB = 1.24rad/s 715 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-83. The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the angular velocity of the link CB at the instant shown, if the link AB is rotating at 4 rad/s. SOLUTION Kinematic Diagram: Since linke AB is rotating about fixed point A, then v B is always directed perpendicular to link AB and its magnitude is Vb ~ u >AB r AB = 4(0.3) = 1.20 m/s. At the instant shown, \ B is directed at an angle 30° with the horizontal. Also, block C is moving horizontally due to the constraint of the guide. Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from \ B and v c . Using law of sines, we have r B/ic _ 0,125 sin 45° ~~ sin 30° r B /ic ~ 0.1768 m r c/ic _ 0,125 sin 105° ~~ sin 30° r c/ic = 0.2415 m The angular velocity of bar BC is given by v B _ 1.20 r B/ic 0.1768 6.79 rad/s Ans. Ans: o) BC = 6.79 rad/s 716 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-84. The conveyor belt is moving to the right at v = 8 ft/s, and at the same instant the cylinder is rolling counterclockwise at co = 2rad/s without slipping. Determine the velocities of the cylinder’s center C and point B at this instant. B A SOLUTION r A ic = | = 4 ft v c = 2(3) = 6.00 ft/s v B = 2(2) = 4.00 ft/s Ans. Ans. ic Ans: Vc = 6.00 ft/s —» v B = 4.00 ft/s - 717 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-85. The conveyor belt is moving to the right at v = 12 ft/s, and at the same instant the cylinder is rolling counterclockwise at w = 6 rad/s while its center has a velocity of 4 ft/s to the left. Determine the velocities of points A and B on the disk at this instant. Does the cylinder slip on the conveyor? o o o 6)o o o o'd 3 SOLUTION r A ic = | = 0-667 ft v A = 6(1 - 0.667) = 2 ft/s -» Ans. v B = 6(1 + 0.667) = 10 ft/s «- Ans. Since v A A 12 ft/s the cylinder slips on the conveyer. Ans. Ans: v A = 2 ft/s —> v B = 10 ft/s *— The cylinder slips. 718 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 86 . As the cord unravels from the wheel’s inner hub, the wheel is rotating at w = 2 rad/s at the instant shown. Determine the velocities of points A and B. SOLUTION r B/ic = 5 + 2 = 7 in. r a/ic ~ V^ 2 + 5 2 = \/7S) in. v B = b)r B / IC = 2(7) = 14 in./s !• v A = to r A /ic = 2( V29) = 10.8 in./s d = tairl Q) = 21 - 8 ° v Ans: v B = 14 in./si v A = 10.8 in./s 6 = 21 . 8 ° ^ Ans. Ans. Ans. Va 719 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-87. If rod CD is rotating with an angular velocity (o CD = 4 rad/s, determine the angular velocities of rods AB and CB at the instant shown. SOLUTION Rotation About A Fixed Axis. For links AB and CD, the magnitudes velocities of C and D are v c = w CD r CD = 4(0.5) = 2.00 m/s t>B — ^AB^AB — w Ab({) And their direction are indicated in Fig. a and b. General Plane Motion. With the results of v c and x B , the IC for link BC can be located as shown in Fig. c. From the geometry of this figure, rc/ic = 0.4 tan 30° = 0.2309 m r B/IC = = 0.4619 m Then, the kinematics gives v c = ‘ABC r c/io 2.00 = u> BC { 0.2309) co BC = 8.6603 rad/s = 8.66 rad/s /) Ans. v B = u >BC r B/ic\ w /tft(l) = 8.6603(0.4619) co AB = 4.00 rad/s /) Ans. Ans: co B c = 8-66 rad/s t) w AB = 4.00 rad/s /) 720 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 88 . If bar AB has an angular velocity oj ab = 6 rad/s, determine the velocity of the slider block C at the instant shown. SOLUTION Kinematic Diagram: Since link AB is rotating about fixed point A, then \ B is always directed perpendicular to link AB and its magnitude is v B = (o AB r AB — 6(0.2) = 1.20 m/s. At the instant shown. \ B is directed with an angle 45° with the horizontal. Also, block C is moving horizontally due to the constraint of the guide. Instantaneous Center: The instantaneous center of zero velocity of bar BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from \ B and Vc ■ Using law of sine, we have r B/ic _ 0.5 sin 60° sin 45° r c/ic _ 0.5 sin 75° ~~ sin 45° r B /ic = 0.6124 m r C /ic — 0.6830 m IC The angular velocity of bar BC is given by v B _ 1.20 r B/ic 0.6124 1.960 rad/s Thus, the velocity of block C is v c = w BC r c/ic = 1-960(0.6830) = 1.34 m/s <— Ans. Ans: v c = 1.34 m/s <— 721 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-89. Show that if the rim of the wheel and its hub maintain contact with the three tracks as the wheel rolls, it is necessary that slipping occurs at the hub A if no slipping occurs at B. Under these conditions, what is the speed at A if the wheel has angular velocity to? SOLUTION Ans: v A = (o ( r 2 ~ tt) 722 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-90. Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point D at this instant. v B = 10 ft/s B SOLUTION 1.6 — x x 5 ~~ 10 5* = 16 — Wx x = 1.06667 ft “ = I^ =9 ' 375rad/S r IC - D = V(0.2667) 2 + (0.8) 2 - 2(0.2667)(0.8)cos 135° = 1.006 ft sin <f> _ sin 135° 0.2667 ~~ 1.006 0 = 10.80° v c = 0.2667(9.375) = 2.50 ft/s <- Ans. v D = 1.006(9.375) = 9.43 ft/s Ans. 6 = 45° + 10.80° = 55.8° M ■ iptXs Ans: v c = 2.50 ft/s <— v D = 9.43 ft/s 0 = 55.8° M 723 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-91. Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point E at this instant. v B = 10 ft/s B SOLUTION 1.6 — x _ x 5 ~ 10 5x = 16 — lOx x = 1.06667 ft 10 co =-= 9.375 rad/s 1.06667 ' Vc = <tt(f/c-c) = 9.375(1.06667 - 0.8) = 2.50 ft/s <— Ans. Ve ~ <*>(ric-E) = 9.375 V(0.8) 2 + (0.26667) 2 = 7.91 ft/s Ans. _ . 0.26667) a 0 = tan - = 18.4° A \ 0.8 / If ft* Ans: v c = 2.50 ft/s <— Vg = 7.91 ft/s e = 18.4° A 724 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-92. Member AB is rotating at co AB = 6 rad/s. Determine the velocity of point D and the angular velocity of members BPD and CD. SOLUTION Rotation About A Fixed Axis. For links AB and CD , the magnitudes of the velocities of B and D are v B = <0ABrAB = 6(0.2) = 1.20 m/s v D = co CD ( 0.2) And their directions are indicated in Figs, a and b. General Plane Motion. With the results of v /; and v fl , the 1C for member BPD can be located as show in Fig. c. From the geometry of this figure. r B/ic ~ r D/ic — 0.4 m Then, the kinematics gives v b 1-20 ■>. <*BPD = -= TwT = 3.00 rad/s fB/ic °- 4 v d = a) BPD r D/ic = (3.00)(0.4) = 1.20 m/s / Thus, v d = w c.d( 0-2); 1.2 = co CD (0.2 ) to C D = 6.00 rad/s *) Ans. Ans. Ans. Ans: w B pd = 3.00 rad/s J v D = 1.20 m/s i/ u>cd = 6.00 rad/s 725 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-93. Member AB is rotating at co AB = 6 rad/s. Determine the velocity of point P, and the angular velocity of member BPD. SOLUTION Rotation About A Fixed Axis. For links AB and CD, the magnitudes of the velocities of B and D are v B = co AB r AB = 6(0.2) = 1.20 m/s v D = (o CD ( 0.2) And their direction are indicated in Fig. a and b General Plane Motion. With the results of x B and x D , the 1C for member BPD can be located as shown in Fig. c. From the geometry of this figure r B /ic = 0.4 m tpjiQ = 0.25 + 0.2 tan 60° = 0.5964 m Then the kinematics give v b 1.20 t»BPD = -= TUT = 3 '°° rad / s ^ r B /lc 0.4 v P = cospoPp/ic = (3.00)(0.5964) = 1.7892 m/s = 1.79 m/s <— Ans. Ans. Ans: co B pd — 3.00 rad/s /) v P = 1.79 m/s <— 726 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-94. The cylinder B rolls on the fixed cylinder A without slipping. If connected bar CD is rotating with an angular velocity to CD = 5 rad/s, determine the angular velocity of cylinder B. Point C is a fixed point. SOLUTION v D = 5(0.4) = 2 m/s 2 a B = 0~3 = 6 ' 67 rad ^ S Ans: (o B = 6.67 rad/s 727 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-95. As the car travels forward at 80 ft/s on a wet road, due to slipping, the rear wheels have an angular velocity a> = 100 rad/s. Determine the speeds of points A, B, and C caused by the motion. SOLUTION v A = 0.6(100) = 60.0 ft/s ->• v c = 2.2(100) = 220 ft/s <- v B = 1.612(100) = 161 ft/s 603 °^ Ans. Ans. Ans. Ans: v A = 60.0 ft/s —» v c = 220 ft/s <— v B = 161 ft/s e = 60.3° 728 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-96. The pinion gear A rolls on the fixed gear rack B with an angular velocity w = 8 rad/s. Determine the velocity of the gear rack C. C SOLUTION Genera! Plane Motion. The location of IC for the gear is at the bottom of the gear where it meshes with gear rack B as shown in Fig. a. Thus, Vq — arc/ic = 8(0.3) = 2.40 m/s <— Ans. Ans: Vq = 2.40 m/s <— 729 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-97. If the hub gear H and ring gear R have angular velocities (o H = 5 rad/s and u> R = 20 rad/s, respectively, determine the angular velocity w s of the spur gear S and the angular velocity of its attached arm OA. SOLUTION 5 _ 0.75 0.1 — x x x = 0.01304 m 0.75 0.01304 57.5 rad/s !) v A = 57.5(0.05 - 0.01304) = 2.125 m/s 2.125 , . o>oa = 02 = 10.6 rad/s 3 Ans. Ans. Ans: co s = 57.5 rad/s/) co 0 a = 10.6 rad/s 5 730 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-98. The IC is at A. If the hub gear H has an angular velocity co H = 5 rad/s, determine the angular velocity of the ring gear R so that the arm OA attached to the spur gear S remains stationary (ioqa = 0)- What is the angular velocity of the spur gear? 0.75 ojc = -= 15.0 rad/s i 0.05 ' SOLUTION Ans. Tftus oh a - 7 £**v Ans: io s = 15.0 rad/s (o R = 3.00 rad/s 731 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-99. The crankshaft AB rotates at u> AB = 50 rad/s about the fixed axis through point A, and the disk at C is held fixed in its support at E. Determine the angular velocity of rod CD at the instant shown. SOLUTION r B/lC r F/1C 0.3 sin 30° 0.3 tan 30° 5 = 0.6 m = 0.5196 m to BF ~ = 8-333 rad/s v F = 8.333(0.5196) = 4.330 m/s 40 mm 300 mm = 50 rad/s Thus, m CD ~ 4.330 0.075 = 57.7 rad/s *) Ans. Ans: co CD = 57.7 rad/s/) 732 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 100 . Cylinder A rolls on the fixed cylinder B without slipping. If bar CD is rotating with an angular velocity of to CD = 3 rad/s, determine the angular velocity of A. SOLUTION Rotation About A Fixed Axis. The magnitude of the velocity of C is Vc = eiCDfDC = 3(0.4) = 1.20 m/s -> General Plane Motion. The IC for cylinder A is located at the bottom of the cylinder where it contacts with cylinder B , since no slipping occurs here, Fig. b. *>c = w a r C /ic\ 1-20 = 01 , 4 ( 0 . 2 ) to a = 6.00 rad/s/) Ans. Ans: co A = 6.00 rad/s /) 733 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 101 . The planet gear A is pin connected to the end of the link BC. If the link rotates about the fixed point B at 4 rad/s, determine the angular velocity of the ring gear R. The sun gear D is fixed from rotating. SOLUTION Gear A: v c = 4(225) = 900mm/s _ 900 _ Vr_ 10A ~ 75 “ 150 v R = 1800 mm/s Ring gear: 1800 "* = ^ = 4rad/S Ans. Ans: (o R = 4 rad/s 734 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 102 . Solve Prob. 16-101 if the sun gear D is rotating clockwise at co D = 5 rad/s while link BC rotates counterclockwise at a>BC = 4 rad/s. SOLUTION Gear A: v P = 5(150) = 750mm/s v c = 4(225) = 900mm/s x _ 75 — x 750 ~~ 900 x = 34.09 mm 750 (o = -= 22.0 rad/s 34.09 ' v R = [75 + (75 - 34.09)] (22) = 2550 mm/s Ring gear: 750 _ 2550 x x + 450 x = 187.5 mm 750 , , . w R = -= 4 rad/s j R 187.5 ' Ans. Ans: co R = 4 rad/s 735 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-103. 736 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-104. At a given instant the bottom A of the ladder has an acceleration a A = 4 ft/s 2 and velocity v A = 6 ft/s, both acting to the left. Determine the acceleration of the top of the ladder, B , and the ladder’s angular acceleration at this same instant. SOLUTION CO ^ = 0.75 rad/s 8 a B ~ a A + ( a B/A)n + ( a B/A)t a B = 4 + (0.75) 2 (16) + a( 16) i ^ 30“ tP 30“ b (<t) 0 = 4 + (0.75) 2 (16 )cos 30° - a(16)sin30° ( + 1) a B = 0 + (0.75) 2 (16)sin30° + a(16)cos30° Solving, a = 1.47 rad/s 2 Ans. ci B = 24.9 ft/s 2 ]. Ans. Also: = a A + “ x r B/A ~ ‘Jib/a — = — 4i + (ctk) X (16cos30°i + 16sin30°j) — (0.75) 2 (16 cos30°i + 16sin30°j) 0 = -4 - 8a - 7.794 — a B = 13.856a — 4.5 a = 1.47 rad/s 2 Ans. a B = 24.9 ft/s 2 i Ans. Ans: a = 1.47 rad/s 2 a B = 24.9 ft/s 2 i a = 1.47 rad/s 2 a B = 24.9 ft/s 2 i 737 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 738 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 106 . Member AB has the angular motions shown. Determine the velocity and acceleration of the slider block C at this instant. SOLUTION Rotation About A Fixed Axis. For member AB, refer to Fig. a. v B = ^ab^ab = 4(2) = 8 m/s «- a B = a AB X r AB — i0 AB r AB = (-5k) X (2j) - 4 2 (2j) = {101 - 32j } m/s 2 General Plane Motion. The IC for member BC can be located using x B and v f as shown in Fig. b. From the geometry of this figure 0 = tan- 1 ^ = 36.87° 0 = 90° - 0 = 53.13° Then r B/IC - 2 ——-= tan 53.13; r B /ic = 2.6667 m — = cos 53.13; r c:lic = 0.8333 m r c/ic The kinematics gives Vb = w BC r B/ic'i 8 = w bc ( 2 - 6667 ) io BC = 3.00 rad/s/) v c = <^BC r c/ic = 3.00(0.8333) = 2.50 m/s i/ Ans. Applying the relative acceleration equation by referring to Fig. c, a c = + a BC x r c/B ~ w BC r C/B -«c(|)i - «c(f)j = (10i - 32j) + a BC k X (-0.51 - 2j) - (3.00 2 ) (-0.51 - 2j) -^a c i - -a c j = (2 a BC + 14.5)1 + (-0.5a BC - 14)j Equating i and j components --a c — 2 a BC + 14.5 (1) 3 ——a c — — 0.5crg C — 14 (2) Solving Eqs. (1) and (2), a c = 12.969 m/s 2 = 13.0 m/s 2 i/ Ans. a BC = —12.4375 rad/s 2 = 12.4 rad/s 2 /) Ans. The negative sign indicates that a BC is directed in the opposite sense from what is shown in Fig. (c). % 6*0 ac = 13.0 m/s 2 / a BC = 12.4 rad/s 2 /) 739 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-107. At a given instant the roller A on the bar has the velocity and acceleration shown. Determine the velocity and acceleration of the roller B, and the bar’s angular velocity and angular acceleration at this instant. SOLUTION General Plane Motion. The IC of the bar can be located using y A and \ B as shown in Fig. a. From the geometry of this figure, r A/ic = r B/ic = 0-6 m Thus, the kinematics give Va = “> r A/ic l 4 = "(°- 6 ) to = 6.667 rad/s = 6.67 rad/s *) Ans. v B = cor B / IC = 6.667(0.6) = 4.00 m/s \ Ans. Applying the relative acceleration equation, by referring to Fig. b , = a A + a X r B/A - 0 / r ll/A a B cos 30°i — a B sin 30°j = —6j + (ak) X (0.6 sin 30°i — 0.6 cos 30°j) - (6.667 2 ) (0.6 sin 30°i - 0.6 cos 30°j) a /n ^ — a B i - -a B j = (0.3' V3a - 13.33)i + (0.3a + 17.09)j Equating i and j components, a B = 0.3V3a - 13.33 1 ~2°b = ^-3“ + 17.09 ( 1 ) ( 2 ) Solving Eqs. (1) and (2) a = -15.66 rad/s 2 = 15.7 rad/s 2 /) a B = -24.79 m/s 2 = 24.8m/s 2 \ The negative signs indicate that a and a B are directed in the senses that opposite to those shown in Fig. b Ans. Ans. it 6 © Ans: to = 6.67 rad/s *) v B = 4.00 m/s \ a = 15.7 rad/s 2 /) a B = 24.8 m/s 2 \ 740 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 108 . The rod is confined to move along the path due to the pins at its ends. At the instant shown, point A has the motion shown. Determine the velocity and acceleration of point B at this instant. SOLUTION Ds = v A + w X r B/A Vb j = 6i + (—tuk) X (—41 - 3j) 0 = 6 — 3co, co = 2 rad/s v B = 4w = 4(2) = 8 ft/s t a B = a A + a X t B/A - (o 1 t B/A 21.331 + (a B ) t j = -31 + «kX (-41 - 3j) - (-2) 2 (-41 - 3j) ( ) 21.33 = -3 + 3a + 16; a = 2.778 rad/s 2 (+T) (a B ), = —(2.778)(4) + 12 = 0.8889 ft/s 2 a B = V(21.33) 2 + (0.8889) 2 = 21.4 ft/s 2 fl = tan- ('= 2.39° A 21.33 ) v A = 6 ft/s a A = 3 ft/s 2 Ans. Ans. Ans. Ans: v B = 8 ft/s t a B = 21.4 ft/s 2 0 = 2.39° ^ 741 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-109. Member AB has the angular motions shown. Determine the angular velocity and angular acceleration of members CB and DC. SOLUTION Rotation About A Fixed Axis. For crank AB, refer to Fig. a. v B = ^AB^AB = 2(0.2) = 0.4 m/s «- a B = a AB X r AB ~ co AB r AB = (4k) X (0.2j) - 2 z (0.2j) = {—0.8i - 0.8j } m/s 2 For link CD, refer to Fig. b. VC = <* >CD r CD = <a C d( 0 - 1 ) 2 a C = a CD X r CD — <ACD r CD = ( _ “C£>k) X (-0.1]) - C0c D (- O.lj) = —0.la CD \ + 0.1co 2 CD j General Plane Motion. The 1C of link CD can be located using x B and \ c of which in this case is at infinity as indicated in Fig. c. Thus, r B n C = r C nc = 00 ■ Thus, kinematics gives Then "sc — Vb r B/IC 00 v c =v B ; co CD (0.1) = 0.4 (o CD = 4.00 rad/s /) Applying the relative acceleration equation by referring to Fig. d, a C = a B + a BC X r C/B ~ Msclc/B -0.1a CD i + 0.1(4.00 2 )j = (—0.8i - 0.8j) + (tr BC k) x (-0.45 sin 60°i —OAotcjji + 1.6j — (0.225ctgc — 0.8)i + (—0.8 — 0.3897ct£c)j m bc — 0 co CD = 4.00 rad/s /) a BC = 6.16 rad/s 2 /) a CD = 21.9 rad/s 2 /) 742 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 110 . The slider block has the motion shown. Determine the angular velocity and angular acceleration of the wheel at this instant. SOLUTION Rotation About A Fixed Axis. For wheel C, refer to Fig. a. t>A = C0 c r c = k>c(0-15) i a A = «C X r c - <4 fc a A = (« c k) X (—0.15i) - <4 ( 0.151) = 0.15 <4 i — 0.15acj General Plane Motion. The 1C for crank AB can be located using y A and \ B as shown in Fig. b. Flere r A /ic = 0.3 m r Bj i C = 0.4 m Then the kinematics gives t>B = t0 AB r B/iC 4 = w ab( 0-4) (o AB =10.0 rad/s t) v a = t0 AB r A/iC g>c(0.15) = 10.0(0.3) io c = 20.0 rad/s *) Ans. Applying the relative acceleration equation by referring to Fig. c, 2 a B = a A T <x AB X r B / A - (o AB r B / A 2i = 0.15(20.0 2 )i - 0.15tr c j + (ctAi?k) X (0.3i - 0.4j) — 10.0 2 (0.3i - 0.4j) 2i = (0.4ct AB + 30)i + (0.3 a AB — 0.15t*c + 40)j Equating i and j components, 2 = 0 .4a AB + 30; a AB = —70.0 rad/s 2 = 70.0 rad/s 2 J 0 = 0.3(—70.0) + 0.15a c + 40; a c = -126.67 rad/s 2 = 127 rad/s 2 Ans. The negative signs indicate that a c and a AB are directed in the sense that those shown in Fig. a and c. 743 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 111 . At a given instant the slider block A is moving to the right with the motion shown. Determine the angular acceleration of link AB and the acceleration of point B at this instant. v A = 4 m/s a A = 6 m/s 2 SOLUTION General Plane Motion. The IC of the link can be located using and v ;i , which in this case is at infinity as shown in Fig. a. Thus r A/IC = r B/IC = 00 Then the kinematics gives v A = oj r A/IC ; 4 = co (°o) &> = 0 v b = v a = 4 m/s Since B moves along a circular path, its acceleration will have tangential and normal Vg 4 2 components. Hence ( a B ) n = — = — = 8 m/s 2 r B 2 Applying the relative acceleration equation by referring to Fig. b , = a A + a X r B/A - io 2 r B/A (a B ) t i — 8j = 6i + (<rk) X (—2 cos 30°i — 2 sin 30°j) — 0 (a B ),i - 8j = (a + 6)i - V3oj Equating i and j componenets, — 8 = — V-?a; a = 8\/3 rad/s 2 = 4.62 rad/s 2 *) Ans. (a B ), = a + 6; (a B ), = 8V3 + 6 = 10.62 m/s 2 Thus, the magnitude of a B is a B = V (a g) 2 + (a B ) 2 = \/l0.62 2 + 8 2 = 13.30 m/s 2 = 13.3 m/s 2 And its direction is defined by 6 = tan - ( a s)t = tan 10.62 = 36.99° = 37.0° ^ Ans. Ans. 744 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 112 . Determine the angular acceleration of link CD if link AB has the angular velocity and angular acceleration shown. SOLUTION Rotation About A Fixed Axis. For link AB. refer to Fig. a. v B = ^ab^ab = 3(1) = 3.00 m/s i a B = a AB X r AB ~ ( °AB' t AB = (-6k) X (li) - 3 2 (li) = { — 9i - 6j}m/s For link CD, refer to Fig. b v c = w CD r DC = m cd(0-5) —* a C = a CD X r DC — bJ CD t DC = (ac/jk) x ( 0-5j) - to CB 2 (-0.5j) = 0.5a CB i + 0.5o>c»j General Plane Motion. The IC of link BC can be located using \ A and \ B as shown in Fig. c. Thus r B /ic = 0-5 m r C /ic = 1 m Then, the kinematics gives v b = C0 BC r B/ic> 3 = tu BC (0.5) co BC = 6.00 rad/s /) v c = w BC r c/tc> w cz)(0-5) = 6.00(1) (o CD = 12.0 rad/s *) Applying the relative acceleration equation by referring to Fig. d, a c = a B + tx B c X r C j B — o ) BC 2 r C / B 0-5« cz? i + 0.5(l2.0 2 )j = (—9i - 6j) + (~a BC k) X (-0.5i + j) —6.00 2 (—0.5i + j) 0-5aczj> + 72j = ( a BC + 9)i + (0.5 q: bc - 42)j 745 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-112. Continued Equating j components, 72 = (0.5 a BC - 42); a BC = 228 rad/s 2 2> Then i component gives 0.5a cz) = 228 + 9; a CD = 474rad/s z 5 Ans. (b) Ans: a CD = 474 rad/s 2 *) 746 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-113. The reel of rope has the angular motion shown. Determine the velocity and acceleration of point A at the instant shown. SOLUTION General Plane Motion. The IC of the reel is located as shown in Fig. a. Here, r A/IC = Vo.l 2 + 0.1 2 = 0.1414 m Then, the Kinematics give v A = cor A /i C = 3(0.1414) = 0.4243 m/s = 0.424 m/s ^5 45 ° Ans. Here a c = ar = 8(0.1) = 0.8 m/s 2 i. Applying the relative acceleration equation by referring to Fig. b. a A = a c + a X r A/c - or r A/c a a = -0-8j + (8k) X (-0.1J) - 3 2 (—O.lj) = {0.8i + O.lj} m/s 2 The magnitude of a A is a A = Vo.8 2 + 0.1 2 = 0.8062 m/s 2 = 0.806 m/s 2 Ans. And its direction is defined by 6 = tan 01 08 = 7.125° = 7.13° Ans. Ans: v A = 0.424 m/s e v = 45° ^ a A = 0.806 m/s 2 d a = 7.13° 747 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-114. The reel of rope has the angular motion shown. Determine the velocity and acceleration of point B at the instant shown. SOLUTION General Plane Motion. The IC of the reel is located as shown in Fig. a. Here, r B/FC = 0.2 m. Then the kinematics gives v B = <i>r B / lc = (3)(0.2) = 0.6 m/s i Ans. Here, a c = otr = 8(0.1) = 0.8 m/s 2 I. Applying the relative acceleration equation, a B = a C + a X r B/C — ^ r B/C sl b = —0.8j + (8k) X (—O.li) - 3 2 (—O.li) = { 0.9i - 1.6j } m/s 2 The magnitude of n B is a B = V0.9 2 + (-1.6) 2 = 1.8358 m/s 2 = 1.84 m/s 2 Ans. And its direction is defined by 6 = tan L6 09 = 60.64° = 60.6° ^ Ans. Ans: v B = 0.6 m/s J. a B = 1.84 m/s 2 0 = 60.6° ^ 748 ) 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-115. A cord is wrapped around the inner spool of the gear. If it is pulled with a constant velocity v, determine the velocities and accelerations of points A and B. The gear rolls on the fixed gear rack. SOLUTION Velocity analysis: v V B = < 0 r BjIC = -(4 r) = 4v —* v A = (or A/IC = -(V(2 r) 2 + (2 r) 2 ) = 2\Zlv ^45° Ans. Ans. Acceleration equation: From Example 16-3, Since a G = 0, a = 0 tB/G = 2r j r A/G = —2r i a B = a G + “ X r B / G — (o 2 r B / G v\ 2 2v 2 = 0 + 0 - I -) (2rj) = j 2v l a B i Ans. *a = a G + a X r A/G \ 2 “> 2 r A /c , v Y 2v l = 0 + 0 - I -) (—2ri) = — 2v 2 a a - Ans. 'V I£ ^•5© Ans: v B = 4v^> v A = 2V2d 6 = 45° 2v 2 | a B — 1 a A 2v 2 749 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 116 . The disk has an angular acceleration a = 8 rad/s 2 and angular velocity a> = 3 rad/s at the instant shown. If it does not slip at A, determine the acceleration of point B. SOLUTION General Plane Motion. Since the disk rolls without slipping, a Q = ar = 8(0.5) = 4 m/s 2 . Applying the relative acceleration equation by referring to Fig. a, = a G + a X r B/0 - co 2 r B/0 a B = (-4i) + (8k) X (0.5 sin 45°i - 0.5cos45°j) - 3 2 (0.5 sin 45°i - 0.5 cos 45°j) a B = { —4.354i + 6.010j } m/s 2 Thus, the magnitude of a B is a B = V(-4.354) 2 + 6.010 2 = 7.4215 m/s 2 = 7.42 m/s 2 Ans. And its direction is given by . / 6 . 010 \ 6 = tan -1 - = 54.08° = 54.1° ^ Ans. \4.354 J w = 3 rad/s a = 8 rad/s 2 (JX) Ans: a B = 7.42 m/s 2 0 = 54.1° 750 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 117 . The disk has an angular acceleration a = 8rad/s 2 and angular velocity <u = 3 rad/s at the instant shown. If it does not slip at A, determine the acceleration of point C. SOLUTION General Plane Motion. Since the disk rolls without slipping, a Q = ar = 8(0.5) = 4 m/s 2 Applying the relative acceleration equation by referring to Fig. a, a c = a 0 + a X r c/0 - co 2 r qo a c = (-4i) + (8k) X (0.5 sin 45°i + 0.5 cos 45°j) — 3 2 (0.5 sin 45°i + 0.5 cos 45°j) a c = { — 10.0104i - 0.3536J } m/s 2 Thus, the magnitude of ac is a c = V(— 10.0104) 2 + (-0.3536) 2 = 10.017 m/s 2 = 10.0 m/s 2 Ans. And its direction is defined by 6 = tair ’ {^4} = 2 -° 23 ° = 2 -° 2 ° ^ AnS ' in = 3 rad/s a = 8 rad/s 2 fa) Ans: a c = 10.0 m/s 2 6 = 2 . 02 ° 5 =" 751 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 118 . A single pulley having both an inner and outer rim is pin- connected to the block at A. As cord CF unwinds from the inner rim of the pulley with the motion shown, cord DE unwinds from the outer rim. Determine the angular acceleration of the pulley and the acceleration of the block at the instant shown. SOLUTION Velocity Analysis: The angular velocity of the pulley can be obtained by using the method of instantaneous center of zero velocity. Since the pulley rotates without slipping about point D , i.e: v D = 0, then point D is the location of the instantaneous center. v F - (or C / IC 2 = w(0.075) to = 26.67 rad/s Acceleration Equation: The angular acceleration of the gear can be obtained by analyzing the angular motion points C and D. Applying Eq. 16-18 with r C /D — {—0.075jj m, we have = 3 B + a X r c/d ~ W ~ r c/D —3i + 17.78j = -35.56j + (-ak) X (-0.075j) - 26.67 2 (-0.075j) —3i + 17.78j = -0.075 ai + 17.78j Equating i and j components, we have —3 = —0.075a a = 40.0 rad/s 2 Ans. 17.78 = 17.78 ( Check !) The acceleration of point A can be obtained by analyzing the angular motion points A and D. Applying Eq. 16-18 with r A / D = {—0.05j) m. we have a A = 3d + a X r a/d ~ M 2 r A /D = —35.56j + (-40.0k) X (-0.05j) - 26.67 2 (-0.05j) = {—2.00i) m/s 2 Thus, a A = 2.00 m/s 2 <— Ans. ~55'5(, m/ ji- radfi 40-0 *■ Ans: a = 40.0 rad/s 2 a a = 2.00 m/s 2 <— 752 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 119 . The wheel rolls without slipping such that at the instant shown it has an angular velocity to and angular acceleration a. Determine the velocity and acceleration of point B on the rod at this instant. SOLUTION a A ~ + @AIO (Pin) t - 1 - (°a)x — om ~ w 2 a ( a A)y = aa a B ~ a A + a BIA(Pin) a B = aa — (o z a + 2 a(a') 0--a, + 2„«'(A=) + 2«(^=) (f a' = 0.577 a - 0.1925oi z a B = 1.58m — l.lloL) 2 a Ans. Ans: v B = 1.58&JO a B = 1.58 aa — 753 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 120 . The collar is moving downward with the motion shown. Determine the angular velocity and angular acceleration of the gear at the instant shown as it rolls along the fixed gear rack. SOLUTION General Plane Motion. For gear C, the location of its IC is indicate in Fig. a. Thus v b = M c r B/(iC) 1 = "cfO-05) I (1) The IC of link AB can be located using V 4 and \ B , which in this case is at infinity. Thus m ab ~ Va T4/(/C) 2 2 _ 00 = 0 Then v b = v a = 2 m/s i Substitute the result of v B into Eq. (1) 2 = co c (0.05) co c = 40.0 rad/s *) Ans. Applying the relative acceleration equation to gear C, Fig. c, with a o = a c r c = “c(0-2) L a fl = a O + a C X r B/0 ~ b) C r B/0 a B = -a c (0.2)j + (a c k) X (0.15i) - 40.0 2 (0.15i) = -240i - 0.05tr c j For link AB, Fig. d , a fl = a A + a AB X r B/A ~ m AB r B/A — 240i — 0.05^^ = (—3j) + (cr^gk) X (0.5 sin 60°i — 0.5 cos 60°j) — 0 —240i — 0.05crc = 0.25^51 + (0.25\/3 a^ B — 3)j Equating i and j components —240 = 0.25 a AB ; a AB = — 960rad/s 2 = 960rad/s 2 /) -0.05rr c = (0.25\/3)(-960) - 3; a c = 8373.84 rad/s 2 = 8374 rad/s 2 !) Ans. Ans: co c = 40.0 rad/s a c = 8374 rad/s 2 *) 754 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 755 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 756 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-122. Continued General Plane Motion. The IC of link BC can be located using \ B and Vc as shown in Fig. c. From the geometry of this figure, r B/ic = 0.5 cos 60° = 0.25 m tqic = 0.5 sin 60° = 0.25\/3 m Then kinematics gives v B = " BC r B/lc\ 0.9 = (o g C (0.25) co BC = 3.60 rad/s Vc — " >BC r c/ic> <*>cd( 0-2) = (3.60) (0.25\/3) co C D = 7.7942 rad/s = 7.79 rad/s *) Ans. Applying the relative acceleration equation by referring to Fig. d, a c = a B + a BC X r C / B — o) BC rc/B -0.2a CD i ~ 0.2(7.7942 z )j = (-2.70i - 2.40j) + (-a sc k) X (-0.5 cos 60°i - 0.5sin60°j) — 3.60 2 (—0.5 cos 60°i — 0.5 sin 60°j) ~0.2a CD i - 12.15j = (0.54 - 0.25\/3a B c)i + (3.2118 + 0.25t* BC )j Equating the j components, -12.15 = 3.2118 + 0.25a BC \ a BC = -61.45 rad/s 2 = 61.45 rad/s 2 *) Then the i component gives -0.2 a CD = 0.54 - 0.25\/3(—61.4474); a CD = -135.74 rad/s 2 = 136 rad/s 2 ^ Ans. Ans: "CD “CD 7.79 rad/s *) 136 rad/s 2 757 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 758 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-123. Continued General Plane Motion. The IC of link BC can be located using \ B and Vc as shown in Fig. c. From the geometry of this figure, r B/ic = 0.5 cos 60° = 0.25 m tqic = 0.5 sin 60° = 0.25A/3 m Then kinematics gives v B = BC r B/lc\ 0.9 = (o g C (0.25) co BC = 3.60 rad/s/) Vc — 0J Bc r cjic:\ “cb(0-2) = (3.60) (0.25\/3) coco = 7.7942 rad/s = 7.79 rad/s *) Ans. Applying the relative acceleration equation by referring to Fig. d, a c = + a BC X r C / B — m B c*c/b -0.2a CD i ~ 0.2(7.7942 z )j = (-2.70i - 2.40j) + (-a sc k) X (-0.5 cos 60°i - 0.5sin60°j) — 3.60 2 (—0.5 cos 60°i — 0.5 sin 60°j) ~0.2a CD i - 12.15j = (0.54 - 0.25V3a BC )i + (3.2118 + 0.25a BC )j Equating the j components, -12.15 = 3.2118 + 0.25 a BC ; a BC = -61.45 rad/s 2 = 61.45 rad/s 2 !) Then the i component gives ~0.2oi cd = 0.54 - 0.25\/3(—61.4474); a CD = -135.74 rad/s 2 = 136 rad/s 2 Ans. From the angular motion of CD. Vc = w cd(®-2) = (7.7942)(0.2) = 1.559 m/s = 1.56 m/s <— Ans. a c = —0.2(—135.74)i - 12.15/ = {27.15i - 12.15/} m/s The magnitude of a c is a c = V27.15 2 + (-12.15) 2 = 29.74 m/s 2 = 29.7 m/s 2 Ans. And its direction is defined by /12 15\ e = tai M ^75) = 24110 = 24 - 10 ^ Ans - Ans: v c = 1-56 m/s «— a c = 29.7 m/s 2 e = 24.1 0 ^ 759 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-124. The disk rolls without slipping such that it has an angular acceleration of a = 4 rad/s 2 and angular velocity of <u = 2 rad/s at the instant shown. Determine the acceleration of points A and B on the link and the link’s angular acceleration at this instant. Assume point A lies on the periphery of the disk, 150 mm from C. SOLUTION The 1C is at so co = 0. *a = a c + a X *A)C - u 2 r A /c *a = 0.6i + (-4k) X (0.15j) - (2) 2 (0.15j) a A = (1.20i — 0.6j)m/s 2 a A = V(1.20) 2 + (-0.6) 2 = 1.34 m/s 2 * - ,an iW - 26 ' 6 "^ A Ans. Ans. Ans: ci A = 1.34 m/s 2 0 = 26.6° 760 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-125. The ends of the bar AB are confined to move along the paths shown. At a given instant, A has a velocity of v A — 4 ft/s and an acceleration of a A = 7 ft/s 2 . Determine the angular velocity and angular acceleration of AB at this instant. SOLUTION Vfl = Va + \ B/A v B = 4+ m(4.788) 30°^ 1 M 5U1 . (-±>) -v B cos 30° = 0 — <u(4.788) sin 51.21° (+ t) «gsin30° = -4 + <n(4.788) cos 51.21° v B = 20.39 ft/s 30° 5 ^ co = 4.73 rad/s *) Ans. a B ~ a A + a B/A a, + 207.9 = 7 + 107.2 + 4.788(a) 30°5^ 60° 2^ 1 ^ 51.21° M 51.21° ( ) a t cos 30° + 207.9 cos 60° = 0 + 107.2 cos 51.21° + 4.788a(sin 51.21°) (+ T) a, sin 30° - 207.9 sin 60° = - 7 - 107.2 sin 51.21° + 4.788a(cos 51.21°) a,(0.866) - 3.732ct = -36.78 a, (0.5) - 3a = 89.49 a t = —607 ft/s 2 a = —131 rad/s 2 = 131 rad/s 2 /) Ans. Also: \ B = \ A = co X r B/A —cos 30°i + yg sin 30°j = —4j + (<uk) X (3i + 3.732j) — v B cos 30° = —tu(3.732) Ugsin30° = —4 + tu(3) co = 4.73 rad/st) Ans. v B = 20.39 ft/s a B - a A ~ M^B/A + a X r B j A (—a t cos30°i + a, sin30°j) + (-207.9 cos 60°i - 207.9 sin 60°j) = -7j - (4.732) 2 (3i + 3.732j) + (ak) X (3i + 3.732j) -a, cos 30° - 207.9 cos 60° = -(4.732) 2 (3) - a(3.732) a, sin30° - 207.9 sin 60° = -7 -(4.732) 2 (3.732) + a(3) Ans: co = 4.73 rad/s /) a = 131 rad/s 2 /) a, = —607 ft/s 2 a = —131 rad/s 2 = 131 rad/s 2 /) Ans. 761 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 126 . The mechanism produces intermittent motion of link AS. If the sprocket S is turning with an angular acceleration a s = 2 rad/s 2 and has an angular velocity a> s = 6 rad/s at the instant shown, determine the angular velocity and angular acceleration of link AB at this instant. The sprocket S is mounted on a shaft which is separate from a collinear shaft attached to AB at A. The pin at C is attached to one of the chain links such that it moves vertically downward. SOLUTION co BC = = 4.950 rad/s 0.2121 ' v B = (4.95)(0.2898) = 1.434 m/s 1.435 , , A a> AB = = 7.1722 rad/s = 7.17 rad/s /> a c = a s r s = 2(0.175) = 0.350 m/s 2 ( a B)/i + ( a fi)( = a C + ( a B/c)n + ( a B/c)f (7.172) 2 (0.2) 3o° ty + 1 , IT O 1 _ — 0.350 _ 1 - + (4.949 ) 2 (0.15) 15° + a BC (0.15) 15°^ ■bo° j -(10.29) cos 30° - (a B ),sin 30° = 0 - (4.949) 2 (0.15) sin 15° - a BC (0.15) cos 15° + T) -(10.29) sin 30° + (a B ),cos30° = -0.350 - (4.949) 2 (0.15) cos 15° + a BC (0.15) sin 15° a BC = 70.8 rad/s 2 . Hence, ( a B ) t = 4.61 m/s 2 a AB ~ fall r B/A 4-61 , = 23.1 rad/s 2 !) Ans. Also, Vc — co s r s = 6(0.175) = 1.05 m/s 1 Tfi = V C + (Osc^Tb/c v B sin30°i — Vgcos30°j = — 1.05j + (— &j BC k)X (0.15 sin 15°i + 0.15 cosl5°j) 3 ^ J ygsin 30° = 0 + cu BC (0.15) cos 15° (+ t) — v B cos 30° = -1.05 - m BC (0.15) sin 15° v B = 1.434 m/s, a> BC = 4.950 rad/s v„ 1.434 w AB — -= -= 7.172 = 7.17 rad/s/) Ans. r B/A 0.2 762 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-126. Continued a B — a c + a B c X r B / C — co’rs/c (a AB k) X (0.2 cos 30°i + 0.2 sin 30°j) - (7.172) 2 (0.2 cos 30°i + 0.2sin30°j) = -(2)(0.175)j + (track) X (0.15 sin 15°i + 0.15cosl5°j) - (4.950) 2 (0.15 sin 15°i + 0.15cosl5°j) -ol ab {0.1) - 8.9108 = -0.1449q: bc - 0.9512 (+t) ^(0.1732) - 5.143 = -0.350+ 0.0388o BC -3.550 ol ab — 23.1 rad/s 2 /) Ans. a BC = 70.8 rad/s 2 Ans: ^ab &AB 7.17 rad/s /) 23.1 rad/s 2 /) 763 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-127. The slider block moves with a velocity of v B = 5 ft/s and an acceleration of a B = 3ft/s 2 . Determine the angular acceleration of rod AB at the instant shown. SOLUTION Angular Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, r B/ic ~ 2 sin 30° = 1 ft >'a/ic = 2 cos 30° = 1.732 ft Thus, mab ~ v B r B/IC 5 I 5 rad/s Then v A = " ABfA/ic = 5(1.732) = 8.660 ft/s Acceleration and Angular Acceleration: Since point A travels along the circular slot, the normal component of its acceleration has a magnitude of ( a A ) n = —— = _ go f t / s 2 an( j j s ejected towards the center of the circular p 1.5 slot. The tangential component is directed along the tangent of the slot. Applying the relative acceleration equation and referring to Fig. b, a A — a B + a AB X r A/B — w AB r A/B 50i — (a A ) t j = 3i + (a AB k) X (—2 cos 30°i + 2 sin 30°j) — 5 2 (—2 cos 30° i + 2 sin 30°j) 50i - (a^), j = (46.30 - a AB )i + {1.112a AB + 25)j Equating the i components. 50 = 46.30 — a AB a AB = —3.70 rad/s 2 = 3.70 rad/s 2 /) Ans. Ans: a AB = 3.70 rad/s 2 /) 764 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 16 - 128 . The slider block moves with a velocity of v B — 5 ft/s and an acceleration of a B = 3 ft/s 2 . Determine the acceleration of A at the instant shown. SOLUTION Angualr Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, r B /ic — 2 sin 30° = 1 ft Thus, Then r A/ic ~ 2 cos 30° = 1.732 ft Vb 5 e ,, w ab = -= t = 5 rad/s r B/IC 1 V A = w ab r a/ic = 5 ( 1 . 732 ) = 8.660 ft/s Acceleration and Angular Acceleration: Since point A travels along the circular slot, the normal component of its acceleration has a magnitude of ( a A\n = ~ = 50 ft/s 2 and is directed towards the center of the circular v ' p 1.5 slot. The tangential component is directed along the tangent of the slot. Applying the relative acceleration equation and referring to Fig. b, a A ~ a B + a AB X r A/B ~ ^AB r A/B 50i — (a A ) t j — 3i + (a^k) X (—2cos30°i + 2 sin 30°j)—5 2 (—2 cos 30°i + 2sin30°j) 50i - (a A ) t j = (46.30 - a AB )i - (1.732a A b + 25)j Equating the i and j components, 50 = 46.30— a AB -{a A )t = -(l-732 a AB + 25) Solving, a AB = —3.70 rad/s 2 (a A ), = 18.59 ft/s 2 i Thus, the magnitude of a A is a A — V (a A )t 2 + ( a A)n 2 ~ \/l8.59 2 + 50 2 = 53.3 ft/s 2 Ans. and its direction is 6 = tan \ a A)t ( ) n = tan -1 ^ 1 ^^ ) = 20.4° ^ Ans. Ans: a A = 53.3 ft/s 2 8 = 20.4° ^ 765 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-129. At the instant shown, ball B is rolling along the slot in the disk with a velocity of 600 mm/s and an acceleration of 150 mm/s 2 , both measured relative to the disk and directed away from O. If at the same instant the disk has the angular velocity and angular acceleration shown, determine the velocity and acceleration of the ball at this instant. SOLUTION Kinematic Equations: v B = v a + 11 X r B /o + ( V B /o)xyz ^ x t b/o + O, X (II X r B / 0 ) + 2H X ( v B /o)xy Z + v 0 = 0 a 0 = 0 fl = { 6k } rad/s fl = {3k} rad/s 2 *b/o = |0.4i) m 0 B/o)xyz = {0.6i}m/s (a B/o)xy Z = {0.15i}m/s 2 Substitute the date into Eqs. (1) and (2) yields: v B = 0 + (6k) X (0.4i) + (0.6i) = {0.6i + 2.4j}m/s Ans. a B = 0 + (3k) X (0.4i) + (6k) X [(6k) X (0.4i)[ + 2(6k) X (0.6i) + (0.151) = { — 14.2i + 8.40j}m/s 2 Ans. Ans: Vg = [0.61 + 2.4j) m/s a B = {—14.2i + 8.40j) m/s 2 z 766 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-130. The crane’s telescopic boom rotates with the angular velocity and angular acceleration shown. At the same instant, the boom is extending with a constant speed of 0.5 ft/s, measured relative to the boom. Determine the magnitudes of the velocity and acceleration of point B at this instant. SOLUTION Reference Frames: The xyz rotating reference frame is attached to boom AB and coincides with the XY fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xy frame with respect to the XY frame is y A ~ a A ~ 0 w ab — [ — 0.02k] rad/s w AB = ol — [—0.01k] rad/s 2 For the motion of point B with respect to the xyz frame, we have *B/A = [60j] ft (Vrel)x^ = [0.5j] ft/s (a, Velocity: Applying the relative velocity equation, \ B = y A + <*AB X *B/A + (v rel )xyz = 0 + (-0.02k) X (60j) + 0.5j = [1.2i + 0.5j] ft / s Thus, the magnitude of v B , Fig. £>, is v B = Vl.2 2 + 0.5 2 = 1.30 ft/s Acceleration: Applying the relative acceleration equation, a fl = a A + <AAB X r B/A + M AB X ( W AB X r B/A) + 2 (A A B X (Uel) = 0 + (-0.01k) X (60j) + (-0.02k) X [(-0.02k) X (60j)] + = [0.62i - 0.024 j] ft/s 2 Thus, the magnitude of a B , Fig. c, is a B = V0.62 2 + (-0.024) 2 = 0.6204 ft/s 2 ib Ans. X- Ans: v B = 1.30 ft/s a B = 0.6204 ft/s 2 767 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 768 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 769 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-133. Water leaves the impeller of the centrifugal pump with a velocity of 25 m/s and acceleration of 30 m/s 2 , both measured relative to the impeller along the blade line AB. Determine the velocity and acceleration of a water particle at A as it leaves the impeller at the instant shown. The impeller rotates with a constant angular velocity of oj = 15 rad/s. SOLUTION Reference Frame: The xyz rotating reference frame is attached to the impeller and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is v G = a 0 = 0 w = [—15k] rad/s io = 0 The motion of point A with respect to the xyz frame is *a/o = [°-3j] m (VreOxyz = (—25cos30°i + 25 sin 30°j) = [—21.65i + 12.5j] m/s ( a rei)x^ = (—30cos30°i + 30sin30°j) = [—25.98i + 15j] m/s 2 Velocity: Applying the relative velocity equation. v A = v 0 + cu X r A/G + (v rel ) x ^ = 0 + (-15k) X (0.3j) + (—21.651 + 12.5j) = [—17.21 + 12.5j] m/s Ans. Acceleration: Applying the relative acceleration equation, y a A = a 0 + (o X r A/G + <o X (a> X r A/G ) + 2 co X (v re i) xyz + (a rel ) xyz = 0 + (-15k) X [(-15k) X (0.3j)] + 2(—15k) X (—21.65i + 12.5j) + (-25.98i + 15j) = [349i + 59711 m/s 2 Ans. Ans: v A = {—17.2i + 12.5j) m/s a A = [349i + 597j] m/s 2 770 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-134. Block A , which is attached to a cord, moves along the slot of a horizontal forked rod. At the instant shown, the cord is pulled down through the hole at O with an acceleration of 4 m/s 2 and its velocity is 2 m/s. Determine the acceleration of the block at this instant. The rod rotates about O with a constant angular velocity a> = 4 rad/s. SOLUTION Motion of moving reference. v Q = 0 a G = 0 n = 4k n = o Motion of A with respect to moving reference. *A/o = O.li Va/o = -2i * a/o = ~4i Thus, »a = a 0 + n X r A/ o + n X (n X t A/ o) + 20 X (v A/0 )^ + (a A/o)xyz = 0 + 0 + (4k) X (4k X O.li) + 2(4k X (-2i)) - 4i a A = {—5.60i — 16j) m/s 2 Ans. Ans: a A = { —5.60i — 16j } m/s 2 771 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-135. Rod AB rotates counterclockwise with a constant angular velocity w = 3 rad/s. Determine the velocity of point C located on the double collar when 0 = 30°. The collar consists of two pin-connected slider blocks which are constrained to move along the circular path and the rod AB. SOLUTION r = 2(0.4 cos 30°) = 0.6928 m r C /A = 0.6928 cos 30°i + 0.6928 sin 30°j = { 0.600i + 0.3464j } m \ c = —0.866u c i + 0.5u c j V C = V A + O X r c/A + (VQA)xyz — 0.866u c i + 0.5u c j = 0 + (3k) X (0.600i + 0.3464j) + {v c / A cos 30°i + v c / A sin 30°j) —0.866w c i + O.Sryj = 0 — 1.039i + 1.80j + Q.%66v c / A i + O.Sric/^j —0.866-y c = -1.039 + 0.866u c/A 0.5w c = 1.80 + 0.5v c / a v c = 2.40 m/s Ans. v c /A = -1.20 m/s Ans: v c = 2.40 m/s 0 = 60° 772 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-136. Rod AB rotates counterclockwise with a constant angular velocity co = 3 rad/s. Determine the velocity and acceleration of point C located on the double collar when 8 = 45°. The collar consists of two pin-connected slider blocks which are constrained to move along the circular path and the rod AB. SOLUTION r C /A = { 0.400i + 0.400j } v>: = ~v c i V C = \ A + flX r c / A + 0 C/A)xyz —u c i = 0 + (3k) X (0.400i + 0.400j) + (v c / A cos 45°i + u c ^sin45°j) —u c i = 0 - 1.20i + 1.20j + 0.101v c / A i + 0.707v c / A j —v c = -1.20 + 0.707v c / A 0 — 1.20 + 0.707v c / a v c = 2.40 m/s Ans. v c / A = -1-697 m/s a c = a A + fi X t C /a + fl X (H X r C / A ) + 2fl X ( Vc/ A )x yz + {*c/A)xyz (2.40) 2 -(a c ) t i - Q4 j = 0 + 0 + 3k X [3k X (0.41 + 0.4j)] + 2(3k) X [0.707(-1.697)i + 0.707(—1.697)j] + OJOlac/A* 0.707flc/yij — (a c )t i — 14.40J = 0 + 0 — 3.60i — 3.60j + 7.20i — 7.20j + 0.707 a C /A^ + 0.707 a c /A j — (^c)i = —3.60 + 7.20 + 0.707 ac/A -14.40 = -3.60 - 7.20 + 0.707 a c/A ac/A = -5.09 m/s 2 («c)r = 0 Thus, (2-40) 2 , a C = (a C )n = Q4 = 14-4 m/s 2 a c = { — 14.4j } m/s 2 Ans. Ans: v c = 2.40 m/s ac = { —14.4j } m/s 2 773 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-137. Particles B and A move along the parabolic and circular paths, respectively. If B has a velocity of 7 m/s in the direction shown and its speed is increasing at 4 m/s 2 , while A has a velocity of 8 m/s in the direction shown and its speed is decreasing at 6 m/s 2 , determine the relative velocity and relative acceleration of B with respect to A. SOLUTION 8 , n = - = 8 rad/s 2 , fi = {—8k} rad/s v B = v A + O, X r B/A + (v B/A ) xyz 7i = —8i + (8k) X (2j) + (\ B/A ) xyz 7i = — 8i - 161 + (y H/A ) xy z (VB/A)x yz = {31.01} m/s n = Y = 6 rad/s 2 , tl = (-6k) rad/s 2 (VA? (8) {a A ) n = —— = — = 64 m/s 2 i y = x dy dx db dx 2 = 2x = 0 c=0 = 2 -IT d~y dx 2 [1 + Op _ 1 ~2 ~ 2 (7) 2 (a B ) n = — = 98 m/s 2 t Ans. a B = a A + h X r B/A + O X (O X r B/A ) + 2 fix (\ B/A ) xyz + (a B /A)xyz 4i + 98j = 6i - 64j + (-6k) X (2j) + (8k) X (8k X 2j) + 2(8k) X (311) + (*, 1/A ) xyz 4i + 98j = 6i - 64j + 121 - 128j + 496j + (a B/A ) xyz (a B /A)xyz = { 14.01 - 206j) m/s 2 Ans. Ans: (VB/A)xyz = {31.01} m/s (* 11 /a )xyz = {—14.01 - 206j) m/s 2 774 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16 - 138 . Collar B moves to the left with a speed of 5 m/s, which is increasing at a constant rate of 1.5 m/s 2 , relative to the hoop, while the hoop rotates with the angular velocity and angular acceleration shown. Determine the magnitudes of the velocity and acceleration of the collar at this instant. SOLUTION Reference Frames: The xyz rotating reference frame is attached to the hoop and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is v a ~ a A ~ 0 oj = [—6k] rad/s co = a = [—3k] rad/s 2 450 mm For the motion of collar B with respect to the xyz frame, *b/a = [—0.45j] m (Vrel)xyz = [~ 5i ] m / s The normal components of (a rel ) xy2 is [{a rt .\} xyx \ n = (areOxyz = [—l-5i + 125j] m/s Velocity: Applying the relative velocity equation, Ij = It + " X r B /A + (Vrel)x^ = 0 + (-6k) X (— 0.45j) + (-51) = [—7.7i] m/s Thus, (^rel)x 5^ 02 v B = 7.7 m/s <— Ans. Acceleration: Applying the relative acceleration equation, “ X tg/ A + w X (w X tg/ A ) + 2(0 X (v re |) t>l j, + (a r el)xj .2 = 0 + (-3k) X (—0.45j) + (-6k) X [(-6k) X (—0.45j)] + 2(-6k) X (-5i) + (—1.5i + 125j) = [-2.851 + 201.2j] m/s 2 Thus, the magnitude of a B is therefore a B = \Z2.85 2 + 201.2 2 = 201 m/s 2 Ans. Ans: v B = 7.7 m/s a B = 201 m/s 2 775 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-139. Block D of the mechanism is confined to move within the slot of member CB. If link AD is rotating at a constant rate of (o AD = 4 rad/s, determine the angular velocity and angular acceleration of member CB at the instant shown. SOLUTION The fixed and rotating X — Y and x — y coordinate system are set to coincide with origin at C as shown in Fig. a. Here the x — y coordinate system is attached to member CB. Thus Motion of moving Reference v c = 0 a c = 0 H = io C u = ojajk a = a ai = Motion of Block D with respect to moving Reference r D /c = { 0.3i} m 0 r D/C.) xyz ( y D/C ) xyz * ( a r>/c) xyz ( a n/c) xyz * The Motions of Block D in the fixed frame are, \ D = to A / D X r D / A = (4k) X (0.2 sin 30°i + 0.2cos30°j) = {— 0.4\/3i + 0.4j } m/s a z) = <x AD X I D/A - io A J(t D/A ) = 0 - 4 2 (0.2 sin 30°i + 0.2 cos 30°j) = { —1.6i - 1.6\/3j} m/s 2 Applying the relative velocity equation, \ D = v c + a X r D/c + (y D/c ) xyz -0.4V3i + 0.4j = 0 + (w CB k) X (0.3i) + (v D / C ) xyz i - 0.4\/3i + 0.4j = ( v D /c) xyz i + 0.3 w CB j Equating i and j components, ( v D/c) xyz = -0.4 V3 m/s 0.4 = 0.3 co CB ; co CB = 1.3333 rad/s = 1.33 rad/s/) Ans. Applying the relative acceleration equation, a Z5 = a C + & X r D/C + a X (fl X t D /c) + 2a X (\D/c) XyZ + ( a D/c) XyZ —1.6i - 1.6V3j = 0 + (a CD k) X (0.3i) + (1.3333k) X (1.3333k X 0.3i) + 2(1.3333k) X (-0.4\/3i) + (a D/c ) xy A 1.6i - 1.6V3J = [(a D/c ) xyz - 0.5333]i + (0.3 a CD - 1.8475)j Equating i and j components 1.6 = [( a D/c ) xyz - 0.5333]; (a D/c ) xyz = 2.1333 m/s 2 -1.6V3 = 0.3 a CD - 1.8475; a CD = -3.0792 rad/s 2 = 3.08 rad/s 2 Ans. Ans: mcb = 1-33 rad/s *) a CD = 3.08 rad/s 2 /) 776 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-140. At the instant shown rod AB has an angular velocity w AB = 4 rad/s and an angular acceleration a AB = 2 rad/s 2 . Determine the angular velocity and angular acceleration of rod CD at this instant. The collar at C is pin connected to CD and slides freely along AB. SOLUTION Coordinate Axes: The origin of both the fixed and moving frames of reference are located at point A. The x, y, z moving frame is attached to and rotate with rod AB since collar C slides along rod AB. Kinematic Equation: Applying Eqs. 16-24 and 16-27, we have v c = V A + n X r c/A + ( Vc/A)xyz (!) a c = 3^4 + fl X r c/A + D X (D X r c/A ) + 2D, X (\ C/A )xyz + (»»C/Ayz ( 2 ) Motion of moving reference v A = 0 *A = 0 fl = 4k rad/s fl = 2k rad/s 2 Motion of C with respect to moving reference r c/A = {0.75i}m (Ac/a) xyz (vc/a) xyz * (a C /a) xyz ( a C/A ) xyz * The velocity and acceleration of collar C can be determined using Eqs. 16-9 and 16-14 with r C /D ~ {—0.5 cos 30°i — 0.5 sin 30°j }m = {—0.4330i — 0.250j) m. v c = "cb x r c/D = ~ co cd^ x (—0.4330i — 0.250j) = —0.250<u Ci ji + 0.4330&) C£ j 2 a c = 01 cd X r c/d ~ mcd^c/d = —a CD k X (—0.4330i - 0.250j) - co 2 CD (-0A330i - 0.250j) = (0.4330&>cd - 0.250 a CD ) i + (0.4330tr CB + 0.250&)^)j Substitute the above data into Eq.(l) yields v C = Va + to X r c/A + (yC/ a) xyz — 0.250 cocd i = 0 4- 4k X 0.75i + ivc/A)xyz * —0.250(u c/) i + 0.4330&) C £ij = (v c ^ A ) xyz i + 3.00j Equating i and j components and solve, we have ( v c/A)xy Z = -1-732 m/s co C D — 6.928 rad/s = 6.93 rad/s Ans. Ill © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-140. Continued Substitute the above data into Eq.(2) yields »C = a A + & x r C/A + O X (O X r c / A ) + 2 fix ( Vc/A)xyz + i a C//dxyz [0.4330 (6.928 2 ) - 0.250 <x CD ]i + [0A330a CD + 0.250(6.928 2 )]j = 0 + 2k X 0.75i + 4k X (4k X 0.75i) + 2 (4k) X (—1.732i) + (a c/A ) xyz i (20.78 - 0.250a cfl )i + (0.4330 a CD + 12)j = [(a c/A )x y z ~ 12.0]i - 12.36j Equating i and j components, we have (■ a c/A)xyz = 46 - 85 m /s 2 a CD = —56.2 rad/s 2 = 56.2 rad/s 2 t) Ans. Ans: oi C d = 6.93 rad/s a C D = 56.2 rad/s 2 778 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-141. The collar C is pinned to rod CD while it slides on rod AB. If rod AB has an angular velocity of 2 rad/s and an angular acceleration of 8 rad/s 2 , both acting counterclockwise, determine the angular velocity and the angular acceleration of rod CD at the instant shown. SOLUTION The fixed and rotating X — Y and x — y coordinate systems are set to coincide with origin at A as shown in Fig. a. Here, the x — y coordinate system is attached to link AC. Thus, Motion of moving Reference Motion of collar C with respect to moving Reference *c/a = {l-5i} m (vc/a) xyz (Vc/a) xyz* (»c/a) xyz ( 8 c/a) xyz * Va = 0 a A = 0 SI = io AB = {2k} rad/s SI = a AB = {8k} rad/s 2 The motions of collar C in the fixed system are v c = "CD x r c/D = ( _<M C£ik) X (—i) = cocdJ &C = a CD x r C/D ~ m cd r c/D = ( _ “c/z)k) x ( _ i) ~ "CZ)( - ') = "coi + “czj Applying the relative velocity equation, Vc = Va + 1lx r c/ A + (\c/A)xyz "Cflj = 0 + (2k) X (1.5i) = (v C / A )xyz i "CZ)j = ( V c/A)xyz i + 3j Equating i and j components ( v C/A)xyz ~ 0 ai C D = 3.00 rad/s {) Ans. Applying the relative acceleration equation, a c = a A + fi X r C /A + SI X (SI X r C /a) + 2H X (v C /a)^^ + ( <d c/A)xyz 3.00 2 i + a CD \ = 0 + (8k) X (1.5i) + (2k) X (2k X 1.5i) + 2(2k) X 0 + ( a c/A ) xyz i 9i + a C D\ — [ {cic/A)xyz - 6 ] i + 12j Equating i and j components, 9 ( < -C'/A)xyz b, (a( a }xyz 15 m/s a C o — 12.0 rad/s 2 {) Ans. Ans: w CD = 3.00 rad/s {) a CD = 12.0 rad/s 2 {) 779 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-142. At the instant shown, the robotic arm AB is rotating counterclockwise at co = 5 rad/s and has an angular acceleration a = 2 rad/s 2 . Simultaneously, the grip BC is rotating counterclockwise at w' = 6 rad/s and a 1 = 2 rad/s 2 , both measured relative to a fixed reference. Determine the velocity and acceleration of the object held at the grip C. SOLUTION v c = \ B + D X r c/B + (v C /b) xyz ~ a fl + X r C/B + n X (D X t C / B ) + 2D, X (yC/Ii)xyz + ( *C/B)xyz ( 2 ) Motion of moving reference D = {6k} rad/s fi = {2k} rad/s 2 Motion of B: Motion of C with respect to moving reference t C /b — {0.125 cos 15°i + 0.125 sin 15°j) m iyc/B)xyz = 0 C a C/B)xyz = 0 \ B = CO X r B/A = (5k) X (0.3 cos 30°i + 0.3 sin 30°j) = {—0.751 + 1.2990j} m/s a B = a X r B / A -co 2 r B j A = (2k) X (0.3 cos 30°i + 0.3 sin 30°j) - (5) 2 (0.3 cos 30°i + 0.3 sin 30°j) = {—6.7952i - 3.2304j) m/s 2 Substitute the data into Eqs. (1) and (2) yields: v c = (—0.75i + 1.2990j) + (6k) X (0.125 cos 15°i + 0.125 sin 15°j) + 0 = {—0.944i + 2.02j) m/s Ans. a c = (-6.79527i - 3.2304j) + (2k) X (0.125 cos 15°i + 0.125 sin 15°j) + (6k) X [(6k) X (0.125 cos 15°i + 0.125 sin 15°j)] + 0 + 0 = {— 11.2i - 4.15j) m/s 2 Ans. Ans: v c = {-0.944i + 2.02j} m/s a c = { —11.2i - 4.15j} m/s 2 780 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-143. Peg B on the gear slides freely along the slot in link AB. If the gear’s center O moves with the velocity and acceleration shown, determine the angular velocity and angular acceleration of the link at this instant. SOLUTION Gear Motion: The 1C of the gear is located at the point where the gear and the gear rack mesh, Fig. a. Thus, 150 mm v 0 = 3 m/s a a = 1.5 m/s' -► Then, „ —— = 20 rad/s r o/ic 0.15 v B = cor B /i C = 20(0.3) = 6m/s- a o 1-5 Since the gear rolls on the gear rack, a = — = ^ ^ = 10 rad/s. By referring to Fig. b, a B ~ + a x t b/o ~ w ~*g/o (a B ) t i - (a B ) n j = 1.5i + (-10k) X 0.15j - 20 2 (0.15j) {a B )t i - (a B )n j = 3i - 60j Thus, (a B ), = 3 m/s 2 (a B ) n = 60m/s 2 015 * Reference Frame: The x'y’z’ rotating reference frame is attached to link AB and coincides with the XYZ fixed reference frame, Figs, c and d. Thus, v /; and a /( with respect to the XYZ frame is \ B = [6 sin 30°i — 6 cos 30° j] = [3i — 5.196j]m/s a B = (3 sin 30° — 60 cos 30°)i + (—3 cos 30° — 60 sin 30°)j = [—50.46i - 32.60j] m/s 2 For motion of the x'y’z 1 frame with reference to the XYZ reference frame, v a = a A ~ 0 co AB = ~w AB \i w AB = —a AB k For the motion of point B with respect to the x'y’z' frame is t B /A [0.6j ]m (v r el) \ ! y’Z (Uel)x'yV j i^relix'y'z' (^rel)x'y'e' J Velocity: Applying the relative velocity equation, V B = + m AB X t B /A + (y T el)x'y'z' 3i - 5.196j = 0 + (-«M B k) X (0.6j) + (v Tel ) x y z ' j 3i - 5.196j = 0.6cu AB i + (v lel ) x y Z 'j Equating the i and j components yields 3 = 0.6co ab (o ab = 5 rad/s Ans. (Vrd)xyz' = -5.196 m/s Acceleration: Applying the relative acceleration equation. a B = a A + m ab X r B / A + co AB X ( io AB X r B / A ) + 2 w AB X (v re i) x y/ + (a re i ) x y z ' -50.46i - 32.60j = 0 + (-a AB k) X (0.6j) + (-5k) X [(-5k) X (0.6j)] + 2(-5k) X (-5.196J) + (u re ,) x yyj -50.46i - 32.60j = (0.6 a AB - 51.96)1 + [(a re] ) x y, - 15]j Equating the i components, -50.46 = 0.6a AB - 51.96 a AB = 2.5 rad/s 2 Ans. Ans: co AB = 5 rad/s /) a AB = 2.5 rad/s 2 /) 781 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-144. The cars on the amusement-park ride rotate around the axle at A with a constant angular velocity co A /f = 2 rad/s, measured relative to the frame AB. At the same time the frame rotates around the main axle support at B with a constant angular velocity cof = 1 rad/s. Determine the velocity and acceleration of the passenger at C at the instant shown. SOLUTION Vc = V A + fi X r c/A + (y C /A)xyz (!) ac = a A + & x r c/A + fix (fix r C / A ) + 2fi X ( v C / A )xy Z + (*c/a) xyz (2) Motion of moving refernce Motion of C with respect to moving reference y X fi = {3k} rad/s fi = 0 Motion of A: \ A = co X r A/B r c/A = {'-8i} ft (yc/jdxyz = o (« C/A)xyz = 0 f = (lk) X (—15cos30°i + 15sin30°]) = {—7.5i - 12.99]} ft/s a a = “ X r A / B - (o 2 t a/b = 0 - (1) 2 (—15cos30°i + 15sin30°j) = {12.991 - 7.5j} ft/s 2 Substitute the data into Eqs.(l) and (2) yields: v c = (—7.51 - 12.99j) + (3k) X (-8i) + 0 = {—7.51 — 37.0J} ft/s Ans. a c = (12.99i - 7.5j) + 0 + (3k) X [(3k) X (-8i) + 0 + 0] = {85.01 — 7.5jj ft/s 2 Ans. Ans: v c = {-7.51 - 37.0]} ft/s a c = {85.Oi - 7.5j} ft/s 2 782 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-145. A ride in an amusement park consists of a rotating arm AB having a constant angular velocity co AB = 2 rad/s about point A and a car mounted at the end of the arm which has a constant angular velocity to' = {—0.5k} rad/s, measured relative to the arm. At the instant shown, determine the velocity and acceleration of the passenger at C. SOLUTION t B /A = (10 cos 30°i + 10 sin 30°j) = {8.661 + 5j} ft Vg = h) AB X t B / A = 2k X (8.66i + 5j) = {—lO.Oi + 17.32j} ft/s &B ~ a AB X r B/A ~ U 2 AB r B/A = 0 - (2) 2 (8.66i + 5j) = {—34.641 - 20j) ft/s 2 n = (2 - 0.5)k = 1.5k v c = v B + n X r c/B + (v C /b) xyz = — 10.0i + 17.32j + 1.5k X (-2j) + 0 = {—7.00i + 17.3j} ft/s a c = a B + O X r c/B + O X (O X t c/B ) + 211 X (v c / B )xyz + (*C/B)xyz = —34.64i - 20j + 0 + (1.5k) X (1.5k) X (-2j) + 0 + 0 = {—34.6i — 15.5j) ft/s 2 Ans. Ans. Ans: v c = {—7.00i + 17.3j) ft/s a c = {—34.6i — 15.5j) ft/s 2 783 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-146. A ride in an amusement park consists of a rotating arm AB that has an angular acceleration of a AB = 1 rad/s 2 when co AB = 2 rad/s at the instant shown. Also at this instant the car mounted at the end of the arm has an angular acceleration of a = {—0.6k} rad/s 2 and angular velocity of w' = {—0.5k} rad/s, measured relative to the arm. Determine the velocity and acceleration of the passenger C at this instant. SOLUTION i B / A = (10 cos 30°i + 10sin30°j) = (8.66i + 5j) ft \ B = w ab x r B/A = 2k X (8.66i + 5j) = {—lO.Oi + 17.32j) ft/s 2 a fl ~ a AB X r B/A ~ ^AB^B/A = (lk) X (8.66i + 5j) - (2) 2 (8.66i + 5j) = {—39.64i - 11.34j) ft/s 2 D = (2-0.5)k = 1.5k fl = (1 - 0.6)k = 0.4k v c = + D X r c/B + ( y cjB ) xyz = — 10.0i + 17.32j + 1.5k X (-2j) + 0 = {—7.00i + 17.3J} ft/s Ans. a C = a B + ^ X r C/B + h X (fl X 1 'c/b) + 2fl X {\c/B)xyz + {^C/B)xyz = —39.64i - 11.34j + (0.4k) X (—2j) + (1.5k) X (1.5k) X (-2j) + 0 + 0 = {— 38.8i — 6.84j) ft/s 2 Ans. Ans: v c = {—7.001 + 17.3j} ft/s a c = {—38.8i - 6.84j) ft/s 2 784 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-147. If the slider block C is fixed to the disk that has a constant counterclockwise angular velocity of 4 rad/s, determine the angular velocity and angular acceleration of the slotted arm AB at the instant shown. SOLUTION v c = ~(4)(60) sin 30°i - 4(60) cos 30°j = -120i - 207.85j a c = (4) 2 (60) sin 60°i - (4) 2 (60) cos 60°j = 831.38i - 480j Thus, V C = + fl X I C / A + (\ C /A)xyz — 120i - 207.85j = 0 + {oj AB k) X (180j) - v c/A \ -120 = -180 o> AB oj A b = 0.667 rad/s t) Ans. -207.85 = -v c/A v c/ A = 207.85 mm/s a c = a A + ft X r c/i + ft X (fl X r c/i ) + 211 X (v c / A )xyz + (a c/ A )x yz 831.38i - 480j = 0 + (o^k) X (180j) + (0.667k) X [(0.667k) X (180j)] + 2(0.667k) X (-207.85j)-a c/A j 831.38i — 480j = —180 cr^i — 80j + 277.13i — a C / A j 831.38 = -180 ou b + 277.13 a AB = -3.08 Thus, a AB = 3.08 rad/s 2 Ans. —480 = —80 — a C j A a c/ A = 400 mm/s 2 Ans: oj ab = 0.667 rad/s 5 a AB = 3.08 rad/s 2 ^ 785 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-148. At the instant shown, car A travels with a speed of 25 m/s, which is decreasing at a constant rate of 2 m/s 2 , while car C travels with a speed of 15 m/s, which is increasing at a constant rate of 3 m/s 2 . Determine the velocity and acceleration of car A with respect to car C. SOLUTION 200 m Reference Frame: The xyz rotating reference frame is attached to car C and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since car C moves along the circular road, its normal component of acceleration is v 2 (a c )„ = -= = 0.9 m/s 2 . Thus, the motion of car C with respect to the XYZ frame is v c = —15 cos 45°i - 15 sin 45°j = [-10.607i - 10.607j] m/s a c = (-0.9 cos 45° - 3 cos 45°)i + (0.9 sin 45° - 3sin45°)j = [—2.7581 - 1.485j] m/s 2 Also, the angular velocity and angular acceleration of the xyz reference frame is v r 15 co = — = -= 0.06 rad/s co = [—0.06k] rad/s p 250 co = ^ = 0.012 rad/s 2 co = [—0.012k] rad/s 2 25 m/s 2 m/s 2 The velocity and accdeleration of car A with respect to the XYZ frame is *a = [25j] m/s a A = [-2j] m/s 2 From the geometry shown in Fig. a, x a/c = —250 sin 45°i - (450 - 250cos45°)j = [—176.781 - 273.22j] m Velocity: Applying the relative velocity equation, VA = V C + CO X r A /c + (Vrel)*^ 25j = (—10.607i - 10.607j) + (-0.06k) X (-176.781 - 273.22j) + (v m] ) xyz 25j = —271 + (v re i) xyz (Vrei)xyz = [27i + 25j] m/s Ans. Acceleration: Applying the relative acceleration equation, a A - a c + co X r a/c + co X (co X r A /c) + 2co X (v re i)x K + (a re i ) xyz —2j = (—2.758i - 1.485j) + (-0.012k) X (-176.78i - 273.22j) + (-0.06k) X [(-0.06k) X (—176.78i - 273.22j)] + 2(-0.06k) X (27i + 25j) + (a re i ) xyz —2j = —2.4i - 1.62j + (a rd )x yz (a re iW = [2.4i - 0.38j] m/s 2 Ans. Ans: (VreOxyz = [27i + 25j] m/s (a re i )xyz = [2-41 - 0.38j] m/s 2 786 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-149. At the instant shown, car B travels with a speed of 15 m/s, which is increasing at a constant rate of 2 m/s 2 , while car C travels with a speed of 15 m/s, which is increasing at a constant rate of 3 m/s 2 . Determine the velocity and acceleration of car B with respect to car C. SOLUTION Reference Frame: The xyz rotating reference frame is attached to C and coincides with the XYZ fixed reference frame at the instant considered. Fig. a. Since B and C move along the circular road, their normal components of acceleration are V 2 152 y 2 152 (a B )n = — = — = 0.9 m/s 2 and ( a c )„ = — = — = 0.9 m/s 2 . Thus, the p 250 p 25U motion of cars B and C with respect to the XYZ frame are D? = [—15i] m/s y c = [— 15 cos 45°i — 15sin45°j] = [— 10.607i — 10.607j] m/s a B = [— 2i + 0.9j] m/s 2 a c = (—0.9 cos 45°—3 cos 45°)i + (0.9 sin 45°-3 sin 45°)j = [— 2.758i - 1.485 j] m/s 2 Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are (*) OJ vc _ 15 p 250 . _ ( a c)t _ 3 " ~~ p ~ 250 0.06 rad/s 0.012 rad/s 2 w = [—0.06k] rad/s ix> = [—0.012k] rad/s 2 From the geometry shown in Fig. a, r B/c = —250 sin 45°i - (250 - 250 cos 45°)j = [—176.78i - 73.22 j] m Velocity: Applying the relative velocity equation, Vs = V C + to X r B/c + (\ Ie l)xyz —15i = (—10.6071 - 10.607j) + (-0.06k) X (—176.781 15i 15i T (v re i) X y. (^rel)x_yz = 0 Acceleration: Applying the relative acceleration equation, = a c + to X r B/c + w X (a, X r B/c ) + 2co X (v rel ) x> , z + (a rel ) xyz —2i + 0.9j = (—2.758i - 1.485J) + (-0.012k) X (-176.78i - 73.22j) + (—0.06k) X [(-0.06k) X (—176.78i - 73.22j)] + 2(-0.06k) X 0 + (a re] ) xyz —2i + 0.9j = —3i + 0.9j + (a rel ) x} , z (a rei)xvz = [!i] m /s 2 An s. - 73.22j) + (v re i) X3 , z Ans. Ans: (Vrel)xyz = 0 (a rel)ryz = {li]m/s 2 787 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 16-150. The two-link mechanism serves to amplify angular motion. Link AB has a pin at B which is confined to move within the slot of link CD. If at the instant shown, AB (input) has an angular velocity of co AB = 2.5 rad/s, determine the angular velocity of CD (output) at this instant. SOLUTION Tba 0-15 m sin 120° ~~ sin 45° i B a — 0.1837 m v c = 0 a c = 0 11 — (U£, c k fl = — ff/xk *b/c = {-0.15 i) m ( v b/c) xyz ( V B/C) xyz* ( a B/c) xyz ( a B/C ) xyz* \ B = co AB X r B / A = (—2.5k) X (—0.1837 cos 15°i + 0.1837 sin 15°j) = |0.1189i + 0.4436j) m/s y B ~ V C + ^ X r B/C + ( v B/c) xyz 0.11891 + 0.4436] = 0 + (-w BC k) X (—0.151) + (v B/c ) xyz i 0.1189i + 0.4436j = (v B/c ) xyz i + 0.15m DC j Solving: (v B /c)xyz = °-H 89 m / s (o DC = 2.96 rad/s /) Ans. Ans: to D c = 2.96 rad/s 788 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 789 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *16-152. The Geneva mechanism is used in a packaging system to convert constant angular motion into intermittent angular motion. The star wheel A makes one sixth of a revolution for each full revolution of the driving wheel B and the attached guide C. To do this, pin P, which is attached to B, slides into one of the radial slots of A, thereby turning wheel A , and then exits the slot. If B has a constant angular velocity of co B = 4 rad/s, determine (o A and a A of wheel A at the instant shown. SOLUTION The circular path of motion of P has a radius of r P = 4 tan 30° = 2.309 in. Thus, \ P = —4(2.309)j = — 9.238j a p = —(4) 2 (2.309)i = —36.951 Thus, \p = \ A + n, X r P/A + (\ P/A ) xyz —9.238j = 0 + (a^k) X (4j) - v P/A j Solving, = 0 Vp/A — 9.238 in./s cog = 4 rad/s Ans. a f — a A + & x r P/A + O X (fl X i 'p/a) + 211 X (\p/ A ) xyz + (ap/ A ) xyz —36.95i = 0 + (a^k) X (4j) + 0 + 0 — a P / A j Solving, —36.95 = —4 a A a A = 9.24 rad/s 2 Ans. a P/A ~ 0 Ans: (o A = 0 a A = 9.24 rad/s 2 *) 790 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-1. Determine the moment of inertia I y for the slender rod. The rod’s density p and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m. SOLUTION Thus, /v = x 2 dm l x 2 (p A dx) = tpA / 3 m = p A l Iy = - m l 2 z Ans. Z- Ans: 791 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-2. The solid cylinder has an outer radius R , height h. and is made from a material having a density that varies from its center as p = k + ar 2 , where k and a are constants. Determine the mass of the cylinder and its moment of inertia about the z axis. SOLUTION Consider a shell element of radius r and mass dm = p dV = p(2tt r dr)h f R m = / (k + «r 2 )(2ir r dr)h „ ,,kR 2 aR\ m = 2irh{— + —) cilv m — tt h R 2 (k H——) dl h h h h r 2 dm = r 2 (p)(2n r dr)h r 2 (k + ar 2 )(2iT r dr) h 2irh / (k r + a r 5 ) dr r k R 4 aR 6 2rr/t[—+ —_ TT hR\, 2 aR 2 ^ —[* + —] Ans. Ans. Ans: m = irhR 2 ^k 4 -— ^ 4 it hR 4 2 k + 2 aR 2 3 792 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 793 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-4. The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration k x . The density of the material is p = 5 Mg/m 3 . SOLUTION dm = p 77 y 1 dx = p it (50x) dx f ^ - OMn 4 = / A >' 2 dm = p 7r P 7T 50/ ~2 50 2 6 1 u 1200 50 x (t t p (50x)} dx .3 (200) 3 0 / /•200 dm = / 77 p (50x) dx 1200 = p 77 (50) 1 - — r z 2 = P ™ ( y )(200) JO ,2 k x = J I -±= J— (200) = 57.7 mm V m V 3 V Ans. Ans: k x = 57.7 mm 794 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-5. Determine the radius of gyration k x of the body. The specific weight of the material is y = 380 lb/ft 3 . SOLUTION dm d I x h m k x pdV = pTry 2 dx 2 (dm) y 2 = -it, py 4 dx 1 — 7 Tpx 413 dx = 86.17p ! irpx 2/3 dx = 60.32p / 86.17p V 60.32p 1.20 in. y Ans. Ans: k x = 1.20 in. 795 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 796 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-7. The frustum is formed by rotating the shaded area around the x axis. Determine the moment of inertia I x and express the result in terms of the total mass m of the frustum. The frustum has a constant density p. SOLUTION , /b 2 , 2 b 2 dm — p dV = pny dx = pul —yxr 4- x + b jdx dl x = — dmy 2 = —pTry 4 dx (■a iA 6 b‘ H-2 4 -X 2 + 4 b 4 -x + b 4 )c lx fl 2 a 4 A 4h 4 o 6 b 4 7 4 b 4 l* + a + — y~X~ + a a 31 64 = —pirab f , f (b 2 2 21,2 u2\j 7 ,2 / dm = pTT / I— yX“ H- x + b )dx — — pirab Jm J0 3 03 I x = — mb 2 x 70 Ans. Ans: 93 I x = — mb 2 x 70 797 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-8. The hemisphere is formed by rotating the shaded area around the y axis. Determine the moment of inertia l y and express the result in terms of the total mass m of the hemisphere. The material has a constant density p. SOLUTION m = p dV = p / 7 t x 2 dy = pir I (r 2 — y 2 )dy Jv Jo Jo p7T r y ~ 3 y -p7T r !y = / l (dm) x z = I 77 x*dy = I ( r z - y l f dy p7T r y - 3 »• r + 5 Jo 4p77 15 Thus, I v = — m r ^ x 2 + y 1 = r 2 1 ^- Ans. V Ans: 798 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-9. Determine the moment of inertia of the homogeneous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x-y plane and having a thickness dz. SOLUTION dV = bx dz — b(a)( 1 — —) dz h dl y = dl y + ( dm)[(^) 2 + z 2 ] 1 7 x 2 = — dm(x 2 ) + dm(—) + dmz 2 = dm( y + z 2 ) = [&(fl)(l-J)<fe](p)[y(l-J) 2 + z 2 } !y = abpj^ [y (~y^) 3 + Z ^ 1 ~ = ab Plftf (h 4 - \ h 4 + ~ \ h ^ + \ (| ; ' 4 “ \ b 4 )] \ = — abhp(a 2 + h 2 ) m = pV = — abhp z X Thus, h m (a 2 + h 2 ) Ans. Ans: YYl / <5 I y = y ( fl2 + b 2 ) 799 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 10 . The pendulum consists of a 4-kg circular plate and a 2-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O. SOLUTION Using the parallel axis theorem by referring to Fig. a, Iq 2(/g + md 2 ) ^( 2 )( 2 2 ) + 2 ( 1 2 ) + i(4)(0.5 2 ) + 4(2.5 2 ) = 28.17 kg-m 2 Thus, the radius of gyration is ko = /~28T7~ V 4 + 2 = 2.167 m = 2.17 m Ans: k 0 = 2.17 m 800 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 11 . The assembly is made of the slender rods that have a mass per unit length of 3 kg/m. Determine the mass moment of inertia of the assembly about an axis perpendicular to the page and passing through point O. 0.8 m O 0.4 m — SOLUTION Using the parallel axis theorem by referring to Fig. a , 4> = S(/ G + md 2 ) = {~[3(1.2)] (l.2 2 ) + [ 3(1.2)] (0.2 2 )} + {^[ 3(0.4)] (0.4 2 ) + [ 3(0.4) ](0.8 2 )} = 1.36 kg • m 2 (a ) Ans: I 0 = 1.36 kg - m 2 801 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 802 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 13 . The wheel consists of a thin ring having a mass of 10 kg and four spokes made from slender rods, each having a mass of 2 kg. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A. SOLUTION 7 a = /„ + md 3 1 ( 4 )( 1) 2 .12 = 7.67 kg • m 2 + 10(0.5) 2 + 18(0.5) 2 Ans. Ans: I A = 7.67 kg -m 2 803 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 14 . If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A. SOLUTION Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 — 1) = 3 ft and a center of mass located at a distance of f 1 + — ^ ft = 2.5 ft from point O can be grouped as segment (2). Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. In — my (4 2 ) + 8 = 84.94 slug • ft 2 1 / 20 U\322 ( 3 2 ) + 20 322 (2.5 2 ) 15 32 2 (l 2 ) The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem I a = I 0 + md 2 , where m = —+ 8) ) 4——— = 8.5404 slug and d = 4 ft. J? ° 32.2 \32.2 / 32.2 5 Thus, I A = 84.94 + 8.5404(4 2 ) = 221.58 slug • ft 2 = 222 slug • ft 2 Ans. (fX) Ans: I A = 222 slug • ft 2 804 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 15 . Determine the moment of inertia about an axis perpendicular to the page and passing through the pin at O. The thin plate has a hole in its center. Its thickness is 50 mm, and the material has a density p = 50 kg/m 3 . SOLUTION I G = ^[50(1.4)(1.4)(0.05)][(1.4) 2 + (1.4) 2 ] - i[50(7r)(0.15) 2 (0.05)](0.15) 2 = 1.5987 kg • m 2 I 0 — Ig + ™i 2 m = 50(1.4)(1.4)(0.05) - 50(tt)(0.15) 2 (0.05) = Iq = 1.5987 + 4.7233(1.4 sin 45°) 2 = 6.23 kg • 4.7233 kg m 2 Ans. Ans: I Q = 6.23 kg • m 2 805 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 16 . Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg/m 2 . SOLUTION Composite Parts: The plate can be subdivided into two segments as shown in Fig. a. Since segment (2) is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated. Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed as m, = -7r(0.2 2 )(20) = 0.8 tt kg and m 2 = (0.2)(0.2)(20) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem. Iq = E/ g + md 2 = | (0.8ir)(0.2 2 ) + 0.8 tt(0.2 2 ) = 0.113 kg-nr 2 Ans. — ( 0 . 8 )( 0 . 2 2 + 0 . 2 2 ) + 0 . 8 ( 0 . 2 2 ) Ans: 1 0 = 0.113 kg • m 2 806 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 17 . Determine the location y of the center of mass G of the assembly and then calculate the moment of inertia about an axis perpendicular to the page and passing through G. The block has a mass of 3 kg and the semicylinder has a mass of 5 kg. SOLUTION Moment inertia of the semicylinder about its center of mass: (Ic)cyc = \mR 2 - Xym Xm 0.2 - 4R\ 2 _ 3-7r / 4(0.2) = 0.3199mfc 3-77 (5) + 0.35(3) 5 + 3 = 0.2032 m = 0.203 m Ans. I G = 0.3199(5)(0.2) 2 + 5 = 0.230 kg • m 2 0.2032 - 0.2 - 4(0.2) \ 377 + -(3)(0.3 2 + 0.4 2 ) + 3(0.35 - 0.2032) 2 Ans. Ans: I G = 0.230 kg-m 2 807 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 18 . Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point O. The block has a mass of 3 kg, and the semicylinder has a mass of 5 kg. SOLUTION {Io)cyi = \' nR2 - = 0.3199 mR 2 ( 4(0 2)\ 2 1 Io = 0.3199(5)(0.2) 2 + 5 0.2 -+ — (3)((0.3) 2 + (0.4) 2 ) + 3(0.350) 2 V 37t / 12 = 0.560 kg • m 2 Ans. Also from the solution to Prob. 17-22, 7 0 = 7 G + md 2 = 0.230 + 8(0.2032) 2 = 0.560 kg • m 2 Ans. Ans: 7 0 = 0.560 kg • m 2 808 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 19 . Determine the moment of inertia of the wheel about an axis which is perpendicular to the page and passes through the center of mass G. The material has a specific weight y = 90 lb/ft 3 . SOLUTION _ l ~~ 2 - 4 90 — (rr)(2) 2 (0.25) 90 (2) 2 + » (2.5) 2 32.2 ( 7r )(2) 2 (l) ( 2) 2 - 4 1/ 90 2\32.2 (tt)(0.25) 2 (0.25) (0.25) 2 90 32.2 (tt)(0.25) 2 (0.25) (l ) 2 = 118.25 = 118 slug-ft 2 Ans. Ans: I G = 118 slug - ft 2 809 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 20 . Determine the moment of inertia of the wheel about an axis which is perpendicular to the page and passes through point O. The material has a specific weight y = 90 lb/ft 3 . SOLUTION 90 m = 322 [ lr( ^ )2(0 ' 25) + 17 {C 2 - 5 )^ 1 ) - ( 2 ) 2 (!)} ~ 4 tt( 0.25) 2 (0.25)] = 27.99 slug From the solution to Prob. 17-18, I G = 118.25 slug • ft 2 I 0 = 118.25 + 27.99(2.5) 2 = 293 slug • ft 2 Ans. 0.25 ft Ans: I 0 = 293 slug • ft 2 810 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 21 . The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. SOLUTION y 2 ym 2 m 1(3) + 2.25(5) 3 + 5 = 1.781m = 1.78 m Ans. Iq — 2/ g + md 2 = ^(3)(2) 2 + 3(1.781 - l) 2 + -^(5)(0.5 2 + l 2 ) + 5(2.25 - 1.781) 2 = 4.45 kg • m 2 Ans. Ans: y = 1.78 m I G = 4.45 kg-m 2 811 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 22 . Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a destiny of p = 7.85 Mg/m 3 . SOLUTION m c = 7.85(10 3 )((0.05)tt( 0.01) 2 ) = 0.1233 kg m p = 7.85(10 3 )((0.03)(0.180)(0.02)) = 0.8478 kg 1 p 1=2 -(0.1233)(0.01) z + (0.1233)(0.06) z ^ (0.8478) ((0.03) 2 + (0.180) 2 ) = 0.00325 kg • m 2 = 3.25 g • m 2 Ans. Ans: 4 = 3.25 g • m 2 812 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 23 . Determine the moment of inertia of the overhung crank about the x' axis. The material is steel having a destiny of p = 7.85 Mg/m 3 . SOLUTION m c = 7.85(10 3 )((0.05)t 7(0.01) 2 ) = 0.1233 kg m p = 7.85(l0 3 )((0.03)(0.180)(0.02)) = 0.8478 kg Ir' = — (0.1233)(0.01) 2 — (0.1233)(0.02) 2 + (0.1233)(0.120) ^ (0.8478)((0.03) 2 + (0.180) 2 ) + (0.8478)(0.06) 2 12 = 0.00719 kg-m 2 = 7.19 g-m 2 2 Ans. Ans: = 7.19 g-m 2 813 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 24 . The door has a weight of 200 lb and a center of gravity at G. Determine how far the door moves in 2 s, starting from rest, if a man pushes on it at C with a horizontal force F = 30 lb. Also, find the vertical reactions at the rollers A and B. SOLUTION 2F* = m(a G ) x \ 30 = (^j)a G a G = 4.83 ft/s 2 C+2M i = N b (12) - 200(6) + 30(9) = (^|)(4.83)(7) N b = 95.0 lb + '\'ZF y = m(a G ) y ; N A + 95.0 - 200 = 0 N a = 105 lb ( ^ ) S = S 0 + V 0 t + |fl G f 2 1 0 s = 0 + 0 + — (4.83)(2) 2 = 9.66 ft Ans. afl- 30V, 300th Ans. Ans. Ans: N b = 95.0 lb N a = 105 lb j = 9.66 ft 814 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 25 . The door has a weight of 200 lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12 ft to the right in 5 s, starting from rest. Also, find the vertical reactions at the rollers A and B. SOLUTION ( )s = s 0 + v 0 t + -I a G t 2 12 = 0 + 0 + ifl G (5) 2 a c = 0.960 ft/s 2 ^ = m(a G ) x ; F = |^(0.96O) F = 5.9627 lb = 5.96 lb Q + ZM a = 2(M*) a ; N b ( 12) - 200(6) + 5.9627(9) = — (0.960)(7) N b = 99.0 lb Ans. + = m(a G ) y \ N A + 99.0 - 200 = 0 N A = 101 lb Ans. Ans: F = 5.96 lb N b = 99.0 lb N a = 101 lb 815 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 26 . The jet aircraft has a total mass of 22 Mg and a center of mass at G. Initially at take-off the engines provide a thrust 2T = 4 kN and T' = 1.5 kN. Determine the acceleration of the plane and the normal reactions on the nose wheel and each of the two wing wheels located at B. Neglect the mass of the wheels and, due to low velocity, neglect any lift caused by the wings. SOLUTION -L 2ZF X = ma x ; 1.5 + 4 = 22 a G + T 2F V = 0 ; 2 B y + A y - 22(9.81) = 0 C +2M, = 2( M k ) b ; 4(2.3) - 1.5(2.5) - 22(9.81)(3) + A y A y = 72.6 kN B y = 71.6 kN a G = 0.250 m/s 2 = - 2200 ( 1 . 2 ) Ans. Ans. Ans. Ans: A y = 72.6 kN B y = 71.6 kN a G = 0.250 m/s 2 816 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 27 . The sports car has a weight of 4500 lb and center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are /jl s = 0.5 and fji k = 0.3, respectively. Neglect the mass of the wheels. SOLUTION C + ^M a = ~2N b (6) + 4500(2) = ^y^c(2.5) Y.F X = m(a G ) x ; 0.3(2 N B ) = + ]lF y = m(a G ) y ; 2N H + 2N A - 4500 = 0 Solving, N a = 1393 lb Ng = 857 lb a G = 3.68 ft/s 2 () v = v Q + a c t 10 = 0 + 3.68 t t = 2.72 s 4 soo Hj Ans. Ans. Ans. Ans: N a = 1393 lb N b = 857 lb t = 2.72 s 817 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 28 . The assembly has a mass of 8 Mg and is hoisted using the boom and pulley system. If the winch at B draws in the cable with an acceleration of 2 m/s 2 , determine the compressive force in the hydraulic cylinder needed to support the boom. The boom has a mass of 2 Mg and mass center at G. SOLUTION SB + 2s l = 1 a B = -2a L 2 = ~2a L a L = — 1 m/s 2 Assembly: + T XF y = ma y \ 2T - 8(l0 3 )(9.81) = 8(l0 3 )(l) T = 43.24 kN Boom: C+SM a = 0; F cd {2) - 2(10 3 )(9.81)(6 cos 60°) - 2(43.24)(l0 3 )(12 cos 60°) = 0 F cd = 289 kN Ans. 1 m XT t/yX Ans: F cd = 289 kN 818 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 29 . The assembly has a mass of 4 Mg and is hoisted using the winch at B. Determine the greatest acceleration of the assembly so that the compressive force in the hydraulic cylinder supporting the boom does not exceed 180 kN. What is the tension in the supporting cable? The boom has a mass of 2 Mg and mass center at G. SOLUTION Boom: C +tM A = 0; 180(10 3 )(2) - 2(l0 3 )(9.81)(6 cos 60°) - 27(12 cos 60°) = 0 T = 25 095 N = 25.1 kN Ans. Assembly: + /£// = ma y ; 2(25 095) - 4(l0 3 )(9.81) = 4(l0 3 ) a a = 2.74 m/s 2 Ans. 1 m ~2 m- Ans: a = 2.74 m/s 2 T = 25.1 kN 819 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 820 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 31 . A car having a weight of 4000 lb begins to skid and turn with the brakes applied to all four wheels. If the coefficient of kinetic friction between the wheels and the road is /x k = 0.8, determine the maximum critical height h of the center of gravity G such that the car does not overturn. Tipping will begin to occur after the car rotates 90° from its original direction of motion and, as shown in the figure, undergoes translation while skidding. Hint: Draw a free-body diagram of the car viewed from the front. When tipping occurs, the normal reactions of the wheels on the right side (or passenger side) are zero. SOLUTION N a represents the reaction for both the front and rear wheels on the left side. XF x = m(a c ) x ; 0.8N A = -yy a G + | %F y = m(a G ) y ; N A - 4000 = 0 C +SM a = t(M k ) A ; 4000(2.5) = |*| (a G )(h) Solving, N a = 4000 lb a G = 25.76 ft/s 2 h = 3.12 ft Ans: h = 3.12 ft 821 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 32 . A force of P = 300 N is applied to the 60-kg cart. Determine the reactions at both the wheels at A and both the wheels at B. Also, what is the acceleration of the cart? The mass center of the cart is at G. SOLUTION Equations of Motions. Referring to the FBD of the cart, Fig. a. *L ~ZF X = m(a G ) x ; 300 cos 30° = 60a a = 4.3301 m/s 2 = 4.33 m/s 2 <— Ans. +T ZF y = m(a c ) y ; N A + N B + 300sin30° - 60(9.81) = 60(0) (1) C+£M g = 0; N b ( 0.2) - A/,(0.3) + 300 cos 30°(0.1) - 300 sin 30°(0.38) = 0 ( 2 ) Solving Eqs. (1) and (2), N a = 113.40 N = 113 N Ans. N b = 325.20 N = 325 N Ans. Ans: a = 4.33 m/s 2 «— N a = 113 N N b = 325 N 822 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 33 . Determine the largest force P that can be applied to the 60-kg cart, without causing one of the wheel reactions, either at A or at B , to be zero. Also, what is the acceleration of the cart? The mass center of the cart is at G. SOLUTION Equations of Motions. Since (0.38 m) tan 30° = 0.22 m > 0.1 m, the line of action of P passes below G. Therefore, P tends to rotate the cart clockwise. The wheels at A will leave the ground before those at B. Then, it is required that N A = 0. Referring, to the FBD of the cart, Fig. a + | tF y = m(a G ) y ; N B + P sin 30° — 60(9.81) = 60(0) ( 1 ) C+ 2M g = 0; P cos 30°(0.1) - P sin 30°(0.38) + N B ( 0.2) = 0 ( 2 ) Solving Eqs. (1) and (2) P = 578.77 N = 579 N Ans. N b = 299.22 N Ans: P = 579 N 823 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 34 . The trailer with its load has a mass of 150 kg and a center of mass at G. If it is subjected to a horizontal force of P = 600 N, determine the trailer’s acceleration and the normal force on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass. SOLUTION Equations of Motion: Writing the force equation of motion -L 2 F x = m(a G ) x ; 600 = 150a a = 4m/s 2 -* Ans. Using this result to write the moment equation about point A, C +ZM a = (M k ) A ; 150(9.81)(1.25) - 600(0.5) - N B ( 2) = -150(4)(1.25) N B = 1144.69 N = 1.14 kN Ans. Using this result to write the force equation of motion along the y axis, + T2-F y = m(a G ) y ; N A + 1144.69 - 150(9.81) = 150(0) N a = 326.81 N = 327 N Ans. / 50(7-80 a/ Ans: a = 4 m/s 2 —* Ng = 1.14 kN N a = 327 N 824 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 35 . The desk has a weight of 75 lb and a center of gravity at G. Determine its initial acceleration if a man pushes on it with a force F = 60 lb. The coefficient of kinetic friction at A and B is /x k = 0.2. SOLUTION -i» Sir = ma x ; + = mciy, C+2M g = 0; Solving, 60 cos 30° - 0.2N a - 0.2 N B = ^a G N a + N b - 75 - 60 sin 30° = 0 60 sin30°(2)— 60 cos30°(l) - N A (2) + N B (2) - Q.2N A (2) - 0.2N b (2) = 0 a G = 13.3 ft/s 2 N a = 44.0 lb N r = 61.0 lb Ans. jo’T'ii 7 m . - £■» ' -‘T-O.Pll k- 4 Mg 7T1 a-> Ans: a G = 13.3 ft/s 2 825 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 36 . The desk has a weight of 75 lb and a center of gravity at G. Determine the initial acceleration of a desk when the man applies enough force F to overcome the static friction at A and B. Also, find the vertical reactions on each of the two legs at A and at B. The coefficients of static and kinetic friction at A and B are /jl s = 0.5 and /r k = 0.2, respectively. SOLUTION Force required to start desk moving; +, XF x = 0; F cos 30° - 0.5 N A - 0.5 N B = 0 + }ZF y = 0; N A + N B - Fsin 30° - 75 = 0 Solving for F by eliminating N A + N B , F = 60.874 lb Desk starts to slide. +> XF Z = m(a G ) x ; 60.874 cos 30° - 0.2 N A - 0.2 N B = 75 322 a G + 12F V = m(a G ) y - N a + N b - 60.874 sin 30° - 75 = 0 Solving for a G by eliminating N A + N B , a G = 13.58 = 13.6 ft/s 2 Ans. C+S34 = %{m k ) a - So that For each leg, -75, N b ( 4) - 75(2) - 60.874 cos 30°(3) = — (13.58)(2) N b = 61.2 lb N a = 44.2 lb N a N b 44.2 2 61.2 2 = 22.1 lb Ans. = 30.6 lb Ans. Ans: a G = 13.6 ft/s 2 N a = 22.1 lb N b = 30.6 lb 826 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 37 . The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to tip on the cart. Slipping does not occur. SOLUTION Equation of Motion. Tipping will occur about edge A. Referring to the FBD and kinetic diagram of the crate, Fig. a, C + £M a = £ (M k ) a ; 150(9.81)(0.25) = (150a)(0.5) a = 4.905 m/s 2 Using the result of a and refer to the FBD of the crate and cart, Fig. b. <L XF x = m(a G ) x P = (150 + 10)(4.905) = 784.8 N = 785 N Ans. a Ans: P = 785 N 827 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 38 . The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to slip or tip on the cart. The coefficient of static friction between the crate and cart is /r s = 0.2. ->— 0.5 m —- SOLUTION Equation of Motion. Assuming that the crate slips before it tips, then Ff = /jl s N = 0.2 N. Referring to the FBD and kinetic diagram of the crate, Fig. a + t2F y = ma y ; N - 150 (9.81) = 150 (0) N = 1471.5 N XF x = m(a G ) x ; 0.2(1471.5) = 150 a a = 1.962 m/s 2 C+SM a = (M k ) A ; 150(9.81)(x) = 150(1.962)(0.5) x = 0.1 m Since x = 0.1m< 0.25 m, the crate indeed slips before it tips. Using the result of a and refer to the FBD of the crate and cart, Fig. b, XF x = m(a G ) x ; P = (150 + 10)(1.962) = 313.92 N = 314 N Ans. a 150MO/J .(&) Ans: P = 314 N 828 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 39 . The bar has a weight per length w and is supported by the smooth collar. If it is released from rest, determine the internal normal force, shear force, and bending moment in the bar as a function of x. SOLUTION Entire bar: S/V = m(a G ) x '\ wl cos 30° wl 8 ( fl c) I a G = g cos 30° Segment: %F X = m{a c ) x ; +i %F y = m(a G ) y ; C + 2 M s = 2(A/fc)s; N = (wx cos 30°) sin 30° = 0.433wa wx - V = wx cos 30°(cos 30°) V = 0.25 wx w - M = wx cos 30°(cos 30°)( M = 0.125wx 2 Ans. Ans: N = 0.433wx V = 0.25wx M = 0.125 wx 2 829 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 40 . The smooth 180-lb pipe has a length of 20 ft and a negligible diameter. It is carried on a truck as shown. Determine the maximum acceleration which the truck can have without causing the normal reaction at A to be zero. Also determine the horizontal and vertical components of force which the truck exerts on the pipe at B. SOLUTION + B 180 * F x = ma x\ B x = — + T2T; = 0; By - 180 = 0 /\ 2 \ 180 / 5 \ C+ZM b = X(M k ) B ; 180(10)^— J = —a T ( 10)(^-J Solving, B x = 432 lb B y = 180 lb a T = 77.3 ft/s 2 Ans. Ans. Ans. Ans: B x = 432 lb By = 180 lb a T = 77.3 ft/s 2 830 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 41 . The smooth 180-lb pipe has a length of 20 ft and a negligible diameter. It is carried on a truck as shown. If the truck accelerates at a = 5 ft/s 2 , determine the normal reaction at A and the horizontal and vertical components of force which the truck exerts on the pipe at B. SOLUTION 'ZFx = ma x \ B x - N J^j = (5) + T2F y = 0; B y - 180 + A a (j|J = 0 C+SM b = t(M k ) B \ -1SO(10)(§) + m - Solving, B x = 73.9 lb B y = 69.7 lb N A = 120 lb 1?0K» Ans. Ans. Ans. Ans: B x = 73.9 lb B y = 69.7 lb N a = 120 lb 831 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 42 . The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and the cart is /jl s = 0.5. SOLUTION Equations of Motion: Assume that the crate slips, then Fj = fj. s N = 0.5/V. C +^M a = 2 (M k ) A ; 50(9.81) cos 15°(x) - 50(9.81) sin 15°(0.5) = 50a cos 15°(0.5) + 50a sin 15°(x) (1) +/21y = m(a G )y ; N - 50(9.81) cos 15° = -50a sin 15° (2) \+2/y = m{a G ) x , ; 50(9.81) sin 15° - 0.5 N = -50a cos 15° ( 3 ) Solving Eqs. (1), (2), and (3) yields N = 447.81 N x = 0.250 m a = 2.01 m/s 2 Ans. Since x < 0.3 m, then crate will not tip. Thus, the crate slips. Ans. Ans: a = 2.01 m/s 2 The crate slips. 832 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-43. Determine the acceleration of the 150-lb cabinet and the normal reaction under the legs A and B if P = 35 lb. The coefficients of static and kinetic friction between the cabinet and the plane are = 0.2 and /r k = 0.15, respectively. The cabinet’s center of gravity is located at G. SOLUTION Equations of Equilibrium: The free-body diagram of the cabinet under the static condition is shown in Fig. a, where P is the unknown minimum force needed to move the cabinet. We will assume that the cabinet slides before it tips. Then, Fa ~ B-sN a ~ 0.2 N A and F B = ^ S N B = Q.2N B . ZF X = 0; P - Q.2N A - 0.2 N b = 0 (1) + T ^F y = 0; N a + N b - 150 = 0 (2) + 2M a = 0; Ng{ 2) - 150(1) - P( 4) = 0 (3) Solving Eqs. (1), (2), and (3) yields P = 30 lb N a = 15 lb N b = 135 lb Since P < 35 lb and is positive, the cabinet will slide. Equations of Motion: Since the cabinet is in motion, F A = /j. k N A = 0.15 N A and F B — b = 0.15N B . Referring to the free-body diagram of the cabinet shown in Fig. b, ^ ZF X = m(a G ) x ; 35 - 0.15 N A - 0.15 N B = Q||)a (4) ^ 2F X = m(a G ) x - N A + N B - 150 = 0 (5) + 2M g = 0; N b ( 1) - 0.15JV b (3.5) - 0.15A a ( 3.5) - N A (1) - 35(0.5) = 0 (6) Solving Eqs. (4), (5), and (6) yields a = 2.68 ft/s 2 Ans. N a = 26.9 lb N b = 123 lb Ans. -1 ft—-+—1 ft- G ,, 3.5 ft P 4ft IT -*| 4 15 Zi y lb c > 3 \ 3 i 3 Vt IH | Ak 'b (*) 35>\\> 4ft F a -o^/Ja 1 I5C \ i A ) I c 7 i □ i □ a 3Sjt Ut if* A/ft H>) Ans: a = 2.68 ft/s 2 N a = 26.9 lb N b = 123 lb 833 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-44. The uniform bar of mass m is pin connected to the collar, which slides along the smooth horizontal rod. If the collar is given a constant acceleration of a, determine the bar’s inclination angle 0. Neglect the collar’s mass. SOLUTION Equations of Motion: Writing the moment equation of motion about point A , + 'ZM a = (M k ) A , mg sin 01 — I = ma cosd 0 = tan ( — vg Ans. Ans: 834 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-45. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the tension in the cable when the truck begins to accelerate at 5 m/s 2 . Also, what are the horizontal and vertical components of reaction at the hinge C? SOLUTION C + SM C = 2(A4) C ; Tsin 30°(2.5) - 12 262.5(1.5 cos 45°) = 1250(5)(1.5 sin 45°) T = 15 708.4 N = 15.7 kN Ans. *L XF x = m(a c ) x ; -C x + 15 708.4 cos 15° = 1250(5) C x = 8.92 kN Ans. +1= m(a G ) y \ C y - 12 262.5 - 15 708.4 sin 15° = 0 C y = 16.3 kN Ans. Ans: T = 15.7 kN C x = 8.92 kN C y = 16.3 kN 835 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-46. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the maximum deceleration of the truck so that the gate does not begin to rotate forward. What are the horizontal and vertical components of reaction at the hinge C? SOLUTION C+SM c = X(M k ) c -, -12 262.5(1.5 cos 45°) = -1250(a)(1.5 sin 45°) a = 9.81 m/s 2 ^ tF x = m{a G ) x - C, = 1250(9.81) C, = 12.3 kN + lXF y = m(a G ) y -, C y - 12 262.5 = 0 C v = 12.3 kN Ans. Ans. Ans. Ans: a = 9.81 m/s 2 C x = 12.3 kN C y = 12.3 kN 836 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-47. The snowmobile has a weight of 250 lb, centered at G\, while the rider has a weight of 150 lb, centered at Go- If the acceleration is a = 20 ft/s 2 , determine the maximum height h of G 2 of the rider so that the snowmobile’s front skid does not lift off the ground. Also, what are the traction (horizontal) force and normal reaction under the rear tracks at A? SOLUTION Equations of Motion: Since the front skid is required to be on the verge of lift off, N B = 0. Writing the moment equation about point A and referring to Fig. a, C +2M A = (M k ) A ■ 250(1.5) + 150(0.5) = ^|(20)(/t max ) + (20)(1) /z max = 3.163 ft = 3.16 ft Ans. 0.5 ft Writing the force equations of motion along the x and y axes, * L '2F x = m(a G ) x ; 150 250 ~ 322 + 322 20 F a = 248.45 lb = 248 lb + T 2 F/ = m(a G ) y ; N a - 250 - 150 = 0 N a = 400 lb Ans. Ans. 150 ft 2501k vn*x * /ft !%(*» *•!*' (a) Ans: h max = 3.16 ft F a = 248 lb N a = 400 lb 837 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-48. The snowmobile has a weight of 250 lb, centered at G 1; while the rider has a weight of 150 lb, centered at G 2 . If h = 3 ft, determine the snowmobile’s maximum permissible acceleration a so that its front skid does not lift off the ground. Also, find the traction (horizontal) force and the normal reaction under the rear tracks at A. SOLUTION Equations of Motion: Since the front skid is required to be on the verge of lift off, N b = 0. Writing the moment equation about point A and referring to Fig. a, C + ^M a = (M k ) A ; 250(1.5) + 150(0.5) = (|^a max )(3) + «ma X )(l) fl max = 20.7 ft/s 2 Ans. Writing the force equations of motion along the x and y axes and using this result, we have ~F X = m(a G ) x ; F --H (2a7)+ ff (2o ' 7) F a = 257.14 lb = 257 lb Ans. + t2F y = m(a G ) y ; N a - 150 - 250 = 0 N a = 400 lb Ans. 0.5 ft Z50 lb 1 50H> 12°.a 3Z.Z /■5ft \ A 250 a 3Z Z (a) Ans: flmax = 20.7 ft/s 2 F a = 257 lb N a = 400 lb 838 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-49. If the cart’s mass is 30 kg and it is subjected to a horizontal force of P = 90 N, determine the tension in cord AB and the horizontal and vertical components of reaction on end C of the uniform 15-kg rod BC. SOLUTION Equations of Motion: The acceleration a of the cart and the rod can be determined by considering the free-body diagram of the cart and rod system shown in Fig. a. %F X = m(a G ) x 90 = (15 + 30)a a = 2 m/s 2 The force in the cord can be obtained directly by writing the moment equation of motion about point C by referring to Fig. b. + SM C = (M k ) c \ Fab sin 30°(1) - 15(9.81) cos 30°(0.5) = -15(2) sin 30°(0.5) Fab = 112.44 N = 112 N Ans. Using this result and applying the force equations of motion along the x and y axes, ■** ~2F X = m(a G ) x ; -C x + 112.44 sin 30° = 15(2) C x = 26.22 N = 26.2 N Ans. + 12^ = m(a G ) y \ C y + 112.44 cos 30° - 15(9.81) = 0 Ans: Far = H2 N C x = 26.2 N C y = 49.8 N 839 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-50. If the cart’s mass is 30 kg, determine the horizontal force P that should be applied to the cart so that the cord AB just becomes slack. The uniform rod BC has a mass of 15 kg. \ / 30°“ 1 m cV #^0 ° P -► SOLUTION Equations of Motion: Since cord AB is required to be on the verge of becoming slack, F AB = O.The corresponding acceleration a of the rod can be obtained directly by writing the moment equation of motion about point C. By referring to Fig. a. + 2M C = 2(M c ) a ; -15(9.81) cos 30°(0.5) = -15asin30°(0.5) a = 16.99 m/s 2 Using this result and writing the force equation of motion along the x axis and referring to the free-body diagram of the cart and rod system shown in Fig. b, ( * )2F x = m{a G ) x - P = (30 + 15)(16.99) = 764.61 N = 765 N Ans. Lb) Ans: P = 765 N 840 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-51. The pipe has a mass of 800 kg and is being towed behind the truck. If the acceleration of the truck is a, = 0.5 m/s 2 , determine the angle 0 and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is /x k = 0.1. SOLUTION ZF X = ma x \ -0.1 N c + T cos 45° = 800(0.5) +1 'ZFy = ma y ; N c ~ 800(9.81) + T sin 45° = 0 C + 2M C = 0; -0.1A C (0.4) + T sin <(>(0.4) = 0 N c = 6770.9 N T = 1523.24 N = 1.52 kN 0.1(6770.9) sin </> = —= 26.39° 1523.24 ^ d = 45° - 0 = 18.6° Ans. Ans: T = 1.52 kN e = 18.6° 841 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-52. The pipe has a mass of 800 kg and is being towed behind a truck. If the angle 0 = 30°, determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is /jL k = 0.1. SOLUTION -L ~ZF r = ma ,; +1 'ZFy = ma y C+2M g = 0; T cos 45° - 0.1N C = 800n N c ~ 800(9.81) + T sin 45° = 0 T sin 15°(0.4) - O.UV c (0.4) = 0 N c = 6161 N T = 2382 N = 2.38 kN a = 1.33 m/s 2 Ans: T = 2.38 kN a = 1.33 m/s 2 842 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-53. The crate C has a weight of 150 lb and rests on the truck elevator for which the coefficient of static friction is /r s = 0.4. Determine the largest initial angular acceleration a, starting from rest, which the parallel links AB and DE can have without causing the crate to slip. No tipping occurs. SOLUTION XF x = ma x \ 0.41V C = (a) cos 30° +1XF V = ma y \ 150 Nc~ 150 = — («)sin30° N c = 195.0 lb a = 19.34 ft/s 2 19.34 = 2 a a = 9.67 rad/s 2 3 Nc f Ans. Ans: a = 9.67 rad/s 2 843 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-54. The crate C has a weight of 150 lb and rests on the truck elevator. Determine the initial friction and normal force of the elevator on the crate if the parallel links are given an angular acceleration a = 2rad/s 2 starting from rest. SOLUTION a = 2 rad/s 2 a = 2a = 4 rad / s 2 tF x = ma x ; F c = — (a) cos 30° +1 XF y = ma y \ Nc — 150 = ( a ) sin 30° F c = 16.1 lb N c = 159 lb Ans. Ans. [+j . i\as? s “ Ans: F c = 16.1 lb N c = 159 lb 844 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-55. The 100-kg uniform crate C rests on the elevator floor where the coefficient of static friction is fx s = 0.4. Determine the largest initial angular acceleration a, starting from rest at 0 = 90°, without causing the crate to slip. No tipping occurs. SOLUTION Equations of Motion. The crate undergoes curvilinear translation. At 0 = 90°, co = 0. Thus, ( a G ) n = co 2 r = 0. However; ( a G ), = ar = a(1.5). Assuming that the crate slides before it tips, then, fy = /jl s N = 0.4 N. = m(a G )„\ 100(9.81) - N = 100(0) N = 981 N Si 7 , = m(a G ) t \ 0.4(981) = 100[a:(1.5)] a = 2.616 rad/s 2 = 2.62 rad/s 2 Ans. C+ 2M g = 0; 0.4(981)(0.6) - 981(x) = 0 x = 0.24 m Since x < 0.3 m, the crate indeed slides before it tips, as assumed. IOOC1-8D W Ans: a = 2.62 rad/s 2 845 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-56. The two uniform 4-kg bars DC and EF are fixed (welded) together at E. Determine the normal force N E , shear force V E , and moment M E , which DC exerts on EF at E if at the instants = 60° BC has an angular velocity a> = 2rad/sand an angular acceleration a = 4rad/s 2 as shown. SOLUTION Equations of Motion. The rod assembly undergoes curvilinear motion. Thus, (a G ), = otr = 4(2) = 8 m/s 2 and {a G ) n = co 2 r = (2 2 )(2) = 8 m/s 2 . Referring to the FBD and kinetic diagram of rod EF, Fig. a <t_ Sir = m ( a G)x\ V E = 4(8) cos 30° + 4(8) cos 60° = 43.71 N = 43.7 N Ans. +1 XF y = m(a G )y, N e - 4(9.81) = 4(8) sin 30° - 4(8) sin 60° N e = 27.53 N = 27.5 N C+SM £ = X(M k ) E , M e = 4(8) cos 30°(0.75) + 4(8) cos 60°(0.75 ) = 32.78 N • m = 32.8 N • m Ans. Ans. Ans: V E = 43.7 N N e = 27.5 N M e = 32.8 N • m 846 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 847 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-58. The uniform 24-kg plate is released from rest at the position shown. Determine its initial angular acceleration and the horizontal and vertical reactions at the pin A. SOLUTION Equations of Motion. The mass moment of inertia of the plate about its center of gravity G is I G = pp(24)(0.5 2 + 0.5 2 ) = 1.00 kg • m 2 . Since the plate is at rest initially to = 0. Thus, (a G ) n = to 2 r G = 0. Here r G = V0.25 2 + 0.25 2 = 0.25 V2 m. Thus, ( a G ) t = ar G = a(o.25V2). Referring to the FBD and kinetic diagram of the plate, Q+XM a = {M k ) A ; —24(9.81)(0.25) = -24[a(0.25V2)] (0.25V2) - 1.00 a a = 14.715 rad/s 2 = 14.7 rad/s 2 Ans. Also, the same result can be obtained by applying %M A = I A a where I A = (24)(0.5 2 + 0.5 2 ) + 24(0.25 V2) 2 = 4.00 kg • m 2 : C+ tM A = I A a\ —24(9.81)(0.25) = -4.00 a a = 14.715 rad/s 2 ^ %F X = m(a G ) x ; A x = 24[l4.715(0.25v^)] cos 45° = 88.29 N = 88.3 N Ans. + ]%F y = m{a G )y, A y - 24(9.81) = -24[l4.715(0.25V2)] sin 45° A y = 147.15 N = 147 N Ans. 4 i A P‘2m^ C * V T 5 4 1 f Ans: a = 14.7 rad/s 2 A x = 88.3 N A y = 147 N 848 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 849 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-60. The bent rod has a mass of 2 kg/m. If it is released from rest in the position shown, determine its initial angular acceleration and the horizontal and vertical components of reaction at A. SOLUTION Equations of Motion. Referring to Fig. a, the location of bent rod is at of gravity G of the - Xxm 2[0.75(1.5)(2)] + 1.5(2)(1.5) „ _ x = —— = - ——— -= 1.00 m Sm 1.5 3(1.5)(2) = 0.75 m The mass moment of inertia of the bent rod about its center of gravity is In = 2 12 (3)( 1.5 2 ) + 3(0.25 2 + 0.75 2 ) ^(3)(l.5 2 ) + 3(0.5 2 ) = 6.1875 kg • m 2 . Here, r G = Vl.OO 2 + 0.75 2 = 1.25 m. Since the bent rod is at rest initially, to = 0. Thus, (a G ) n = oo 2 r G = 0. Also, ( a G ), = ar G = a( 1.25). Referring to the FBD and kinetic diagram of the plate, C+ XM a = (M k ) A ; 9(9.81)(1) = 9[a(1.25)](1.25) + 6.1875 a a = 4.36 rad/s 2 *) Ans. Also, the same result can be obtained by applying %M A = I A a where 4 = j^(3)(l.5 2 ) + 3(0.75 2 ) + ^(3)(1.5 2 ) + 3(l.5 2 + 0.75 2 ) C+ = I A a, 24 = m(a G ) x ; + 124 = m{a G )y, + — (3)(l.5 2 ) + 3(l.5 2 + 0.75 2 ) = 20.25 kg• m 2 Yr 9(9.81 )(1) = 20.25 a a = 4.36 rad/s 2 2 3 a A x = 9(4.36(1.25)] = 29.43 N = 29.4 N A, - 9(9.81) = -9(4.36(1.25)] - A, = 49.05 N = 49.1 N Ans. Ans. /•5/n. ft 7 I. X 1 -‘4 &■) Ans: a = 4.36 rad/s 2 !) A x = 29.4 N A y = 49.1 N 850 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 61 . If a horizontal force of P= 100 N is applied to the 300-kg reel of cable, determine its initial angular acceleration. The reel rests on rollers at A and B and has a radius of gyration of k a = 0.6 m. SOLUTION Equations of Motions. The mass moment of inertia of the reel about O I Q = Mk 2 0 = 300(0.6 2 ) = 108 kg • m 2 . Referring to the FBD of the reel, Fig. a. C+ 2M 0 = I Q a; -100(0.75) = 108(-a) a = 0.6944 rad/s 2 = 0.694 rad/s 2 Ans. 300CW a / Ans: a = 0.694 rad/s 2 851 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 62 . The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb • ft/rad, so that the torque developed is M = (50) lb • ft, where 8 is in radians. If the bar is released from rest when it is vertical at 8 = 90°, determine its angular velocity at the instant 8 = 0°. SOLUTION 1 10 C+2M 0 = I 0 a; -58= [^(^)(2) 2 ]« - 48.3 8 = a a dd = (o dco 48.3 8 dd co dco 48.3 n 2 2 1 2 co = 10.9 rad/s ioU. Ans. Ans: co = 10.9 rad/s 852 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 63 . The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k — 5 lb • ft/rad, so that the torque developed is M — (50) lb • ft, where 8 is in radians. If the bar is released from rest when it is vertical at 0 = 90°, determine its angular velocity at the instant 0 = 45°. SOLUTION C +2M 0 = I„a- 5 8 = [^(^(Z) 2 ]^ a = - 48.30 a d8 = w dto co = 9.45 rad/s Ans. Ans: to = 9.45 rad/s 853 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 64 . A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (MS 2 ) N, where 6 is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revolutions. The disk has an initial angular velocity of oj 0 = 1 rad/s. SOLUTION Equations of Motion. The mass moment inertia of the disk about O is lo = \ mrl = \ (8)(0.3 2 ) = 0.36 kg • m 2 . Referring to the FBD of the disk, Fig. a, Q+ tM 0 = 1o a ; Q 6 2 j(0.3) = 0.36 a F a = (0.2083 6 2 ) rad/s 2 Kinematics. Using the result of a, integrate codco = add with the initial condition (o = 0 when 6 = 0, r 5(2ir) codco = 0.2083 6 2 d6 0(fi> 2 - 1) = 0.06944 6 2 5(2tt) 0 u> = 65.63 rad/s = 65.6 rad/s Ans. Ans: w = 65.6 rad/s 854 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 65 . Disk A has a weight of 5 lb and disk B has a weight of 10 lb. If no slipping occurs between them, determine the couple moment M which must be applied to disk A to give it an angular acceleration of 4 rad/s 2 . SOLUTION Disk A: t+'2.M A = I A a A \ M - F d ( 0.5) |_2\32.2 (0.5) 2 (4) Disk B: +^M B = I B a B - F d (0.75) 1 2 “B r A <XA = r B a H 0.5(4) = 0.75a B Solving: a B = 2.67 rad/s 2 ; F D = 0.311 lb M = 0.233 lb • ft a = 4 rad/s 2 A 5U Ans. Ans: M = 0.233 lb • ft 855 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 856 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 67 . If the cord at B suddenly fails, determine the horizontal and vertical components of the initial reaction at the pin A, and the angular acceleration of the 120-kg beam. Treat the beam as a uniform slender rod. SOLUTION Equations of Motion. The mass moment of inertia of the beam about A is I A = ^(120)(4 2 ) + 120(2 2 ) = 640 kg-m 2 . Initially, the beam is at rest, w = 0. Thus, ( a G ) n = o) 2 r = 0. Also, (a G ), = ar G = a(2) = 2a. Referring to the FBD of the beam. Fig. a C+ = I A a\ 800(4) + 120(9.81 )(2) = 640 a a = 8.67875 rad/s 2 = 8.68 rad/s 2 Ans. %F n = m(a G ) n ; o II Ans. = m(a G ) t ; 800 + 120(9.81) + A t = 120[2(8.67875)] A. = 105.7 N = 106 N Ans. Ans: a = 8.68 rad/s 2 An = 0 A, = 106 N 857 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 68 . The device acts as a pop-up barrier to prevent the passage of a vehicle. It consists of a 100-kg steel plate AC and a 200-kg counterweight solid concrete block located as shown. Determine the moment of inertia of the plate and block about the hinged axis through A. Neglect the mass of the supporting arms AB. Also, determine the initial angular acceleration of the assembly when it is released from rest at 0 = 45°. 0.3 m 1.25 m SOLUTION Mass Moment of Inertia: I A = — (100)(l.25 2 ) + 100(0.625 2 ) + ^(200)(0.5 2 + 0.3 2 ) + 200(V0.75 2 + 0.15 2 ) 2 = 174.75 kg • m 2 = 175 kg • m 2 Ans. Equation of Motion: Applying Eq. 17-16, we have C +^M a = I A a-, 100(9.81)(0.625) + 200(9.81) sin 45°(0.15) -200(9.81) cos 45°(0.75) = -174.75a a = 1.25 rad/s 2 Ans. 200(1-60 r! A # V 1 ^ oL — . A Ans: I A = 175 kg • m 2 a = 1.25 rad/s 2 858 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 859 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 70 . The 20-kg roll of paper has a radius of gyration k A = 90 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. If the roll rests against a wall for which the coefficient of kinetic friction is fJ k = 0.2, determine the constant vertical force F that must be applied to the roll to pull off 1 m of paper in t = 3 s starting from rest. Neglect the mass of paper that is removed. SOLUTION ( + i ) 5 = S 0 + V Q t + 1 = 0 + 0 + —flc(3) 2 a c = 0.222 m/s 2 a = “ C ' = 1.778 rad/s 2 0.125 ' +. 1,F X = m(a Gx ); Nc ~ T AB cos 67.38° = 0 +t tF y = m(a G )y, T ab sin 67.38° - 0.2 N c ~ 20(9.81) - F = 0 C+ = I A a; -0.2fV c (0.125) + F(0.125) = 20(0.09) 2 ( 1.778) Solving: N c = 99.3 N T ab = 258 N F = 22.1 N Ans: F = 22.1 N Ans. 860 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 71 . The reel of cable has a mass of 400 kg and a radius of gyration of k A = 0.75 m. Determine its angular velocity whenf = 2 s, starting from rest, if the force P = (20 1 2 + 80) N, when t is in seconds. Neglect the mass of the unwound cable, and assume it is always at a radius of 0.5 m. SOLUTION Equations of Motion. The mass moment of inertia of the reel about A is I A = Mk A = 400(0.75 2 ) = 225 kg • m 2 . Referring to the FBD of the reel, Fig. a Q+XM a = I A a ; —(20 1 2 + 80)(0.5) = 225(-a) a = —(t 1 + 4) rad/s 2 45 Kinematics. Using the result of a , integrate dco = adt , with the initial condition o> = 0 at t = 0, pco />2s n S <,! + 4) ‘" w = 0.4741 rad/s = 0.474 rad/s Ans. 400 WO *J ^zoi^+ao ol Ans: co = 0.474 rad/s 861 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 72 . The 30-kg disk is originally spinning at co = 125 rad/ s. If it is placed on the ground, for which the coefficient of kinetic friction is p c = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of force which the member AB exerts on the pin at A during this time? Neglect the mass of AB. SOLUTION Equations of Motion. The mass moment of inertia of the disk about B is 1 i I B = —mr 2 = — (30)(0.3 2 ) = 1.35 kg - m 2 . Since it is required to slip at C, Ff = iJ-(A'c = 0-5 N c . Referring to the FBD of the disk, Fig. a, 125 rad/s + XF x = m(a G ) x ; 0.5 N c - F ab cos 45° = 30(0) (1) +1 SF V = m{a G ) y : N c - F ab sin 45° - 30(9.81) = 30(0) (2) Solving Eqs. (1) and (2), N c = 588.6 N F ab = 416.20 N Subsequently, C + 'M B = I B a\ 0.5(588.6)(0.3) = 1.35a a = 65.4 rad/s 2 *) Referring to the FBD of pin A, Fig. b, + . XF x = 0; 416.20 cos 45° - A x = 0 A x = 294.3 N = 294 N Ans. + | SF y = 0; 416.20 sin 45° - A y = 0 Kinematic. Using the result of a , A y = 294.3 N = 294 N Ans. + CO — (Oq + OLt\ 0 = 125 + (-65.4)1 t = 1.911s = 1.91s Ans. 3o(9-ddd Ox) fa 4& tori A x = 294 N Ay = 294 N t = 1.91 s 862 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 73 . Cable is unwound from a spool supported on small rollers at A and B by exerting a force T = 300 N on the cable. Compute the time needed to unravel 5 m of cable from the spool if the spool and cable have a total mass of 600 kg and a radius of gyration of k Q = 1.2 m. For the calculation, neglect the mass of the cable being unwound and the mass of the rollers at A and B. The rollers turn with no friction. SOLUTION I 0 = mk 2 0 = 600(1.2) 2 = 864 kg -m 2 C + tM 0 = I 0 a\ 300(0.8) = 864(a) s 5 The angular displacement 6 = - = —— = y 0.8 a = 0.2778 rad/s 2 6.25 rad. 1 7 e = e 0 + co 0 r+ -a c t- 6.25 = 0 + 0 + ^(0.27778)t 2 t = 6.71 s Ans. Ans: t = 6.71 s 863 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 74 . The 5-kg cylinder is initially at rest when it is placed in contact with the wall B and the rotor at A. If the rotor always maintains a constant clockwise angular velocity w = 6 rad/s, determine the initial angular acceleration of the cylinder. The coefficient of kinetic friction at the contacting surfaces B and C is fi k = 0.2. SOLUTION Equations of Motion: The mass moment of inertia of the cylinder about point O is given by I 0 = \ mr 1 — — (5)(0.125 2 ) = 0.0390625 kg • m 2 . Applying Eq. 17-16, we have ^J.F X = m(a G ) x ; N B + 0.2 N A cos 45° - N A sin 45° = 0 (1) + t = m(a G ) y ; 0.2 N B + 0.2 N A sin 45° + N A cos 45° - 5(9.81) = 0 ( 2 ) C +2M 0 = I 0 a\ 0.2 N a (0.125) - 0.2 N B (0.125) = 0.0390625a ( 3 ) Solving Eqs. (1), (2), and (3) yields; N a = 51.01 N N b = 28.85 N a = 14.2 rad/s 2 Ans. Ans: a = 14.2 rad/s 2 864 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 75 . The wheel has a mass of 25 kg and a radius of gyration k B = 0.15 m. It is originally spinning at w = 40 rad/s. If it is placed on the ground, for which the coefficient of kinetic friction is /jl c — 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of reaction which the pin at A exerts on AB during this time? Neglect the mass of AB. SOLUTION I B = mk B = 25(0.15) 2 = 0.5625 kg-m 2 + t ^F y = m(a G ) y ; (§) F AB + N c ~ 25(9.81) = 0 ^F x = m(a G ) x ; 0.5 N c - (f)F AB = 0 C + ~2M B = I B a; 0.5N C (0.2) = 0.5625(—a) Solvings Eqs. (1),(2) and (3) yields: F ab = 111.48 N N c = 178.4 N a = —31.71 rad/s 2 A = \F AB = 0.8(111.48) = 89.2 N A y = | F ab = 0.6(111.48) = 66.9 N co = (o 0 + a c t 0 = 40 + (—31.71) t t = 1.26 s Ans: A x = 89.2 N Ay = 66.9 N t = 1.25 s Ans. Ans. 865 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 76 . The 20-kg roll of paper has a radius of gyration k A = 120 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. The roll rests on the floor, for which the coefficient of kinetic friction is H k = 0.2. If a horizontal force F= 60 N is applied to the end of the paper, determine the initial angular acceleration of the roll as the paper unrolls. F SOLUTION Equations of Motion. The mass moment of inertia of the paper roll about A is I A = mk A = 20(0.12 2 ) = 0.288 kg - m 2 . Since it is required to slip at C, the friction is Ff = fx k N = 0.2 N. Referring to the FBD of the paper roll, Fig. a i %F X = m(a c ) x \ 0.2 N - F AB Q + 60 = 20(0) (1) + t = m(a G ) y ; N - “ 20(9.81) = 20(0) (2) Solving Eqs. (1) and (2) Fab = 145.94 N N = 283.76 N Subsequently C+ 1M a = l A a\ 0.2(283.76)(0.3) - 60(0.3) = 0.288(-<z) a = 3.3824 rad/s 2 = 3.38 rad/s 2 Ans. A) ( 0 .) Ans: a = 3.38 rad/s 2 866 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 77 . Disk D turns with a constant clockwise angular velocity of 30 rad/s. Disk E has a weight of 60 lb and is initially at rest when it is brought into contact with D. Determine the time required for disk E to attain the same angular velocity as disk D. The coefficient of kinetic friction between the two disks is fji k = 0.3. Neglect the weight of bar BC. SOLUTION 30 rad/s Equations of Motion: The mass moment of inertia of disk E about point B is given by I B = \ mr 2 = 2 ( ^+^(1 2 ) = 0.9317 slug • ft 2 . Applying Eq. 17-16, we have ^F x = m(a G ) x ; 0.3 N - F bc cos 45° = 0 ( 1 ) +1 2 = m(a G ) y ; N - F bc sin 45° - 60 = 0 ( 2 ) Q +J,M 0 = Iq a\ 0.3A(1) = 0.9317a ( 3 ) Solving Eqs. (1), (2) and (3) yields: F bc = 36.37 lb N = 85.71 lb a = 27.60 rad/s 2 Kinematics: Applying equation to = to 0 + a t , we have (C+) 30 = 0 + 27.60? t = 1.09 s Ans. Go iy tV. Ans: t = 1.09 s 867 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 78 . Two cylinders A and B , having a weight of 10 lb and 5 lb, respectively, are attached to the ends of a cord which passes over a 3-lb pulley (disk). If the cylinders are released from rest, determine their speed in t = 0.5 s. The cord does not slip on the pulley. Neglect the mass of the cord. Suggestion: Analyze the “system” consisting of both the cylinders and the pulley. SOLUTION Equation of Motion: The mass moment of inertia of the pulley (disk) about point O is given by I Q = ^mr 2 = ^ (^f/>)( 0 - 752 ) a a a = — = Applying Eq. 17-16, we have 0.02620 slug • ft 2 . Here, a = ar or C+2M 0 = I 0 a; 5(0.75) - 10(0.75) = -0.02620^^^) (0.75) - 3H ,32.2 a = 9.758 ft/s 2 Kinematic: Applying equation v = v 0 + at , we have v = 0 T 9.758(0.5) = 4.88 ft/s 10 322 (0.75) I 0 tg-o.ozi,zo(~j Ans. 10 lb Ans: v = 4.88 ft/s 868 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 79 . The two blocks A and B have a mass of 5 kg and 10 kg, respectively. If the pulley can be treated as a disk of mass 3 kg and radius 0.15 m, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. SOLUTION Kinematics: Since the pulley rotates about a fixed axis passes through point O , its angular acceleration is a a a = — =-= 6.6667a r 0.15 The mass moment of inertia of the pulley about point O is I„ = ^Mr 2 = |(3)(0.15 2 ) = 0.03375 kg-rn 2 Equation of Motion: Write the moment equation of motion about point O by referring to the free-body and kinetic diagram of the system shown in Fig. a, C +2M„ = 2(M*) 0 ; 5(9.81)(0.15) - 10(9.81)(0.15) = —0.03375(6.6667a) - 5a(0.15) - 10a(0.15) a = 2.973 m/s 2 = 2.97 m/s 2 Ans. tomotl I0O. (a.) Ans: a = 2.97 m/s 2 869 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 80 . The two blocks A and B have a mass m A and m B , respectively, where m B > m A . If the pulley can be treated as a disk of mass M, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. SOLUTION C +2M C - 2(M it ) c ; m B g( r ) ~ m Ag( r ) ~ ( —Mr 2 Jot + m B r 2 a + m A r 2 a g{m B - m A ) r( + m B + m A g(m B - m A ) 1 -M + m B + m A Ans. ij) 1,13 ij ■v ( V\ 6 0 Ans: a g(m B - m A ) —M + m B + m A 870 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 81 . Determine the angular acceleration of the 25-kg diving board and the horizontal and vertical components of reaction at the pin A the instant the man jumps off Assume that the board is uniform and rigid, and that at the instant he jumps off the spring is compressed a maximum amount of 200 mm, w = 0, and the board is horizontal. Take k = 7 kN/m. SOLUTION C + ~~ +12 F t = m ( a c) t ; ^ 2 F n = m(a G ) n ■ 1.5(1400 - 245.25) = - 1400 - 245.25 - A y = A x — 0 (25)(3) 2 25(1.5a) Solving, A, = 0 A y = 289 N a = 23.1 rad/s 2 Ans. Ans. Ans. 245.25 N 0 \J A a ¥ 1.5 25(1.5 a) Ans: A x = 0 A y = 289 N a = 23.1 rad/s 2 871 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 82 . The lightweight turbine consists of a rotor which is powered from a torque applied at its center. At the instant the rotor is horizontal it has an angular velocity of 15 rad/s and a clockwise angular acceleration of 8 rad/s 2 . Determine the internal normal force, shear force, and moment at a section through A. Assume the rotor is a 50-m-long slender rod, having a mass of 3 kg/m. SOLUTION 2F„ = m(a G ) n ; N A = 45(15) 2 (17.5) = 177kN + iSF f = m(a G ) t ; V A + 45(9.81) = 45(8)(17.5) V A = 5.86 kN C+2M, = X(Af k ) A ; M a + 45(9.81)(7.5) ^(45)(15) 2 (8) + [45(8)(17.5)](7.5) M a = 50.7 kN • m Ans. Ans: N a = 111 kN V A = 5.86 kN M a = 50.7 kN • m 872 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 83 . The two-bar assembly is released from rest in the position shown. Determine the initial bending moment at the fixed joint B. Each bar has a mass m and length /. SOLUTION Assembly: I a = \ml 2 + ^ 2 (m)(lf + m{l 2 + (^) 2 ) = 1.667 ml 2 C+ZM a = I A a- mg(-) + mg(l) = (1.661 ml 2 )a 0.9 g Segment BC: C+2M b = ^(M k ) B - M = 12 -ml 2 a + m(l 2 + 6 2 ) 1/2 «( 1/2 , )(g) „ 1 ,2 1 o ,°- 9 S, M = —ml a = — ml (—-—) M = 0.3 gml Ans. C H />vt b - rw\ Ans: M = 0.3gml 873 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-84. The armature (slender rod) AB has a mass of 0.2 kg and can pivot about the pin at A. Movement is controlled by the electromagnet E , which exerts a horizontal attractive force on the armature at B of F B — (0.2(10~ 3 )/~ 2 ) N, where / in meters is the gap between the armature and the magnet at any instant. If the armature lies in the horizontal plane, and is originally at rest, determine the speed of the contact at B the instant / = 0.01 m. Originally / = 0.02 m. SOLUTION Equation of Motion: The mass moment of inertia of the armature about point A is given by I A = I G + mr 2 G = (0.2) (0.15 2 ) + 0.2(0.075 2 ) = 1.50(l0~ 3 )kg • m 2 Applying Eq. 17-16, we have „ 0 . 2 ( 10 ~ 3 ) , Q + 2M a = I A a; - j 2 —- (0.15) = 1.50(l0~ 3 ) a 0.02 Kinematic: From the geometry, l = 0.02 — 0.150. Then dl — —0.15 dd or dl v dv dd =-. Also, co =-hence dco =-. Substitute into equation codco = add , 0.15 0.15 0.15 4 we have v ( dv\ _ ( dl\ 015^015 ) ~ \ 015 ) vdv = —0.15 adl -0.15 v = 0.548 m/s Ans. K- o-ZOO "t> 015m 6 -- Ans: v = 0.548 m/s 874 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 85 . The bar has a weight per length of w. If it is rotating in the vertical plane at a constant rate w about point O, determine the internal normal force, shear force, and moment as a function of x and 6. SOLUTION L--U z, Forces: ■—~to 2 ^ | + S + wx J. (1) Moments: O = M -\sx ( 2 ) Solving (1) and (2), N = V = M = wx w~ g L wx sin 6 1 wx 2 sind -) 2 > + cos 6 Ans. Ans. Ans. Ans: + cos 6 V = wx sin 0 M = 1 2 wx 2 sin 6 875 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 86 . The 4-kg slender rod is initially supported horizontally by a spring at B and pin at A. Determine the angular acceleration of the rod and the acceleration of the rod’s mass center at the instant the 100-N force is applied. 100 N SOLUTION Equation of Motion. The mass moment of inertia of the rod about A is I A = —(4)(3 2 ) + 4(1.5 2 ) = 12.0 kg • m 2 . Initially, the beam is at rest, co = 0. Thus, ( a G ) n = w 2 r = 0. Also, (a G ), = ar G = <*(1-5). The force developed in the spring before the application of the 4(9.81)N 100 N force is F sp = ---= 19.62 N. Referring to the FBD of the rod, Fig. a, C + M a = I A a; 19.62(3) - 100(1.5) - 4(9.81)(1.5) = 12.0(-a) a = 12.5rad/s^) Ans. Then (a G ) t = 12.5(1.5) = 18.75 m/s 2 i Since ( a G )„ = 0. Then a G = ( a G) t = 18.75 m/s 2 I Ans. lOOti (&) Ans: a = 12.5 rad/s/) a G = 18.75 m/s 2 J. 876 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 87 . The 100-kg pendulum has a center of mass at G and a radius of gyration about G of k G = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant 0 = 90° when the pendulum is rotating at &) = 8 rad/s. Neglect the weight of the beam and the support. SOLUTION Equations of Motion: Since the pendulum rotates about the fixed axis passing through point C, (a G ), = ar G = a(0.75) and ( a G )„ = (o 2 r G = 8 2 (0.75) = 48 m/s 2 . Here, the mass moment of inertia of the pendulum about this axis is I c = 100(0.25) 2 + 100(0.75 2 ) = 62.5 kg • m 2 . Writing the moment equation of motion about point C and referring to the free-body diagram of the pendulum. Fig. a, we have C + 2 M c = f c a; 0 = 62.5 a a = 0 Using this result to write the force equations of motion along the n and t axes, = m(a G ) t ; -C, = 100[0(0.75)] C, = 0 + T2F„ = m(a G ) n ; C„ - 100(9.81) = 100(48) C„ = 5781 N Equilibrium: Writing the moment equation of equilibrium about point A and using the free-body diagram of the beam in Fig. b , we have + SM ,4 = 0; N b (1.2) - 5781(0.6) = 0 N B = 2890.5 N = 2.89 kN Ans. Using this result to write the force equations of equilibrium along the x and y axes, we have '2,F X = 0; A x = 0 Ans. + T2Fj, = 0; A y + 2890.5 - 5781 = 0 A y = 2890.5 N = 2.89 kN Ans. (a.) C** *731 d ---- ^ _ r.=, > > i '•t “ lm A* ' r - O'btY) Av ^6 { (b) Ans: N b = 2.89 kN A x = 0 A y = 2.89 kN 877 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 88 . The 100-kg pendulum has a center of mass at G and a radius of gyration about G of k G = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant 0 = 0° when the pendulum is rotating at <o = 4 rad/s. Neglect the weight of the beam and the support. SOLUTION Equations of Motion: Since the pendulum rotates about the fixed axis passing through point C, ( a G ) t = ar G = a(0.75) and (a G ) n = w 2 r G = 4 2 (0.75) = 12 m/s 2 . Here, the mass moment of inertia of the pendulum about this axis is I c = 100(0.25 2 ) + 100(0.75) 2 = 62.5 kg • m 2 . Writing the moment equation of motion about point C and referring to the free-body diagram shown in Fig. a, C+2M c = / c a; ~100(9.81)(0.75) =-62.5a a = 11.772 rad/s 2 Using this result to write the force equations of motion along the n and t axes, we have + '['ZF t = m{a G ), ; C, - 100(9.81) = -100[11.772(0.75)] C, = 98.1 N = m(a G ) n ; C„ = 100(12) C n = 1200 N Equilibrium: Writing the moment equation of equilibrium about point A and using the free-body diagram of the beam in Fig. b. + 2M a = 0; N b (1.2) - 98.1(0.6) - 1200(1) = 0 N B = 1049.05 N = 1.05 kN Ans. Using this result to write the force equations of equilibrium along the x and y axes, we have ^ 2F X = 0; 1200 - A x = 0 A x = 1200 N = 1.20 kN Ans. + T = 0; 1049.05 - 98.1 ~ A y = 0 A y = 950.95 N = 951 N Ans. tootf-BO C a ) Ans: N b = 1.05 kN A x = 1.20 kN Ay = 951N 878 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 89 . The “Catherine wheel" is a firework that consists of a coiled tube of powder which is pinned at its center. If the powder burns at a constant rate of 20 g/s such as that the exhaust gases always exert a force having a constant magnitude of 0.3 N, directed tangent to the wheel, determine the angular velocity of the wheel when 75% of the mass is burned off. Initially, the wheel is at rest and has a mass of 100 g and a radius of r = 75 mm. For the calculation, consider the wheel to always be a thin disk. SOLUTION Mass of wheel when 75% of the powder is burned = 0.025 kg 0.075 kg Time to burn off 75 % = -— = 3.75 s 0.02 kg/s m(t) = 0.1 — 0.02 1 Mass of disk per unit area is Po = m A 0-1 kg 77(0.075 m) 2 5.6588 kg/m 2 At any time t, „ _ 0.1 - 0 . 02 / 5.6588 =-^- r(t) = yj- 0.1 - 0 . 02 / r(5.6588) + 1,Mc — Ic a '< 0.3r = —mr 2 a 0.6 0.6 (0.1 - 0.02r). 0.1 - 0 . 02 1 77(5.6588) a = 0.6(Vt 7(5.6588)) [0.1 - 0.02r] i a = 2.530[0.1 - 0.02r] I db) = a dt p(0 ft / da) = 2.530 / [0.1 - 0.02r] i dt Jo Jo w = 253[(0.1 - 0.02/) i - 3.162] For t = 3.75 s, oj = 800 rad/s Ans. Ans: co = 800 rad/s 879 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 90 . If the disk in Fig. 17—21« rolls without slipping , show that when moments are summed about the instantaneous center of zero velocity, 1C. it is possible to use the moment equation 2M /C = l/ G o, where I IC represents the moment of inertia of the disk calculated about the instantaneous axis of zero velocity. SOLUTION C + 2M /c = E(M X ), C ; 2M )C = I G a + (ma G )r Since there is no slipping, a G = ar Thus, 2M /C = (/ G + rar 2 )o: By the parallel-axis thoerem, the term in parenthesis represents I IC . Thus, 2M /C — I [(-a Q.E.D. To Ans: 2M /C = I ic a 880 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 91 . The 20-kg punching bag has a radius of gyration about its center of mass G of k G = 0.4 m. If it is initially at rest and is subjected to a horizontal force F = 30 N, determine the initial angular acceleration of the bag and the tension in the supporting cable AB. SOLUTION = m(a G ) x ; 30 = 20 (a G ) x + t2F y = m(a c )y\ T - 196.2 = 20 (a G ) y C+2M G = / G a; 30(0.6) = 20(0.4) 2 ct a = 5.62 rad/s 2 (a G )x = 1-5 m/s 2 a B = a G + a B/G a B i = («c)>'j + («g).T “ “(0.3)i (+T) (a G )y = 0 Thus, T = 196 N Ans. Ans. T (ObVO t (06)j . o< tO= 0 a- ^ Ca 6 ) t = a B Ans: a = 5.62 rad/s 2 T = 196 N 881 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. * 17 - 92 . The uniform 150-lb beam is initially at rest when the forces are applied to the cables. Determine the magnitude of the acceleration of the mass center and the angular acceleration of the beam at this instant. F a = 100 lb 12 ft SOLUTION Equations of Motion: The mass moment of inertia of the beam about its mass center is I G = — ml 2 = — f^^-Vl2 2 ) = 55.90 slug • ft 2 . G 12 12 V32.2 ’ 5 ^F x = m(a G ) x \ 200 cos 60° = (a G ) x (a G ) x = 21.47 ft/s 2 1 + \'LF y = m(a G ) y \ 100 + 200 sin 60° — 150 = ff^( a G)y ( a G )y = 26.45 ft/s 2 + 2M g = I G a\ 200 sin 60°(6) - 100(6) = 55.90a: a = 7.857 rad/s 2 = 7.86 rad/s 2 Thus, the magnitude of a G is a G = V(fl G ) x 2 + (a G ) y 2 = V21.47 2 + 26.45 2 = 34.1 ft/s 2 Ans: a = 7.86 rad/s 2 a G = 34.1 ft/s 2 882 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 - 93 . The slender 12-kg bar has a clockwise angular velocity of tu = 2 rad/s when it is in the position shown. Determine its angular acceleration and the normal reactions of the smooth surface A and B at this instant. SOLUTION Equations of Motion. The mass moment of inertia of the rod about its center of gravity G is I G = — ml 2 = p^(12)(3 2 ) = 9.00 kg • m 2 . Referring to the FBD and kinetic diagram of the rod, Fig. a = m(a G ) x ; N b ~ 12 (a c ) x ( 1 ) +1 = m(a G ) y ; N a - 12(9.81) = -12(«g), ( 2 ) Q+XM 0 = ( M k ) 0 ; —12(9.81)(1.5 cos 60°) = -12(00^(1.5 sin 60°) — 12(a G )).(l-5 cos 60°) — 9.00a V3 (a G ) x + {a G ) y + a = 9.81 ( 3 ) Kinematics. Applying the relative acceleration equation relating a G and a fl by referring to Fig. b, a G = a s + a X t G/B - art G/B —(a G ) x i ~ ( a e))j = ~ a B) + (—ctk) X (— 1.5 cos 60°i — 1.5sin60°j) —2 Z (—1.5 cos 60°i - 1.5 sin 60°j) -(aG),! - (a G ) y j = (3 - 0.75V3a)i + (0.75a - a B + 3\/3)j 883 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-93. Continued Equating i and j components, -(og)x = 3 - 0.75 V3« ( 4 ) ~(a G ) y = 0.15a - a B + 3V3 (5) Also, relate a B and a A , + a X t A/B - 0 ) 2 r A/B —a A i = —a B j + (—ak) X (—3 cos 60°i — 3 sin 60°j) —2 2 (—3 cos 60°i — 3 sin 60°j) —= (6 — 1.5\/3a)i + (1.5ct — a B + 6\/3)j Equating j components, 0 = 1.5a ~ a B + 6V3; a B = 1.5a + 6\/3 (6) Substituting Eq. (6) into (5) (a G ) y = 0.75a + 3V3 ( 7 ) Substituting Eq. (4) and (7) into (3) V3(0.75V3a - 3 ) + 0.75a + 3V3 + a = 9.81 a = 2.4525 rad/s z = 2.45^) rad/s 2 Ans. Substituting this result into Eqs. (4) and (7) ~(a G ) x = 3 - (0.75V3)(2.4525); (a G ) x = 0.1859 m/s 2 (a G ) y = 0.75(2.4525) + 3V3; (a G ) y = 7.0355 m/s 2 Substituting these results into Eqs. (1) and (2) N b = 12(0.1859); N B = 2.2307 N = 2.23 N Ans. N a — 12(9.81) — —12(7.0355); N A = 33.2937 N = 33.3 N Ans. Ans: a = 2.45 rad/s 2 N b = 2.23 N N a = 33.3 N 884 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-94. The tire has a weight of 30 lb and a radius of gyration of k G = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are /jl s = 0.2 and /jL k = 0.15, determine the tire’s angular acceleration as it rolls down the incline. Set 6 = 12°. SOLUTION +/^F X = m(a G ) x ; 30 sin 12° - F = (^^j a G +\2Fy = m(a G )y ; N - 30 cos 12° = 0 Q +SM g = I G a; F(1.25) (0.6) 2 a Assume the wheel does not slip. a G = (1.25)a Solving: F = 1.171b N = 29.34 lb a G = 5.44 ft/s 2 a = 4.35 rad/s 2 F max = 0.2(29.34) = 5.87 lb > 1.17 lb 1.25 Ans. OK Ans: a = 4.32 rad/s 2 885 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-95. The tire has a weight of 30 lb and a radius of gyration of k G = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are yu s = 0.2 and = 0.15, determine the maximum angle 9 of the inclined plane so that the tire rolls without slipping. SOLUTION Since wheel is on the verge of slipping: +/Sf, = m(a G ) x ; 30 sin 9 — 0.2 N = +\hF y = m(a G ) y ; TV — 30 cos 0 = 0 Y 30 C+2M c = 7 G a; 0.2A(1.25) = f — 30 3Z2 (0.6) 2 (1.25ct) Substituting Eqs.(2) and (3) into Eq. (1), 30 sin 9 — 6 cos 9 = 26.042 cos 6 30 sin 9 = 32.042 cos 6 tan 6 = 1.068 9 = 46.9° ( 1 ) ( 2 ) (3) Ans. 1.25 Ans: 9 = 46.9° 886 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-96. The spool has a mass of 100 kg and a radius of gyration of k G = 0.3 m. If the coefficients of static and kinetic friction atAareyii s = 0.2 and = 0.15, respectively, determine the angular acceleration of the spool if P = 50 N. SOLUTION -^^F x = m{a G ) x ; 50 + F a = 100a G + = m(a G ) y ; N a - 100(9.81) = 0 C +2M C = I G a; 50(0.25) - F a (0.4) = [100(0.3) 2 ]a Assume no slipping: a G = 0.4a a = 1.30 rad/s 2 a G = 0.520 m/s 2 N A = 981 N F A = 2.00 N Since (F A ) max = 0.2(981) = 196.2 N > 2.00 N OK Ans: a = 1.30 rad/s 2 887 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-97. Solve Prob. 17-96 if the cord and force P = 50 N are directed vertically upwards. SOLUTION -L = m(a G )x; F A = 100a G +1 'ZFy = m(a G ) y \ N A + 50 - 100(9.81) = 0 C + 2M g = I G a; 50(0.25) - F a (0.4) = [100(0.3) 2 ]a Assume no slipping: a G = 0.4 a a = 0.500 rad/s 2 a G = 0.2 m/s 2 N A = 931 N F A = 20 N Since (F^) max = 0.2(931) = 186.2 N > 20 N Ans: a = 0.500 rad/s 2 888 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-98. The spool has a mass of 100 kg and a radius of gyration k G = 0.3 m. If the coefficients of static and kinetic friction at A are yii s = 0.2 and = 0.15, respectively, determine the angular acceleration of the spool if P = 600 N. SOLUTION -L 1,F X = m(a G ) x ; 600 + F A = 100 a G + T 2F y = m(a G ) y ; N A - 100(9.81) = 0 C+ZM g = I G a\ 600(0.25) - F a (0A) = [100(0.3) 2 ]t* Assume no slipping: a G = 0.4 a a = 15.6 rad/s 2 a G = 6.24 m/s 2 N A = 981 N F A = 24.0 N Since (F A ) max = 0.2(981) = 196.2 N > 24.0 N Ans: a = 15.6 rad/s 2 889 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-99. The 12-kg uniform bar is supported by a roller at A. If a horizontal force of F = 80 N is applied to the roller, determine the acceleration of the center of the roller at the instant the force is applied. Neglect the weight and the size of the roller. SOLUTION Equations of Motion. The mass moment of inertia of the bar about its center of gravity G \ i is I G = — ml 2 = — (12)(2 2 ) = 4.00 kg • m 2 . Referring to the FBD and kinetic diagram of the bar, Fig. a i -%F X = m(a G ) x ; 80 = 12 (a G ) x (a G ) x = 6.6667 m/s 2 -» Q+1M a = (jjL k ) A ; 0 = 12(6.6667)(1) - 4.00 a a = 20.0 rad/s 2 /> Kinematic. Since the bar is initially at rest, co = 0. Applying the relative acceleration equation by referring to Fig. b, ae = a a + a X r G/A - io 2 r G/A 6.6667i + («G)vj = a A x + (—20.0k) X (— j) — 0 6.6667i + (a G ) y \ = (a A — 20)i Equating i and j components, 6.6667 = a A — 20; a A = 26.67 m/s 2 = 26.7 m/s 2 —» Ans. (' a G)y = 0 Ch) Ans: a A = 26.7 m/s 2 —* 890 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *17-100. A force of F = 10 N is applied to the 10-kg ring as shown. If slipping does not occur, determine the ring’s initial angular acceleration, and the acceleration of its mass center, G. Neglect the thickness of the ring. SOLUTION Equations of Motion. The mass moment of inertia of the ring about its center of gravity G is I G = mr 2 = 10(0.4 2 ) = 1.60 kg • m 2 . Referring to the FBD and kinetic diagram of the ring, Fig. a, C + SM C = O k ) c \ (10 sin 45°)(0.4 cos 30°) - (10 cos 45°)[0.4(1 + sin 30°)] = —(10« g )( 0.4) - 1.60a 4 a G + 1.60 a = 1.7932 (1) Kinematics. Since the ring rolls without slipping, a G = ar = a(0.4) (2) Solving Eqs. (1) and (2) a = 0.5604 rad/s 2 = 0.560 rad/s 2 /> Ans. a G = 0.2241 m/s 2 = 0.224 m/s 2 —» Ans. Ans: a = 0.560 rad/s 2 a G = 0.224 m/s 2 —> 891 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-101. If the coefficient of static friction at C is = 0.3, determine the largest force F that can be applied to the 5-kg ring, without causing it to slip. Neglect the thickness of the ring. SOLUTION Equations of Motion: The mass moment of inertia of the ring about its center of gravity G is I G = mr 2 = 10(0.4 2 ) = 1.60 kg -m 2 . Here, it is required that the ring is on the verge of slipping at C, fy = /jl s N = 0.3 N. Referring to the FBD and kinetic diagram of the ring, Fig. a + t2Fj, = m(a G )y, F sin 45° + N - 10(9.81) = 10(0) (1) i XF x = m{a G ) x \ F cos 45° - 0.3 N = 10 a G (2) C + £M g = I G a; F sin 15°(0.4) - 0.3 N(0.4) = -1.60a (3) Kinematics. Since the ring rolls without slipping, a G = ar = a(0.4) (4) Solving Eqs. (1) to (4), F = 42.34 N = 42.3 N Ans. N = 68.16 N a = 2.373 rad/s 2 2 a G = 0.9490 m/s 2 -» Ans: F = 42.3 N 892 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17-102. The 25-lb slender rod has a length of 6 ft. Using a collar of negligible mass, its end A is confined to move along the smooth circular bar of radius 3\/2 ft. End B rests on the floor, for which the coefficient of kinetic friction is /jl b = 0.4. If the bar is released from rest when 8 = 30°, determine the angular acceleration of the bar at this instant. SOLUTION i = m(a G ) x ; -0.4 N B + N A cos 45° = (a G ) x + t SF V = m{a G ) y \ N B - 25 - N A sin 45° = (a G ) y C + tM G = I G a; N b ( 3 cos 30°) - 0.4 N B (3 sin 30°) + ^sinl5°(3)= i 1 2 ( 3 2 2 5 2 )(6)^ a fl = a A + a B/A a B = a A + 6“ (+$$ 0 = —a A sin 45° + 6a(cos 30°)
a A = 7.34847a
a c = + a GM

( ci G ) x + ( a G ) y = 7.34847a + 3a
<- J, ^45° M 30°

(i) ( a G ) x = -5.196a + 1.5a = -3.696a

(+ i) (a G ) y = 5.196a - 2.598a = 2.598a

Solving Eqs. (l)-