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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 







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21 






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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 



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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 



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* 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 





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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 












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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 








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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 








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* 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 












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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 










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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 



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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 









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* 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 












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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 









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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 


















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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 






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* 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 


















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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 





















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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 











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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 
























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* 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 



















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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 




























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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 
















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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 





















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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 




















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* 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 




















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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 




























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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 


v dv = ads 


/»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 


















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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 






















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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 










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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 













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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 


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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 


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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 


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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 


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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') 



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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 


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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 . 


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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 


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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 


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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 


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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 . 


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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 


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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° 


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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 


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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) 


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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 


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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) 


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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 


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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 


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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 


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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 







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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 







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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 








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* 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 


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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 










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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 


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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 


























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*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 


















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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 










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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 


















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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 






















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* 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° ^ 


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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 















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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 










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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 










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*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 


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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 






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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 







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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 



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* 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 



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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 


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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 


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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 


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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 


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* 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 


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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 


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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 


















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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° ^ 


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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° 


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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° ^ 


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*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 


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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 



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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 








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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 °^ 


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* 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 








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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 


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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 




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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 



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* 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 


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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 


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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 



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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 


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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. 


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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 



















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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 







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* 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 






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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 


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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 


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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 


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* 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 


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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 


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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 


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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 


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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 


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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 


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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 


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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 



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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 


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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 


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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 


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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 


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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 


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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° 


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* 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 


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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 


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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 


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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 


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* 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° 


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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 






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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 


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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 







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* 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 


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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 



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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 


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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 


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* 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 










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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 



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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 





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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 








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* 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 






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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 



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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 





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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 




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* 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 










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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 




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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 






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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 














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* 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 














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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 














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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. 


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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 


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* 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 






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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 


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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 


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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 


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* 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 


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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 



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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 

















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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 


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* 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 


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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 






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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 






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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. 


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* 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 


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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 












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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 


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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 










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* 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 










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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 


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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. 


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* 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 —» 


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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 


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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 


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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 -» 


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* 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 


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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 


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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 


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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 


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* 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' 


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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. 


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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 


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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 


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* 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 | 


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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 


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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 \ 


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* 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. 


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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 


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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° ^ 


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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 


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* 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+ 




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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 ] 


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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 


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* 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° 


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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 


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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° 


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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 


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* 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 


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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 


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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° ^ 


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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^ 


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*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°?" 


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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 


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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 


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* 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 



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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. 


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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° 


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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 


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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 


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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 


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* 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 


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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 


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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 


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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 


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*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 


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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 


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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 



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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 


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*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 


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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 —» 


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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 


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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 


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* 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 


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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 


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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 


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* 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 


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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 ° 


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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 | 


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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 


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* 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 


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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 


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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 


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* 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 


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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 


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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 


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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 


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* 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 


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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/ 


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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 


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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 


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* 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 


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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 


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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 


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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 



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* 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 




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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 


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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. 


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*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 


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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 


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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 


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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 


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*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 


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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 


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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 






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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 


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*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 


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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 


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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 


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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 


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* 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 


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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 







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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 


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* 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 


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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 


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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. [1] 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 


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* 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. [1] 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 


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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 


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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 


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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° 


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* 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 


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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 


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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 


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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 



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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° 


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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 


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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 


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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 


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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=6 rad/s 
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 


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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 


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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 
0 = 1.3043 rad/s 

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 


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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° 

The positive sign indicates that ip is measured from extended radial line in positive 
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 <■> 

6 = — rad/s 2 
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 


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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° 

The negative sign indicates that if> is measured from extended radial line in the 
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 


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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 


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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 


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* 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 


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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 


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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 

















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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 


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* 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 


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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 




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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 


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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 


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* 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 


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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 


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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 


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* 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 


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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 


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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 


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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 


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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 


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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 


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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 


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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: 



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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 


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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 


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*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 


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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 
















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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 


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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 


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* 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 


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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 


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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 


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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 


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*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 


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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 


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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 


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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 


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* 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 


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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 


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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 


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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 


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* 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 


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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 



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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. 



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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 























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* 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 


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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 


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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 


















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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 




















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* 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 


















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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 























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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 










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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 


















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* 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 










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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 

















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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 





































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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 























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* 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 









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397 







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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 
















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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 




















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* 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. [1], (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. [2]). Substituting these 
values into Eq. [3] 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 



































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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 


















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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 





























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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 





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* 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 


















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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 
















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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 













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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 














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* 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 



















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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 












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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 















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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 


















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412 






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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 
































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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 











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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 


























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* 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 























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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 







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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 




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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 





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* 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 










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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 




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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 











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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 














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* 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 













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425 








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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 


























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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 




























































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* 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 


























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429 









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430 








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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 






























































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* 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 



















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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 


















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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 











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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 











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* 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 










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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 









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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 










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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 













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* 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 







































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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 























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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 









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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 

































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* 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 


























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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 


























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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 














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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 
















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* 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 
















































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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 






















































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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 


































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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 

































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* 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 









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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 































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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 

















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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 





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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 












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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 
















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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 














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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 




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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 



















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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 







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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 













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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 
















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* 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 























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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 


















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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 



















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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 















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*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 

































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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 





























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470 


















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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 















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*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 





















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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 




































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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 





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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 




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476 

















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* 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 




















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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 








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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 






























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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 







































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* 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 



















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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 








































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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 





































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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 











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* 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 



















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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 






















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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 


















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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 



































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* 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 



































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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 


































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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 




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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 












































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


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


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 



















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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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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. 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 













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*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 











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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 










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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 

















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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 










































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* 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 





































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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 




































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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 











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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 





























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* 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 






















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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 






















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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 














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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 





















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* 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 













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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 















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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 



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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 
















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* 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 














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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 











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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 











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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 
















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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. [1] and [2] yields 


v t = 8.62 m/s Ans. 

v b = 6.619 m/s 


Ans: 

v t = 8.62 m/s 


519 












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520 

















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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 




















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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 





































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* 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 










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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 












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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 











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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 








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* 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 



















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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 









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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 







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530 







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* 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 















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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 








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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 




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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 





























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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 




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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 












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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 



































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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 








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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 







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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 










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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 












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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 











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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 











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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 













































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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 
















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546 



























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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 













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



548 










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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 










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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 











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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 









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* 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 





















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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 














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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 
























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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 





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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 














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*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 




























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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 





























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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 



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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 














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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 






































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562 























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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 























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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 









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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 












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*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 









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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 












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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 




















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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 



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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 
















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*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 









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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 















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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 

























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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 






















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* 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 





















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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 





















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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 


































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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 


































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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 















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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 











































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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 














































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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 









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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 










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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 







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585 





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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 








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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 























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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 







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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 














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590 





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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 












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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 










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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 









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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 


















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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 


































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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 























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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 













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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 






























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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 


































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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 







































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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 































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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 













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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 






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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 








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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 

























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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 







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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 












































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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 






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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 


























































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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 


















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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 























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612 







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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 














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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 



















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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 









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* 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 










































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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 









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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 





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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 

















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620 













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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 


















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622 

















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623 





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* 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 





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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 
co = 3.25 rad/s 
a = 5 rad/s 2 

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 




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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 = 


’ 12 rad/s 


(3 1 1 + 12 )dt 


co - 12 = (t 3 + 12 1) 


2 s 

0 


co = 44.0 rad/s 

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 






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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] 

co = 25.43 rad/s 

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 





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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 







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* 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 




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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 


5 rad/s 
(0 A 

5 rad/s 

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) 

to c = 21.982 rad/s 

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] 
w A = 13.2588 rad/s 

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

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

(joq = 16.5735 rad/s 
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 

{10 1 2 } rad/s 


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 



rad 


0 = y(3 3 ) = 90.0 rad 


Ans. 


a = (10f) m/s 2 



Ans: 

a = 60 rad / s 2 
co = 90.0 rad/s 
0 = 90.0 rad 


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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 


= 300 rad/s 


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 


^ = 600 rad/s 


Ans. 


Ans: 

co B = 300 rad/s 
<o c = 600 rad/s 


641 




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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: 

co = 42.7 rad/s 
6 = 42.7 rad 


643 






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* 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 
co = 3.584 rad/s 

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 










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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 — 




156 rad/s 


Ans. 


Ans: 

w B = 156 rad/s 


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, 
w A = 36 rad/s 


Wa'a = w Br B 


36(0.05) = w B (0.15) 

w B = 12 rad/s Ans. 


Ans: 

w B = 12 rad/s 


649 



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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: 

co D = 4.00 rad/s 
a D = 0.400 rad/s 2 


651 













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* 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 


dco = adt 
/» 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, 

|(506.25) = 211 rad/s Ans. 


rA 

<*>b = — W A 
r R 


05 

L2 


Ans: 

co B = 211 rad/s 


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 


da> = adt 
/■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) 


13.18 rad/s 


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


70.31 rad/s 2 


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 
d A = 6 rad 

<°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: 

co c = 1.68 rad/s 
6 C = 1-68 rad 


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: 

co = 148 rad/s 


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 


dco = adt 



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) 


64.95 rad/s 


‘ (u) (25, - sl) 


129.9 rad/s 2 


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(^) 

Mmax = 8.49 rad/s Ans. 

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

v c = 0.6 m/s Ans. 


Ans: 

(mb )max = 8.49 rad/s 
(«c)max = 0.6 m/s 


659 












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662 













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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 
8 = 6 rad. 


SOLUTION 


i dco = a dd 


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

co = 20 
At 0 = 6 rad, 

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 


co = 28.87 rad/s = 28.9 rad/s 

Ans. 

Subsequently, Eq. (2) gives 


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


a = -469.57 rad/s 2 = 470 rad/s 2 *) 

Ans. 


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




Ca> 


Ans: 

to = 28.9 rad/s /) 
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: 

t<> = 8.70 rad/s 
a = —50.5 rad/s 2 


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 
link CD at this instant. 


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) 

a = —183 rad/s 2 



Ans. 


Ans. 


Ans: 

co = —19.2 rad/s 
a = -183 rad/s 2 


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. [1], 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. [3], 
dt dt 

4> = 60.0°. Substitute these values into Eq. [2] 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 




















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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 



























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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 










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* 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 





















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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 











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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 
























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680 







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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 





























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* 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 





































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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 























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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 





























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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 









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* 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 





























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687 








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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 














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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 
























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* 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 
































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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 





























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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 














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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 


















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*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 








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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 








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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 





















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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 


























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* 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 







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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 




















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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 

























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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 






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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 















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* 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 
















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*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 



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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 




















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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 

















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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 












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*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 












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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 









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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 











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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 ) [1] 

0.500 v c = 14.14 - 0.2598 at AC [2] 


Solving Eqs. [1] and [2] 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. [3] and [4] yields 

wcd — 157.74 rad/s 

v D = 7.07 m/s Ans. 


Ans: 

v D = 7.07 m/s 


711 













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* 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 













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*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 





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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 











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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 













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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 


















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*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 










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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 






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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 














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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 































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* 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 

















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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 





















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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 















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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 













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*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 




































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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 


































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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 







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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 











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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 












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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 










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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 










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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 










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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 



















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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 















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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 


















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16-103. 



736 




















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*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 







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738 








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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 

































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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 











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* 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 


















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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 

























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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 






























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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 

























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* 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 























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*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 










































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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 
















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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 




















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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 












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* 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 












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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 











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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 















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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 








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* 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 *) 


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755 











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756 





































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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 


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758 






































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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 





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*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 













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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 








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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 


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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 



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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 


























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* 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 




































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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 









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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 





















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768 







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769 







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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 












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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 






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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 








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*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 








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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 























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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 



















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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 












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*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 














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*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 


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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 {) 


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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 


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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 





























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*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 


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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 







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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 







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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 











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*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 


















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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 













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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 






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789 








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*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 








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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 






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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 





















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793 







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*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 


















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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 






















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796 




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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 













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*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 










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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 








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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 






























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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 




























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802 











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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 















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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 






















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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 





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* 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 


















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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 





















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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 














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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 





















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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 












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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 


















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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 
































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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 


































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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 

































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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 
































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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 














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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 














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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 
























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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 






















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



820 










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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 





































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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 



































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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 

























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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 






















































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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 






















































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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 












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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 


















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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 



















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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 












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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 




















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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 











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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 


























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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 























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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 
















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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 















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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 




















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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 





















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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 




























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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 














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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 







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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 





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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 











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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 












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



856 









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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 


























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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 































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859 












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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 













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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 











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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 





















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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 












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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 










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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 


















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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 















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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 





















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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 




































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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 












































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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 
























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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 




















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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 






























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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 













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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 





















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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 




































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* 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 
































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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 

















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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 





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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 



























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* 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 























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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 




















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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 



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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 








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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 








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*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 











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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 














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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 












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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 



































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*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 



















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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 














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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)-