CHAPTER 11
PROBLEM 11.1
A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS
coordinates are used to determine his displacement as a function of time: x= 0.5t3 + t2 + 2t where x and t are
expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder
when t = 5 seconds.
SOLUTION
Position:
Velocity:
Acceleration:
At t  5 s,
x  0.5t 3  t 2  2t
v
dx
 1.5t 2  2t  2
dt
a
dv
 3t  2
dt
x  0.5  5  52  2  5
x  97.5 ft 
v  1.5 5  2  5  2
v  49.5 ft/s 
3
2
a  3  5  2 
a  17 ft/s 2 
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PROBLEM 11.2
The motion of a particle is defined by the relation x  2t 3  9t 2  12t  10, where x and t are expressed in feet
and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v  0.
SOLUTION
x  2t 3  9t 2  12t  10
Differentiating,
v
dx
 6t 2  18t  12  6(t 2  3t  2)
dt
 6(t  2)(t  1)
a
dv
 12t  18
dt
So v  0 at t  1 s and t  2 s.
At t  1 s,
x1  2  9  12  10  15
a1  12  18  6
t  1.000 s 
x1  15.00 ft 
a1  6.00 ft/s 2 
At t  2 s,
x2  2(2)3  9(2)2  12(2)  10  14
t  2.00 s 
x2  14.00 ft 
a2  (12)(2)  18  6
a2  6.00 ft/s 2 
PROBLEM 11.3
The vertical motion of mass A is defined by the relation x  10 sin 2t  15cos 2t  100,
where x and t are expressed in mm and seconds, respectively. Determine (a) the position,
velocity and acceleration of A when t  1 s, (b) the maximum velocity and acceleration of A.
SOLUTION
x  10sin 2t  15cos 2t  100
v
dx
 20cos 2t  30sin 2t
dt
a
dv
 40sin 2t  60 cos 2t
dt
For trigonometric functions set calculator to radians:
(a) At t  1 s.
x1  10sin 2  15cos 2  100  102.9
v1  20cos 2  30sin 2  35.6
a1  40sin 2  60cos 2  11.40
x1  102.9 mm 
v1  35.6 mm/s 
a1  11.40 mm/s2 
(b) Maximum velocity occurs when a  0.
40sin 2t  60 cos 2t  0
tan 2t  
60
 1.5
40
2t  tan 1 (1.5)  0.9828 and 0.9828  
Reject the negative value. 2t  2.1588
t  1.0794 s
t  1.0794 s for vmax
so
vmax  20 cos(2.1588)  30sin(2.1588)
 36.056
vmax  36.1 mm/s 
Note that we could have also used
vmax  202  302  36.056
by combining the sine and cosine terms.
For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms.
amax  402  602  72.1 mm/s2
amax  72.1 mm/s 2 
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PROBLEM 11.4
A loaded railroad car is rolling at a constant velocity
when it couples with a spring and dashpot bumper
system. After the coupling, the motion of the car is
defined by the relation x  60e4.8t sin16t where x and t
are expressed in mm and seconds, respectively.
Determine the position, the velocity and the
acceleration of the railroad car when (a) t  0,
(b) t  0.3 s.
SOLUTION
x  60e4.8t sin16t
dx
 60( 4.8)e 4.8t sin16t  60(16)e 4.8t cos16t
dt
v  288e 4.8t sin16t  960e 4.8t cos16t
v
a
dv
 1382.4e4.8t sin16t  4608e4.8t cos16t
dt
 4608e4.8t cos16t  15360e4.8t sin16t
a  13977.6e4.8t sin16t  9216e4.8 cos16t
(a) At t  0,
x0  0
v0  960 mm/s
a0  9216 mm/s 2
(b) At t  0.3 s,
x0  0 mm 
v0  960 mm/s
a0  9220 mm/s2


e 4.8t  e 1.44  0.23692
sin16t  sin 4.8  0.99616
cos16t  cos 4.8  0.08750
x0.3  (60)(0.23692)(0.99616)  14.16
x0.3  14.16 mm

v0.3  87.9 mm/s

v0.3  (288)(0.23692)(0.99616)
 (960)(0.23692)(0.08750)  87.9
a0.3  (13977.6)(0.23692)(0.99616)
 (9216)(0.23692)(0.08750)  3108
a0.3  3110 mm/s 2 
or 3.11 m/s2

PROBLEM 11.5
The motion of a particle is defined by the relation x  6t 4  2t 3  12t 2  3t  3, where x and t are expressed in
meters and seconds, respectively. Determine the time, the position, and the velocity when a  0.
SOLUTION
We have
x  6t 4  2t 3  12t 2  3t  3
Then
v
dx
 24t 3  6t 2  24t  3
dt
and
a
dv
 72t 2  12t  24
dt
When a  0:
72t 2  12t  24  12(6t 2  t  2)  0
(3t  2)(2t  1)  0
or
t
or
At t 
2
s:
3
2
1
s and t   s (Reject)
3
2
4
3
2
2
2
2
2
x2/3  6    2    12    3    3
3
3
3
3
3
v2/3
t  0.667 s 
2
2
2
2
 24    6    24    3
3
3
3
or
x2/3  0.259 m 
or v2/3  8.56 m/s 
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PROBLEM 11.6
The motion of a particle is defined by the relation x  t 3  9t 2  24t  8, where x and t are expressed in inches
and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance
traveled when the acceleration is zero.
SOLUTION
We have
x  t 3  9t 2  24t  8
Then
v
dx
 3t 2  18t  24
dt
and
a
dv
 6 t  18
dt
(a)
When v  0:
3 t 2  18t  24  3(t 2  6t  8)  0
(t  2)(t  4)  0
t  2.00 s and t  4.00 s 
(b)
When a  0:
6t  18  0 or t  3 s
x3  (3)3  9(3)2  24(3)  8
At t  3 s:
First observe that 0  t  2 s:
v0
2 s  t  3 s:
v0
or
x3  10.00 in. 
Now
At t  0:
x0  8 in.
At t  2 s:
x2  (2)3  9(2)2  24(2)  8  12 in.
Then
x2  x0  12  ( 8)  20 in.
| x3  x2 |  |10  12|  2 in.
Total distance traveled  (20  2) in.
Total distance  22.0 in. 
PROBLEM 11.7
A girl operates a radio-controlled model car in a vacant
parking lot. The girl’s position is at the origin of the xy
coordinate axes, and the surface of the parking lot lies in the
x-y plane. She drives the car in a straight line so that the x
coordinate is defined by the relation x(t) = 0.5t3 - 3t2 + 3t +
2, where x and t are expressed in meters and seconds,
respectively. Determine (a) when the velocity is zero, (b)
the position and total distance travelled when the
acceleration is zero.
SOLUTION
Position:
x  t   0.5t 3  3t 2  3t  2
Velocity:
v t  
dx
dt
v  t   1.5t 2  6t  3
(a) Time when v=0
0  1.5t 2  6t  3
t
Acceleration:
6  62  4 1.5  3
a t  
t  0.586 s and t  3.414 s 
2  1.5
dv
dt
a  t   3t  6
Time when a=0
0  3t  6
t2s
(b) Position at t=2 s
x  2   0.5  2   3  2   3  2  2
3
2
x  2  0 m 
To find total distance note that car changes direction at t=0.586 s
Position at t=0 s
x  0   0.5  0   3  0   3  0  2
3
2
x  0  2
Position at t=0.586 s
x  0.586   0.5  0.586   3  0.586   3  0.586  2
3
2
x  0.586  2.828 m
Distances traveled:
From t=0 to t=0.586 s:
x  0.586   x  0   0.828 m
From t=0.586 to t=2 s:
x  2   x  0.586   2.828 m
Total distance traveled  0.828 m  2.828 m
Total distance  3.656 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.8
The motion of a particle is defined by the relation x  t   t  2 , where x and t are expressed in feet and
seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance
traveled by the particle from t  0 to t  4 s..
3
2
SOLUTION
Position:
x  t   t 2   t  2
Velocity:
v t  
3
dx
dt
v  t   2t  3  t  2 
2

v  t   2t  3 t 2  4t  4

 3t 2  14t  12
Time when v(t)=0
0  3t 2  14t  12
t
14  142  4  3 12 
2  3
t  1.131 s and t  3.535 s
(a) Position at t=1.131 s
x 1.131  1.131  1.1.31  2 
2
3
x 1.131  1.935 ft. 
Position at t=3.535 s
x  3.535    3.535    3.535  2 
2
3
x  3.531  8.879 ft. 
To find total distance traveled note that the particle changes direction at t=1.131 s and again at t=3.535 s.
Position at t=0 s
x 0  0  0  2
2
3
x  4   4   4  2
3
x  0  8 ft
Position at t=4 s
2
x  4  8 ft
(b) Distances traveled:
From t=0 to t=1.131 s:
x 1.131  x  0   6.065 ft.
From t=1.131 to t=3.531 s:
x  3.535   x 1.131  6.944 ft.
From t=3.531 to t=4 s:
x  4   x  3.535   0.879 ft.
Total distance traveled  6.065 ft  6.944 ft  0.879 ft
Total distance  13.888 ft. 
PROBLEM 11.9
The brakes of a car are applied, causing it to slow down at a rate
of 10 ft/s2. Knowing that the car stops in 100 ft, determine
(a) how fast the car was traveling immediately before the brakes
were applied, (b) the time required for the car to stop.
SOLUTION
a  10 ft/s 2
(a)
Velocity at x  0.
v

0
v0
0
(b)
dv
 a  10
dx
vdv  

xf
0
(10)dx
v02
 10 x f  (10)(300)
2
v02  6000
v0  77.5 ft/s2 
Time to stop.
dv
 a  10
dx

0
dv  
v0

tf
0
10dt
0  v0  10t f
tf 
v0 77.5

10
10
t f  7.75 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.10
The acceleration of a particle is defined by the relation a  3e  0.2t , where a and t are expressed in ft/s 2 and
seconds, respectively. Knowing that x 0 and v 0 at t 0, determine the velocity and position of the
particle when t 0.5 s.
SOLUTION
Acceleration:
a  3e0.2t ft/s2
Given :
v0  0 ft/s, x0  0 ft
Velocity:
a
dv
 dv  adt
dt
v
t
v0
0
 dv   adt
t
v  vo 
 3e
0.2t
0

t
dt  v  0=  15 e0.2t
0

v  15 1  e0.2t ft/s
Position:
v
dx
 dx  vdt
dt
x

t

dx  vdt
0
x0
t
 


x  xo  15 1  e 0.2t dt  x  0=15 t  5 e 0.2t

0

t
0

x  15 t  5 e 0.2t  75 ft
Velocity at t=0.5 s
Position at t=0.5 s


v  15 1  e0.20.5 ft/s

v  0.5  1.427 ft/s 

x  15 0.5  5 e 0.20.5  75 ft
x  0.5  0.363 ft. 
PROBLEM 11.11
The acceleration of a particle is directly proportional to the square of the time t. When t  0, the particle is
at x  24 m. Knowing that at t  6 s, x  96 m and v  18 m/s, express x and v in terms of t.
SOLUTION
a  kt 2
We have
k  constant
dv
 a  kt 2
dt
Now
v

or
1
v  18  k (t 3  216)
3
18
dv 

t
At t  6 s, v  18 m/s:
6
kt 2 dt
1
v  18  k (t 3  216)(m/s)
3
or
dx
1
 v  18  k (t 3  216)
dt
3
Also
x
1

18  k (t 3  216)  dt
3


or
1 1

x  24  18t  k  t 4  216t 
3 4

24
dx 

t
At t  0, x  24 m:
0

Now
At t  6 s, x  96 m:
or
Then
or
and
or
1 1

96  24  18(6)  k  (6)4  216(6) 
3 4

k
1
m/s 4
9
1  1  1

x  24  18t    t 4  216t 
3  9  4

x(t ) 
1 4
t  10t  24
108

11
v  18    (t 3  216)
3 9 
v(t ) 
1 3
t  10
27
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
PROBLEM 11.12
The acceleration of a particle is defined by the relation a  kt 2 . (a) Knowing that v  8 m/s when t  0
and that v  8 m/s when t  2 s, determine the constant k. (b) Write the equations of motion, knowing also
that x  0 when t  2 s.
SOLUTION
a  kt 2
dv
 a  kt 2
dt
(1)
t  0, v  8 m/s and t  2 s, v  8 ft/s

(a)
8
8
dv 

2
0
kt 2 dt
1
8  ( 8)  k (2)3
3
(b)
k  6.00 m/s 4 
Substituting k  6 m/s4 into (1)
dv
 a  6t 2
dt
t  0, v  8 m/s:

v
8
dv 

t
0
a  6t 2 
6t 2 dt
1
v  (8)  6(t )3
3
v  2t 3  8 
dx
 v  2t 3  8
dt
t  2 s, x  0:

x
0
dx 

t
2
(2t 3  8) dt; x 
1 4
t  8t
2
1
 1

x   t 4  8t    (2)4  8(2) 
2
 2

1
x  t 4  8t  8  16
2
t
2
x
1 4
t  8t  8 
2
PROBLEM 11.13
A Scotch yoke is a mechanism that transforms the circular motion of a
crank into the reciprocating motion of a shaft (or vice versa). It has been
used in a number of different internal combustion engines and in control
valves. In the Scotch yoke shown, the acceleration of Point A is defined by
the relation a 1.8sin kt, where a and t are expressed in m/s2 and seconds,
respectively, and k 3 rad/s. Knowing that x 0 and v 0.6 m/s when t 0,
determine the velocity and position of Point A when t 0.5 s.
SOLUTION
Acceleration:
a  1.8sin kt m/s2
Given :
v0  0.6 m/s, x0  0,
Velocity:
dv
a
 dv  adt 
dt
t
k  3 rad/s
v
t
v0
0
 dv   adt
t
v  v0   0 a dt  1.8  0 sin kt dt 
v  0.6 
1.8
cos kt
k
t
0
1.8
(cos kt  1)  0.6 cos kt  0.6
3
v  0.6cos kt m/s
Position:
v
dx
 dx  vdt 
dt
x  x0 
x0
x

t

dx  vdt
x0
0
0.6
t
t
 0 v dt  0.6  0 cos kt dt  k sin kt
t
0
0.6
(sin kt  0)  0.2sin kt
3
x  0.2sin kt m
When t =0.5 s,
kt  (3)(0.5)  1.5 rad
v  0.6 cos1.5  0.0424 m/s
v  42.4 mm/s 
x  0.2sin1.5  0.1995 m
x  199.5 mm 
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PROBLEM 11.14
For the scotch yoke mechanism shown, the acceleration of Point A is defined by the
relation a  1.08sin kt  1.44cos kt , where a and t are expressed in m/s2 and
seconds, respectively, and k = 3 rad/s. Knowing that x = 0.16 m and v = 0.36 m/s
when t = 0, determine the velocity and position of Point A when t = 0.5 s.
SOLUTION
Acceleration:
a  1.08sin kt  1.44cos kt m/s2
Given :
v0  0.36 m/s, x0  0.16,
Velocity:
dv
a
 dv  adt 
dt
Integrate:
k  3 rad/s
v
t
v0
0
 dv   adt
t
t
v  v0  1.08  0 sin kt dt  1.44  0 cos kt dt
v  0.36 

1.08
cos kt
k
t
0

1.44
sin kt
k
t
0
1.08
1.44
(cos 3t  1) 
(sin 3t  0)
3
3
 0.36 cos 3t  0.36  0.48sin 3t
v  0.36 cos 3t  0.48sin 3t m/s
Evaluate at t  0.5 s
 0.36 cos1.5  0.36  0.48sin1.5
Position:
v
dx
 dx  vdt 
dt
x

t

dx  vdt
0
x0
t
v  453 mm/s 
t
t
x  x0   0 v dt  0.36  0 cos kt dt  0.48  0 sin kt dt
x  0.16 
0.36
sin kt
k
t
0

0.48
cos kt
k
t
0
0.36
0.48
(sin 3t  0) 
(cos 3t  1)
3
3
 0.12sin 3t  0.16 cos 3t  0.16

x  0.12sin 3t  0.16 cos 3t m
Evaluate at t  0.5 s
x  0.12sin1.5  0.16 cos1.5  0.1310 m
x  131.0 mm 
PROBLEM 11.15
A piece of electronic equipment that is surrounded by
packing material is dropped so that it hits the ground with a
speed of 4 m/s. After contact the equipment experiences an
acceleration of a  kx, where k is a constant and x is the
compression of the packing material. If the packing material
experiences a maximum compression of 20 mm, determine
the maximum acceleration of the equipment.
SOLUTION
a
vdv
 k x
dx
Separate and integrate.

vf
v0
vdv  

xf
0
k x dx
1 2 1 2
1
v f  v0   kx 2
2
2
2
xf
0
1
  k x 2f
2
Use v0  4 m/s, x f  0.02 m, and v f  0. Solve for k.
0
1 2
1
(4)   k (0.02) 2
2
2
k  40, 000 s 2
Maximum acceleration.
amax  kxmax : (40,000)(0.02)  800 m/s2
a  800 m/s2 
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PROBLEM 11.16
A projectile enters a resisting medium at x  0 with an initial velocity
v0  900 ft/s and travels 4 in. before coming to rest. Assuming that the
velocity of the projectile is defined by the relation v  v0  kx, where v is
expressed in ft/s and x is in feet, determine (a) the initial acceleration of the
projectile, (b) the time required for the projectile to penetrate 3.9 in. into the
resisting medium.
SOLUTION
First note
When x 
 4 
0  (900 ft/s)  k  ft 
 12 
4
ft, v  0:
12
k  2700
or
(a)
1
s
We have
v  v0  kx
Then
a
or
a  k (v0  kx)
At t  0:
1
a  2700 (900 ft/s  0)
s
dv d
 (v0  kx)   kv
dt dt
a0  2.43  106 ft/s2 
or
(b)
dx
 v  v0  kx
dt
We have
At t  0, x  0:
or

x
0
dx

v0  kx
t
 dt
0
1
 [ln(v0  kx )]0x  t
k
or
t
1  v0  1  1 

ln 
  ln 
k  v0  kx  k  1  vk x 
0


When x  3.9 in.:
t

1
ln 
1
2700 s 1 
or
1
2700 1/s
900 ft/s

3.9
12

ft  
t  1.366  103 s 
PROBLEM 11.17
The acceleration of a particle is defined by the relation a   k/x. It has been experimentally determined that
v  15 ft/s when x  0.6 ft and that v  9 ft/s when x  1.2 ft. Determine (a) the velocity of the particle
when x  1.5 ft, (b) the position of the particle at which its velocity is zero.
SOLUTION
a
vdv  k

dx
x
Separate and integrate using x  0.6 ft, v  15 ft/s.

v
15
vdv   k

x
0.6
v
dx
x
1 2
 k ln x
v
2 15
x
0.6
1 2 1
 x 
v  (15) 2  k ln 

2
2
 0.6 
(1)
When v  9 ft/s, x  1.2 ft
1 2 1
 1.2 
(9)  (15)2  k ln 

2
2
 0.6 
Solve for k.
k  103.874 ft 2 /s 2
(a)
Velocity when x  65 ft.
Substitute
k  103.874 ft 2 /s2
and x  1.5 ft into (1).
1 2 1
 1.5 
v  (15)2  103.874 ln 

2
2
 0.6 
v  5.89 ft/s 
(b)
Position when for v  0,
1
 x 
0  (15)2  103.874 ln 

2
 0.6 
 x 
ln 
  1.083
 0.6 


 x  1.772 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.18
A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass
tube under the magnetic repelling force of another steel magnet C located at a
distance x  0.004 m from B. The force is inversely proportional to the square of
the distance between B and C. If block A is suddenly removed, the acceleration
of block B is a  9.81  k /x 2 , where a and x are expressed in m/s2 and m,
respectively, and k  4  104 m3 /s2 . Determine the maximum velocity and
acceleration of B.
SOLUTION
0  9.81 
The maximum velocity occurs when a  0.
xm2 
k
4  104

 40.775  106 m 2
9.81
9.81
k
xm2
xm  0.0063855 m
The acceleration is given as a function of x.
v
dv
k
 a  9.81  2
dx
x
Separate variables and integrate:
vdv  9.81dx 

v
0

vdv  9.81
x
x0
k dx
x2
dx  k

x
x0
dx
x2
1 1 
1 2
v  9.81( x  x0 )  k   
2
 x x0 
 1
1 2
1 
vm  9.81( xm  x0 )  k 
 
2
 xm x0 
1
1 

 9.81(0.0063855  0.004)  (4  104 ) 


 0.0063855 0.004 
 0.023402  0.037358  0.013956 m 2 /s 2
Maximum velocity:
vm  0.1671 m/s
vm  167.1 mm/s 
The maximum acceleration occurs when x is smallest, that is, x  0.004 m.
am  9.81 
4  104
(0.004)2
am  15.19 m/s2

PROBLEM 11.19
Based on experimental observations, the acceleration of a particle is defined by the relation a  (0.1  sin x/b),
where a and x are expressed in m/s2 and meters, respectively. Knowing that b  0.8 m and that v  1 m/s
when x  0, determine (a) the velocity of the particle when x  1 m, (b) the position where the velocity is
maximum, (c) the maximum velocity.
SOLUTION
We have
When x  0, v  1 m/s:
v

v
1
dv
x 

 a    0.1  sin
dx
0.8 

vdv 

x 

  0.1  sin
dx
0 
0.8 
x
x
1 2
x 

(v  1)   0.1x  0.8 cos
2
0.8  0

or
1 2
x
v  0.1x  0.8 cos
 0.3
2
0.8
or
(a)
When x  1 m:
1 2
1
 0.3
v  0.1(1)  0.8 cos
2
0.8
v   0.323 m/s 
or
(b)
When v  vmax, a  0:
or
(c)
x 

  0.1  sin
0
0.8 

x  0.080 134 m
x  0.0801 m 
When x  0.080 134 m:
1 2
0.080 134
vmax  0.1(0.080 134)  0.8 cos
 0.3
2
0.8
 0.504 m 2 /s 2
or
vmax  1.004 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.20
A spring AB is attached to a support at A and to a collar. The
unstretched length of the spring is l. Knowing that the collar is
released from rest at x  x0 and has an acceleration defined by
the relation a  100( x  lx / l 2  x 2 ) , determine the velocity of
the collar as it passes through Point C.
SOLUTION
av
Since a is function of x,

dv
lx
 100  x 

dx
l 2  x2





Separate variables and integrate:

vf
v0
vdv  100


lx
x
2
x0 
l  x2

0
 x2
1 2 1 2
v f  v0  100 
 l l 2  x2
2
2
 2

 dx





0
x0
 x2
1 2
v f  0  100   0  l 2  l l 2  x02
 2
2




1 2 100 2
(l  x02  l 2  2l l 2  x02 )
vf 
2
2
100
( l 2  x02  l ) 2

2
v f  10( l 2  x02  l ) 
PROBLEM 11.21


The acceleration of a particle is defined by the relation a  k 1  e  x , where k is a constant. Knowing that
the velocity of the particle is v   9 m/s when x   3 m and that the particle comes to rest at the origin,
determine (a) the value of k, (b) the velocity of the particle when x   2 m.
SOLUTION


Acceleration:
a  k 1  e x
Given :
at x   3 m, v  9 m/s
at x  0 m, v  0 m/s
adx  vdv

k 1 e
x
 dx  vdv
Integrate using x=-3 m and v=9 m/s as the lower limits of the integrals
x
 
v


k 1  e x dx  vdv
3
9

k x  e x
Velocity:


x
v

3
1 2
v
2 9

k x  e  x  3  e3
  12 v
2

1 2
(9)
2
(1)
(a) Now substitute v=0 m/s and x=0 m into (1) and solve for k


k 0  e 0  3  e3
  12 0
2

1 2
(9)
2
k  2.52 m2 /s 2 
(b) Find velocity when x= -2 m using the equation (1) and the value of k


2.518 2  e 2  3  e3
  12 v
2
1
 (9) 2
2
v  4.70 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.22
2
Starting from x  0 with no initial velocity, a particle is given an acceleration a  0.1 v  16, where a
2
and v are expressed in ft/s and ft/s, respectively. Determine (a) the position of the particle when v 
3ft/s, (b) the speed and acceleration of the particle when x  4 ft.
SOLUTION
a
vdv
 0.1(v 2  16)1/ 2
dx
(1)
Separate and integrate.

v
0
vdv
v 2  16


x
0
0.1 dx
v
(v 2  16)1/2  0.1 x
0
2
1/2
(v  16)
(a)
(b)
 4  0.1 x
x  10[(v2  16)1/2  4]
(2)
x  10[(32  16)1/2  4]
x  10.00 ft 
v  3 ft/s.
x  4 ft.
From (2),
(v 2  16)1/2  4  0.1x  4  (0.1)(4)  4.4
v 2  16  19.36
v 2  3.36ft 2 /s 2
From (1),
a  0.1(1.8332  16)1/2
v  1.833 ft/s 
a  0.440 ft/s2 
PROBLEM 11.23
A ball is dropped from a boat so that it strikes the surface of a
lake with a speed of 16.5 ft/s. While in the water the ball
experiences an acceleration of a  10  0.8v, where a and v
are expressed in ft/s2 and ft/s, respectively. Knowing the
ball takes 3 s to reach the bottom of the lake, determine (a) the
depth of the lake, (b) the speed of the ball when it hits the
bottom of the lake.
SOLUTION
a
dv
 10  0.8v
dt
Separate and integrate:

dv

v0 10  0.8v
v

t
0
dt
v

1
ln(10  0.8v)  t
0.8
v0
 10  0.8v
ln 
 10  0.8v0

  0.8t

10  0.8v  (10  0.8v0 )e0.8t
0.8v  10  (10  0.8v0 )e0.8t
or
v  12.5  (12.5  v0 )e0.8t
v  12.5  4e0.8t
With v0  16.5ft/s
Integrate to determine x as a function of t.
dx
 12.5  4e0.8t
v
dt

x
0
dx 

t
0
(12.5  4e 0.8t )dt
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.23 (Continued)
t
x  12.5t  5e0.8t  12.5t  5e0.8t  5
0
(a)
At t  35 s,
x  12.5(3)  5e2.4  5  42.046 ft
(b)
v  12.5  4e2.4  12.863 ft/s
x  42.0 ft 
v  12.86 ft/s 
PROBLEM 11.24
The acceleration of a particle is defined by the relation a  k v , where k is a constant. Knowing that x  0
and v  81 m/s at t  0 and that v  36 m/s when x  18 m, determine (a) the velocity of the particle when
x  20 m, (b) the time required for the particle to come to rest.
SOLUTION
(a)
We have
dv
 a  k v
dx
v
v dv  k dx
so that
v

or
2 3/2 v
[v ]81   kx
3
or
2 3/2
[v  729]   kx
3
When x  18 m, v  36 m/s:
v dv 
81

x
When x  0, v  81 m/s:
0
 k dx
2
(363/2  729)  k (18)
3
k  19 m/s 2
or
Finally
When x  20 m:
2 3/2
(v  729)  19(20)
3
v3/2  159
or
(b)
We have
dv
 a  19 v
dt
At t  0, v  81 m/s:

v
dv
81
v


t
0
19dt
or
v
2[ v ]81
 19t
or
2( v  9)  19t
When v  0:
or
v  29.3 m/s 
2(9)  19t
t  0.947 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.25
2.5
The acceleration of a particle is defined by the relation a  kv , where k is a constant. The particle starts at
x  0 with a velocity of 16 mm/s, and when x  6 mm the velocity is observed to be 4 mm/s. Determine
(a) the velocity of the particle when x  5 mm, (b) the time at which the velocity of the particle is 9 mm/s.
SOLUTION
Acceleration:
a  kv 2.5
Given :
at t=0, x  0 mm, v  16 mm/s
at x  6 mm, v  4 mm/s
adx  vdv
2.5
kv dx  vdv
Separate variables
 kdx  v  3 2 dv
Integrate using x=-0 m and v=16 mm/s as the lower limits of the integrals
x

v

-kdx  v 3 2 dv
0
16
x
 2v 1 2
 kx
0
Velocity and position:
kx  2v 1 2 
v
16
1
2
(1)
Now substitute v=4 mm/s and x=6 mm into (1) and solve for k
1
2
k  0.0833 mm 3 2 s1 2
k  6   41 2 
(a) Find velocity when x= 5 mm using the equation (1) and the value of k
k  5  2v 1 2 
1
2
v  4.76 mm/s
(b)
Separate variables
a
dv
or adt  dv
dt
a
dv
or adt  dv
dt
 kdt  v 2.5 dv
PROBLEM 11.25 (Continued)
Integrate using t=0 and v=16 mm/s as the lower limits of the integrals
t
v
 -kdt   v
0
v
kt
0
kt 
dv
16
t
Velocity and time:
5 2
2
  v 3 2
3
16
2 3 2 1

v
3
96
(2)
Find time when v= 9 mm/s using the equation (2) and the value of k
kt 
2 3 2 1
9 
3
96
t  0.171 s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.26
A human powered vehicle (HPV) team wants to model the
acceleration during the 260 m sprint race (the first 60 m is called a
flying start) using a = A – Cv2, where a is acceleration in m/s2 and v
is the velocity in m/s. From wind tunnel testing, they found that
C = 0.0012 m-1. Knowing that the cyclist is going 100 km/h at the
260 meter mark, what is the value of A?
SOLUTION
Acceleration:
a  A  Cv2 m/s2
Given :
C  0.0012 m1 , v f  100 km/hr when x f  260 m
Note: 100 km/hr = 27.78 m/s
adx  vdv
 A  Cv  dx  vdv
2
Separate variables
dx 
vdv
A  Cv 2
Integrate starting from rest and traveling a distance xf with a final velocity vf.
xf
vf
0
0
vdv
 dx   A  Cv
x
x 0f  
2

1
ln A  Cv 2
2C

1
xf  
ln A  v 2f
2C
A
1 
xf 
ln 
2C  A  Cv 2f
Next solve for A
A
Cv 2f e
1 e

vf
0
  21C ln  A




2Cx f
2Cx f
Now substitute values of C, xf and vf and solve for A
A  1.995 m/s2 
PROBLEM 11.27
Experimental data indicate that in a region downstream of a given
louvered supply vent the velocity of the emitted air is defined by
v  0.18v0 /x, where v and x are expressed in m/s and meters,
respectively, and v0 is the initial discharge velocity of the air. For
v0  3.6 m/s, determine (a) the acceleration of the air at x  2 m,
(b) the time required for the air to flow from x  1 to x  3 m.
SOLUTION
(a)
dv
dx
0.18v0 d  0.18v0 

x dx  x 
av
We have

a
When x  2 m:
0.0324v02
x3
0.0324(3.6)2
(2)3
a  0.0525 m/s2 
or
(b)
0.18v0
dx
v
dt
x
We have
From x  1 m to x  3 m:

3
1
xdx 

t3
t1
0.18v0 dt
3
or
1 2
 2 x   0.18v0 (t3  t1 )

1
or
(t3  t1 ) 
or
1
2
(9  1)
0.18(3.6)
t3  t1  6.17 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.28
Based on observations, the speed of a jogger can be approximated by the relation
v  7.5(1  0.04 x)0.3 , where v and x are expressed in mi/h and miles, respectively.
Knowing that x  0 at t  0, determine (a) the distance the jogger has run when t  1 h,
(b) the jogger’s acceleration in ft/s2 at t  0, (c) the time required for the jogger to run 6 mi.
SOLUTION
(a)
dx
 v  7.5(1  0.04 x )0.3
dt
We have

At t  0, x  0:
or
x
0
dx

(1  0.04 x)0.3
t
 7.5dt
0
1  1 
[(1  0.04 x)0.7 ]0x  7.5t



0.7  0.04 
1  (1  0.04 x)0.7  0.21t
or
or
x
1
[1  (1  0.21t )1/0.7 ]
0.04
At t  1 h:
x
1
{1  [1  0.21(1)]1/0.7 }
0.04
(1)
x  7.15 mi 
or
(b)
We have
av
dv
dx
d
[7.5(1  0.04 x)0.3 ]
dx
 7.52 (1  0.04 x)0.3 [(0.3)(0.04)(1  0.04 x)0.7 ]
 7.5(1  0.04 x)0.3
 0.675(1  0.04 x)0.4
At t  0, x  0:
a0  0.675 mi/h 2 
5280 ft  1 h 


1 mi
 3600 s 
a0  275  106 ft/s 2 
or
(c)
From Eq. (1)
t
When x  6 mi:
t
or
2
1
[1  (1  0.04 x)0.7 ]
0.21
1
{1  [1  0.04(6)]0.7 }
0.21
 0.83229 h
t  49.9 min 
PROBLEM 11.29
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
32.2
[1  ( y/20.9  106 )]2
a
where a and y are expressed in ft/s2 and feet, respectively. Using this expression, compute
the height reached by a projectile fired vertically upward from the surface of the earth if
its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s.
SOLUTION
We have
v
dv
a
dy
1 
32.2
y
20.9  106
When
y  0,
provided that v does reduce to zero,
y  ymax , v  0

Then
0
v0
v dv 
v  v0
ymax
0
1 
32.2
y
20.9  106

2
dy

1
 1345.96  10 1 
 1  ymax
20.9  106

or
v02
or
ymax 
v0  1800 ft/s:
ymax 
6
v0  3000 ft/s:
or
y
 max


0




v02
64.4 
v02
20.9  106
(1800) 2
64.4 
(1800) 2
20.9  106
ymax  50.4  103 ft 
or
(b)
2

1
1
 v02  32.2  20.9  106
y

2
1
20.9  106

or
(a)


ymax 
(3000) 2
64.4 
(3000)2
20.9  106
ymax  140.7  103 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.29 (Continued)
(c)
v0  36,700 ft/s:
ymax 
(36, 700)2
64.4 
(36,700)2
20.9  106
 3.03  1010 ft
This solution is invalid since the velocity does not reduce to zero. The velocity 36,700 ft/s is above the
escape velocity vR from the earth. For vR and above.
ymax

PROBLEM 11.30
The acceleration due to gravity of a particle falling toward the earth is a   gR 2 /r 2, where r
is the distance from the center of the earth to the particle, R is the radius of the earth, and g
is the acceleration due to gravity at the surface of the earth. If R  3960 mi, calculate the
escape velocity, that is, the minimum velocity with which a particle must be projected
vertically upward from the surface of the earth if it is not to return to the earth. (Hint:
v  0 for r  .)
SOLUTION
We have
v
dv
gR 2
a 2
dr
r
r  R , v  ve
When
r  , v  0
then

0
ve
vdv 


R

gR 2
dr
r2

or
1
1 
 ve2  gR 2  
2
 r R
or
ve  2 gR
1/ 2
5280 ft 

  2  32.2 ft/s 2  3960 mi 
1 mi 

or
ve  36.7  103 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.31
The velocity of a particle is v  v0 [1  sin ( t/T )]. Knowing that the particle starts from the origin with an
initial velocity v0 , determine (a) its position and its acceleration at t  3T , (b) its average velocity during the
interval t  0 to t  T .
SOLUTION
(a)

dx
  t 
 v  v0 1  sin   
dt
 T 

We have
At t  0, x  0:

x
0
dx 


  t 
v0 1  sin    dt
0
 T 

t
t
 T
 T
  t 
 t  T 
x  v0 t  cos     v0 t  cos    
 T 0
T  
 
 
At t  3T :

T
2T 
   3T  T 

x3T  v0 3T  cos 
   v0  3T 


 
 T  


a
At t  3T :
(b)
(1)
x3T  2.36 v0T 

t
dv d  
  t   
 v0 1  sin      v0 cos
dt dt  
T
T
 T   
a3T  v0

T
cos
  3T
a3T 
T
 v0
T

Using Eq. (1)
At t  0:
At t  T :
Now
T
T

x0  v0 0  cos(0)    0





T
 T
xT  v0 T  cos 

 T

vave 
2T
 T

     v0  T  



xT  x0 0.363v0T  0

t
T 0

  0.363v0T

vave  0.363v0 
Problem 11.32
An eccentric circular cam, which serves a similar function as the
Scotch yoke mechanism in Problem 11.13, is used in
conjunction with a flat face follower to control motion in pumps
and in steam engine valves. Knowing that the eccentricity is
denoted by e, the maximum range of the displacement of the
follower is dmax, and the maximum velocity of the follower is
vmax, determine the displacement, velocity, and acceleration of
the follower.
SOLUTION
Constraint:
y  r  e cos 
(1)
Differentiate:
y  e sin 
(2)
Differentiate again:

y  esin   e2 cos
(3)
ymax occurs when cosθ =1 and ymin occurs when cosθ = -1
d max  ymax  ymin
d max  r  e   r  e 
d max  2e
e
dmax
2
(4)
yr
Substitute (4) into (1) to get Position
d max
cos  
2
Max Velocity occurs when sin   1
vmax  e
 
vmax
e
(5)
y  vmax sin 
Substitute (5) into (2) to get velocity and assume cw rotation.
Substitute (4) and (5) into (3)

y 
Acceleration:
 4v 2
d max 
d
 sin   max  2max
2
2  d max

 cos 


y 
d max 
2v 2
 sin   max cos  
d max
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.33
An airplane begins its take-off run at A with zero
velocity and a constant acceleration a. Knowing that
it becomes airborne 30 s later at B and that the
distance AB is 900 m, determine (a) the acceleration
a, (b) the take-off velocity v B .
SOLUTION
Given :
x A  0 , xB  900 m , v A  0 , t  30 s
Uniform Acceleration:
xB  xA  vAt 
1 2
at
2
Substitute known values:
900 m  0  0 
1
2
a  30 s 
2
a  2.0 m/s 2 
(a) Solve for acceleration
Uniform Acceleration:
vB  vA  at
Substitute known values:
vB  0  2 m/s 2  30 s 
(b) Velocity at takeoff (point B)


vB  60.0 m/s
PROBLEM 11.34
A motorist is traveling at 54 km/h when she
observes that a traffic light 240 m ahead of her
turns red. The traffic light is timed to stay red for
24 s. If the motorist wishes to pass the light
without stopping just as it turns green again,
determine (a) the required uniform deceleration of
the car, (b) the speed of the car as it passes the
light.
SOLUTION
Uniformly accelerated motion:
x0  0
(a)
x  x0  v0 t 
v0  54 km/h  15 m/s
1 2
at
2
when t  24s, x  240 m:
240 m  0  (15 m/s)(24 s) 
1
a (24 s) 2
2
a  0.4167 m/s 2
(b)
a  0.417 m/s2 
v  v0  a t
when t  24s:
v  (15 m/s)  (0.4167 m/s)(24 s)
v  5.00 m/s
v  18.00 km/h
v  18.00 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.35
Steep safety ramps are built beside mountain highways to
enable vehicles with defective brakes to stop safely. A truck
enters a 750-ft ramp at a high speed v0 and travels 540 ft in 6 s
at constant deceleration before its speed is reduced to v0 / 2.
Assuming the same constant deceleration, determine (a) the
additional time required for the truck to stop, (b) the additional
distance traveled by the truck.
SOLUTION
Given :
xo  0 , x A  540 m , t A  6 s , vA 
Uniform Acceleration:
v A  vo  at A
Substitute known values:
1
vo  vo   a  6 
2
a
1
vo , Lramp  750 ft
2
1
vo
12
1 2
at A
2
Uniform Acceleration:
xA  xo  vot A 
Substitute known values:
540  0  vo  6  
1
2
vo  6 
24
vo  120 ft/s and a  10 ft/s 2
(a)
vB  vo  at B
Substitute known values:
0  120  10tB
Solve for tB
tB  12 s
tB  t A  6.0 s 
Additional time to stop
1 2
atB
2
(b)
xB  xo  votB 
Substitute known values:
1
2
xB  0  120 12   10 12 
2
Solve for xB
xB  720 ft
Additional time to stop
xB  x A  180.0 ft 
PROBLEM 11.36
A group of students launches a model rocket in the vertical direction. Based on tracking
data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered
portion of the flight and that the rocket landed 16 s later. Knowing that the descent
parachute failed to deploy so that the rocket fell freely to the ground after reaching its
maximum altitude and assuming that g  32.2 ft/s 2 , determine (a) the speed v1 of the
rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.
SOLUTION
(a)
1 2
at
2
We have
y  y1  v1t 
At tland ,
y0
Then
0  89.6 ft  v1 (16 s)

1
(32.2 ft/s 2 )(16 s) 2
2
v1  252 ft/s 
or
(b)
We have
v 2  v12  2a( y  y1 )
At
y  y max ,
Then
0  (252 ft/s)2  2(32.2 ft/s 2 )( ymax  89.6) ft
or
v0
y max  1076 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.37
A small package is released from rest at A and
moves along the skate wheel conveyor ABCD.
The package has a uniform acceleration of 4.8
m/s 2 as it moves down sections AB and CD,
and its velocity is constant between B and C. If
the velocity of the package at D is 7.2 m/s,
determine (a) the distance d between C and D,
(b) the time required for the package to reach D.
SOLUTION
(a)
For A
B and C
D we have
v 2  v02  2a( x  x0 )
2
vBC
 0  2(4.8 m/s2 )(3  0) m
Then, at B
 28.8 m2 /s 2
(vBC  5.3666 m/s)
2
vD2  vBC
 2aCD ( xD  xC )
and at D
d  x D  xC
(7.2 m/s) 2  (28.8 m 2 /s 2 )  2(4.8 m/s 2 )d
or
d  2.40 m 
or
(b)
For A
B and C
D we have
v  v0  at
Then A
B
5.3666 m/s  0  (4.8 m/s2 )t AB
t AB  1.118 04 s
or
and C
7.2 m/s  5.3666 m/s  (4.8 m/s 2 )tCD
D
tCD  0.38196 s
or
Now, for B
C, we have
xC  x B  v BC t BC
or
3 m  (5.3666 m/s)t BC
or
t BC  0.55901 s
Finally,
or
t D  t AB  t BC  tCD  (1.11804  0.55901  0.38196) s
t D  2.06 s 
PROBLEM 11.38
A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with
constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine (a) his
acceleration, (b) his final velocity, (c) his time for the race.
SOLUTION
0  x  35 m, a  constant
Given:
35 m  x  100 m, v  constant
At t  0, v  0 when
x  35 m, t  5.4 s
Find:
(a)
a
(b)
v when x  100 m
(c)
t when x  100 m
(a)
We have
At t  5.4 s:
or
x  0  0t 
35 m 
1 2
at
2
for
0  x  35 m
1
a (5.4 s) 2
2
a  2.4005 m/s 2
a  2.40 m/s 2 
(b)
First note that v  vmax for 35 m  x  100 m.
Now
(c)
v 2  0  2a( x  0)
for
0  x  35 m
When x  35 m:
2
vmax
 2(2.4005 m/s 2 )(35 m)
or
vmax  12.9628 m/s
We have
When x  100 m:
or
x  x1  v0 (t  t1 )
vmax  12.96 m/s 
for
35 m  x  100 m
100 m  35 m  (12.9628 m/s)(t 2  5.4) s
t 2  10.41 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.39
Automobile A starts from O and
accelerates at the constant rate of
0.75 m/s2. A short time later it is
passed by bus B which is traveling
in the opposite direction at a
constant speed of 6 m/s. Knowing
that bus B passes point O 20 s after
automobile A started from there,
determine when and where the
vehicles passed each other.
SOLUTION
Place origin at 0.
Motion of auto.
 xA 0  0,  vA 0  0,
x A   x A 0   v A 0 t 
aA  0.75 m/s2
1
1
a At 2  0  0     0.75  t 2
2
2
xA  0.375t 2 m
Motion of bus.
 xB 0  ?,  vB 0   6 m/s,
aB  0
xB   xB 0   vB 0 t   xB 0  6t m
At t  20 s , x B  0.
0   xB 0   6  20 
Hence,
 xB 0  120 m
x B  120  6 t
When the vehicles pass each other, x B  x A .
120  6t  0.375 t 2
0.375 t 2  6 t  120  0
t
t
 6  (6) 2   4  0.375  120 
 2  0.375 
 6  14.697
 11.596 s
0.75
and  27.6 s
t  11.60 s 
Reject the negative root.
Corresponding values of xA and xB.
xA   0.37511.596  50.4 m
2
xB  120   6 11.596   50.4 m
x  50.4 m 
PROBLEM 11.40
In a boat race, boat A is leading boat B by 50
m and both boats are traveling at a constant
speed of 180 km/h. At t  0, the boats
accelerate at constant rates. Knowing that
when B passes A, t  8 s and v A  225 km/h,
determine (a) the acceleration of A, (b) the
acceleration of B.
SOLUTION
(a)
We have
v A  (v A ) 0  a At
(v A ) 0  180 km/h  50 m/s
At t  8 s:
Then
v A  225 km/h  62.5 m/s
62.5 m/s  50 m/s  a A (8 s)
a A  1.563 m/s2 
or
(b)
We have
1
1
x A  ( x A )0  (v A )0 t  a At 2  50 m  (50 m/s)(8 s)  (1.5625 m/s 2 )(8 s)2  500 m
2
2
and
1
xB  0  (vB ) 0 t  a B t 2
2
At t  8 s:
x A  xB
( v B ) 0  50 m/s
1
500 m  (50 m/s)(8 s)  aB (8 s)2
2
or
aB  3.13 m/s2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.41
As relay runner A enters the 65-ft-long exchange zone with a
speed of 30 ft/s, he begins to slow down. He hands the baton to
runner B 2.5 s later as they leave the exchange zone with the
same velocity. Determine (a) the uniform acceleration of each
of the runners, (b) when runner B should begin to run.
SOLUTION
Let x  0 at the start of the exchange zone, and t  0 as runner A enters the exchange zone.
Then,
 v A 0
Motion of runner A:
v A   v A 0  a At
 30 ft/s.
x A   v A 0 t 
1
a At 2
2
(a) Solving for aA , and noting that xA  65 ft when t  2.5 s
aA 
2  x A   v A 0 t 
t
2

2  65   30  2.5  
a A  3.20 ft/s2 
2.52
 vA  f
At t  2.5 s,
 30   3.20  2.5  22.0 ft/s
Motion of runner B:  vB 0  0 and xB  0 at the starting time tB of runner B.
vB2   vB 0  2aB xB
2
Then,
Solving for aB , and noting that vB   vA  f  22.0 ft/s
When xB  65 ft,
aB 
 vB 2   vB 02
2 xB

22.02  0
 2  65
aB  3.723 ft/s2 
v B   v B 0  a B  t  t B   a B  t  t B 
t  tB 
vB
aB
(b) Solving for t B , and noting that vB  22.0 ft/s when t  2.5 s
tB  t 
vB
22.0
 2.5 
 3.41 s
aB
3.723
Runner B should begin 3.41 s before runner A reaches the exchange zone. 
PROBLEM 11.42
Automobiles A and B are traveling in adjacent highway lanes
and at t  0 have the positions and speeds shown. Knowing
that automobile A has a constant acceleration of 1.8 ft/s 2 and
that B has a constant deceleration of 1.2 ft/s 2 , determine
(a) when and where A will overtake B, (b) the speed of each
automobile at that time.
SOLUTION
a A  1.8 ft/s 2
aB  1.2 ft/s 2
(v A )0  24 mi/h  35.2 ft/s
A overtakes B
(vB )0  36 mi/h  52.8 ft/s
Motion of auto A:
v A  ( v A ) 0  a A t  35.2  1.8t
x A  ( x A ) 0  (v A ) 0 t 
(1)
1
1
a At 2  0  35.2t  (1.8)t 2
2
2
(2)
Motion of auto B:
v B  ( v B ) 0  a B t  52.8  1.2t
xB  ( xB ) 0  ( v B ) 0 t 
(a)
(3)
1
1
aB t 2  75  52.8t  (1.2)t 2
2
2
(4)
A overtakes B at t  t1 .
xA  xB : 35.2t  0.9t12  75  52.8t1  0.6t12
1.5t12  17.6t1  75  0
t1  3.22 s
Eq. (2):
and
t1  15.0546
xA  35.2(15.05)  0.9(15.05)2
t1  15.05 s 
x A  734 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.42 (Continued)
(b)
Velocities when t1  15.05 s
Eq. (1):
v A  35.2  1.8(15.05)
v A  62.29 ft/s
Eq. (3):
v A  42.5 mi/h

v B  23.7 mi/h

v B  52.8  1.2(15.05)
v B  34.74 ft/s
PROBLEM 11.43
Two automobiles A and B are approaching each other in adjacent highway lanes. At t  0, A and B are 3200 ft
apart, their speeds are v A  65 mi/h and v B  40 mi/h, and they are at Points P and Q, respectively. Knowing
that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the
uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time.
SOLUTION
(a)
x A  0  (v A ) 0 t 
We have
(x is positive
3200 m  (95.333 m/s)(40 s) 
Also, xB  0  (vB )0 t 
1
aB t 2
2
Then (95.333 ft/s)t AB 
(c)
1
aB (42 s) 2
2
aB  0.83447 ft/s 2
When the cars pass each other
Solving
aA  0.767 ft/s2 
( v B ) 0  40 mi/h  58.667 ft/s
3200 ft  (58.667 ft/s)(42 s) 
or
or
1
a A (40 s) 2
2
; origin at Q.)
At t  42 s:
(b)
(v A )0  65 mi/h  95.33 ft/s
; origin at P.)
At t  40 s:
(xB is positive
1
a At 2
2
aB  0.834 ft/s2 
x A  x B  3200 ft
1
1
2
2
 (58.667 ft/s)t AB  (0.83447 ft/s 2 )t AB
 3200 ft
( 0.76667 ft/s)t AB
2
2
2
0.03390t AB
 154t AB  3200  0
t  20.685 s and t  4563 s
We have
v B  (v B ) 0  a B t
At t  t AB :
vB  58.667 ft/s  (0.83447 ft/s2 )(20.685 s)
 75.927 ft/s
t 0
t AB  20.7 s 
v B  51.8 mi/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.44
An elevator is moving upward at a constant speed of 4 m/s. A man standing 10 m
above the top of the elevator throws a ball upward with a speed of 3 m/s. Determine
(a) when the ball will hit the elevator, (b) where the ball will hit the elevator with
respect to the location of the man.
SOLUTION
Place the origin of the position coordinate at the level of the standing man, the positive direction being up.
The ball undergoes uniformly accelerated motion.
y B  ( y B ) 0  ( vB ) 0 t 
1 2
gt
2
with ( yB )0  0, (vB )0  3 m/s, and g  9.81 m/s 2 .
yB  3t  4.905t 2
The elevator undergoes uniform motion.
y E  ( y E )0  vE t
with ( y E ) 0   10 m and v E  4 m/s.
(a)
Set y B  y E
Time of impact.
3t  4.905t 2  10  4t
4.905t 2  t  10  0
t  1.330 s 
t  1.3295 and 1.5334
(b)
Location of impact.
y B  (3)(1.3295)  (4.905)(1.3295) 2  4.68 m
y E  10  (4)(1.3295)  4.68 m
(checks)
4.68 m below the man 
PROBLEM 11.45
Two rockets are launched at a fireworks display. Rocket A is launched
with an initial velocity v0  100 m/s and rocket B is launched t1 seconds
later with the same initial velocity. The two rockets are timed to explode
simultaneously at a height of 300 m as A is falling and B is rising.
Assuming a constant acceleration g  9.81 m/s2 , determine (a) the time t1,
(b) the velocity of B relative to A at the time of the explosion.
SOLUTION
Place origin at ground level. The motion of rockets A and B is
Rocket A:
v A  ( v A ) 0  gt  100  9.81t
y A  ( y A ) 0  (v A ) 0 t 
Rocket B:
(1)
1 2
gt  100t  4.905t 2
2
(2)
v B  ( v B ) 0  g (t  t1 )  100  9.81(t  t1 )
(3)
1
g (t  t1 ) 2
2
 100(t  t1 )  4.905(t  t1 ) 2
y B  ( y B )0  (vB )0 (t  t1 ) 
(4)
Time of explosion of rockets A and B. y A  y B  300 ft
From (2),
300  100t  4.905t 2
4.905t 2  100t  300  0
t  16.732 s and 3.655 s
From (4),
300  100(t  t1 )  4.905(t  t12 )
t  t1  16.732 s
Since rocket A is falling,
Since rocket B is rising,
(a)
Time t1:
(b)
Relative velocity at explosion.
and
3.655 s
t  16.732 s
t  t1  3.655 s
t1  t  (t  t1 )
From (1),
v A  100  (9.81)(16.732)   64.15 m/s
From (3),
v B  100  (9.81)(16.732  13.08)  64.15 m/s
Relative velocity:
vB/A  vB  v A
t1  13.08 s 
vB/A  128.3 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.46
Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a
constant speed of 60 mi/h. At t  0, A starts and accelerates at a constant rate a A , while at t  5 s, B begins to
slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other
x  294 ft and v A  v B , determine (a) the acceleration a A , (b) when the vehicles pass each other, (c) the
distance d between the vehicles at t  0.
SOLUTION
For t  0:
v A  0  a At
xA  0  0 
1
a At 2
2
0  t  5 s:
xB  0  (vB )0 t
At t  5 s:
x B  (88 ft/s)(5 s)  440 ft
For t  5 s:
vB  (vB )0  aB (t  5)
1
aB   a A
6
xB  ( xB ) S  (vB )0 (t  5) 
1
aB (t  5) 2
2
( v B ) 0  60 mi/h  88 ft/s
Assume t  5 s when the cars pass each other.
At that time (t AB ),
v A  vB :
a At AB  (88 ft/s) 
x A  294 ft :
294 ft 
Then
or
a A 76 t AB  65 
1
2
2
a At AB

1
2
a At AB
2
88
294
2
44t AB
 343t AB  245  0
aA
(t AB  5)
6
PROBLEM 11.46 (Continued)
Solving
(a)
With t AB  5 s,
t AB  0.795 s
294 ft 
and
t AB  7.00 s
1
a A (7.00 s) 2
2
a A  12.00 ft/s2 
or
(b)
t AB  7.00 s 
From above
Note: An acceptable solution cannot be found if it is assumed that t AB  5 s.
(c)
We have
d  x  ( xB )t AB
 294 ft  440 ft  (88 ft/s)(2.00 s)
1 1

    12.00 ft/s 2  (2.00 s) 2
2 6

or
d  906 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.47
The elevator shown in the figure moves downward with a constant velocity of
4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the
counterweight W, (c) the relative velocity of the cable C with respect to the
elevator, (d ) the relative velocity of the counterweight W with respect to the
elevator.
SOLUTION
Choose the positive direction downward.
(a)
Velocity of cable C.
yC  2 y E  constant
vC  2 v E  0
v E  4 m/s
But,
or
(b)
vC   2 v E   8 m/s
v C  8.00 m/s 
Velocity of counterweight W.
yW  y E  constant
vW  v E  0
(c)
vW   v E   4 m/s
vW  4.00 m/s 
Relative velocity of C with respect to E.
vC/E  vC  vE  (8 m/s)  (4 m/s)  12 m/s
vC/E  12.00 m/s 
(d )
Relative velocity of W with respect to E.
vW /E  vW  vE  (4 m/s)  (4 m/s)  8 m/s
vW /E  8.00 m/s 
PROBLEM 11.48
The elevator shown starts from rest and moves upward with a constant
acceleration. If the counterweight W moves through 30 ft in 5 s, determine
(a) the acceleration of the elevator and the cable C, (b) the velocity of the
elevator after 5 s.
SOLUTION
We choose positive direction downward for motion of counterweight.
yW 
At t  5 s,
1
aW t 2
2
yW  30 ft
30 ft 
1
aW (5 s) 2
2
aW  2.4 ft/s2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW  2.4 ft/s 2
yW  y E  constant
vW  v E  0,
and
aW  a E  0
aE  aW  (2.4 ft/s2 ),
yC  2 y E  constant,
vC  2 v E  0,
a E  2.40 ft/s2 
and
aC  2 a E  0
aC  2aE  2(2.4 ft/s2 )  4.8 ft/s2 ,
aC  4.80 ft/s 2 
Velocity of elevator after 5 s.
vE  (vE )0  aE t  0  (2.4 ft/s 2 )(5 s)  12 ft/s
( v E ) 5  12.00 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.49
An athlete pulls handle A to the left with a
constant velocity of 0.5 m/s. Determine (a)
the velocity of the weight B, (b) the
relative velocity of weight B with respect
to the handle A.
SOLUTION
x A  4 y B  constant
From the diagram:
(a)
v A  4vB  0
Differentiate
Sketch:
(1)
x A  0.5 m/s
v B   0.125 m/s
Solve (1) for vB
vB  0.125 m/s  
(b) Finding the relative velocity:

vB  0.125  m/s

v A  0.5  m/s



v B / A  vB  v A
 0.125    0.5   m/s
or
v B / A  0.5  0.125  m/s
v B / A  0.5154 m/s
14 
Problem 11.50
An athlete pulls handle A to the left with a constant
acceleration. Knowing that after the weight B has been
lifted 4 in. its velocity is 2 ft/s, determine (a) the
accelerations of handle A and weight B (b) the velocity and
change in position of handle A after 0.5 sec.
SOLUTION
(a)
From the diagram:
Sketch:
x A  3 y B  constant
v A  3v B  0
(1)
a A  3a B  0
(2)
Uniform Acceleration of weight B

v B 2   v B o +2aB yB   y B o
2
Substitute in known values

 2 ft/s 2 =02  2aB  
4

ft  0 
12


aB  6 ft/s 2
Using (2)
a A  3(6 ft/s 2 )  0
aA  18 ft/s2  
aB  6 ft/s2  
(b) Uniform Acceleration of Handle A
vA   vA o  aAt


vA  0  18 ft/s 2  0.5 s 
v A  9 ft/s  
or
Also
1
y A   y A  o   v A  o t  a At 2
2
y A   y A o  0 
or


1
2
18 ft/s2  0.5 s 
2
y A  2.25 ft  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.51
Slider block B moves to the right
with a constant velocity of 300
mm/s. Determine (a) the velocity
of slider block A, (b) the velocity
of portion C of the cable, (c) the
velocity of portion D of the cable,
(d ) the relative velocity of portion
C of the cable with respect to slider
block A.
SOLUTION
x B  ( x B  x A )  2 x A  constant
From the diagram
Then
2 v B  3v A  0
(1)
and
2 a B  3a A  0
(2)
 x D  x A  constant
Also, we have
vD  v A  0
Then
(a)
Substituting into Eq. (1)
2(300 mm/s)  3v A  0
or
(b)
From the diagram
Then
Substituting
From the diagram
Then
Substituting
or
v A  200 mm/s

v C  600 mm/s

v D  200 mm/s

x B  ( x B  xC )  constant
2 v B  vC  0
2(300 mm/s)  vC  0
or
(c)
(3)
( xC  x A )  ( x D  x A )  constant
vC  2 v A  v D  0
600 mm/s  2(200 mm/s)  v D  0
PROBLEM 11.51 (Continued)
(d)
We have
vC/A  vC  v A
 600 mm/s  200 mm/s
or
vC/A  400 mm/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.52
At the instant shown, slider block
B is moving with a constant
acceleration, and its speed is 150
mm/s. Knowing that after slider
block A has moved 240 mm to the
right its velocity is 60 mm/s,
determine (a) the accelerations of
A and B, (b) the acceleration of
portion D of the cable, (c) the
velocity and change in position of
slider block B after 4 s.
SOLUTION
x B  ( x B  x A )  2 x A  constant
From the diagram
Then
2 v B  3v A  0
(1)
and
2 a B  3a A  0
(2)
(a)
First observe that if block A moves to the right, v A  and Eq. (1)  v B  . Then, using
Eq. (1) at t  0
2(150 mm/s)  3( v A ) 0  0
(v A ) 0  100 mm/s
or
Also, Eq. (2) and a B  constant  a A  constant
vA2  (vA )02  2aA [ xA  ( xA )0 ]
Then
When x A  ( x A ) 0  240 mm:
(60 mm/s)2  (100 mm/s)2  2a A (240 mm)
or
or
aA  
40
mm/s 2
3
a A  13.33 mm/s2

PROBLEM 11.52 (Continued)
Then, substituting into Eq. (2)
 40

2aB  3  
mm/s2   0
 3

aB  20 mm/s2
or
(b)
a B  20.0 mm/s2

From the diagram,  x D  x A  constant
vD  v A  0
Then
Substituting
aD  a A  0
 40

mm/s2   0
aD   
 3

or
(c)
We have
vB  (vB )0  a B t
At t  4 s:
vB  150 mm/s  (20.0 mm/s2 )(4 s)
or
Also
At t  4 s:
xB  ( x B ) 0  ( v B ) 0 t 

v B  70.0 mm/s

1
aB t 2
2
xB  ( xB )0  (150 mm/s)(4 s)

or
a D  13.33 mm/s2
1
(20.0 mm/s 2 )(4 s)2
2
x B  ( x B ) 0  440 mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.53
A farmer wants to get his hay bales into the top loft of
his barn by walking his horse forward with a constant
velocity of 1 ft/s. Determine the velocity and
acceleration of the hay bale when the horse is 10 ft
away from the barn.
SOLUTION
From the diagram, we can write an expression for the total length of rope:
L  20  yB  xH2  (20)2
Differentiate
dL
0
dt
xH x H
0   y B 
Velocities
vB  y B and vH  xH
vB 
Differentiate again:
x  (20) 2
2
H
aB 
aB 
or
aB 
at xH  10
vB 
xH vH
x  (20) 2
2
H
dvB
dt
xH vH
xH2  (20) 2
vH2
xH2  (20) 2
xH vH

xH2  (20) 2
xH a H

xH2  (20) 2
aB 

xH vH ( 12 )(2 xH ) xH
x
x
2
H
 (20) 2 
3/2
xH2 vH2
2
H
 (20) 2 
3/2
10(1)
10    20 
2
vB  0.447 ft/s 
2
1
2
2
10    20 
2
and


10  0 
2
2
10    20 
aB  0.0447  0  0.000894
10  1

3/ 2
102   202 
2
2
aB  0.0358 ft/s 2 
PROBLEM 11.54
The motor M reels in the cable at a constant rate of 100 mm/s. Determine
(a) the velocity of load L, (b) the velocity of pulley B with respect to load L.
SOLUTION
Let xB and xL be the positions, respectively, of pulley B and load L measured downward from a fixed elevation
above both. Let xM be the position of a point on the cable about to enter the reel driven by the motor. Then,
considering the lengths of the two cables,
xM  3xB  constant
vM  3vB  0
xL  ( xL  xB )  constant
2vL  vB  0
vM  100 mm/s
with
vB  
vL 
vM
 33.333 m/s
3
vB
 16.667 mm/s
2
v L  16.67 mm/s 
(a)
Velocity of load L.
(b)
Velocity of pulley B with respect to load L. vB/L  vB  vL  33.333  (16.667)  16.667
v B/L  16.67 mm/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.55
Collar A starts from rest at t = 0 and moves upward with a constant acceleration of
3.6 in./s2. Knowing that collar B moves downward with a constant velocity of 18 in./s,
determine (a) the time at which the velocity of block C is zero, (b) the corresponding
position of block C.
SOLUTION
Define positions as positive downward from a fixed level.
Constraint of cable.
 yB  y A    yC
 y A   2  yC  yB   constant
3 yC  y B  2 y A  constant
Differentiate
3vC  v B  2v A  0
(1)
Differentiate again
3aC  a B  2 a A  0
(2)
Motion of block C:
 vA 0  0,
aA   3.6 in./s2, vB   vB 0  18 in./s, aB  0
1
 vB   2  v A    6 in./s
0
0
3
Using (1)
 vC 0 
Using (2)
aC 
Constant acceleration
vC   vC 0  aC t
Constant acceleration
xC   xC 0   vC 0 t 
1
1
 aB  2a A   0   2   3.6    2.4 in./s2
3
3
(3)
 6  1.2t
 6t  0.6t
1
aC t 2
2
(4)
2
(a) Time at vC  0 :
Using (3)
0  6  2.4t
t  2.5 s 
(b) Corresponding position of block C:
Using (4)
2
1
xC   xC 0   6  2.5      2.4  2.5 
2
 
xC   xC 0  7.5 in.  
PROBLEM 11.56
Block A starts from rest at t  0 and moves downward with a constant acceleration
of 6 in./s2. Knowing that block B moves up with a constant velocity of 3 in./s,
determine (a) the time when the velocity of block C is zero, (b) the corresponding
position of block C.
SOLUTION
The cable lengths are constant.
L1  2 yC  2 yD  constant
L 2  y A  yB  ( yB  yD )  constant
Eliminate yD.
L1  2 L 2  2 yC  2 yD  2 y A  2 yB  2( yB  yD )  constant
2( yC  y A  2 yB )  constant
Differentiate to obtain relationships for velocities and accelerations, positive downward.
vC  v A  2 v B  0
(1)
aC  a A  2 a B  0
(2)
Use units of inches and seconds.
Motion of block A:
v A  a A t  6t
y A 
Motion of block B:
1
1
a At 2  (6)t 2  3t 2
2
2
v B  3 in./s
v B   3 in./s
 y B  v B t   3t
Motion of block C:
From (1),
vC  v A  2vB  6t  2(3)  6  6t
yC 
(a)
Time when vC is zero.
(b)
Corresponding position.

t
0
vC dt  6t  3t 2
6  6t  0
yC  (6)(1)  (3)(1)2  3 in.
t  1.000 s 
 yC  3.00 in. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.57
Block B starts from rest, block A moves with a constant
acceleration, and slider block C moves to the right with a
constant acceleration of 75 mm/s 2 . Knowing that at t  2 s the
velocities of B and C are 480 mm/s downward and 280 mm/s to
the right, respectively, determine (a) the accelerations of A and
B, (b) the initial velocities of A and C, (c) the change in position
of slider block C after 3 s.
SOLUTION
From the diagram
3 y A  4 y B  xC  constant
Then
3v A  4 v B  vC  0
(1)
and
3 a A  4 a B  aC  0
(2)
( v B )  0,
Given:
a A  constent
(a C )  75 mm/s 2
At t  2 s,
v B  480 mm/s
v C  280 mm/s
(a)
Eq. (2) and a A  constant and aC  constant  a B  constant
vB  0  a B t
Then
At t  2 s:
480 mm/s  a B (2 s)
aB  240 mm/s 2

or
a B  240 mm/s2 
or
a A  345 mm/s2 
Substituting into Eq. (2)
3a A  4(240 mm/s 2 )  (75 mm/s 2 )  0
a A   345 mm/s



PROBLEM 11.57 (Continued)
(b)
vC  ( vC ) 0  aC t
We have
At t  2 s:
280 mm/s  (vC ) 0  (75 mm/s)(2 s)
vC  130 mm/s

or
( v C ) 0  130.0 mm/s

Then, substituting into Eq. (1) at t  0
3( v A ) 0  4(0)  (130 mm/s)  0
v A  43.3 mm/s
(c)
We have
At t  3 s:
xC  ( xC )0  (vC )0 t 
or
1
aC t 2
2
xC  ( xC )0  (130 mm/s)(3 s) 
 728 mm
(v A ) 0  43.3 mm/s 
1
(75 mm/s 2 )(3 s)2
2
or
x C  ( x C ) 0  728 mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.58
Block B moves downward with a constant velocity of 20 mm/s.
At t  0, block A is moving upward with a constant acceleration,
and its velocity is 30 mm/s. Knowing that at t  3 s slider block
C has moved 57 mm to the right, determine (a) the velocity of
slider block C at t  0, (b) the accelerations of A and C, (c) the
change in position of block A after 5 s.
SOLUTION
From the diagram
3 y A  4 y B  xC  constant
Then
3v A  4 v B  vC  0
(1)
and
3 a A  4 a B  aC  0
(2)
v B  20 mm/s ;
Given:
( v A ) 0  30 mm/s
(a)
Substituting into Eq. (1) at t  0
3(  30 mm/s)  4(20 mm/s)  ( vC ) 0  0
( vC ) 0  10 mm/s
(b)
We have
At t  3 s:
( v C ) 0  10.00 mm/s
or
xC  ( xC )0  (vC )0 t 
1
aC t 2
2
57 mm  (10 mm/s)(3 s) 
1
aC (3 s) 2
2
aC  6 mm/s 2

Now

v B  constant  a B  0
or
aC  6.00 mm/s 2

PROBLEM 11.58 (Continued)
Then, substituting into Eq. (2)
3a A  4(0)  (6 mm/s 2 )  0
a A  2 mm/s2
(c)
We have
At t  5 s:
y A  ( y A ) 0  (v A ) 0 t 
or
a A  2.00 mm/s2 
1
a At 2
2
y A  ( y A )0  (30 mm/s)(5 s) 
1
(2 mm/s 2 )(5 s)2
2
 175 mm
or
y A  ( y A ) 0  175.0 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.59
The system shown starts from rest, and each component moves with a constant
acceleration. If the relative acceleration of block C with respect to collar B is 60
mm/s 2 upward and the relative acceleration of block D with respect to block A is
110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the
change in position of block D after 5 s.
SOLUTION
From the diagram
2 y A  2 y B  yC  constant
Cable 1:
Then
2v A  2v B  vC  0
(1)
and
2 a A  2 a B  aC  0
(2)
( y D  y A )  ( y D  y B )  constant
Cable 2:
Then
and
(3)
 a A  aB  2aD  0
(4)
At t  0, v  0; all accelerations constant;
aC/B  60 mm/s2 , aD /A  110 mm/s2
Given:
(a)
 v A  vB  2vD  0
We have
aC /B  aC  a B   60
or
a B  aC  60
and
a D /A  a D  a A  110
or
a A  a D  110
Substituting into Eqs. (2) and (4)
Eq. (2):
or
Eq. (4):
or
2( a D  110)  2( aC  60)  aC  0
3aC  2 a D  100
(5)
 ( a D  110)  ( aC  60)  2 a D  0
 aC  a D   50
(6)
PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and a D
aC  40 mm/s2
aD  10 mm/s2
Now
vC  0  aC t
At t  3 s:
vC  (40 mm/s 2 )(3 s)
v C  120.0 mm/s 
or
(b)
We have
At t  5 s:
or
yD  ( yD )0  (0)t 
yD  ( yD )0 
1
aD t 2
2
1
( 10 mm/s 2 )(5 s)2
2
y D  ( y D ) 0  125.0 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.60*
The system shown starts from rest, and the length of the upper cord is adjusted so
that A, B, and C are initially at the same level. Each component moves with a
constant acceleration, and after 2 s the relative change in position of block C with
respect to block A is 280 mm upward. Knowing that when the relative velocity of
collar B with respect to block A is 80 mm/s downward, the displacements of A
and B are 160 mm downward and 320 mm downward, respectively, determine
(a) the accelerations of A and B if aB  10 mm/s2 , (b) the change in position of
block D when the velocity of block C is 600 mm/s upward.
SOLUTION
From the diagram
Cable 1:
2 y A  2 y B  yC  constant
Then
2v A  2v B  vC  0
(1)
and
2 a A  2 a B  aC  0
(2)
Cable 2: ( y D  y A )  ( y D  y B )  constant
Then
 v A  vB  2vD  0
(3)
and
 a A  aB  2aD  0
(4)
Given: At
t0
v0
( y A )0  ( yB )0  ( yC )0
All accelerations constant.
At t  2 s
yC /A  280 mm
When
v B /A  80 mm/s
y A  ( y A ) 0  160 mm
y B  ( y B ) 0  320 mm
aB  10 mm/s2
PROBLEM 11.60* (Continued)
(a)
We have
y A  ( y A )0  (0)t 
1
a At 2
2
and
yC  ( yC )0  (0)t 
1
aC t 2
2
Then
yC/A  yC  y A 
1
(aC  a A )t 2
2
At t  2 s, yC/A  280 mm:
280 mm 
or
1
(aC  a A )(2 s)2
2
aC  a A  140
(5)
Substituting into Eq. (2)
2 a A  2 a B  ( a A  140)  0
or
1
a A  (140  2aB )
3
Now
vB  0  a B t
(6)
v A  0  a At
v B/A  v B  v A  ( a B  a A )t
Also
When

yB  ( yB )0  (0)t 
v B /A  80 mm/s :
1
aB t 2
2
80  ( a B  a A )t
 y A  160 mm :
160 
1
a At 2
2
 y B  320 mm :
320 
1
aB t 2
2
(7)
1
(aB  a A )t 2
2
Then
160 
Using Eq. (7)
320  (80)t or t  4 s
Then
160 
1
a A (4)2
2
or
a A  20.0 mm/s2 
and
320 
1
aB (4)2
2
or
a B  40.0 mm/s2 
Note that Eq. (6) is not used; thus, the problem is over-determined.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.60* (Continued)
(b)
Substituting into Eq. (5)
aC  20  140  120 mm/s2
and into Eq. (4)
(20 mm/s2 )  (40 mm/s2 )  2aD  0
or
aD  30 mm/s2
Now
vC  0  aC t
When vC  600 mm/s:
or
Also
At t  5 s:
or
600 mm/s  (120 mm/s2 )t
t 5s
yD  ( yD )0  (0)t 
yD  ( yD )0 
1
aD t 2
2
1
(30 mm/s 2 )(5 s)2
2
y D  ( y D ) 0  375 mm 
PROBLEM 11.61
A particle moves in a straight line with a constant acceleration
of 4 ft/s 2 for 6 s, zero acceleration for the next 4 s, and a
constant acceleration of 4 ft/s 2 for the next 4 s. Knowing that
the particle starts from the origin and that its velocity is 8 ft/s
during the zero acceleration time interval, (a) construct the
v  t and x  t curves for 0  t  14 s, (b) determine the
position and the velocity of the particle and the total distance
traveled when t  14 s.
SOLUTION
From the a  t curve
A1   24 ft/s, A2  16 ft/s
(a) Construct the v  t curve :
v6   8 ft/s
v0  v6  A1
 8   24 
 16 ft/s
v10   8 ft/s
v14  v10  A2
 8  16

 8 ft/s
From the v  t curve
A3  32 ft, A4   8 ft
A5   32 ft, A6   8 ft , A7  8 ft
(b) Construct the x  t curve
x0  0
x4  x0  A3  32 ft
x6  x4  A4  24 ft
x10  x6  A5   8 ft
x12  x10  A6   16 ft

x14  x12  A7   8 ft
Total Distance Traveled
0  t  4 s,
d1  32  0  32 ft
4 s  t  12 s,
d 2  16  32  48 ft
12 s  t  14 s,
d3  8   16  8 ft
d  32  48  8
d  88 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.62
A particle moves in a straight line with a constant acceleration
of 4 ft/s 2 for 6 s, zero acceleration for the next 4 s, and a
constant acceleration of 4 ft/s 2 for the next 4 s. Knowing that
the particle starts from the origin with v0  16 ft/s,
(a) construct the v  t and x  t curves for 0  t  14 s,
(b) determine the amount of time during which the particle is
further than 16 ft from the origin.
SOLUTION
From the a  t curve
A1   24 ft/s, A2  16 ft/s
(b) Construct the v  t curve :
v0  16 ft/s
v6  v0  A1
 16   24 
 8 ft/s
v10  v6   8 ft/s
v14  v10  A2
 8  16

 8 ft/s
From the v  t curve
A3  32 ft, A4   8 ft
A5   32 ft, A6   8 ft , A7  8 ft
Construct the x  t curve
x0  0
x4  x0  A3  32 ft
x6  x4  A4  24 ft
x10  x6  A5   8 ft
x12  x10  A6   16 ft

x14  x12  A7   8 ft
(b)
Time for x  16 ft.
From the x  t diagram, this is time interval t1 to t 2 .
From 0  t  6 s,
dx
 v  16  4t
dt
Integrating, using limits x  0 when t  0 and x  16 ft when t  t1
 x160
t
 16t  2t 2 
0
or
16  16t1  2t12
PROBLEM 11.62 (Continued)
t12  8t1  8  0
or
Using the quadratic formula:
t1 
8
82   4 18 
 2 1
 4  2.828  1.172 s
and
6.828 s
The larger root is out of range, thus t1  1.172 s
From 6  t  10,
x  24  8  t  6   72  8t
Setting x  16,
16  72  8t 2
Time Interval:
t  t2  t1
or
t2  7 s
t  5.83 s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.63
A particle moves in a straight line with the
velocity shown in the figure. Knowing that
x  540 m at t  0, (a) construct the at
and xt curves for 0  t  50 s, and
determine (b) the total distance traveled by
the particle when t  50 s, (c) the two times
at which x  0.
SOLUTION
(a)
at  slope of v  t curve at time t
From t  0 to t  10 s:
v  constant  a  0
20  60
 5 m/s 2
26  10
t  10 s to t  26 s:
a
t  26 s to t  41 s:
v  constant  a  0
t  41 s to t  46 s:
a
t  46 s:
5  (20)
 3 m/s 2
46  41
v  constant  a  0
x2  x1  (area under v  t curve from t1 to t 2 )
At t  10 s:
x10   540  10(60)  60 m
Next, find time at which v  0. Using similar triangles
tv  0  10
60
At
t  22 s:
t  26 s:
t  41 s:
t  46 s:
t  50 s:

26  10
80
1
x22  60  (12)(60)  420 m
2
1
x26  420  (4)(20)  380 m
2
x41  380  15(20)  80 m
 20  5 
x46  80  5 
  17.5 m
 2 
x50  17.5  4(5)  2.5 m
or
tv  0  22 s
PROBLEM 11.63 (Continued)
(b)
From t  0 to t  22 s: Distance traveled  420  (540)
 960 m 
t  22 s to t  50 s: Distance traveled  | 2.5  420|
 422.5 m
Total distance traveled  (960  422.5) ft  1382.5 m
Total distance traveled  1383 m 
(c)
Using similar triangles
Between 0 and 10 s:
(t x 0 )1  0 10

540
600
(t x  0 )1  9.00 s 
Between 46 s and 50 s:
(t x 0 ) 2  46 4

17.5
20
(t x  0 ) 2  49.5 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.64
A particle moves in a straight line with the
velocity shown in the figure. Knowing that
x  540 m at t  0, (a) construct the a t
and x t curves for 0  t  50 s, and
determine (b) the maximum value of the
position coordinate of the particle, (c) the
values of t for which the particle is at
x  100 m.
SOLUTION
(a)
at  slope of v  t curve at time t
From t  0 to t  10 s:
v  constant  a  0
20  60
 5 m/s 2
26  10
t  10 s to t  26 s:
a
t  26 s to t  41 s:
v  constant  a  0
t  41 s to t  46 s:
a
t  46 s:
5  (20)
 3 m/s 2
46  41
v  constant  a  0
x2  x1  (area under v  t curve from t1 to t 2 )
At t  10 s:
x10   540  10(60)  60 m
Next, find time at which v  0. Using similar triangles
tv  0  10
60
At
t  22 s:
t  26 s:
t  41 s:
t  46 s:
t  50 s:

26  10
80
or
1
x22  60  (12)(60)  420 m
2
1
x26  420  (4)(20)  380 m
2
x41  380  15(20)  80 m
 20  5 
x46  80  5 
  17.5 m
 2 
x50  17.5  4(5)  2.5 m
tv  0  22 s
PROBLEM 11.64 (Continued)
(b)
Reading from the x t curve
(c)
Between 10 s and 22 s
xmax  420 m 
100 m  420 m  (area under v t curve from t , to 22 s) m
1
100  420  (22  t1 )(v1 )
2
(22  t1 )( v1 )  640
Using similar triangles
v1
60

22  t1 12
Then
v1  5(22  t1 )
or
(22  t1 )[5(22  t1 )]  640
t1  10.69 s
and
t1  33.3 s
10 s  t1  22 s 
We have
t1  10.69 s 
Between 26 s and 41 s:
Using similar triangles
41  t2 15

20
300
t 2  40.0 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.65
A particle moves in a straight line with the velocity shown in the
figure. Knowing that x = – 48 ft at t = 0, draw the a–t and x–t
curves for 0 < t < 40 s and determine (a) the maximum value of
the position coordinate of the particle, (b) the values of t for
which the particle is at a distance of 108 ft from the origin.
SOLUTION
The a  t curve is the slope of the v  t curve
0  t  10 s,
a0
10 s < t  18 s,
a
18  6
 1.5 ft/s 2
18  10
18 s < t  30 s,
a
18  18
  3 ft/s 2
30  18
30 s < t  40 s
a0

Points on the x–t curve may be calculated using areas of the v–t curve.
A1  (10)(6)  60 ft
A2 
1
(6  18)(18  10)  96 ft
2
A3 
1
(18)(24  18)  54 ft
2
A4 
1
(18)(30  24)   54 ft
2
A5  (  18)(40  30)   180 ft
Value of x at specific times:
x0   48 ft
x10  x0  A1  12 ft
x18  x10  A2  108 ft
x24  x18  A3  162 ft
x30  x24  A4  108 ft
x40  x30  A5   72 ft

PROBLEM 11.65 (Continued)
(a) Maximum value of x occurs when: v  0, i.e. t  24 s.
xmax  162 ft 
(b) Times when x  108 ft.
From the x  t curve
t  18 s and t  30 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.66
A parachutist is in free fall at a rate of 200 km/h when he opens his parachute at
an altitude of 600 m. Following a rapid and constant deceleration, he then
descends at a constant rate of 50 km/h from 586 m to 30 m, where he maneuvers
the parachute into the wind to further slow his descent. Knowing that the
parachutist lands with a negligible downward velocity, determine (a) the time
required for the parachutist to land after opening his parachute, (b) the initial
deceleration.
SOLUTION
Assume second deceleration is constant. Also, note that
200 km/h  55.555 m/s,
50 km/h  13.888 m/s
PROBLEM 11.66 (Continued)
(a)
Now  x  area under v t curve for given time interval
Then
 55.555  13.888 
(586  600) m  t1 
 m/s
2


t1  0.4032 s
(30  586) m  t2 (13.888 m/s)
t2  40.0346 s
1
(0  30) m   (t3 )(13.888 m/s)
2
t3  4.3203 s
t total  (0.4032  40.0346  4.3203) s
(b)
We have
ainitial 

t total  44.8 s 
 vinitial
t1
[13.888  (55.555)] m/s
0.4032 s
 103.3 m/s 2
ainitial  103.3 m/s2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.67
A commuter train traveling at 40
mi/h is 3 mi from a station. The
train then decelerates so that its
speed is 20 mi/h when it is 0.5 mi
from the station. Knowing that the
train arrives at the station 7.5 min
after beginning to decelerate and
assuming constant decelerations,
determine (a) the time required for
the train to travel the first 2.5 mi,
(b) the speed of the train as it
arrives at the station, (c) the final
constant deceleration of the train.
SOLUTION
Given: At t  0, v  40 mi/h, x  0; when x  2.5 mi, v  20 mi/h;
at t  7.5 min, x  3 mi; constant decelerations.
The v  t curve is first drawn as shown.
(a)
A1  2.5 mi
We have
1h
 40  20 
(t1 min) 
mi/h 
 2.5 mi

60 min
 2 
t1  5.00 min 
(b)
A2  0.5 mi
We have
 20  v2
(7.5  5) min  
 2
1h

 mi/h  60 min  0.5 mi

v2  4.00 mi/h 
(c)
We have
afinal  a12

(4  20) mi/h 5280 ft 1 min
1h



(7.5  5) min
mi
60 s 3600 s
afinal  0.1564 ft/s2 
PROBLEM 11.68
A temperature sensor is attached to slider AB which moves
back and forth through 60 in. The maximum velocities of the
slider are 12 in./s to the right and 30 in./s to the left. When the
slider is moving to the right, it accelerates and decelerates at a
constant rate of 6 in./s2; when moving to the left, the slider
accelerates and decelerates at a constant rate of 20 in./s2.
Determine the time required for the slider to complete a full
cycle, and construct the v – t and x t curves of its motion.
SOLUTION
The v  t curve is first drawn as shown. Then
ta 
vright
aright

12 in./s
2s
6 in./s 2
vleft 30 in./s

aleft 20 in./s
 1.5 s
td 
A1  60 in.
Now
[(t1  2) s](12 in./s)  60 in.
or
t1  7 s
or
A2  60 in.
and
or
{[(t 2  7)  1.5] s}(30 in./s)  60 in.
or
t 2  10.5 s
tcycle  t2
Now
tcycle  10.5 s 
We have xii  xi  (area under v  t curve from ti to tii )
At
t  2 s:
t  5 s:
1
(2) (12)  12 in.
2
x5  12  (5  2)(12)
x2 
 48 in.
t  7 s:
t  8.5 s:
t  9 s:
x7  60 in.
1
x8.5  60  (1.5)(30)
2
 37.5 in.
x9  37.5  (0.5)(30)
 22.5 in.
t  10.5 s:
x10.5  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.69
In a water-tank test involving the launching of a small model
boat, the model’s initial horizontal velocity is 6 m/s, and its
horizontal acceleration varies linearly from 12 m/s 2 at t  0
to 2 m/s 2 at t  t1 and then remains equal to 2 m/s 2 until
t  1.4 s. Knowing that v  1.8 m/s when t  t1 , determine
(a) the value of t1 , (b) the velocity and the position of the
model at t  1.4 s.
SOLUTION
Given:
v0  6 m/s; for 0  t  t1 ,
for
t1  t  1.4 s a  2 m/s2 ;
at
t  0 a  12 m/s2 ;
at t  t1
a  2 m/s 2 , v  1.8 m/s 2
The at and vt curves are first drawn as shown. The time axis is not
drawn to scale.
(a)
We have
vt1  v0  A1
 12  2 
2
1.8 m/s  6 m/s  (t1 s) 
 m/s
 2 
t1  0.6 s 
(b)
We have
v1.4  vt1  A2
v1.4  1.8 m/s  (1.4  0.6) s  2 m/s 2
v1.4  0.20 m/s 
Now x1.4  A3  A4 , where A3 is most easily determined using
integration. Thus,
for 0  t  t1 :
Now
a
2  (12)
50
t  12  t  12
0.6
3
dv
50
 a  t  12
dt
3
PROBLEM 11.69 (Continued)
At t  0, v  6 m/s:

v
6
dv 

t
50

t  12  dt

0 3

v 6
or
25 2
t  12t
3
We have
dx
25
 v  6  12t  t 2
dt
3
Then
A3 

xt1
0
dx 

0.6
0
(6  12t 
25 2
t )dt
3
0.6
25 

 6t  6t 2  t 3   2.04 m
9 0

Also
Then
or
 1.8  0.2 
A4  (1.4  0.6) 
  0.8 m
2


x1.4  (2.04  0.8) m
x1.4  2.84 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.70
The acceleration record shown was obtained for a small
airplane traveling along a straight course. Knowing that x  0
and v  60 m/s when t  0, determine (a) the velocity and
position of the plane at t  20 s, (b) its average velocity
during the interval 6 s  t  14 s.
SOLUTION
Geometry of “bell-shaped” portion of vt curve
The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of vt diagram is:
= x = 6 m
(a)
When t  20 s:
v20  60 m/s 
x20  (60 m/s) (20 s)  (shaded area)
 1200 m  6 m
(b)
From t  6 s to t  14 s:
x20  1194 m 
t  8 s
 x  (60 m/s)(14 s  6 s)  (shaded area)
 (60 m/s)(8 s)  6 m  480 m  6 m  474 m
vaverage 
 x 474 m

t
8s
vaverage  59.25 m/s 
PROBLEM 11.71
In a 400-m race, runner A reaches her maximum velocity v A in
4 s with constant acceleration and maintains that velocity until
she reaches the half-way point with a split time of 25 s. Runner
B reaches her maximum velocity v B in 5 s with constant
acceleration and maintains that velocity until she reaches the
half-way point with a split time of 25.2 s. Both runners then
run the second half of the race with the same constant
deceleration of 0.1 m/s 2 . Determine (a) the race times for both
runners, (b) the position of the winner relative to the loser
when the winner reaches the finish line.
SOLUTION
Sketch v  t curves for first 200 m.
Runner A:
t1  4 s, t 2  25  4  21 s
A1 
1
(4)(v A )max  2(v A )max
2
A2  21(v A ) max
A1  A2   x  200 m
23(v A ) max  200
Runner B:
or
t1  5 s,
A1 
(v A ) max  8.6957 m/s
t 2  25.2  5  20.2 s
1
(5)(vB ) max  2.5(vB )max
2
A2  20.2(v B ) max
A1  A2   x  200 m
22.7(v B ) max  200
Sketch v  t curve for second 200 m.
A3  vmax t3 
t3 
Runner A:
or
(v B ) max  8.8106 m/s
 v  | a |t3  0.1t3
1
vt3  200
2
or
vmax  (vmax )2  (4)(0.05)(200)
(vmax ) A  8.6957,
Reject the larger root. Then total time
(2)(0.05)
(t3 ) A  146.64 s
0.05t32  vmax t3  200  0

 10 vmax  (vmax )2  40
and

27.279 s
t A  25  27.279  52.279 s
t A  52.2 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.71 (Continued)
Runner B: (vmax ) B  8.8106, (t3 ) B  149.45 s
Reject the larger root. Then total time
and
26.765 s
t B  25.2  26.765  51.965 s
t B  52.0 s 
Velocity of A at t  51.965 s:
v1  8.6957  (0.1)(51.965  25)  5.999 m/s
Velocity of A at t  51.279 s:
v2  8.6957  (0.1)(52.279  25)  5.968 m/s
Over 51.965 s  t  52.965 s, runner A covers a distance  x
 x  vave (t ) 
1
(5.999  5.968)(52.279  51.965)
2
 x  1.879 m 
PROBLEM 11.72
A car and a truck are both traveling at the
constant speed of 35 mi/h; the car is 40 ft
behind the truck. The driver of the car
wants to pass the truck, i.e., he wishes to
place his car at B, 40 ft in front of the truck,
and then resume the speed of 35 mi/h. The
maximum acceleration of the car is 5 ft/s 2
and the maximum deceleration obtained by
applying the brakes is 20 ft/s 2 . What is the
shortest time in which the driver of the car
can complete the passing operation if he
does not at any time exceed a speed of
50 mi/h? Draw the vt curve.
SOLUTION
Relative to truck, car must move a distance:
Allowable increase in speed:
 x  16  40  50  40  146 ft
 vm  50  35  15 mi/h  22 ft/s
Acceleration Phase:
t1  22/5  4.4 s
A1 
1
(22)(4.4)  48.4 ft
2
Deceleration Phase:
t3  22/20  1.1 s
A3 
1
(22)(1.1)  12.1 ft
2
But:  x  A1  A2  A3 :
146 ft  48.4  (22)t 2  12.1
t total  t1  t 2  t3  4.4 s  3.89 s  1.1 s  9.39 s
t 2  3.89 s
t B  9.39 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.73
Solve Problem 11.72, assuming that the driver of the car does not pay any attention to the speed limit while
passing and concentrates on reaching position B and resuming a speed of 35 mi/h in the shortest possible time.
What is the maximum speed reached? Draw the vt curve.
PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft
behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in
front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s 2 and
the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the
driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h?
Draw the vt curve.
SOLUTION
Relative to truck, car must move a distance:
 x  16  40  50  40  146 ft
vm  5t1  20t2 ;
 x  A1  A2 :
t2 
146 ft 
1
(vm )(t1  t2 )
2
146 ft 
1
1 

(5t1 )  t1  t1 
2
4 

t12  46.72
1
t1
4
t1  6.835 s
t2 
1
t1  1.709
4
t total  t1  t 2  6.835  1.709
t B  8.54 s 
 vm  5t1  5(6.835)  34.18 ft/s  23.3 mi/h
Speed vtotal  35 mi/h, vm  35 mi/h  23.3 mi/h
vm  58.3 mi/h 
PROBLEM 11.74
Car A is traveling on a highway at a
constant speed ( v A ) 0  60 mi/h,
and is 380 ft from the entrance of
an access ramp when car B enters
the acceleration lane at that point
at a speed ( v B ) 0  15 mi/h. Car B
accelerates uniformly and enters
the main traffic lane after traveling
200 ft in 5 s. It then continues to
accelerate at the same rate until it
reaches a speed of 60 mi/h, which
it then maintains. Determine the
final distance between the two
cars.
SOLUTION
(v A )0  60 mi/h, (vB )0  1.5 mi/h; at t  0,
Given:
( x A )0  380 ft, ( xB )0  0; at t  5 s,
xB  200 ft; for 15 mi/h  vB  60 mi/h,
aB  constant; for vB  60 mi/h,
First note
aB  0
60 mi/h  88 ft/s
15 mi/h  22 ft/s
The vt curves of the two cars are then drawn as shown.
Using the coordinate system shown, we have
at t  5 s, x B  200 ft :
 22 + (vB )5 
(5 s) 
 ft/s  200 ft
2


( v B ) 5  58 ft/s
or
Then, using similar triangles, we have
(88  22) ft/s (58  22) ft/s
(  aB )

5s
t1
t1  9.16 67 s
or
Finally, at t  t1


 22 + 88 
xB/A  xB  x A  (9.1667 s) 
ft/s 

 2 


 [380 ft  (9.1667 s) (88 ft/s)]
or
xB/A  77.5 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.75
An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2
until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the
elevator begins to move, a man standing 12 m above the initial position of the top of
the elevator throws a ball upward with an initial velocity of 20 m/s. Determine when
the ball will hit the elevator.
SOLUTION
Given:
At t  0 v E  0; For 0  v E  7.8 m/s, aE  1.2 m/s 2 ;
For v E  7.8 m/s, a E  0;
At t  2 s, vB  20m/s 
The vt curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve
for the elevator is 1.2 m/s 2 , while the slope of the curve for the ball is  g (9.81 m/s 2 ).
The time t1 is the time when v E reaches 7.8 m/s.
Thus,
or
or
v E  (0)  a E t
7.8 m/s  (1.2 m/s 2 )t1
t1  6.5 s
The time ttop is the time at which the ball reaches the top of its trajectory.
Thus,
or
or
vB  (vB ) 0  g (t  2)
0  20 m/s  (9.81 m/s2 )(ttop  2) s
ttop  4.0387 s
PROBLEM 11.75 (Continued)
Using the coordinate system shown, we have
0  t  t1 :
1

yE  12 m   aE t 2  m
2

At t  ttop :
yB 
and
yE  12 m 
1
(4.0387  2) s  (20 m/s)
2
 20.387 m
1
(1.2 m/s 2 )(4.0387 s) 2
2
 2.213 m
t  [2  2(4.0387  2)] s  6.0774 s, y B  0
At
and at t  t1 ,
yE  12 m 
1
(6.5 s) (7.8 m/s)  13.35 m
2
The ball hits the elevator ( y B  y E ) when ttop  t  t1.
For t  ttop :
1

yB  20.387 m   g (t  ttop )2  m
2


Then,
yB  yE
when
1
(9.81 m/s 2 ) (t  4.0387) 2
2
1
 12 m  (1.2 m/s 2 ) (t s) 2
2
20.387 m 
5.505t 2  39.6196t  47.619  0
or
Solving
Since 1.525 s is less than 2 s,
t  1.525 s and t  5.67 s
t  5.67 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.76
Car A is traveling at 40 mi/h when it
enters a 30 mi/h speed zone. The driver
of car A decelerates at a rate of 16 ft/s2
until reaching a speed of 30 mi/h, which
she then maintains. When car B, which
was initially 60 ft behind car A and
traveling at a constant speed of 45 mi/h,
enters the speed zone, its driver
decelerates at a rate of 20 ft/s2 until
reaching a speed of 28 mi/h. Knowing
that the driver of car B maintains a speed
of 28 mi/h, determine (a) the closest that
car B comes to car A, (b) the time at
which car A is 70 ft in front of car B.
SOLUTION
(v A )0  40 mi/h; For 30 mi/h  v A  40 mi/h, a A  16 ft/s 2 ; For v A  30 mi/h, a A  0;
Given:
( x A /B )0  60 ft; (vB )0  45 mi/h;
When xB  0, aB  20 ft/s 2 ;
For vB  28 mi/h, aB  0
First note
40 mi/h  58.667 ft/s 30 mi/h  44 ft/s
45 mi/h  66 ft/s
28 mi/h  41.067 ft/s
At t  0
The vt curves of the two cars are as shown.
At
t  0:
Car A enters the speed zone.
t  (t B )1:
Car B enters the speed zone.
t  tA:
Car A reaches its final speed.
t  tmin :
vA  vB
t  (tB )2 :
Car B reaches its final speed.
PROBLEM 11.76 (Continued)
(a)
(v A )final  (v A )0
tA
aA 
We have
16 ft/s 2 
or
(44  58.667) ft/s
tA
t A  0.91669 s
or
Also
60 ft  (t B )1 (vB )0
or
60 ft  (t B )1 (66 ft/s)
aB 
and
20 ft/s 2 
or
or
(t B )1  0.90909 s
(vB )final  (vB )0
(t B ) 2  (t B )1
(41.067  66) ft/s
[(t B )2  0.90909] s
Car B will continue to overtake car A while vB  vA . Therefore, ( xA/B )min will occur when v A  vB ,
which occurs for
(t B )1  tmin  (tB )2
For this time interval
v A  44 ft/s
vB  (vB )0  aB [t  (t B )1 ]
Then
at t  tmin :
44 ft/s  66 ft/s  (20 ft/s2 )(tmin  0.90909) s
tmin  2.00909 s
or
Finally ( x A/B ) min  ( x A )t  ( xB )t
min
min
  (v )  (v A )final 

 t A  A 0
 (tmin  t A )(v A )final 

2

 


 (v )  (v A )final  
 ( xB )0  (t B )1 (vB )0  [tmin  (t B )1 ]  B 0

2





 58.667  44 
ft/s  (2.00909  0.91669) s  (44 ft/s) 
 (0.91669 s) 

2





 66  44  
  60 ft  (0.90909 s)(66 ft/s)  (2.00909  0.90909) s  
 ft/s 
 2  

 (47.057  48.066) ft  (60  60.000  60.500) ft
 34.623 ft
or ( xA/B )min  34.6 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.76 (Continued)
(b)
Since ( xA/B )  60 ft for t  tmin , it follows that xA/B  70 ft for t  (tB )2
[Note (t B )2  tmin ]. Then, for t  (tB )2
x A/B  ( x A/B )min  [(t  tmin )(vA )final ]


 (vB )final 
 (v )
 [(t B )2  (tmin )]  A final
 [t  (tB )2 ](vB )final 

2




or
70 ft  34.623 ft  [(t  2.00909) s  (44 ft/s)]


 44  41.06 
 (2.15574  2.00909) s  
 ft/s  (t  2.15574) s  (41.067) ft/s 
2




or
t  14.14 s 
PROBLEM 11.77
An accelerometer record for the motion of a given part of a
mechanism is approximated by an arc of a parabola for 0.2 s and a
straight line for the next 0.2 s as shown in the figure. Knowing that
v  0 when t  0 and x  0.8 ft when t  0.4 s, (a) construct the v t
curve for 0  t  0.4 s, (b) determine the position of the part at t 
0.3 s and t  0.2 s.
SOLUTION
Divide the area of the a t curve into the four areas A1, A2 , A3 and A4.
2
(8)(0.2)  1.0667 ft/s
3
A2  (16)(0.2)  3.2 ft/s
A1 
1
(16  8)(0.1)  1.2 ft/s
2
1
A4  (8)(0.1)  0.4 ft/s
2
A3 
Velocities: v0  0
v0.2  v0  A1  A2
v0.2  4.27 ft/s 
v0.3  v0.2  A3
v0.3  5.47 ft/s 
v0.4  v0.3  A4
v0.4  5.87 ft/s 
Sketch the v  t curve and divide its area into A5, A6 , and A7 as shown.
0.8
0.4
 x dx  0.8  x   t vdt
At t  0.3 s,
With A5 
With A5  A6 
0.4
x  0.8   t vdt
x0.3  0.8  A5  (5.47)(0.1)
2
(0.4)(0.1)  0.0267 ft,
3
At t  0.2 s,
and
or
x0.3  0.227 ft 
x0.2  0.8  ( A5  A6 )  A7
2
(1.6)(0.2)  0.2133 ft,
3
A7  (4.27)(0.2)  0.8533 ft
x0.2  0.8  0.2133  0.8533
x0.2  0.267 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.78
A car is traveling at a constant speed of
54 km/h when its driver sees a child run into
the road. The driver applies her brakes until
the child returns to the sidewalk and then
accelerates to resume her original speed of
54 km/h; the acceleration record of the car is
shown in the figure. Assuming x  0 when
t  0, determine (a) the time t1 at which the
velocity is again 54 km/h, (b) the position of
the car at that time, (c) the average velocity of
the car during the interval 1 s  t  t1.
SOLUTION
Given:
At
t  0, x  0, v  54 km/h;
For t  t1 ,
v  54 km/h
First note
(a)
54 km/h  15 m/s
vb  va  (area under at curve from ta to tb )
We have
Then
at
t  2 s:
v  15  (1)(6)  9 m/s
t  4.5 s:
1
v  9  (2.5)(6)  1.5 m/s
2
t  t1:
15  1.5 
1
(t1  4.5)(2)
2
or
(b)
Using the above values of the velocities, the vt curve is drawn as shown.
t1  18.00 s 
PROBLEM 11.78 (Continued)
Now
x at t  18 s
x18  0   (area under the vt curve from t  0 to t  18 s)
 15  9 
 (1 s)(15 m/s)  (1 s) 
 m/s
 2 
1


+  (2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s) 
3


2


 (13.5 s)(1.5 m/s)  (13.5 s)(13.5 m/s) 
3


 [15  12  (3.75  6.25)  (20.25  121.50)] m
 178.75 m
(c)
First note
or
x18  178.8 m 
x1  15 m
x18  178.75 m
Now
or
vave 
 x (178.75  15) m

 9.6324 m/s
t
(18  1) s
vave  34.7 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.79
An airport shuttle train travels between two terminals that are 1.6 mi apart. To maintain passenger comfort,
the acceleration of the train is limited to 4 ft/s2, and the jerk, or rate of change of acceleration, is limited to
0.8 ft/s2 per second. If the shuttle has a maximum speed of 20 mi/h, determine (a) the shortest time for the
shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle.
SOLUTION
xmax  1.6 mi; | amax|  4 ft/s2
Given:
 da 
 0.8 ft/s2 /s; vmax  20 mi/h
 dt 
 max
First note
20 mi/h  29.333 ft/s
1.6 mi  8448 ft
(a)
To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the
time for which v  vmax . The time  t required for the train to have an acceleration of 4 ft/s2 is found
from
a
 da 
 max
 dt 
t
 max
4 ft/s 2
0.8 ft/s 2 /s
or
t 
or
t  5 s
Now,
da


 since dt  constant 


after 5 s, the speed of the train is
v5 
1
( t )( amax )
2
or
v5 
1
(5 s)(4 ft/s 2 )  10 ft/s
2
Then, since v5  vmax , the train will continue to accelerate at 4 ft/s2 until v  vmax . The at curve
must then have the shape shown. Note that the magnitude of the slope of each inclined portion of the
curve is 0.8 ft/s2/s.
PROBLEM 11.79 (Continued)
Now
at t  (10  t1 ) s, v  vmax :
1

2  (5 s)(4 ft/s 2 )   (t1 )(4 ft/s 2 )  29.333 ft/s
2

or
t1  2.3333 s
Then
at t  5 s:
1
(5)(4)  10 ft/s
2
t  7.3333 s: v  10  (2.3333)(4)  19.3332 ft/s
1
t  12.3333 s: v  19.3332  (5)(4)  29.3332 ft/s
2
v0
Using symmetry, the vt curve is then drawn as shown.
Noting that A1  A2  A3  A4 and that the area under the vt curve is equal to xmax, we have

 10  19.3332  
2  (2.3333 s) 
 ft/s 
2

 

 (10  t2 ) s  (29.3332 ft/s)  8448 ft
or
t2  275.67 s
Then
tmin  4(5 s)  2(2.3333 s)  275.67 s
 300.34 s
tmin  5.01 min 
or
(b)
We have
or
vave 
x
1.6 mi 3600 s


t 300.34 s
1h
vave  19.18 mi/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.80
During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before
temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to 4.8 ft/s2 per
second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average
values of the velocity of the belt during that time.
SOLUTION
Given:
(a)
At
t  0, x  0, v  0; xmax  1.2 ft;
when
 da 
x  xmax , v  0;  
 4.8 ft/s 2
 dt max
Observing that vmax must occur at t  12 tmin , the at curve must have the shape shown. Note that the
magnitude of the slope of each portion of the curve is 4.8 ft/s2/s.
We have
at
t  t :
1
1
v  0  (t )(amax )  amax t
2
2
t  2 t :
vmax 
1
1
amax t  (t )( amax )  amax t
2
2
Using symmetry, the vt is then drawn as shown.
Noting that A1  A2  A3  A4 and that the area under the vt curve is equal to xmax, we have
(2t )(vmax )  xmax
vmax  amax t  2amax t 2  xmax
PROBLEM 11.80 (Continued)
Now
amax
 4.8 ft/s2 /s so that
t
2(4.8t ft/s3 )t 2  1.2 ft
or
Then
t  0.5 s
tmin  4t
tmin  2.00 s 
or
(b)
We have
vmax  amax t
 (4.8 ft/s 2 /s  t)t
 4.8 ft/s 2 /s  (0.5 s) 2
vmax  1.2 ft/s 
or
Also
or
vave 
x
1.2 ft

ttotal 2.00 s
vave  0.6 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.81
Two seconds are required to bring the piston
rod of an air cylinder to rest; the acceleration
record of the piston rod during the 2 s is as
shown. Determine by approximate means (a)
the initial velocity of the piston rod, (b) the
distance traveled by the piston rod as it is
brought to rest.
SOLUTION
at curve; at
Given:
1.
t  2 s, v  0
The at curve is first approximated with a series of rectangles, each of width t  0.25 s. The area
(t)(aave) of each rectangle is approximately equal to the change in velocity v for the specified
interval of time. Thus,
v  aave t
where the values of aave and v are given in columns 1 and 2, respectively, of the following table.
2.
v(2)  v0 
Now
and approximating the area

2
0

2
0
a dt  0
a dt under the at curve by aave t  v, the initial velocity is then
equal to
v0  v
Finally, using
v2  v1  v12
where v12 is the change in velocity between times t1 and t2, the velocity at the end of each
0.25 interval can be computed; see column 3 of the table and the vt curve.
3.
The vt curve is then approximated with a series of rectangles, each of width 0.25 s. The area
(t )(vave ) of each rectangle is approximately equal to the change in position  x for the specified
interval of time. Thus
 x  vave t
where vave and  x are given in columns 4 and 5, respectively, of the table.
PROBLEM 11.81 (Continued)
4.
With x0  0 and noting that
x2  x1   x12
where  x12 is the change in position between times t1 and t2, the position at the end of each 0.25 s
interval can be computed; see column 6 of the table and the xt curve.
(a)
We had found
(b)
At t  2 s
v0  1.914 m/s 
x  0.840 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.82
The acceleration record shown was obtained
during the speed trials of a sports car.
Knowing that the car starts from rest,
determine by approximate means (a) the
velocity of the car at t  8 s, (b) the distance
the car has traveled at t  20 s.
SOLUTION
Given: at curve; at
1.
t  0, x  0, v  0
The at curve is first approximated with a series of rectangles, each of width t  2 s. The
area (t )(aave ) of each rectangle is approximately equal to the change in velocity v for the
specified interval of time. Thus,
v  aave t
where the values of aave and v are given in columns 1 and 2, respectively, of the following table.
2.
Noting that v0  0 and that
v2  v1  v12
where v12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s
interval can be computed; see column 3 of the table and the vt curve.
3.
The vt curve is next approximated with a series of rectangles, each of width t  2 s. The
area (t )(vave ) of each rectangle is approximately equal to the change in position x for the
specified interval of time.
Thus,
x  vave t
where vave and x are given in columns 4 and 5, respectively, of the table.
4.
With x0  0 and noting that
x2  x1   x12
where  x12 is the change in position between times t1 and t2, the position at the end of each 2 s
interval can be computed; see column 6 of the table and the xt curve.
PROBLEM 11.82 (Continued)
(a)
At t  8 s, v  32.58 m/s
(b)
At t  20 s
or
v  117.3 km/h 
x  660 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.83
A training airplane has a velocity of 126 ft/s when
it lands on an aircraft carrier. As the arresting gear
of the carrier brings the airplane to rest, the
velocity and the acceleration of the airplane are
recorded; the results are shown (solid curve) in the
figure. Determine by approximate means (a) the
time required for the airplane to come to rest,
(b) the distance traveled in that time.
SOLUTION
Given: a  v curve:
v0  126 ft/s
The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on
the figure).
For uniformly accelerated motion
v22  v12  2a ( x2  x1 )
v2  v1  a (t2  t1 )
v22  v12
2a
v v
t  2 1
a
x 
or
For the five regions shown above, we have
Region
v1 , ft/s
v2 , ft/s
a, ft/s 2
x, ft
t , s
1
126
120
12.5
59.0
0.480
2
120
100
33
66.7
0.606
3
100
80
45.5
39.6
0.440
4
80
40
54
44.4
0.741
5
40
0
58
13.8
0.690
223.5
2.957

(a)
From the table, when v  0
t  2.96 s 
(b)
From the table and assuming x0  0, when v  0
x  224 ft 
PROBLEM 11.84
Shown in the figure is a portion of the experimentally
determined v  x curve for a shuttle cart. Determine by
approximate means the acceleration of the cart (a) when
x  10 in., (b) when v  80 in./s.
SOLUTION
Given: v  x curve
First note that the slope of the above curve is
dv
.
dx
av
(a)
Now
dv
dx
When
x  10 in., v  55 in./s
Then
 40 in./s 
a  55 in./s 

 13.5 in. 
a  163.0 in./s2 
or
(b)
When v  80 in./s, we have
 40 in./s 
a  80 in./s 

 28 in. 
or
a  114.3 in./s2 
Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary
that the same scale be used for the x and v axes (e.g., 1 in.  50 in., 1 in.  50 in./s). In the above
solution, v and  x were measured directly, so different scales could be used.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.85
An elevator starts from rest and rises 40 m to its maximum
velocity in T s with the acceleration record shown in the figure.
Determine (a) the required time T, (b) the maximum velocity,
(c) the velocity and position of the elevator at t = T/2.
SOLUTION
Find the position of the elevator from the a  t curve using the moment-area method.
First, find the areas A1 and A2 on the a  t curve
A1 
1
T
 0.6   0.1T m/s
2
3
A2 
1
2T
 0.6   0.2T m/s
2
3
(a) Apply the moment-area formula: x  x0  v0t 
T
0 T  t  adt


 T 1  2T   
2  T 
x  v0t   A1   T      A2  T    
  
3  3 

 3 3  3 

7 2
8 2
40  0 
T 
T
90
90
15 2
T

90
1
 T2
6
T 2   40  6 
 240 s 2
T  15.49 s 
(b)
vmax  v0  A1  A2  0  0.1T  0.2T  0.3T
vmax  4.65 m/s 
(c)
To find the position at t=T/2, first find the areas A1 and A2 on the a  t curve
A1  0.1T
A4 
A3 
T
1
 0.6   0.05T
2
6
1
T
 0.45  0.0375T
2
6
Apply the moment-area formula: x  x0  v0t 
T 2

0  2  t  adt


T
PROBLEM 11.85 (Continued)
x  v0
T
 T 2T
 A1  
2
9
2

2 T 
1 T 
  A3     A4   

3 6
3 6
T
 5T 
T 
 0   0.1T      0.05T      0.0375T   0.035417T 2
18
 18 
9
  0.035417 15.49 
2
x  8.50 m 
v  v0  A1  A3  A4  0.1875T
v  2.90 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.86
Two road rally checkpoints A and B are located
on the same highway and are 8 mi apart. The
speed limits for the first 5 mi and the last 3 mi
are 60 mi/h and 35 mi/h, respectively. Drivers
must stop at each checkpoint, and the specified
time between points A and B is 10 min 20 s.
Knowing that a driver accelerates and
decelerates at the same constant rate, determine
the magnitude of her acceleration if she travels
at the speed limit as much as possible.
SOLUTION
Given
10
20

60 3600
 0.1722 h
10 min 20 s 
Sketch the v  t curve and note ta 
60
25
35
, tb 
, tc 
a
a
a
Find Areas A1 and A2
1
1
 60   ta    25 tb
2
2
1
1
 60 t1  1800  312.5
a
a
(1)
1
 35 tc
2
1
 6.0278  35t1  612.5
a
(2)
A1  60t1 
A2  35  0.1722  t1  
Also given
A1  5 mi and A2  3 mi
Setting (1) equal to 5 mi:
60t1  2112.5
Setting (2) equal to 3 mi:
35t1  612.5
Solving equations (3) and (4) for t1 and
1
5
a
(3)
1
 3.0278
a
(4)
1
:
a
t1  85.45  103 h  5.13 min
1
 60.23  106 h 2 / mi
a
16.616  10   5280

3
3
a  16.616  10 mi/h
2
 3600 
2
a  6.77 ft/s 2 
PROBLEM 11.87
As shown in the figure, from t  0 to t  4 s, the acceleration of
a given particle is represented by a parabola. Knowing that x  0
and v  8 m/s when t  0, (a) construct the vt and x–t curves
for 0  t  4 s, (b) determine the position of the particle at t  3 s.
(Hint: Use table inside the front cover.)
SOLUTION
Given
At
t  0, x  0, v  8 m/s
(a) We have
v2  v1  (area under at curve from t1 to t2)
and
x2  x1  (area under vt curve from t1 to t2)
Then, using the formula for the area of a parabolic spandrel, we have
1
at t  2 s: v  8  (2)(12)  0
3
1
t  4 s: v  0  (2)(12)  8 m/s
3
The vt curve is then drawn as shown.
Note: The area under each portion of the curve is a spandrel of Order no. 3.
(2)(8)
4m
3 1
(2)(8)
0
t  4 s: x  4 
3 1
Now at t  2 s: x  0 
The xt curve is then drawn as shown.
(b) We have
or

At t  3 s: a  3(3  2)2  3 m/s 2
1
v  0  (1)(3)  1 m/s
3
(1)(1)
x 4
3 1
x3  3.75 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.88
A particle moves in a straight line with the acceleration
shown in the figure. Knowing that the particle starts from
the origin with v0  2 m/s, (a) construct the v  t and
x  t curves for 0  t  18 s, (b) determine the position
and the velocity of the particle and the total distance
traveled when t  18 s.
SOLUTION
Find the indicated areas under the a t curve:
A1   0.758  6 m/s
A2   2 4  8 m/s
A3   6 6  36 m/s
Find the values of velocity at times indicated:
v0  2 m/s
v8  v0  A1  8 m/s
v12  v8  A2  0
v18  v12  A3  36 m/s
(a) Sketch v  t curve using straight line portions over the constant acceleration periods.
Find the indicated areas under the v  t curve:
1
 2  88  40 m
2
1
A5   8 4  16 m
2
1
A6   36 6  108 m
2
A4 
Find the values of position at times indicated:
x0  0
x8  x0  A4  40 m
x12  x8  A5  56 m
x18  x12  A6  52 m
Sketch the x  t curve using the positions and times calculated:
(b) Position at t=18s
Velocity at t-18s
Total Distance Traveled:  56  108
x18  52 m 
v18  36 m/s 
d  164 m 
PROBLEM 11.89
A ball is thrown so that the motion is defined by
the equations x  5t and y  2  6t  4.9t 2 , where
x and y are expressed in meters and t is expressed
in seconds. Determine (a) the velocity at t  1 s,
(b) the horizontal distance the ball travels before
hitting the ground.
SOLUTION
Units are meters and seconds.
Horizontal motion:
vx 
dx
5
dt
Vertical motion:
vy 
dy
 6  9.8t
dt
(a)
Velocity at t  1 s.
vx  5
v y  6  9.8  3.8
v  vx2  v 2y  52  3.82  6.28 m/s
tan  
(b)
vy
vx

3.8
5
Horizontal distance:
  37.2
v  6.28 m/s
37.2 
( y  0)
y  2  6t  4.9t 2
t  1.4971 s
x  (5)(1.4971)  7.4856 m
x  7.49 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.90
The motion of a vibrating particle is defined by the position vector
r  10(1  e3t )i  (4e2t sin15t ) j, where r and t are expressed in
millimeters and seconds, respectively. Determine the velocity and
acceleration when (a) t  0, (b) t  0.5 s.
SOLUTION
r  10(1  e3t )i  (4e2t sin15t ) j
Then
v
and
a
(a)
dr
 30e 3t i  [60e 2t cos15t  8e 2t sin15t ]j
dt
dv
 90e 3t i  [120e2t cos15t  900e2t sin15t  120e 2t cos15t  16e2t sin15t ]j
dt
 90e 3t i  [240e2t cos15t  884e2t sin15t ]j
When t  0:
v  30i  60 j mm/s
v  67.1 mm/s
63.4 
a  90i  240 j mm/s 2
a  256 mm/s2
69.4 
v  8.29 mm/s
36.2 
a  336 mm/s2
86.6 
When t  0.5 s:
v  30e1.5 i  [60e1 cos 7.5  8e1 sin 7.5]
 6.694i  4.8906 j mm/s
a  90e1.5 i  [240e1 cos 7.5  884e1 sin 7.5 j]
 20.08i  335.65 j mm/s 2
PROBLEM 11.91
The motion of a vibrating particle is defined by the position vector
r  (4sin  t )i  (cos 2 t ) j, where r is expressed in inches and t in
seconds. (a) Determine the velocity and acceleration when t  1 s.
(b) Show that the path of the particle is parabolic.
SOLUTION
r  (4sin  t )i  (cos 2 t ) j
v  (4 cos  t )i  (2 sin 2 t ) j
a  (4 2 sin  t )i  (4 2 cos 2 t ) j
(a)
When t  1 s:
v  (4 cos  )i  (2 sin 2 ) j
v  (4 in/s)i 
a  (4 2 sin  )i  (4 2 cos  ) j
(b)
a  (4 2in/s2 ) j 
Path of particle:
Since r  xi  y j ;
x  4sin  t ,
y   cos 2 t
Recall that cos 2  1  2sin 2  and write
y   cos 2 t  (1  2sin 2  t )
But since x  4sin  t or sin  t 
(1)
1
x, Eq.(1) yields
4
2

1  
y   1  2  x  
 4  

y
1 2
x  1 (Parabola) 
8
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.92
The motion of a particle is defined by the equations x  10t  5sin t and y  10  5cos t , where x and y are
expresed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval 0  t  2 ,
and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the
corresponding times, positions, and directions of the velocities.
SOLUTION
Sketch the path of the particle, i.e., plot of y versus x.
Using x  10t  5sin t , and y  10  5cos t obtain the values in the table below. Plot as shown.
t(s)
x(ft)
y(ft)
0
0.00
5
10.71
10
31.41
15
52.12
10
62.83
5

2

3

2
2
PROBLEM 11.92 (Continued)
(a)
Differentiate with respect to t to obtain velocity components.
vx 
dx
 10  5cos t and v y  5sin t
dt
v 2  vx2  v 2y  (10  5cos t )2  25sin 2 t  125  100cos t
d (v ) 2
 100sin t  0 t  0.   .  2   N 
dt
When t  2 N  .
cos t  1.
and
v2 is minimum.
When t  (2 N  1) .
cos t  1.
and
v2 is maximum.
(v 2 )min  125  100  25(ft/s)2 vmin  5 ft/s 
(v 2 )max  125  100  225(ft/s)2
(b)
vmax  15 ft/s 
When v  vmin.
When N  0,1, 2,
x  10(2 N )  5sin(2 N )
x  20 N ft 
y  10  5cos(2 N )
y  5 ft 
vx  10  5cos(2 N )
vx  5 ft/s 
v y  5sin(2 N )
tan  
vy
vx
 0,
vy  0 
 0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.92 (Continued)
t  (2 N  1) s 
When v  vmax .
x  10[2 ( N  1)]  5sin[2 ( N  1)]
x  20 ( N  1) ft 
y  10  5cos[2 ( N  1)]
y  15 ft 
vx  10  5cos[2 ( N  1)]
vx  15 ft/s 
v y  5sin[2 ( N  1)]
tan  
vy
vx
 0,
vy  0 
 0 
PROBLEM 11.93
The damped motion of a vibrating particle is defined by the
position vector r  x1[1  1/(t  1)]i  ( y1e t/2 cos 2 t ) j, where
t is expressed in seconds. For x1  30 mm and y1  20 mm,
determine the position, the velocity, and the acceleration of the
particle when (a) t  0, (b) t  1.5 s.
SOLUTION
We have
1 

r  30  1 
i  20(e t/ 2 cos 2 t ) j

 t 1
Then
v
 30
1
 

i  20   e t/ 2 cos 2 t  2 e t/ 2 sin 2 t  j
2
(t  1)
 2

 30

1
1

i  20 e t/ 2  cos 2 t  2 sin 2 t   j
2
2
(t  1)



a
and
dr
dt
dv
dt
 30

(a)
At t  0:
 

2
1

i  20   e t/ 2  cos 2 t  2 sin 2 t   e t/ 2 ( sin 2 t  4 cos 2 t )  j
3
(t  1)
2

 2

60
i  10 2 e t/ 2 (4 sin 2 t  7.5 cos 2 t ) j
(t  1)3
 1
r  30 1   i  20(1) j
 1
r  20 mm 
or
 1
1

v  30   i  20 (1)   0   j
1
2
 

 
v  43.4 mm/s
or
46.3° 
60
a   i  10 2 (1)(0  7.5) j
(1)
or
a  743 mm/s 2
85.4° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.93 (Continued)
(b)
At t  1.5 s:
1 

0.75
r  30 1 
(cos 3 ) j
 i  20e
2.5


 (18 mm)i  (1.8956 mm) j
or
r  18.10 mm
6.01° 
v  5.65 mm/s
31.8° 
a  70.3 mm/s2
86.9° 
30
1

i  20 e0.75  cos 3  0  j
2
2
(2.5)


 (4.80 mm/s)i  (2.9778 mm/s) j
v
or
a
60
i  10 2 e 0.75 (0  7.5 cos 3 ) j
(2.5)3
 (3.84 mm/s 2 )i  (70.1582 mm/s 2 ) j
or
Problem 11.94
A girl operates a radio-controlled model car in a vacant
parking lot. The girl’s position is at the origin of the xy
coordinate axes, and the surface of the parking lot lies in
the x-y plane. The motion of the car is defined by the

position vector r  (2  2t 2 )iˆ  (6  t 3 ) ˆj where r and t
are expressed in meters and seconds, respectively.
Determine (a) the distance between the car and the girl
when t = 2 s, (b) the distance the car traveled in the interval
from t = 0 to t = 2 s, (c) the speed and direction of the car’s
velocity at t = 2 s, (d) the magnitude of the car’s
acceleration at t = 2 s.
SOLUTION
Given:
r  (2  2t 2 )i  (6  t 3 ) j
(a) At t=2s
r  2   10i  14 j m
r  2   10 2  14 2 m
r  2  =17.20 m 
(b) The car is traveling on a curved path. The distance that the car travels during any infinitesimal interval
is given by:
ds  dx 2  dy 2
where:
dx  4tdt and dy  3t 2 dt
Substituting
ds  16t 2  9t 4 dt
Integrating:

(c) Velocity can be found by
s
0
2
ds   t 16  9t 2 dt
0
s
3/2 2
1
16  9t 2  |

0
27
v
dr
dt
s  11.52 m 
v  4ti  3t 2 j m/s
At t=2 s
v  8i  12 j m/s
(d) Acceleration can be found by a 
v  14.42 m/s
56.31 
dv
dt
a  4i  6tj m/s 2
At t = 2 s
a  4i  12 j m/s 2
a  12.65 m/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r  (Rt cos nt)i  ctj 
(Rt sin nt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve
described by the particle is a conic helix.)
SOLUTION
We have
r  ( Rt cos n t )i  ctj  ( Rt sin n t )k
Then
v
and
a
dr
 R(cos n t  n t sin n t )i  cj  R(sin n t  n t cos n t )k
dt
dv
dt
 R (n sin n t  n sin n t  n2t cos n t )i
 R(n cos n t  n cos n t  n2t sin n t )k
 R (2n sin n t  n2t cos n t )i  R(2n cos n t  n2t sin n t )k
Now
v 2  vx2  v 2y  vz2
 [ R (cos n t  n t sin n t )]2  (c) 2  [ R(sin n t  n t cos n t )]2


 R 2  cos 2 n t  2n t sin n t cos n t  n2 t 2 sin 2 n t

 sin 2 n t  2n t sin n t cos n t  n2 t 2 cos 2 n t   c 2





 R 2 1  n2t 2  c 2

Also,

v  R 2 1  n2 t 2  c 2 
or
a 2  ax2  a 2y  az2


2
  R 2n sin n t  n2 t cos n t   (0) 2





2
  R 2n cos n t  n2 t sin n t 


2
2
2
3
 R 4n sin n t  4n t sin n t cos n t  n4 t 2 cos 2 n t




 4n2 cos 2 n t  4n3t sin n t cos n t  n4 t 2 sin 2 n t 


 R 2 4n2  n4 t 2
or

a  Rn 4  n2t 2 
PROBLEM 11.96
The three-dimensional motion of a particle is defined by the position
vector r  ( At cos t )i  ( A t 2  1) j  ( Bt sin t )k , where r and t are
expressed in feet and seconds, respectively. Show that the curve
described by the particle lies on the hyperboloid (y/A)2  (x/A)2 
(z/B)2  1. For A  3 and B  1, determine (a) the magnitudes of the
velocity and acceleration when t  0, (b) the smallest nonzero value
of t for which the position vector and the velocity are perpendicular to
each other.
SOLUTION
We have
r  ( At cos t )i  ( A t 2  1) j  ( Bt sin t )k
or
x  At cos t
cos t 
Then
x
At
y  A t 2  1 z  Bt sin t
sin t 
2
 y
t2    1
 A
z
Bt
2
2
2
2
2
x z
t2      
 A  B
or
2
Then
 y
x z
 A  1   A    B 
 
   
or
 y x  z
 A    A   B  1
     
2
(a)
2
 x   z 
cos 2 t  sin 2 t  1        1
 At   Bt 
Now
2
2

Q.E.D.
With A  3 and B  1, we have
v
and
dr
t
 3(cos t  t sin t )i  3
j  (sin t  t cos t )k
2
dt
t 1
 j
t2  1  t
dv
a
 3( sin t  sin t  t cos t )i  3
dt
(t 2  1)
 (cos t  cos t  t sin t )k
 3(2 sin t  t cos t )i  3
t
t 2 1
1
j  (2 cos t  t sin t )k
(t  1)3/2
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.96 (Continued)
At t  0:
v  3(1  0)i  (0) j  (0)k
v  vx2  v 2y  vz2
v  3 ft/s 
or
and
Then
a  3(0)i  3(1) j  (2  0)k
a 2  (0)2  (3)2  (2)2  13
a  3.61 ft/s2 
or
(b)
If r and v are perpendicular, r  v  0
t


[(3t cos t )i  (3 t 2  1) j  (t sin t )k ]  [3(cos t  t sin t )i   3
 j  (sin t  t cos t )k ]  0
2
 t 1 
or
Expanding
t


(3t cos t )[3(cos t  t sin t )]  (3 t 2  1)  3
  (t sin t )(sin t  t cos t )  0
2
 t 1 
(9t cos2 t  9t 2 sin t cos t )  (9t )  (t sin 2 t  t 2 sin t cos t )  0
or (with t  0)
10  8 cos2 t  8t sin t cos t  0
or
Using “trial and error” or numerical methods, the smallest root is
Note: The next root is t  4.38 s. 
7  2 cos 2t  2t sin 2t  0
t  3.82 s 
PROBLEM 11.97
An airplane used to drop water on brushfires is flying
horizontally in a straight line at 180 mi/h at an altitude of
300 ft. Determine the distance d at which the pilot should
release the water so that it will hit the fire at B.
SOLUTION
First note
v0  180 km/h  264 ft/s
Place origin of coordinates at Point A.
Vertical motion. (Uniformly accelerated motion)
y  0  (0)t 
1 2
gt
2
1
300 ft   (32.2 ft/s 2 )t 2
2
At B:
tB  4.31666 s
or
Horizontal motion. (Uniform)
x  0  (v x ) 0 t
At B:
or
d  (264 ft/s)(4.31666 s)
d  1140 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.98
A ski jumper starts with a horizontal take-off velocity of 25
m/s and lands on a straight landing hill inclined at 30o.
Determine (a) the time between take-off and landing, (b) the
length d of the jump, (c) the maximum vertical distance
between the jumper and the landing hill.
SOLUTION
y   x tan 30
(a) At the landing point,
x  x0   v x 0 t  v0t
Horizontal motion:
 0 t  12 gt 2   12 gt 2
Vertical motion:
y  y0  v y
from which
t2  
2y
2 x tan 30 2v0t tan 30


g
g
g
t 
Rejecting the t  0 solution gives
d 
(b) Landing distance:
2v0 tan 30  2  25  tan 30

g
9.81
 25 2.94 
x
v0t


cos 30 cos 30
cos 30
t  2.94 s 
d  84.9 m 
h  x tan 30  y
(c) Vertical distance:
h  v0t tan 30 
or
1 2
gt
2
Differentiating and setting equal to zero,
dh
 v0 tan 30  gt  0
dt
Then,
hmax 
or
t 
vo tan 30
g
 v0  v0 tan 30 tan 30  1 g  v0 tan 30 

2 
g
v 2 tan 2 30  25  tan 30 
 0

2g
 2 9.81
2
g
2


2
hmax  10.62 m 
PROBLEM 11.99
A baseball pitching machine
“throws”
baseballs
with
a
horizontal velocity v0. Knowing
that height h varies between 788
mm and 1068 mm, determine
(a) the range of values of v0, (b) the
values of  corresponding to
h  788 mm and h  1068 mm.
SOLUTION
(a)
y0  1.5 m, (v y )0  0
Vertical motion:
t 
2( y0  y)
g
or
tB 
2( y0  h)
g
When h  788 mm  0.788 m,
tB 
(2)(1.5  0.788)
 0.3810 s
9.81
When h  1068 mm  1.068 m,
tB 
(2)(1.5  1.068)
 0.2968 s
9.81
y  y0  (v y )0 t 
At Point B,
Horizontal motion:
1 2
gt
2
or
y h
x0  0, (vx )0  v0 ,
x  v0t
or
v0 
x
x
 B
t
tB
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.99 (Continued)
v0 
12.2
 32.02 m/s
0.3810
and
v0 
12.2
 41.11 m/s
0.2968
32.02 m/s  v0  41.11 m/s
or
Vertical motion:
v y  (v y )0  gt   gt
Horizontal motion:
vx  v0
With xB  12.2 m,
(b)
we get
tan   
115.3 km/h  v0  148.0 km/h 
(v y ) B
dy
gt

 B
(vx ) B
dx
v0
For h  0.788 m,
tan  
(9.81)(0.3810)
 0.11673,
32.02
  6.66 
For h  1.068 m,
tan  
(9.81)(0.2968)
 0.07082,
41.11
  4.05 
PROBLEM 11.100
While delivering newspapers, a
girl throws a newspaper with a
horizontal velocity v0. Determine
the range of values of v0 if the
newspaper is to land between
Points B and C.
SOLUTION
Vertical motion. (Uniformly accelerated motion)
y  0  (0)t 
1 2
gt
2
Horizontal motion. (Uniform)
x  0  (vx )0 t  v0t
At B:
y:
tB  0.455016 s
or
Then
x:
y:
or
1
2 ft   (32.2 ft/s 2 )t 2
2
tC  0.352454 s
or
Then
7 ft  (v0 ) B (0.455016 s)
(v0 ) B  15.38 ft/s
or
At C:
1
1
3 ft   (32.2 ft/s 2 )t 2
3
2
x:
1
12 ft  (v0 )C (0.352454 s)
3
(v0 )C  35.0 ft/s
15.38 ft/s  v0  35.0 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.101
Water flows from a drain spout with an initial velocity of
2.5 ft/s at an angle of 15° with the horizontal. Determine the
range of values of the distance d for which the water will
enter the trough BC.
SOLUTION
First note
(vx )0  (2.5 ft/s) cos 15  2.4148 ft/s
(v y )0  (2.5 ft/s) sin 15  0.64705 ft/s
Vertical motion. (Uniformly accelerated motion)
y  0  (v y ) 0 t 
1 2
gt
2
At the top of the trough
8.8 ft  ( 0.64705 ft/s) t 
or
tBC  0.719491 s
1
(32.2 ft/s 2 ) t 2
2
(the other root is negative)
Horizontal motion. (Uniform)
x  0  (vx ) 0 t
In time t BC
xBC  (2.4148 ft/s)(0.719491 s)  1.737 ft
Thus, the trough must be placed so that
xB  1.737 ft or xC  1.737 ft
Since the trough is 2 ft wide, it then follows that
0  d  1.737 ft 
PROBLEM 11.102
In slow pitch softball the
underhand pitch must reach a
maximum height of between 1.8
m and 3.7 m above the ground. A
pitch is made with an initial
velocity v 0 of magnitude 13 m/s
at an angle of 33 with the
horizontal. Determine (a) if the
pitch meets the maximum height
requirement, (b) the height of the
ball as it reaches the batter.
SOLUTION
v0  13 m/s,   33, x0  0, y0  0.6 m
v y  v0 sin   gt
Vertical motion:
y  y0   v0 sin   t 
At maximum height,
vy  0
(a)
t 
or
t 
1 2
gt
2
v0 sin 
g
13sin 33
 0.7217 s
9.81
ymax  0.6  13sin 33  0.7217  
1
 9.81 0.7217 2
2
ymax  3.16 m 
1.8 m  3.16 m  3.7 m
Horizontal motion:
x  x0   v0 cos   t
At x  15.2 m,
t 
(b) Corresponding value of y :
yes 
or
t 
x  x0
v0 cos 
15.2  0
 1.3941 s
13cos 33
y  0.6  13sin 33 1.3941 
1
 9.811.39412
2
y  0.937 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.103
A volleyball player serves the
ball with an initial velocity v0 of
magnitude 13.40 m/s at an angle
of 20° with the horizontal.
Determine (a) if the ball will
clear the top of the net, (b) how
far from the net the ball will land.
SOLUTION
First note
(vx )0  (13.40 m/s) cos 20  12.5919 m/s
(v y )0  (13.40 m/s) sin 20  4.5831 m/s
(a)
Horizontal motion. (Uniform)
x  0  (vx ) 0 t
At C
9 m  (12.5919 m/s) t or tC  0.71475 s
Vertical motion. (Uniformly accelerated motion)
y  y0  (v y )0 t 
At C:
1 2
gt
2
yC  2.1 m  (4.5831 m/s)(0.71475 s)
1
(9.81 m/s 2 )(0.71475 s) 2
2
 2.87 m

yC  2.43 m (height of net)  ball clears net 
(b)
At B, y  0:
0  2.1 m  (4.5831 m/s)t 
1
(9.81 m/s 2 )t 2
2
Solving
t B  1.271175 s (the other root is negative)
Then
d  (vx )0 t B  (12.5919 m/s)(1.271175 s)
 16.01 m
The ball lands
b  (16.01  9.00) m  7.01 m from the net 
PROBLEM 11.104
A golfer hits a golf ball with an initial velocity of 160 ft/s at an angle of 25° with the horizontal. Knowing that
the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B
where the ball first lands.
SOLUTION
(vx )0  (160 ft/s) cos 25
First note
(v y )0  (160 ft/s) sin 25
xB  d cos 5 yB  d sin 5
and at B
Now
Horizontal motion. (Uniform)
x  0  (vx ) 0 t
d cos 5  (160 cos 25)t or t B 
At B
cos 5
d
160 cos 25
Vertical motion. (Uniformly accelerated motion)
y  0  (v y ) 0 t 
1 2
gt
2
( g  32.2 ft/s 2 )
1 2
gt B
2
At B:
 d sin 5  (160 sin 25)t B 
Substituting for tB
 cos 5 
1  cos 5  2
d sin 5  (160 sin 25) 
d  g 
 d
2  160 cos 25 
 160 cos 25 
2
or
or
2
(160 cos 25)2 (tan 5  tan 25)
32.2 cos 5
 726.06 ft
d
d  242 yd 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.105
A homeowner uses a snowblower to clear his
driveway. Knowing that the snow is discharged
at an average angle of 40° with the horizontal,
determine the initial velocity v0 of the snow.
SOLUTION
First note
(vx )0  v0 cos 40
(v y )0  v0 sin 40
Horizontal motion. (Uniform)
x  0  (vx ) 0 t
At B:
14  (v0 cos 40) t or t B 
14
v0 cos 40
Vertical motion. (Uniformly accelerated motion)
y  0  (v y ) 0 t 
At B:
1 2
gt
2
1.5  (v0 sin 40) t B 
( g  32.2 ft/s 2 )
1 2
gt B
2
Substituting for tB

 1 

14
14
1.5  (v0 sin 40) 
 g

 v0 cos 40  2  v0 cos 40 
or
or
v02 
1
2
2
(32.2)(196)/ cos 2 40
1.5  14 tan 40
v0  22.9 ft/s 
PROBLEM 11.106
At halftime of a football game souvenir balls are
thrown to the spectators with a velocity v0.
Determine the range of values of v0 if the balls are
to land between Points B and C.
SOLUTION
The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below Point A.
The coordinates of Point A are x0  0, y0  2m. The components of initial velocity are (vx )0  v0 cos 40 m/s
and (v y )0  v0 sin 40.
Horizontal motion:
x  x0  (vx )0 t  (v0 cos 40)t
Vertical motion:
y  y0  (v y )0 t 
(1)
1 2
gt
2
1
 2  (v0 sin 40)   (9.81)t 2
2
From (1),
v0t 
(2)
x
cos 40
(3)
y  2  x tan 40  4.905t 2
Then
t2 
Point B:
2  x tan 40  y
4.905
(4)
x  8  10cos 35  16.1915 m
y  1.5  10sin 35  7.2358 m
16.1915
 21.1365 m
cos 40
2  16.1915 tan 40  7.2358
t2 
4.905
21.1365
v0 
1.3048
v0 t 
t  1.3048 s
v0  16.199 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.106 (Continued)
Point C:
x  8  (10  7) cos35  21.9256 m
y  1.5  (10  7)sin 35  11.2508 m
Range of values of v0.
v0t 
21.9256
 28.622 m
cos 40
t2 
2  21.9256 tan 40  11.2508
4.905
v0 
28.622
1.3656
t  1.3656 s
v0  20.96 m/s
16.20 m/s  v0  21.0 m/s 
PROBLEM 11.107
A basketball player shoots when she is 16 ft from the
backboard. Knowing that the ball has an initial velocity v0 at an
angle of 30° with the horizontal, determine the value of v0 when
d is equal to (a) 9 in., (b) 17 in.
SOLUTION
First note
(vx )0  v0 cos 30
Horizontal motion. (Uniform)
(v y )0  v0 sin 30
x  0  (vx ) 0 t
(16  d )  (v0 cos 30) t or tB 
At B:
16  d
v0 cos 30
y  0  (v y ) 0 t 
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1 2
gt B
2
At B:
3.2  (v0 sin 30) t B 
Substituting for tB
 16  d  1  16  d 
3.2  (v0 sin 30) 
 g

 v0 cos 30  2  v0 cos 30 
or
v02 
(a)
d  9 in.:
v02
d  17 in.:
v02
(b)


( g  32.2 ft/s 2 )
2
2 g (16  d )2
3

1
3
(16  d )  3.2 

2(32.2) 16  129 
3

1
3
16  129   3.2
2(32.2) 16  17
12 
3

1
3
2
v0  29.8 ft/s 
2
16  1712   3.2
v0  29.6 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.108
A tennis player serves the ball at a height h  2.5 m with an initial velocity of v0 at an angle of 5° with the
horizontal. Determine the range for which of v0 for which the ball will land in the service area which extends
to 6.4 m beyond the net.
SOLUTION
The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below the
point where the racket impacts the ball. The coordinates of this impact point are x0  0, y0  h  2.5 m. The
components of initial velocity are (vx )0  v0 cos5 and (v y )0  v0 sin 5.
Horizontal motion:
x  x0  (vx )0 t  (v0 cos5)t
Vertical motion:
y  y0  (v y )0 t 
(1)
1 2
gt
2
1
 2.5  (v0 sin 5)t   (9.81)t 2
2
From (1),
Then
v0t 
x
cos 5
(2)
(3)
y  2.5  x tan 5  4.905t 2
t2 
2.5  x tan 5  y
4.905
(4)
At the minimum speed the ball just clears the net.
x  12.2 m,
y  0.914 m
12.2
 12.2466 m
cos 5
2.5  12.2 tan 5  0.914
t2 
4.905
12.2466
v0 
0.32517
v0t 
t  0.32517 s
v0  37.66 m/s
PROBLEM 11.108 (Continued)
At the maximum speed the ball lands 6.4 m beyond the net.
x  12.2  6.4  18.6 m
y0
18.6
v0t 
 18.6710 m
cos 5
2.5  18.6 tan 5  0
t2 
t  0.42181 s
4.905
18.6710
v0 
v0  44.26 m/s
0.42181
Range for v0.
37.7 m/s  v0  44.3 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.109
The nozzle at A discharges cooling water with an initial
velocity v0 at an angle of 6° with the horizontal onto a grinding
wheel 350 mm in diameter. Determine the range of values of
the initial velocity for which the water will land on the
grinding wheel between Points B and C.
SOLUTION
First note
(vx )0  v0 cos 6
(v y )0  v0 sin 6
Horizontal motion. (Uniform)
x  x0  (vx )0 t
Vertical motion. (Uniformly accelerated motion)
y  y0  (v y )0 t 
1 2
gt
2
( g  9.81 m/s 2 )
x  (0.175 m) sin 10
At Point B:
y  (0.175 m) cos 10
x : 0.175 sin 10  0.020  (v0 cos 6)t
tB 
or
0.050388
v0 cos 6
y : 0.175 cos 10  0.205  (v0 sin 6)t B 
1 2
gt B
2
Substituting for tB
 0.050388  1
 0.050388 
0.032659  (v0 sin 6) 
  (9.81) 

 v0 cos 6  2
 v0 cos 6 
2
PROBLEM 11.109 (Continued)
v02
or

1
2
(9.81)(0.050388) 2
cos 2 6(0.032659  0.050388 tan 6)
(v0 ) B  0.678 m/s
or
x  (0.175 m) cos 30
At Point C:
y  (0.175 m) sin 30
x : 0.175 cos 30  0.020  (v0 cos 6)t
tC 
or
0.171554
v0 cos 6
y : 0.175 sin 30  0.205  (v0 sin 6)tC 
1 2
gtC
2
Substituting for tC
 0.171554  1
 0.171554 
0.117500  (v0 sin 6) 
  (9.81) 

 v0 cos 6  2
 v0 cos 6 
or
or

v02 
1
2
2
(9.81)(0.171554) 2
cos 2 6(0.117500  0.171554 tan 6)
(v0 )C  1.211 m/s
0.678 m/s  v0  1.211 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.110
While holding one of its ends, a worker lobs a coil of rope over the
lowest limb of a tree. If he throws the rope with an initial velocity v0
at an angle of 65° with the horizontal, determine the range of values
of v0 for which the rope will go over only the lowest limb.
SOLUTION
First note
(v x )0  v0 cos 65
(v y )0  v0 sin 65
Horizontal motion. (Uniform)
x  0  (vx ) 0 t
At either B or C, x  5 m
s  (v0 cos 65)t B ,C
or
t B ,C 
5
(v0 cos 65)
Vertical motion. (Uniformly accelerated motion)
y  0  (v y ) 0 t 
1 2
gt
2
( g  9.81 m/s 2 )
At the tree limbs, t  t B ,C

 1 

5
5
yB ,C  (v0 sin 65) 
 g

 v0 cos 65  2  v0 cos 65 
2
PROBLEM 11.110 (Continued)
v02 
or

1
2
(9.81)(25)
2
cos 65(5 tan 65  yB , C )
686.566
5 tan 65  yB , C
At Point B:
v02 
686.566
5 tan 65  5
or
(v0 ) B  10.95 m/s
At Point C:
v02 
686.566
5 tan 65  5.9
or
(v0 )C  11.93 m/s

10.95 m/s  v0  11.93 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.111
The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle  with the
horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle , (b) the angle  that the
velocity of the ball at Point B forms with the horizontal.
SOLUTION
First note
v0  72 km/h  20 m/s
(vx )0  v0 cos   (20 m/s) cos 
and
(v y )0  v0 sin   (20 m/s) sin 
(a)
Horizontal motion. (Uniform)
x  0  (vx )0 t  (20 cos  ) t
At Point B:
14  (20 cos  )t or t B 
7
10 cos 
Vertical motion. (Uniformly accelerated motion)
y  0  (v y ) 0 t 
At Point B:
1 2
1
gt  (20 sin  )t  gt 2
2
2
0.08  (20 sin  )t B 
( g  9.81 m/s 2 )
1 2
gt B
2
Substituting for tB

 1 

7
7
0.08  (20 sin  ) 
 g

 10 cos   2  10 cos  
or
Now
8  1400 tan  
1
49
g
2 cos 2 
1
 sec2   1  tan 2 
2
cos 
2
PROBLEM 11.111 (Continued)
Then
or
Solving
8  1400 tan   24.5 g (1  tan 2  )
240.345 tan 2   1400 tan   248.345  0
  10.3786 and   79.949
Rejecting the second root because it is not physically reasonable, we have
  10.38 
(b)
We have
vx  (vx )0  20 cos 
and
v y  (v y )0  gt  20 sin   gt
At Point B:
(v y ) B  20 sin   gt B
 20 sin  
7g
10 cos 
Noting that at Point B, v y  0, we have
tan  


or
|(v y ) B |
vx
7g
10 cos 
 20 sin 
20 cos 
7
9.81
200 cos 10.3786
 sin 10.3786
cos 10.3786
  9.74 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.112
A model rocket is launched from Point A with an initial velocity v0 of
75 m/s. If the rocket’s descent parachute does not deploy and the
rocket lands a distance d  100 m from A, determine (a) the angle 
that v0 forms with the vertical, (b) the maximum height above Point A
reached by the rocket, and (c) the duration of the flight.
SOLUTION
Set the origin at Point A.
x0  0,
y0  0
Horizontal motion:
x  v0t sin 
Vertical motion:
y  v0 t cos  
cos  
sin 2   cos2  
sin  
x
v0t
1 2
gt
2
1 
1

y  gt 2 

v0 t 
2

(2)
2
1  2 
1 2 


x
y
gt


  1
2
(v0t )2 

 
x 2  y 2  gyt 2 

1 24
g t  v02t 2
4

1 24
g t  v02  gy t 2  ( x 2  y 2 )  0
4
At Point B,
(1)
x 2  y 2  100 m,
x  100 cos 30 m
y  100 sin 30  50 m
1
(9.81) 2 t 4  [752  (9.81)(50)] t 2  1002  0
4
24.0590 t 4  6115.5 t 2  10000  0
t 2  252.54 s 2
and 1.6458 s 2
t  15.8916 s
and 1.2829 s
(3)
PROBLEM 11.112 (Continued)
Restrictions on  :
0    120
tan  
x
y  gt
1
2
2

100 cos 30
 0.0729
50  (4.905)(15.8916) 2
  4.1669
100 cos 30
 2.0655
50  (4.905)(1.2829) 2
  115.8331
and
Use   4.1669 corresponding to the steeper possible trajectory.
(a)
Angle  .
(b)
Maximum height.
  4.17 
v y  0 at
y  ymax
v y  v0 cos   gt  0
t
v0 cos 
g
ymax  v0 t cos  

(c)

Duration of the flight.
v 2 cos 2 
1
gt  0
2
2g
(75)2 cos 2 4.1669
(2)(9.81)
ymax  285 m 
(time to reach B)
t  15.89 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.113
The initial velocity v0 of a hockey puck is 105 mi/h. Determine (a) the largest value (less than 45°) of
the angle  for which the puck will enter the net, (b) the corresponding time required for the puck to reach
the net.
SOLUTION
First note
v0  105 mi/h  154 ft/s
(vx )0  v0 cos   (154 ft/s) cos 
and
(v y )0  v0 sin   (154 ft/s) sin 
(a)
Horizontal motion. (Uniform)
x  0  (vx )0 t  (154 cos  )t
At the front of the net,
Then
or
x  16 ft
16  (154 cos  )t
tenter 
8
77 cos 
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
 (154 sin  ) t  gt 2
2
y  0  (v y ) 0 t 
( g  32.2 ft/s 2 )
At the front of the net,
yfront  (154 sin  ) tenter 
1 2
gtenter
2

 1 

8
8
 (154 sin  ) 
 g

 77 cos   2  77 cos  
32 g
 16 tan  
5929 cos 2 
2
PROBLEM 11.113 (Continued)
Now
Then
or
1
 sec2   1  tan 2 
cos 2 
yfront  16 tan  
tan 2  
32 g
(1  tan 2  )
5929
 5929

5929
tan   1 
yfront   0
2g
32 g


5929
2g


   5929
2g


2


1/ 2

y
 4 1  5929
32 g front 

2
Then
tan  
or
 5929 2 
5929
5929

yfront  
tan  
  
  1 
4  32.2  4  32.2   32  32.2
 
or
tan   46.0326  [(46.0326)2  (1  5.7541 yfront )]1/2
1/ 2
Now 0  yfront  4 ft so that the positive root will yield values of   45 for all values of yfront.
When the negative root is selected,  increases as yfront is increased. Therefore, for  max , set
yfront  yC  4 ft
(b)
Then
tan   46.0326  [(46.0326)2  (1  5.7541  4)]1/2
or
 max  14.6604
We had found
tenter 

or
 max  14.66 
8
77 cos 
8
77 cos 14.6604
tenter  0.1074 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.114
A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an
initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the
worker can wash from his position at A, (b) the corresponding angle .
SOLUTION
First note
(vx )0  v0 cos   (11.5 m/s) cos 
(v y )0  v0 sin   (11.5 m/s) sin 
By observation, dmax occurs when
ymax  1.1 m.
Vertical motion. (Uniformly accelerated motion)
v y  (v y )0  gt
 (11.5 sin  )  gt
y  ymax
When
Then
1 2
gt
2
1
 (11.5 sin  ) t  gt 2
2
y  0  (v y ) 0 t 
at B, (v y ) B  0
(v y ) B  0  (11.5 sin  )  gt
11.5 sin 
g
( g  9.81 m/s 2 )
or
tB 
and
yB  (11.5 sin  ) t B 
1 2
gt B
2
Substituting for tB and noting yB  1.1 m
 11.5 sin   1  11.5 sin  
1.1  (11.5 sin  ) 
 g

g
g

 2 

1
(11.5) 2 sin 2 

2g
or
sin 2  
2.2  9.81
11.52
  23.8265
2
PROBLEM 11.114 (Continued)
(a)
Horizontal motion. (Uniform)
x  0  (vx )0 t  (11.5 cos  ) t
At Point B:
where
Then
or
(b)
From above
x  d max
tB 
and t  t B
11.5
sin 23.8265  0.47356 s
9.81
d max  (11.5)(cos 23.8265)(0.47356)
d max  4.98 m 
  23.8 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.115
An oscillating garden sprinkler which
discharges water with an initial velocity
v0 of 8 m/s is used to water a vegetable
garden. Determine the distance d to the
farthest Point B that will be watered and
the corresponding angle  when (a) the
vegetables are just beginning to grow,
(b) the height h of the corn is 1.8 m.
SOLUTION
First note
(vx )0  v0 cos   (8 m/s) cos 
(v y )0  v0 sin   (8 m/s) sin 
Horizontal motion. (Uniform)
x  0  (vx )0 t  (8 cos  ) t
At Point B:
or
x  d : d  (8 cos  ) t
d
tB 
8 cos 
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
 (8 sin  ) t  gt 2 ( g  9.81 m/s 2 )
2
1
0  (8 sin  ) t B  gt B2
2
y  0  (v y ) 0 t 
At Point B:
Simplifying and substituting for tB
0  8 sin  
d
or
(a)
1  d 
g

2  8 cos  
64
sin 2
g
(1)
When h  0, the water can follow any physically possible trajectory. It then follows from
Eq. (1) that d is maximum when 2  90
  45 
or
Then
or
d
64
sin (2  45)
9.81
d max  6.52 m 
PROBLEM 11.115 (Continued)
(b)
Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as  increases
in value from 0 to 45° and then d decreases as  is further increased. Thus, dmax occurs for the value
of  closest to 45° and for which the water just passes over the first row of corn plants. At this row,
xcom  1.5 m
tcorn 
so that
1.5
8 cos 
Also, with ycorn  h, we have
h  (8 sin  ) tcorn 
1 2
gtcorn
2
Substituting for tcorn and noting h  1.8 m,
 1.5  1  1.5 
1.8  (8 sin  ) 
 g

 8 cos   2  8 cos  
or
Now
Then
or
1.8  1.5 tan  
2
2.25 g
128 cos 2 
1
 sec2   1  tan 2 
2
cos 
1.8  1.5 tan  
2.25(9.81)
(1  tan 2  )
128
0.172441 tan 2   1.5 tan   1.972441  0
  58.229 and   81.965
Solving
From the above discussion, it follows that d  dmax when
  58.2 
Finally, using Eq. (1)
d
or
64
sin (2  58.229)
9.81
d max  5.84 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.116*
A nozzle at A discharges water
with an initial velocity of 36 ft/s at
an angle  with the horizontal.
Determine (a) the distance d to the
farthest point B on the roof that the
water can reach, (b) the
corresponding angle  . Check
that the stream will clear the edge
of the roof.
SOLUTION
Horizontal motion:
x   v0 cos   t
Vertical motion:
y  y0   v0 sin   t 
y  y0  x tan  
Let
u  x tan 
x2 
du
 xmax 
2

2
2  36 


32.2
(a) Maximum distance:

y  y0  u 

g
x2  u 2
2v02

   0.
d x2
du
   2v
d x2
v0  36 ft/s,
1 2
1
gx 2
gt  y0  x tan  
2
2  v0 cos  2
2v02
 u  y0  y   u 2
g
The maximum value of x 2 is required:
Data:
x
v0 cos 
gx 2
1  tan 2 
2v02
so that
Solving for x 2 :
t
or
2
0
g
y0  3.6 ft,
 2u  0
or
yB  18 ft
u
v02
g
2
36 

u
32.2
 40.2484  3.6  18   40.2484 2  460.78 ft 2
d  xmax  13.5
 40.2484 ft
xmax  21.466 ft
d  7.97 ft 
PROBLEM 11.116* (Continued)
(b) Angle  .
tan  
xmax tan 
u
40.2484


 1.875
xmax
xmax
21.466
  61.9 
  61.93
Check the edge.
y  y0  x tan  
gx 2
2  v0 cos  
y  3.6  13.5 1.875  
2
 32.2 13.5 2
2
2 36 cos 61.93
y  18.69 ft 
Since y  18 ft, the stream clears the edge.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.117
The velocities of skiers A and B are
as shown. Determine the velocity
of A with respect to B.
SOLUTION
We have
vA  vB  vA/B
The graphical representation of this equation is then as shown.
Then
v 2A/B  302  452  2(30)(45) cos 15
or
vA/B  17.80450 ft/s
and
or
30
17.80450

sin 
sin 15
  25.8554°
  25  50.8554°
vA/B  17.8 ft/s
Alternative solution.
vA/B  vA  vB
 30 cos 10 i  30 sin 10 j  (45 cos 25 i  45 sin 25 j)
 11.2396i  13.8084 j
 5.05 m/s  17.8 ft/s
50.9°
50.9° 
PROBLEM 11.118
The three blocks shown move with constant velocities. Find the velocity of each
block, knowing that the relative velocity of A with respect to C is 300 mm/s upward
and that the relative velocity of B with respect to A is 200 mm/s downward.
SOLUTION
From the diagram
Cable 1:
y A  yD  constant
Then
vA  vD  0
Cable 2:
(1)
( yB  yD )  ( yC  yD )  constant
vB  vC  2vD  0
Then
(2)
Combining Eqs. (1) and (2) to eliminate vD ,
2vA  vB  vC  0
(3)
Now
vA/C  vA  vC  300 mm/s
(4)
and
vB/A  vB  vA  200 mm/s
(5)
(3)  (4)  (5) 
Then
(2vA  vB  vC )  (v A  vC )  (vB  v A )  (300)  (200)
v A  125 mm/s 
or
and using Eq. (5)
vB  (125)  200
v B  75 mm/s 
or
Eq. (4)
or
125  vC  300
vC  175 mm/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.119
Three seconds after automobile B passes through the intersection shown,
automobile A passes through the same intersection. Knowing that the speed
of each automobile is constant, determine (a) the relative velocity of B with
respect to A, (b) the change in position of B with respect to A during a 4-s
interval, (c) the distance between the two automobiles 2 s after A has
passed through the intersection.
SOLUTION
v A  45 mi/h  66 ft/s
vB  30 mi/h  44 ft/s
Law of cosines
vB2/A  662  442  2(66)(44) cos110
vB/A  90.99 ft/s
vB  v A  vB/A
Law of sines
sin  sin110

66
90.99
  42.97
  90    90  42.97  47.03
v B/A  91.0 ft/s
(a)
Relative velocity:
(b)
Change in position for t  4 s.
rB/A  vB/A t  (91.0 ft/s)(4 s)
(c)
Distance between autos 2 seconds after auto A has passed intersection.
Auto A travels for 2 s.
vA  66 ft/s
20°
rA  vAt  (66 ft/s)(2 s)  132 ft
rA  132 ft
20°
rB/A  364 ft
47.0° 
47.0° 
PROBLEM 11.119 (Continued)
Auto B
v B  44 ft/s
rB  v B t  (44 ft/s)(5 s)  220 ft
rB  rA  rB/A
Law of cosines
rB2/A  (132)2  (220) 2  2(132)(220) cos110
rB/A  292.7 ft
Distance between autos  293 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.120
Shore-based radar indicates that a ferry leaves its slip with a
velocity v  18 km/h
70°, while instruments aboard the
ferry indicate a speed of 18.4 km/h and a heading of 30° west
of south relative to the river. Determine the velocity of the
river.
SOLUTION
We have
v F  v R  v F/R
or v F  v F/R  v R
The graphical representation of the second equation is then as shown.
We have
vR2  182  18.42  2(18)(18.4) cos 10
or
vR  3.1974 km/h
and
or
18
3.1974

sin  sin 10
  77.84
Noting that
v R  3.20 km/h
Alternatively one could use vector algebra.
17.8° 
PROBLEM 11.121
Airplanes A and B are flying at the same altitude and are
tracking the eye of hurricane C. The relative velocity of C
with respect to A is vC/A  350 km/h
75°, and the relative
velocity of C with respect to B is vC/B  400 km/h
40°.
Determine (a) the relative velocity of B with respect to A,
(b) the velocity of A if ground-based radar indicates that the
hurricane is moving at a speed of 30 km/h due north, (c) the
change in position of C with respect to B during a 15-min
interval.
SOLUTION
(a)
We have
vC  v A  vC/A
and
vC  v B  vC/B
Then
v A  vC/A  v B  vC/B
or
v B  v A  vC/A  vC/B
Now
v B  v A  v B/A
so that
v B/A  vC/A  vC/B
or
vC/A  vC/B  v B/A
The graphical representation of the last equation is then as shown.
We have
vB2/A  3502  4002  2(350)(400) cos 65
or
vB/A  405.175 km/h
and
or
400 405.175

sin 
sin 65
  63.474
75    11.526
v B/A  405 km/h
11.53° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.121 (Continued)
(b)
We have
vC  v A  vC/A
or
v A  (30 km/h) j  (350 km/h)( cos 75i  sin 75 j )
v A  (90.587 km/h)i  (368.07 km/h) j
v A  379 km/h
or
(c)
76.17° 
Noting that the velocities of B and C are constant, we have
rB  (rB )0  v B t
Now
rC  (rC )0  vC t
rC/B  rC  rB  [(rC )0  (rB )0 ]  ( vC  v B )t
 [(rC )0  (rB )0 ]  vC/B t
Then
rC/B  (rC/B )t2  (rC/B )t1  vC/B (t2  t1 )  vC/B t
For t  15 min:
1 
rC/B  (400 km/h)  h   100 km
4 
rC/B  100 km
40° 
PROBLEM 11.122
Instruments in an airplane which is in level flight indicate that the velocity relative
to the air (airspeed) is 120 km/h and the direction of the relative velocity vector
(heading) is 70° east of north. Instruments on the ground indicate that the velocity of
the airplane (ground speed) is 110 km/h and the direction of flight (course) is 60°
east of north. Determine the wind speed and direction.
SOLUTION
Let i and j be unit vectors in directions east and north respectively.
Velocity of plane relative to air.
v P/ A  120 km/h  cos 20 i  sin 20 j
Velocity of plane.
v P  110 km/h  cos30 i  sin 30j
But
v P  v A  v P/ A
Velocity of air.
v A  v P  v P/ A
v A  110cos30  120cos 20  i  110sin 30  120sin 20  j
   17.50 km/h  i  13.96 km/h  j
vA 
17.50 2  13.96 2
tan  
17.50
13.96
 22.4 km/h
  51.4
v A  22.4 km/h at 51.4 west of north 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.123
Knowing that at the instant shown
block B has a velocity of 2 ft/s to
the right and an acceleration of 3
ft/s2 to the left, determine (a) the
velocity of block A, (b) the
acceleration of block A.
SOLUTION
Given:
vB  2 ft/s, aB  3 ft/s
Length of Cable
L  y A  2 xB  constants
Differentiate Length Equation
0  vA  2vB
(1)
Differentiate again
0  a A  2aB
(2)
(e) Rearrange (1)
vA  2vB
 2  2  ft/s
vA  4 ft/s
(f) Rearrange (2)
v A  4 ft/s  
a A  2aB
 2  3 ft/s 2
a A  6 ft/s 2
a A  6 ft/s 2  
PROBLEM 11.124
Knowing that at the instant shown block A has a velocity of 8
in./s and an acceleration of 6 in./s2 both directed down the
incline, determine (a) the velocity of block B, (b) the
acceleration of block B.
SOLUTION
From the diagram
2 x A  xB/A  constant
Then
2vA  vB/A  0
| vB/A |  16 in./s
or
2a A  aB/A  0
and
| aB/A |  12 in./s2
or
Note that v B/A and a B/A must be parallel to the top surface of block A.
(a)
We have
v B  v A  v B/A
The graphical representation of this equation is then as shown. Note that because A is moving
downward, B must be moving upward relative to A.
We have
vB2  82  162  2(8)(16) cos 15
or
vB  8.5278 in./s
and
or
8
8.5278

sin  sin 15
  14.05
v B  8.53 in./s
(b)
54.1° 
The same technique that was used to determine v B can be used to determine a B . An alternative method
is as follows.
We have
a B  a A  a B/A
 (6i)  12( cos 15i  sin 15 j)*
 (5.5911 in./s 2 )i  (3.1058 in./s 2 ) j
or
a B  6.40 in./s2
54.1° 
* Note the orientation of the coordinate axes on the sketch of the system.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.125
A boat is moving to the right with a constant
deceleration of 0.3 m/s2 when a boy standing on the
deck D throws a ball with an initial velocity relative to
the deck which is vertical. The ball rises to a maximum
height of 8 m above the release point and the boy must
step forward a distance d to catch it at the same height
as the release point. Determine (a) the distance d,
(b) the relative velocity of the ball with respect to the
deck when the ball is caught.
SOLUTION
Horizontal motion of the ball:
vx  (vx )0 ,
Vertical motion of the ball:
v y  (v y )0  gt
yB  (v y )0 t 
At maximum height,
xball  (vx )0 t
1 2
gt , (v y ) 2  (v y )02  2 gy
2
vy  0
and
y  ymax
(v y )2  2 gymax  (2)(9.81)(8)  156.96 m 2 /s 2
(v y )0  12.528 m/s
At time of catch,
tcatch  2.554 s
or
Motion of the deck:
1
(9.81)t 2
2
and
v y  12.528 m/s
y  0  12.528 
vx  (vx )0  aDt ,
xdeck  (vx )0 t 
1
aDt 2
2
Motion of the ball relative to the deck:
(vB/D ) x  (vx )0  [(vx )0  aDt ]  aDt
1
1


xB/D  (vx )0 t  (vx )0 t  aDt 2    aDt 2
2
2


(vB/D ) y  (v y )0  gt , yB/D  yB
(a)
(b)
At time of catch,
1
d  xD/B   ( 0.3)(2.554) 2
2
(vB/D ) x  ( 0.3)(2.554)   0.766 m/s
(vB/D ) y  12.528 m/s
d  0.979 m 
or 0.766 m/s
v B/D  12.55 m/s
86.5 
PROBLEM 11.126
The assembly of rod A and wedge B starts from rest and moves
to the right with a constant acceleration of 2 mm/s2. Determine
(a) the acceleration of wedge C, (b) the velocity of wedge C
when t  10 s.
SOLUTION
(a)
We have
aC  a B  aC/B
The graphical representation of this equation is then as shown.
First note
  180  (20  105)
 55
Then
aC
2

sin 20 sin 55
aC  0.83506 mm/s 2
(b)
aC  0.835 mm/s2
75° 
vC  8.35 mm/s
75° 
For uniformly accelerated motion
vC  0  aC t
At t  10 s:
vC  (0.83506 mm/s 2 )(10 s)
 8.3506 mm/s
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.127
Determine the required velocity of the belt B if the
relative velocity with which the sand hits belt B is to be
(a) vertical, (b) as small as possible.
SOLUTION
A grain of sand will undergo projectile motion.
vsx  vsx  constant  5 ft/s
0
vsy  2 gh  (2)(32.2 ft/s2 )(3 ft)  13.90 ft/s 
y-direction.
v S/B  v S  v B
Relative velocity.
(a)
(1)
If vS/B is vertical,
vS /B j  5i  13.9 j  (vB cos 15i  vB sin 15 j)
 5i  13.9 j  vB cos 15i  vB sin 15 j
Equate components.
i : 0  5  vB cos 15
vB 
5
 5.176 ft/s
cos 15
v B  5.18 ft/s
(b)
15° 
vS/C is as small as possible, so make vS/B  to vB into (1).
vS/B sin 15i  vS/B cos 15 j   5i  13.9 j  vB cos 15i  vB sin 15 j
Equate components and transpose terms.
(sin 15) vS/B  (cos 15) vB  5
(cos 15) vS/B  (sin 15) vB  13.90
Solving,
vS/B  14.72 ft/s
vB  1.232 ft/s
v B  1.232 ft/s
15° 
PROBLEM 11.128
Conveyor belt A, which forms a 20° angle
with the horizontal, moves at a constant
speed of 4 ft/s and is used to load an
airplane. Knowing that a worker tosses
duffel bag B with an initial velocity of 2.5
ft/s at an angle of 30° with the horizontal,
determine the velocity of the bag relative to
the belt as it lands on the belt.
SOLUTION
First determine the velocity of the bag as it lands on the belt. Now
[(vB ) x ]0  (vB )0 cos 30
 (2.5 ft/s) cos 30
[(vB ) y ]0  (vB )0 sin 30
 (2.5 ft/s)sin 30
Horizontal motion. (Uniform)
x  0  [(vB ) x ]0 t
(vB ) x  [(vB ) x ]0
 (2.5 cos 30)t
 2.5 cos 30
Vertical motion. (Uniformly accelerated motion)
y  y0  [(vB ) y ]0 t 
1 2
gt
2
 1.5  (2.5 sin 30) t 
(vB ) y  [(vB ) y ]0  gt
1 2
gt
2
 2.5 sin 30  gt
The equation of the line collinear with the top surface of the belt is
y  x tan 20
Thus, when the bag reaches the belt
1.5  (2.5 sin 30) t 
1 2
gt  [(2.5 cos 30) t ]tan 20
2
or
1
(32.2) t 2  2.5(cos 30 tan 20  sin 30) t  1.5  0
2
or
16.1t 2  0.46198t  1.5  0
Solving
t  0.31992 s and t  0.29122 s (Reject)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.128 (Continued)
The velocity v B of the bag as it lands on the belt is then
v B  (2.5 cos 30)i  [2.5 sin 30  32.2(0.319 92)]j
 (2.1651 ft/s)i  (9.0514 ft/s) j
Finally
or
v B  v A  v B/A
v B/A  (2.1651i  9.0514 j)  4(cos 20i  sin 20 j)
 (1.59367 ft/s)i  (10.4195 ft/s) j
or
v B/A  10.54 ft/s
81.3° 
PROBLEM 11.129
During a rainstorm the paths of the raindrops appear to form an angle of 30° with the vertical and to be directed
to the left when observed from a side window of a train moving at a speed of 15 km/h. A short time later, after
the speed of the train has increased to 24 km/h, the angle between the vertical and the paths of the drops
appears to be 45°. If the train were stopped, at what angle and with what velocity would the drops be observed
to fall?
SOLUTION
vrain  vtrain  vrain/train
Case :
vT  15 km/h
;
vR /T
30°
Case :
vT  24 km/h
;
vR /T
45°
Case :
(vR ) y tan 30  15  (vR ) x
(1)
Case :
(vR ) y tan 45  24  (vR ) x
(2)
Substract (1) from (2)
(vR ) y (tan 45  tan 30)  9
(vR ) y  21.294 km/h
Eq. (2):
21.294 tan 45  25  (vR ) x
(vR ) x  2.706 km/h
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.129 (Continued)
3.706
21.294
  7.24
tan  
vR 
21.294
 21.47 km/h  5.96 m/s
cos 7.24
vR  5.96 m/s
82.8° 
Alternate solution
Alternate, vector equation
v R  vT  v R /T
For first case,
v R  15i  vR /T 1 ( sin 30i  cos30 j)
For second case,
v R  24i  vR /T  2 ( sin 45i  cos 45 j)
Set equal
15i  vR /T 1 ( sin 30i  cos30 j)  24i  vR /T  2 ( sin 45i  cos 45 j)
Separate into components:
i:
15  vR /T 1 sin 30  24  vR /T  2 sin 45
vR /T 1 sin 30  vR /T  2 sin 45  9
(3)
vR /T 1 cos30  vR /T  2 cos 45
j:
vR /T 1 cos30  vR /T 2 cos 45  0
(4)
Solving Eqs. (3) and (4) simultaneously,
vR /T 1  24.5885 km/h
vR /T  2  30.1146 km/h
Substitute v R /T 2 back into equation for v R .
v R  24i  30.1146( sin 45i  cos 45 j)
v R  2.71i  21.29 j
 21.29 
  82.7585
 2.71 
  tan 1 
v R  21.4654 km/hr  5.96 m/s
v R  5.96 m/s
82.8° 
PROBLEM 11.130
Instruments in airplane A indicate that with respect to the air the plane is headed 30o
north of east with an airspeed of 300 mi/h. At the same time radar on ship B indicates
that the relative velocity of the plane with respect to the ship is 280 mi/h in the
direction 33o north of east. Knowing that the ship is steaming due south at 12 mi/h,
determine (a) the velocity of the airplane, (b) the wind speed and direction.
SOLUTION
Let the x-axis be directed east, and the y-axis be directed north.
Airspeed:
30  300  cos 30i  sin 30j mi/h
v A/W  300 mi/h
v A/B  280  cos 33i  sin 33 j mi/h
Plane relative to ship:
v B  12 mi/h
Ship:
 12 j
(a) Velocity of airplane.
v A  v B  v A/B  12 j  280  cos 33i  sin 33 j
 234.83i  140.50 j
v A  274 mi/h
30.9 
vW  26.7 mi/h
20.8 
(b) Wind velocity.
vW  v A  vW / A  v A  v AW
/
 234.83i  140.50 j  300  cos30i  sin 30j
 24.98i  9.50 j
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.131
When a small boat travels north at 5 km/h, a flag mounted on
its stern forms an angle   50 with the centerline of the boat
as shown. A short time later, when the boat travels east at 20
km/h, angle  is again 50°. Determine the speed and the
direction of the wind.
SOLUTION
We have
vW  v B  vW/B
Using this equation, the two cases are then graphically represented as shown.
With vW now defined, the above diagram is redrawn for the two cases for clarity.
Noting that
  180  (50  90   )
 40  
We have
vW
5

sin 50 sin (40   )
  180  (50   )
 130  
vW
20

sin 50 sin (130   )
PROBLEM 11.131 (Continued)
Therefore
or
or
5
20

sin (40   ) sin (130   )
sin 130 cos   cos 130 sin   4(sin 40 cos   cos 40 sin  )
tan  
sin 130  4 sin 40
cos 130  4 cos 40
or
  25.964
Then
vW 
5 sin 50
 15.79 km/h
sin (40  25.964)
vW  15.79 km/h
26.0° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.132
As part of a department store
display, a model train D runs on a
slight incline between the store’s
up and down escalators. When the
train and shoppers pass Point A,
the train appears to a shopper on
the up escalator B to move
downward at an angle of 22° with
the horizontal, and to a shopper
on the down escalator C to move
upward at an angle of 23° with the
horizontal and to travel to the left.
Knowing that the speed of the
escalators is 3 ft/s, determine the
speed and the direction of the
train.
SOLUTION
v D  v B  v D/B
We have
v D  vC  v D/C
The graphical representations of these equations are then as shown.
Then
vD
3

sin 8 sin (22   )
Equating the expressions for
vD
3

sin 7 sin (23   )
vD
3
sin 8
sin 7

sin (22   ) sin (23   )
or
or
or
Then
sin 8 (sin 23 cos   cos 23 sin  )
 sin 7 (sin 22 cos   cos 22 sin  )
tan  
sin 8 sin 23  sin 7 sin 22
sin 8 cos 23  sin 7 cos 22
  2.0728
vD 
3 sin 8
 1.024 ft/s
sin (22  2.0728)
v D  1.024 ft/s
2.07° 
PROBLEM 11.132 (Continued)
Alternate solution using components.
v B  (3 ft/s)
30  (2.5981 ft/s)i  (1.5 ft/s) j
vC  (3 ft/s)
30  (2.5981 ft/s)i  (1.5 ft/s) j
v D/B  u1
22  (u1 cos 22)i  (u1 sin 22) j
v D/C  u2
23  (u2 cos 23)i  (u2 sin 23) j
v D  vD
  (vD cos  )i  (vD sin  ) j
v D  v B  v D/B  vC  v D/C
2.5981i  1.5 j  (u1 cos 22)i  (u1 sin 22) j  2.5981i  1.5 j  (u2 cos 23)i  (u2 sin 23) j
Separate into components, transpose, and change signs.
u1 cos 22  u2 cos 23  0
u1 sin 22  u1 sin 23  3
Solving for u1 and u2 ,
u1  3.9054 ft/s
u2  3.9337 ft/s
v D  2.5981i  1.5 j  (3.9054 cos 22)i  (3.9054 sin 22) j
 (1.0229 ft/s)i  (0.0370 ft/s) j
or
v D  2.5981i  1.5 j  (3.9337 cos 23)i  (3.9337 sin 23) j
 (1.0229 ft/s)i  (0.0370 ft/s) j
v D  1.024 ft/s
2.07° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.133
Determine the smallest radius that should be used for a
highway if the normal component of the acceleration of a car
traveling at 72 km/h is not to exceed 0.8 m/s2 .
SOLUTION
an 
v2
an  0.8 m/s 2

v  72 km/h  20 m/s
0.8 m/s 2 
(20 m/s) 2

  500 m  
PROBLEM 11.134
Determine the maximum speed that the cars of the roller-coaster
can reach along the circular portion AB of the track if  is 25 m
and the normal component of their acceleration cannot exceed 3 g.
SOLUTION
We have
an 
v2

Then
(vmax )2AB  (3  9.81 m/s2 )(25 m)
or
(vmax ) AB  27.124 m/s
or
(vmax ) AB  97.6 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 11.135
Human centrifuges are often used to simulate different
acceleration levels for pilots and astronauts. Space shuttle
pilots typically face inwards towards the center of the gondola
in order to experience a simulated 3g forward acceleration.
Knowing that the astronaut sits 5 m from the axis of rotation
and experiences 3 g’s inward, determine her velocity.
SOLUTION
Given:
an  3g ,   5 m
an 
Rearrange
v2

v   an
v   3g
Substitute in known values
v  5*3*9.81 m/s
vA  12.13 m/s 
PROBLEM 11.136
The diameter of the eye of a stationary hurricane is 20 mi and the
maximum wind speed is 100 mi/h at the eye wall, r  10 mi.
Assuming that the wind speed is constant for constant r and
decreases uniformly with increasing r to 40 mi/h at r  110 mi,
determine the magnitude of the acceleration of the air at (a) r  10
mi, (b) r  60 mi, (c) r  110 mi.
SOLUTION
(a)
r  10 mi  52800 ft,
a
(b)
r  60 mi  316800 ft,
v  100 mi/h  146.67 ft/s
146.67 
v2

r
52800
v  100 
2
40  100
 60  10   70 mi/h  102.67 ft/s
110  10
102.67 
v2
a

r
316800
(c)
r  110 mi  580800 ft,
a  0.407 ft/s2 
2
a  0.0333 ft/s2 
v  40 mi/h  58.67 ft/s
 58.67 
v2
a

r
580800
2
a  0.00593 ft/s2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.137
The peripheral speed of the tooth of a 10-in.-diameter circular saw blade is 150 ft/s when the power to the saw
is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in 9 s. Determine
the time at which the total acceleration of the tooth is 130 ft/s2.
SOLUTION
v  v0  at t
For uniformly decelerated motion:
At t  9 s,
0  150  at  9  ,
at  16.667 ft/s 2
a 2  at2  an2
Total acceleration:
1/2
an   a 2  at2 
Normal acceleration:
or
an 
1/2
2
2
 130    16.667  


v2

,
where
 
 5
v 2   an    128.93  53.72 ft 2/s 2 ,
 12 
Time:
t 
v  v0
7.329  150

16.667
at
 128.93 ft/s 2
1
5
diameter 
ft
2
12
v  7.329 ft/s
t  8.56 s 
PROBLEM 11.138
A robot arm moves so that P travels in a circle about Point B,
which is not moving. Knowing that P starts from rest, and its speed
increases at a constant rate of 10 mm/s2, determine (a) the
magnitude of the acceleration when t  4 s, (b) the time for the
magnitude of the acceleration to be 80 mm/s2.
SOLUTION
Tangential acceleration:
at  10 mm/s2
v  at t
Speed:
v2
an 
where
  0.8 m  800 mm
(a)
When t  4 s

v  (10)(4)  40 mm/s
an 
Acceleration:


at2t 2
Normal acceleration:
(40) 2
 2 mm/s 2
800
a  at2  an2  (10)2  (2)2
a  10.20 mm/s2 
(b)
Time when a  80 mm/s 2
a 2  an2  at2
2
 (10) 2 t 2 
2
(80)  
  10
800


2
t 4  403200 s 4
t  25.2 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.139
A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the
maximum total acceleration of the train must not exceed 1.5 m/s 2 , determine (a) the shortest distance in
which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at .
SOLUTION
When v  72 km/h  20 m/s and   400 m,
an 
v2


(20)2
 1.000 m/s 2
400
a  an2  at2
But
at  a2  an2  (1.5)2  (1.000)2  1.11803 m/s2
Since the train is accelerating, reject the negative value.
(a)
Distance to reach the speed.
v0  0
Let
x0  0
v12  v02  2at ( x1  x0 )  2at x1
x1 
(b)

v12
(20)2

2at (2)(1.11803)
x1  178.9 m 
Corresponding tangential acceleration.
at  1.118 m/s2 
PROBLEM 11.140
A motorist starts from rest at Point A on a circular entrance ramp
when t  0, increases the speed of her automobile at a constant
rate and enters the highway at Point B. Knowing that her speed
continues to increase at the same rate until it reaches 100 km/h
at Point C, determine (a) the speed at Point B, (b) the
magnitude of the total acceleration when t  20 s.
SOLUTION
v0  0
Speeds:
v1  100 km/h  27.78 m/s
s 
Distance:

2
(150)  100  335.6 m
v12  v02  2at s
Tangential component of acceleration:
at 
At Point B,
v12  v02
(27.78) 2  0

 1.1495 m/s 2
2s
(2)(335.6)
vB2  v02  2at sB
where
sB 

2
(150)  235.6 m
vB2  0  (2)(1.1495)(235.6)  541.69 m2 /s 2
vB  23.27 m/s
(a)
At t  20 s,
vB  83.8 km/h 
v  v0  at t  0  (1.1495)(20)  22.99 m/s
  150 m
Since v  vB , the car is still on the curve.
(b)
Normal component of acceleration:
an 
Magnitude of total acceleration:
|a| 
v2


(22.99)2
 3.524 m/s 2
150
at2  an2  (1.1495)2  (3.524)2
| a |  3.71 m/s2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.141
Racecar A is traveling on a straight portion of the track while
racecar B is traveling on a circular portion of the track. At the
instant shown, the speed of A is increasing at the rate of 10
m/s2, and the speed of B is decreasing at the rate of 6 m/s2. For
the position shown, determine (a) the velocity of B relative to
A, (b) the acceleration of B relative to A.
SOLUTION
v A  240 km/h  66.67 m/s
Speeds:
vB  200 km/h  55.56 m/s
vA  66.67 m/s
Velocities:
vB  55.56 m/s
(a)
50°
vB/A  vB  vA
Relative velocity:
vB/A  (55.56 cos 50)   55.56 sin 50   66.67
 30.96   42.56
 52.63 m/s
53.96°
vB /A  189.5 km/h
Tangential accelerations:
(aA )t  10 m/s2
(aB )t  6 m/s 2
an 
Normal accelerations:

(   )
Car B:
(   300 m)
Acceleration of B relative to A:
50°
v2
Car A:
(aB )n 
(b)
54.0° 
(aA )n  0
(55.56) 2
 10.288
300
(aB )n  10.288 m/s2
40°
aB/A  aB  aA
a B/A  (a B )t  (a B )n  (a A )t  (a A )n
6
50  10.288
40  10   0
 (6 cos 50  10.288cos 40  10)
 (6sin 50  10.288sin 40)
 21.738   2.017
a B/A  21.8 m/s 2
5.3° 
PROBLEM 11.142
At a given instant in an airplane race, airplane A is flying
horizontally in a straight line, and its speed is being increased
at the rate of 8 m/s 2 . Airplane B is flying at the same altitude
as airplane A and, as it rounds a pylon, is following a circular
path of 300-m radius. Knowing that at the given instant the
speed of B is being decreased at the rate of 3 m/s 2 , determine,
for the positions shown, (a) the velocity of B relative to A,
(b) the acceleration of B relative to A.
SOLUTION
First note
v A  450 km/h vB  540 km/h  150 m/s
(a)
v B  v A  v B/A
We have
The graphical representation of this equation is then as shown.
We have
vB2/A  4502  5402  2(450)(540) cos 60°
vB/A  501.10 km/h
and
540 501.10

sin  sin 60
  68.9
v B/A  501 km/h
(b)
First note
Now
a A  8 m/s2
(aB ) n 
v 2B
B

(a B )t  3 m/s2
Then
60
(150 m/s) 2
300 m
(a B )n  75 m/s2

68.9 
30 
a B  (a B )t + (a B ) n
 3( cos 60 i  sin 60 j) + 75(cos 30 i  sin 30 j)
= (66.452 m/s 2 )i  (34.902 m/s 2 ) j
Finally
a B  a A  a B/A
a B/A  ( 66.452i  34.902 j)  (8i )
 (74.452 m/s 2 )i  (34.902 m/s 2 ) j
a B/A  82.2 m/s2
25.1 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.143
A race car enters the circular portion of a track that has a radius
of 70 m. When the car enters the curve at point P, it is
travelling with a speed of 120 km/h that is increasing at 5 m/s2.
Three seconds later, determine (a) the total acceleration of the
car in xy components, (b) the linear velocity of the car in xy
components.
SOLUTION
Speeds:
vto  120 km/h  33.33 m/s
vt  vto  at t
vt  33.33  5(3)  48.33 m/s
Tangential Acceleration:
1
s  so  vto t  at t 2
2
1
s  33.33(3)  (5)(3)2  122.5 m
2
s  R
s 122.5
 1.75 rad
 
R
70
  100.3
at  5 m/s 2
Normal Acceleration:
an 
Distance:
Location:
an 
(a)
Accelerations:
v2

48.332
 33.37 m/s 2
70
ax  5cos(10.3)  33.37sin(10.3)
a x  1.047 m/s 2 
a y  5 sin(10.3)  33.37 cos(10.3)
a y  33.726 m/s 2 
(b)
Velocities:
vx  48.33cos(10.3)
vx  47.55 m/s 
v y  48.33sin(10.3)
v y  8.64 m/s 
PROBLEM 11.144
An airplane flying at a constant speed of 240 m/s makes a banked horizontal turn.
What is the minimum allowable radius of the turn if the structural specifications
require that the acceleration of the airplane shall never exceed 4g?
SOLUTION
Given:
v  240 m/s, amax  4 g
Acceleration:
an 
Rearrange:

v2
an
Substitute in known values:

2402
m
4*9.81
v2

  1467.9 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.145
A golfer hits a golf ball from Point A with an initial velocity of 50 m/s at an angle
of 25° with the horizontal. Determine the radius of curvature of the trajectory
described by the ball (a) at Point A, (b) at the highest point of the trajectory.
SOLUTION
(a)
We have
or
(a A )n 
A 
v A2
A
(50 m/s)2
(9.81 m/s 2 ) cos 25
 A  281 m 
or
(b)
We have
( aB ) n 
vB2
B
where Point B is the highest point of the trajectory, so that
vB  (vA ) x  vA cos 25
Then
or
B 
[(50 m/s) cos 25°]2
9.81 m/s 2
 B  209 m 
PROBLEM 11.146
Three children are throwing
snowballs at each other. Child A
throws a snowball with a horizontal
velocity v0. If the snowball just
passes over the head of child B and
hits child C, determine the radius of
curvature
of
the
trajectory
described by the snowball (a) at
Point B, (b) at Point C.
SOLUTION
The motion is projectile motion. Place the origin at Point A.
Horizontal motion:
v x  v0
x  v0t
Vertical motion:
y0  0,
(v y )  0
v y   gt
1
y   gt 2
2
2h
,
g
t
where h is the vertical distance fallen.
| v y|  2 gh
v 2  vx2  v 2y  v02  2 gh
Speed:
Direction of velocity.
cos  
v0
v
Direction of normal acceleration.
an  g cos  
Radius of curvature:
At Point B,

gv0 v 2


v
v3
gv0
hB  1 m; xB  7 m
tB 
(2)(1 m)
 0.45152 s
9.81 m/s 2
xB  v0 t B
v0 
xB
7m

 15.504 m/s
t B 0.45152 s
vB2  (15.504) 2  (2)(9.81)(1)  259.97 m 2 /s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.146 (Continued)
(a)
Radius of curvature at Point B.
B 
At Point C
(259.97 m 2 /s 2 )3/2
(9.81 m/s 2 )(15.504 m/s)
 B  27.6 m 
hC  1 m  2 m  3 m
vC2  (15.504)2  (2)(9.81)(3)  299.23 m2 /s2
(b)
Radius of curvature at Point C.
C 
(299.23 m 2 /s 2 )3/2
(9.81 m/s2 )(15.504 m/s)
C  34.0 m 
PROBLEM 11.147
Coal is discharged from the tailgate A of a dump truck with an
initial velocity v A  2 m/s 50 . Determine the radius of
curvature of the trajectory described by the coal (a) at Point A,
(b) at the point of the trajectory 1 m below Point A.
SOLUTION
a A  g  9.81 m/s2
(a) At Point A.
Sketch tangential and normal components of acceleration at A.
(a A )n  g cos50
A 
v A2
(2)2

(a A ) n
9.81cos50
 A  0.634 m 
(b) At Point B, 1 meter below Point A.
Horizontal motion: (vB ) x  (vA ) x  2cos50  1.286 m/s
(vB )2y  (vA )2y  2a y ( yB  y A )
Vertical motion:
 (2cos 40) 2  (2)(9.81)(1)
 21.97 m 2 /s 2
(vB ) y  4.687 m/s
tan  
(vB ) y
(vB ) x

4.687
,
1.286
or
  74.6
aB  g cos 74.6
B 

(vB )2x  (vB ) 2y
vB 2

(a B ) n
g cos 74.6
(1.286) 2  21.97
9.81cos 74.6
 B  9.07 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.148
From measurements of a photograph, it has been found that as the
stream of water shown left the nozzle at A, it had a radius of
curvature of 25 m. Determine (a) the initial velocity vA of the
stream, (b) the radius of curvature of the stream as it reaches its
maximum height at B.
SOLUTION
(a)
We have
(a A )n 
v A2
A
or
4

v A2   (9.81 m/s 2 )  (25 m)
5

or
vA  14.0071 m/s
vA  14.01 m/s
(b)
We have
Where
Then
or
( aB ) n 
vB2
B
vB  (v A ) x 
B
36.9° 
4
vA
5
4
 14.0071 m/s 

5

2
9.81 m/s2
 B  12.80 m 
PROBLEM 11.149
A child throws a ball from point A with an initial
velocity v0 at an angle of 3 with the horizontal.
Knowing that the ball hits a wall at point B,
determine (a) the magnitude of the initial velocity,
(b) the minimum radius of curvature of the
trajectory.
SOLUTION
Horizontal motion.
vx  v0 cos
x  v0 t cos 
Vertical motion.
v y  v0 sin   gt
y  y0  v0 t sin  
y  y0  x tan  
Eliminate t.
1 2
gt
2
gx 2
2 v02 cos 2 
(1)
Solving (1) for v0 and applying result at point B
v0 
gx 2

2  y0  x tan   y  cos 2 
 2 1.5  6 tan 3  0.97   cos 2 3 
v0  14.48 m/s 
(a)
Magnitude of initial velocity.
(b)
Minimum radius of curvature of trajectory.
an  g 
 9.81 6 2
v2


v2
v2

an g cos
(2)
where  is the slope angle of the trajectory.
The minimum value of  occurs at the highest point of the trajectory where cos   1
and v  vx  v0 cos 
Then
2
v 2 cos 2  14.48  cos 3
 0

9.81
g
2
 min
min  21.3 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.150
A projectile is fired from Point A
with an initial velocity v 0 . (a)
Show that the radius of curvature
of the trajectory of the projectile
reaches its minimum value at the
highest Point B of the trajectory.
(b) Denoting by  the angle
formed by the trajectory and the
horizontal at a given Point C, show
that the radius of curvature of the
trajectory at C is   min /cos3  .
SOLUTION
For the arbitrary Point C, we have
(aC ) n 
or
C 
vC2
C
vC2
g cos 
Noting that the horizontal motion is uniform, we have
(v A ) x  (vC ) x
where
Then
(vA ) x  v0 cos 
v0 cos   vC cos 
cos 
v0
cos 
or
vC 
so that
1
C 
g cos 
(a)
2
 cos  
v 2 cos 2 
v0   0

g cos3 
 cos  
In the expression for C , v0 ,  , and g are constants, so that C is minimum where cos  is
maximum. By observation, this occurs at Point B where   0.
 min   B 
(b)
(vC ) x  vC cos
C 
1
cos3 
C 
min
cos3 
v02 cos 2 
g
 v02 cos 2 

g

Q.E.D.

Q.E.D.




PROBLEM 11.151*
Determine the radius of curvature of the path described by the particle of Problem 11.95 when t  0.
PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector
r  (Rt cos nt)i  ctj  (Rt sin nt)k. Determine the magnitudes of the velocity and acceleration of the
particle. (The space curve described by the particle is a conic helix.)
SOLUTION
We have
v
dr
 R (cos n t  n t sin n t )i  cj  R (sin n t  n t cos n t )k
dt
and
a
dv
 R  n sin n t  n sin n t  n2t cos n t i
dt




 R n cos n t  n cos n t  n2t sin n t k
a  n R [(2 sin n t  n t cos n t )i  (2cos n t  nt sin nt ) k ]
or
v 2  R 2 (cos n t  n t sin n t ) 2  c 2  R 2 (sin n t  n t cos n t ) 2
Now


 R 2 1  n2t 2  c 2


1/ 2
v   R 2 1  n2 t 2  c 2 


Then
R 2n2t
dv

1/ 2
dt  2
R 1  n2t 2  c 2 


and


2
2
 dv   v
a 2  at2  an2     
 dt   
Now
At t  0:



2
dv
0
dt
a  n R(2 k ) or a  2n R
v2  R2  c2
Then, with
we have
or
dv
 0,
dt
a
2n R 
v2

R2  c2


R2  c2

2n R
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.152*
Determine the radius of curvature of the path described by the particle
of Problem 11.96 when t  0, A  3, and B  1.
PROBLEM 11.96
The three-dimensional motion of a particle is defined by the position
vector r  ( At cos t )i  ( A t 2  1) j  ( Bt sin t )k , where r and t are
expressed in feet and seconds, respectively. Show that the curve
described by the particle lies on the hyperboloid (y/A)2  (x/A)2 
(z/B)2  1. For A  3 and B  1, determine (a) the magnitudes of the
velocity and acceleration when t  0, (b) the smallest nonzero value
of t for which the position vector and the velocity are perpendicular to
each other.
SOLUTION
With
A  3,
B 1
we have
r  (3t cos t )i  3 t 2  1 j  (t sin t )k
Now
v
and
a


 3t 
dr
 3(cos t  t sin t )i   2
j  (sin t  t cos t )k
 t  1 
dt




2
dv
 3( sin t  sin t  t cos t )i  3  t  1  t 

dt
t2  1

 (cos t  cos t  t sin t )k
  3(2sin t  t cos t )i  3
t
t2  1

 j


1
j
(t  1)1/ 2
2
 (2 cos t  t sin t )k
v 2  9(cos t  t sin t )2  9
Then
t2
 (sin t  t cos t )2
2
t 1
Expanding and simplifying yields
v 2  t 4  19t 2  1  8(cos2 t  t 4 sin 2 t )  8(t 3  t )sin 2t
Then
and
v  [t 4  19t 2  1  8(cos 2 t  t 4 sin 2 t )  8(t 3  t )sin 2t ]1/ 2
dv 4t 3  38t  8(2cos t sin t  4t 3sin 2 t  2t 4sin t cos t )  8[(3t 2  1)sin 2t  2(t 3  t ) cos 2t ]

dt
2[t 4  19t 2  1  8(cos 2 t  t 4 sin 2 t )  8(t 3  t )sin 2t ]1/ 2
PROBLEM 11.152* (Continued)
2
a 
Now
at2

an2
2
2
 dv   v 
     
 dt    
2
At t  0:
a  3 j  2k
or
a  13 ft/s 2
dv
0
dt
v 2  9 (ft/s) 2
Then, with
dv
 0,
dt
we have
a
or

v2

9 ft 2 /s 2
13 ft/s 2
  2.50 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.153
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g ( R /r )2 , where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing
that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s2 ,
determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular.
Earth: (umean )orbit  107 Mm/h.
SOLUTION
g  274 m/s2 ,
For the sun,
R
and
Given that an 
gR 2
v2
a

and
that
for
a
circular
orbit
n
r
r2
Eliminating an and solving for r,
r 
gR 2
v2
v  107  106 m/h  29.72  103 m/s
For the planet Earth,
Then
1
1
D    (1.39  109 )  0.695  109 m
2
2
r 
(274)(0.695  109 ) 2
 149.8  109 m
(29.72  103 ) 2
r  149.8 Gm 
PROBLEM 11.154
A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the
acceleration of the satellite is equal to g ( R/r )2 , where g is the acceleration of gravity at the surface of the
planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing
that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s 2 ,
determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular.
Saturn: (umean )orbit  34.7 Mm/h.
SOLUTION
g  274 m/s2
For the sun,
R
and
Given that an 
1
1
D    (1.39  109 )  0.695  109 m
2
2
gR 2
v2
a

.
and
that
for
a
circular
orbit:
n
r
r2
gR 2
v2
Eliminating an and solving for r,
r 
For the planet Saturn,
v  34.7  106 m/h  9.639  103 m/s
Then,
r 
(274)(0.695  109 ) 2
 1.425  1012 m
(9.639  103 )2
r  1425 Gm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.155
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Venus: g  29.20 ft/s2 , R  3761 mi.
SOLUTION
From Problems 11.153 and 11.154,
an 
gR 2
r2
For a circular orbit,
an 
v2
r
v R
Eliminating an and solving for v,
For Venus,
g
r
g  29.20 ft/s2
R  3761 mi  19.858  106 ft.
r  3761  100  3861 mi  20.386  106 ft
Then,
v  19.858  106
29.20
 23.766  103 ft/s
20.386  106
v  16200 mi/h 
PROBLEM 11.156
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Mars: g  12.17 ft/s2 , R  2102 mi.
SOLUTION
From Problems 11.153 and 11.154,
an 
gR 2
r2
For a circular orbit,
an 
v2
r
v R
Eliminating an and solving for v,
For Mars,
g
r
g  12.17 ft/s2
R  2102 mi  11.0986  106 ft
r  2102  100  2202 mi  11.6266  106 ft
Then,
v  11.0986  106
12.17
 11.35  103 ft/s
6
11.6266  10
v  7740 mi/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.157
Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a
circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154).
Jupiter: g  75.35 ft/s2 , R  44, 432 mi.
SOLUTION
From Problems 11.153 and 11.154,
an 
gR 2
r2
For a circular orbit,
an 
v2
r
v R
Eliminating an and solving for v,
For Jupiter,
g
r
g  75.35 ft/s2
R  44432 mi  234.60  106 ft
r  44432  100  44532 mi  235.13  106 ft
Then,
v  (234.60  106 )
75.35
 132.8  103 ft/s
235.13  106
v  90600 mi/h 
PROBLEM 11.158
A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its
2
acceleration is equal to g  R / r  , where g  9.81 m/s 2 , R = radius of the earth = 6370 km, and r = distance
from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle of radius
384  103 km , determine the speed of the moon relative to the earth.
SOLUTION
gR 2
r2
Normal acceleration:
an 
Solve for v2:
v 2  ran 
Data:
g  9.81 m/s 2 ,
an 
and
v2


v2
r
gR 2
r
R  6370 km = 6.370  106 m
r  384  103 km = 384  106 m
2
v 
 9.81  6.370  106 
384  10
v = 1.018 m/s
6
2
 1.0366  106 m 2 /s 2
v  3670 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.159
Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space
Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth.
(See information given in Problems 11.153–11.155.)
SOLUTION
an  g
We have
Then
or
g
R2
r2
and an 
v2
r
R2 v2

r
r2
vR
g
r
where
r Rh
The circumference s of the circular orbit is equal to
s  2 r
Assuming that the speed of the telescope is constant, we have
s  vtorbit
Substituting for s and v
2 r  R
or
torbit 

or
g
torbit
r
2 r 3/2
R g
2
[(6370  590) km]3/2
1h

3
2 1/2
6370 km [9.81  10 km/s ]
3600 s
torbit  1.606 h 
PROBLEM 11.160
Satellites A and B are traveling in the same plane in circular orbits around the
earth at altitudes of 120 and 200 mi, respectively. If at t  0 the satellites are
aligned as shown and knowing that the radius of the earth is R  3960 mi,
determine when the satellites will next be radially aligned. (See information
given in Problems 11.153–11.155.)
SOLUTION
an  g
We have
Then
g
R2
r2
R2 v2

r
r2
and an 
or
v2
r
vR
g
r
r Rh
where
The circumference s of a circular orbit is
equal to
s  2 r
Assuming that the speeds of the satellites are constant, we have
s  vT
Substituting for s and v
2 r  R
or
Now
T
g
T
r
2 r 3/ 2 2 ( R  h)3/ 2

R g
R
g
hB  hA  (T ) B  (T ) A
Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B
completes N orbits, satellite A must complete ( N  1) orbits.
Thus,
TC  N (T ) B  ( N  1)(T ) A
or
 2 ( R  hB )3/ 2 
 2 ( R  hA )3/ 2 
N
  ( N  1) 

g
g
 R

 R

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PROBLEM 11.160 (Continued)
N
or

Then
( R  hA )3/ 2
( R  hB )

1
3960  200
3960 120
3/ 2

 ( R  hA )
3/ 2
TC  N (T ) B  N
3/ 2


1
R  hB
R  hA

3/ 2
1
 33.835 orbits
1
2 ( R  hB )3/ 2
R
g
(3960  200) mi  1 h
2
 33.835
3960 mi 32.2 ft/s 2  1 mi 1/ 2 3600s
5280 ft
3/ 2


TC  51.2 h 
or
Alternative solution
From above, we have (T ) B  (T ) A . Thus, when the satellites are next radially aligned, the angles  A and  B
swept out by radial lines drawn to the satellites must differ by 2 . That is,
 A   B  2
For a circular orbit
s  r
From above
s  vt and v  R
Then

At time TC :
or
R g
( R  hA )
3/ 2
TC 
TC 

R g
R g
s vt 1 
g
t
  R
 t  3/ 2 t 

r r r
r 
r
( R  h)3/2
R g
( R  hB )3/ 2
TC  2
2


1
1
R g

 ( R  hA )3/ 2 ( R  hB )3/ 2 
2

1 mi
(3960 mi) 32.2 ft/s 2  5280
ft


or
g
r
1
(3960  120) mi3/ 2
1


1/ 2
1
(3960  200) mi3/ 2
1h
3600 s
TC  51.2 h 
PROBLEM 11.161
The oscillation of rod OA about O is defined by the relation   3/ sin  t 
where and t are expressed in radians and seconds, respectively. Collar B slides
along the rod so that its distance from O is r  6(1  e2t ) where r and t
are expressed in inches and seconds, respectively. When t  1 s, determine (a) the
velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the
collar relative to the rod.
SOLUTION
Calculate the derivatives with respect to time.
3
r  6  6e 2t in.

r  12e 2t in/s
  3cos t rad/s

r  24e 2t in/s 2
  3 sin  t rad/s 2

sin  t rad
At t  1 s,
(a)
3
r  6  6e 2  5.1880 in.

r  12e 2  1.6240 in/s
  3cos   3 rad/s

r  24e 2  3.2480 in/s 2
  3 sin   0

sin   0
Velocity of the collar.
v  re r  re  1.6240 e r  (5.1880)(3)e
v  (1.624 in/s)er  (15.56 in/s)e 
(b)
Acceleration of the collar.
a  (
r  r2 )er  (r  2r)e
 [3.2480  (5.1880)(3)2 ]er  (5.1880)(0)  (2)(1.6240)(3)]e
(49.9 in/s2 )er  (9.74 in/s2 )e 
(c)
Acceleration of the collar relative to the rod.
a B /OA  
rer
a B /OA  (3.25 in/s2 )er 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.162
The path of a particle P is a limaçon. The motion of the particle is
defined by the relations r  b  2  cos  t  and    t , where t and
 are expressed in seconds and radians, respectively. Determine
(a) the velocity and the acceleration of the particle when t  2 s,
(b) the value of  for which the magnitude of the velocity is
maximum.
SOLUTION
Differentiate the expressions for r and  with respect to time.
r  b  2  cos  t  ,
r    b sin  t ,

r    2b cos  t
   t,
   ,
  0
sin t  0,
(a) At t  2 s,
r  3b,
r  0,
cos t  1

r    2b,
  2 rad,
   rad/s
v  r  3 b,
vr  r  0 ,
v  3 be 
ar  
r  r 2    2b   3b   2   4 2b
a  r  2r  0,
a   4 2ber 
(b) Values of  for which v is maximum.
vr  r    b sin  t
v  r   b  2  cos  t  

2
v 2  vr2  v 2   2b 2 sin 2  t   2  cos  t  


  2b2 sin 2  t  4  4cos  t  cos 2  t 
  2b2  5  4cos  t 
v
is maximum when
ut
cos  t  1
   t,
or
hence
 t  0,
2 ,
4 ,
6 , etc
  2N , N  0, 1, 2,  
PROBLEM 11.163
During a parasailing ride, the boat is traveling at
a constant 30 km/hr with a 200 m long tow line.
At the instant shown, the angle between the line
and the water is 30º and is increasing at a
constant rate of 2º/s. Determine the velocity and
acceleration of the parasailer at this instant.
SOLUTION
Given:
v B  30i km/hr  8.333i m/s
aB  0
ê
eˆr
r  200 m, r  0, 
r 0
  30
r
P
  2 / s  0.0349 rad/s

=0
Relative Motion relations:
B
vP  vB  vP/ B
aP  aB  aP / B
Using Radial and Transverse components:
v P / B  rer  re
r  r2 er  r  2r e
a P / B  

Substitute in known values:



v P / B  6.981e m/s
a P / B  0.2437er m/s2
Change to rectangular coordinates:
v P / B  6.981*sin 30i  6.981*cos 30 j m/s
=3.491i  6.046 j m/s
a P / B  0.2437 *   cos 30  i  0.2437 *sin 30 j m/s 2
 0.2111i  0.1219 j m/s 2
Substitute into Relative Motion relations:
v P  8.33i  3.491i  6.046 j m/s
=11.824i  6.046 j m/s
a P  0  0.2111i  0.1219 j m/s 2
 0.2111i  0.1219 j m/s 2
Velocity:
Acceleration:
vP  13.280 m/s
27.08 
aP  0.2437 m/s
30.00 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.164
Some parasailing systems use a winch to pull the rider back to the boat. During the interval when  is
between 20º and 40º, (where t = 0 at  = 20º) the angle increases at the constant rate of 2 º/s. During this time,
/
600
, where r and t are expressed in ft and s,
the length of the rope is defined by the relationship
respectively. Knowing that the boat is travelling at a constant rate of 15 knots (where 1 knot = 1.15 mi/h),
(a) plot the magnitude of the velocity of the parasailer as a function of time (b) determine the magnitude of
the acceleration of the parasailer when t = 5 s.
SOLUTION
Given:
v B  15i knots*
1.15mph 5280 ft / mi
*
 25.30i ft/s
knot
3600s / hr
aB  0
ê
eˆr
1 5
r  600  t 2 m
8
20    40
  2 / s  0.0349 rad/s
r
P

B
=0
Take Derivatives of r(t):
Relative Motion relations:
5 32
t m/s
16
15 1

r   t 2 m/s 2
32
r  
vP  vB  vP/ B
aP  aB  aP / B
Using Radial and Transverse components:
v P / B  rer  re
a P / B  
r  r2 er  r  2r e




Change velocity to rectangular coordinates:
v P / B  r( cos  i + sin  j)  r  sin  i  cos  j
Substitute into Relative Motion relations:

 
  
r  r  e   r  2r  e

v P  vB  r cos   r sin  i  r sin   r cos  j
aP
Note that:
2
v P  vPx i  vPy j
a P  ar er  a e
r

PROBLEM 11.164 (Continued)
vPx  vB  r cos   r sin 
v  r sin   r cos 
Where:
Py
ar  
r  r 2
a  r  2r

2
2
v P  vPx
 vPy
Magnitude of velocity:
a P  ar2  a2
(a) Plot of Velocity as a function of theta:
Magnitude of Velocity as Theta Changes
47
46
45
Velocity, (ft/s)
44
43
42
41
40
39
38
37
20
22
24
26
28
30
32
theta, (deg)
34
36
38
40
(b) Magnitude of Acceleration at t=5s:
r  600 
1 52
 5  593.012 m
8
3
5
 5 2  -3.494 m/s
16
1
15

r    5 2  -1.048 m/s 2
32
r  


  r  2r   0.2439 m/s
a r  
r  r 2  1.771 m/s 2
a
2
a P  1.7712  0.24392
aP  1.787 m/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.165
As rod OA rotates, pin P moves along the parabola BCD. Knowing that the
equation of this parabola is r  2b /(1  cos  ) and that   kt , determine the
velocity and acceleration of P when (a)   0, (b)   90.
SOLUTION
2b
  kt
1  cos kt
2bk sin kt
r 
  k   0
(1  cos kt ) 2
2bk

r
[(1  cos kt ) 2 k cos kt  (sin kt )2(1  cos kt )(k sin kt )]
4
(1  cos kt )
r
(a)
When   kt  0:
2bk
1
[(2) 2 k (1)  0]  bk 2
4
2
(2)
r b
r  0

r
 0
  k
  0
v  r  bk
vr  r  0
1
1

ar  
r  r2  bk 2  bk 2   bk 2 
2
2


a  r  2r  b(0)  2(0)  0

(b)
v  bk e 
1
a   bk 2e r 
2
When   kt  90°:
r  2b
r  2bk
  90
  k
vr  r  2bk
2bk
[0  2k ]  4bk 2
19
  0

r
v  r  2bk
ar  
r  r2  4bk 2  2bk 2  2bk 2
a  r  2r  2b(0)  2(2bk )k  4bk 2

v  2bk er  2bk e 
a  2bk 2er  4bk 2e 
PROBLEM 11.166
The pin at B is free to slide along the circular slot DE
and along the rotating rod OC. Assuming that the rod
OC rotates at a constant rate  , (a) show that the
acceleration of pin B is of constant magnitude,
(b) determine the direction of the acceleration of pin B.
SOLUTION
From the sketch:
r  2b cos 
r  2b sin  
Since   constant,   0

r  2b cos  2
ar  
r  r2  2b cos   2  (2b cos  )2
a  4b cos   2
r
a  r  2r  (2b cos  )(0)  2(2b sin  ) 2
a  4b sin   2

a  ar2  a2  4b 2 ( cos  ) 2  (  sin  )2
a  4b 2
Since both b and  are constant, we find that
a  constant 
  tan 1
 4b sin   2
a
 tan 1 
2
ar
 4b cos  



  tan 1 (tan  )
 
Thus, a is directed toward A 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.167
To study the performance of a racecar, a high-speed camera is
positioned at Point A. The camera is mounted on a mechanism
which permits it to record the motion of the car as the car
travels on straightway BC. Determine (a) the speed of the car in
terms of b,  , and , (b) the magnitude of the acceleration in
terms of b,  , , and .
SOLUTION
(a)
We have
r
b
cos 
Then
r 
b sin 
cos 2 
We have
v 2  vr2  v2  (r) 2  (r)2
2
or
 b sin    b 2
 
  

2
 cos    cos  
 b2 2
b 2 2  sin 2


1

 
4
cos 2  cos 2
 cos 
b
v
cos 2
For the position of the car shown,  is decreasing; thus, the negative root is chosen.
v
b

cos 2
Alternative solution.
From the diagram
or
or




r  v sin 
b sin 
 v sin 
cos 2
v
b

cos 2
PROBLEM 11.167 (Continued)
(b)
For rectilinear motion
a
dv
dt
Using the answer from Part a
v
Then
a
b
cos2
d 
b
 
dt  cos 2
 b



 cos2  (2 cos  sin  )
cos 4
a
or
b
(  22 tan  ) 
cos 2
Alternative solution
From above
Then
Now
where
and
b
b sin 
r 
cos 
cos 2 
( sin    2 cos  )(cos2 )  ( sin  )(2 cos  sin  )

r b
cos 4
  sin   2 (1  sin 2  ) 
 b


2
cos3
 cos 

r
a 2  ar2  a2
  sin   2 (1  sin 2  )  b 2

ar  
r  r 2  b 

2
cos 2
 cos 
 cos 
2 2 sin 2  
b  


sin




cos  
cos 2 
b sin  
ar 
(  22 tan  )
cos 2
b
b 2 sin 
a  r  2r 
2
cos 
cos 2

Then
b cos  
(  2 tan  )
cos 2
a
b
(  22 tan  )[(sin  )2  (cos  )2 ]1/ 2
2
cos 
For the position of the car shown,  is negative; for a to be positive, the negative root is chosen.
b
(  2 2 tan  ) 
a
cos 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.168
After taking off, a helicopter climbs
in a straight line at a constant angle
 . Its flight is tracked by radar from
Point A. Determine the speed of the
helicopter in terms of d,  ,  , and .
SOLUTION
From the diagram
r
d

sin (180   ) sin (    )
or
d sin   r (sin  cos   cos  sin  )
tan 
tan  cos   sin 
or
rd
Then
r  d tan 
( tan  sin   cos  ) 

(tan  cos   sin  )2
tan  sin   cos 
 d tan 
(tan  cos   sin  )2
From the diagram
vr  v cos (   )
where
vr  r
Then
tan  sin   cos 
d tan 
 v(cos  cos   sin  sin  )
(tan  cos   sin  )2
 v cos  (tan  sin   cos  )
v
or
Alternative solution.
We have
v 2  vr2  v2  (r)2  (r)2
d tan  sec 

(tan  cos  sin  )2
PROBLEM 11.168 (Continued)
Using the expressions for r and r from above

tan  sin   cos  
v   d tan 

(tan  cos   sin  )2 

2
1/ 2
or
 (tan  sin   cos  )2

d tan 
v
 1

2
(tan  cos   sin  )  (tan  cos   sin  )

1/ 2


d tan 
tan 2   1


2
(tan  cos   sin  )  (tan  cos   sin  ) 
Note that as  increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v
d tan  sec 

(tan  cos   sin  )2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.169
At the bottom of a loop in the vertical plane,
an airplane has a horizontal velocity of 315
mi/h and is speeding up at a rate of 10 ft/s2.
The radius of curvature of the loop is 1 mi.
The plane is being tracked by radar at O.
r ,  and
What are the recorded values of r, 
 for this instant?
SOLUTION
Geometry. The polar coordinates are
r  (2400)2  (1800)2  3000 ft
Velocity Analysis.
 1800 
  36.87
 2400 
  tan 1 
v  315 mi/h  462 ft/s
vr  462 cos   369.6 ft/s
v  462sin   277.2 ft/s
vr  r
v  r
 
r  370 ft/s 
v
277.2

r
3000
  0.0924 rad/s 
Acceleration analysis.
at  10 ft/s 2
an 
v2


(462) 2
 40.425 ft/s 2
5280
PROBLEM 11.169 (Continued)
ar  at cos   an sin   10 cos 36.87  40.425 sin 36.87  32.255 ft/s 2
a  at sin   an cos   10 sin 36.87  40.425 cos 36.87  26.34 ft/s 2
ar  
r  r 2 
r  ar  r 2

r  32.255  (3000)(0.0924)2
a  r  2r
a
2r
   
r
r
26.34 (2)(369.6)(0.0924)


3000
3000

r  57.9 ft/s 2 
  0.0315 rad/s2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.170
Pin C is attached to rod BC and slides freely in the slot of rod OA
which rotates at the constant rate  . At the instant when   60,
r and . Express your answers in terms
determine (a) r and , (b) 
of d and  .
SOLUTION
Looking at d and  as polar coordinates with d  0,
v  d   d ,
vd  d  0
a  d   2d   0,
(a)
Velocity analysis:
ad  d  d  2  d  2
r  d 3 for angles shown.
Geometry analysis:
Sketch the directions of v, er and e.
vr  r  v  er  d  cos120
1
r   d  
2
v  r  v  e  d  cos30
 
(b)
Acceleration analysis:
d  23
d  cos 30

r
d 3
 
1

2
Sketch the directions of a, er and e.
ar  a  e r  a cos150  

r  r2  

r  
3
d 2
2
3
d 2
2
3
3
1 
d  2  r 2  
d 2  d 3   
2
2
2 
2

r 
3
d 2 
4
1
a  a  e  d  2 cos120   d  2
2



a  r  2r

 
1
(a  2r) 
r
1  1
 1
 1  
2
  2 d   (2)   2 d   2   
3d 



  0 
PROBLEM 11.171
For the racecar of Problem 11.167, it was found that it took 0.5
s for the car to travel from the position   60 to the position
  35. Knowing that b  25 m, determine the average speed
of the car during the 0.5-s interval.
PROBLEM 11.167 To study the performance of a racecar, a
high-speed camera is positioned at Point A. The camera is
mounted on a mechanism which permits it to record the motion
of the car as the car travels on straightway BC. Determine
(a) the speed of the car in terms of b,  , and , (b) the
magnitude of the acceleration in terms of b,  , , and .
SOLUTION
From the diagram:
r12  25 tan 60  25 tan 35
 25.796 m
Now
vave 

r12
t12
25.796 m
0.5 s
 51.592 m/s
or
vave  185.7 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.172
For the helicopter of Problem 11.168, it was found that when the helicopter was at B, the distance and the
angle of elevation of the helicopter were r  3000 ft and   20, respectively. Four seconds later, the radar
station sighted the helicopter at r  3320 ft and   23.1. Determine the average speed and the angle of
climb  of the helicopter during the 4-s interval.
PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle  . Its flight is
tracked by radar from Point A. Determine the speed of the helicopter in terms of d,  ,  , and .
SOLUTION
We have
r0  3000 ft
r4  3320 ft
 0  20
 4  23.1
From the diagram:
r 2  30002  33202
 2(3000)(3320) cos (23.1  20)
or
r  362.70 ft
Now
vave 
r
t
362.70 ft

4s
 90.675 ft/s
vave  61.8 mi/h 
or
Also,
or
or
r cos   r4 cos  4  r0 cos 0
cos  
3320 cos 23.1  3000 cos 20
362.70
  49.7 
PROBLEM 11.173
A particle moves along the spiral shown. Determine the magnitude
of the velocity of the particle in terms of b,  , and .
SOLUTION
1 2

Given:
r  be 2
Take time derivative
r  be 2 
Radial and Transverse Components
1 2

1 2

vr  r  be 2 
1 2

v  r  be 2 
Magnitude of Velocity
v 2  vr2  v2
2
 1 2 
  be 2   2  1  2






1 2

v  be 2

2

1
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1/2
 
PROBLEM 11.174
A particle moves along the spiral shown. Determine the magnitude of the
velocity of the particle in terms of b,  , and .
SOLUTION
b
Given:
r 
Take time derivative
r  
Radial and Transverse Components
vr  r  
2
2b 

3
v  r 
Magnitude of Velocity
2b 

3
b 

2
v 2  vr2  v2


4b 2  2 b 2  2
  4
6

b
2
6

 4   
2
2
v
b
3
4   
2
12
 
PROBLEM 11.175
A particle moves along the spiral shown. Knowing that  is
constant and denoting this constant by , determine the
magnitude of the acceleration of the particle in terms of b,
 , and  .
SOLUTION
1 2

r  be 2
  
Given:
Take time derivatives
1 2

r  be 2 
1 2


r  be 2
Radial Component of Acceleration
 
 

 2  

2
ar  
r  r2
1 2

 be 2
1 2

 be 2
Transverse Component of Acceleration
  0
and
 
2
  2     2 

 
2
 

 

 

a  r  2r
1 2

1 2

 be 2   2be 2  2

Substitute Given Values
1 2

be 2
  
  0
and
1 2

ar  be 2
Magnitude of Acceleration
  2 2 


 2
and
1 2

a  be 2
 2 
2
a 2  ar2  a2
2
 1 2 
  be 2   4  4 2  4






1 2


a  be 2   2  4

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1
2  2
PROBLEM 11.176
A particle moves along the spiral shown. Knowing that  is constant and
denoting this constant by , determine the magnitude of the acceleration of
the particle in terms of b,  , and  .
SOLUTION
b
Given:
r
Take time derivatives
r  

r 
Radial Component of Acceleration
2b 

3
2b  6b  2
  4
3


ar  
r  r 2


Transverse Component of Acceleration
,    and   0
2
b
2b  6b  2
  4   2  2
3

b

4


 2  6
2
  2 2

a  r  2r
b 
 2b 
   2    3   2
2

  
b
 3   4 2



&&&a d


0
Substitute Given Values
ar 
Magnitude of Acceleration
b

4
 6   
2
2
and
a  
4b
3
2
a 2  ar2  a2


b2
8
b2
8
36  12
36  4
2
2

  4 4 
16b 2
6
4

  4 4
a
b

4
36  4
2
 4

12
 2
PROBLEM 11.177
The motion of a particle on the surface of a right circular cylinder is
defined by the relations R  A,   2 t , and z  B sin 2 nt , where A and
B are constants and n is an integer. Determine the magnitudes of the
velocity and acceleration of the particle at any time t.
SOLUTION
RA
R  0
  2 t
z  B sin 2 nt
  2
z  2 n B cos 2 nt
  0
R
  0

z  4 2 n 2 B sin 2 nt
Velocity (Eq. 11.49)
v  R e R  Re  zk
v
 A(2 )e  2 n B cos 2 nt k
v  2 A2  n 2 B 2 cos 2 2 nt 
Acceleration (Eq. 11.50)
 - R 2 )e  ( R  2 R)e  
a  (R
zk
R

a  4 2 Aek  4 2 n2 B sin 2 nt k
a  4 2 A2  n 4 B 2 sin 2 2 nt 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.178
Show that r  h sin  knowing that at the instant shown, step
AB of the step exerciser is rotating counterclockwise at a
constant rate .
SOLUTION
From the diagram
r 2  d 2  h2  2dh cos 
Then
Now
or
2rr  2dh sin 
r
d

sin  sin 
r
d sin 
sin 
Substituting for r in the expression for r
 d sin  

 sin   r  dh sin 


or
r  h sin 
Q.E.D.
Alternative solution.
First note
  180  (   )
Now
v  v r  v  re r  re
With B as the origin
vP  d
(d  constant  d  0)

PROBLEM 11.178 (Continued)
With O as the origin
(vP )r  r
where
(vP )r  vP sin 
r  d sin 
Then
h
d

sin  sin 
Now
d sin   h sin 
or
substituting
r  h sin 
Q.E.D.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.179
The three-dimensional motion of a particle is defined by the relations R  A(1  et ),   2 t , and
z  B(1  et ). Determine the magnitudes of the velocity and acceleration when (a) t  0, (b) t  .
SOLUTION
R  A(1  et )
R  Aet
   Aet
R
  2 t
z  B(1  et )
  2
z  Bet
  0

z   Bet
Velocity (Eq. 11.49)
v  R e R  Re  zk
v  Aet e R  2 A(1  et )e  Bet k
(a)
When
t  0: et  e0  1;
(b)
When
t   : et  e  0
v  A2  B 2 
v  Ae R  Bk
v  2 Ae
v  2 A 
Acceleration (Eq. 11.50)
  R1 )e  ( R  2 R)e  
a  (R
zk
R

 [ Aet  A(1  et )4 2 ]e R  [0  2 Aet (2 )]e  Bet k
(a)
When
t  0: et  e0  1
a   Ae R  4 Ae  B k
a  A2  (4 A)2  B2
(b)
When
a  (1  16 2 ) A2  B2 
t   : et  e  0
a  4 2 Ae R
a  4 2 A 
PROBLEM 11.180*
For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis.
PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r 
(Rt cos nt)i  ctj  (Rt sin nt)k. Determine the magnitudes of the velocity and acceleration of the particle.
(The space curve described by the particle is a conic helix.)
SOLUTION
First note that the vectors v and a lie in the osculating plane.
Now
r  ( Rt cos n t )i  ctj  ( Rt sin n t )k
Then
v
dr
 R (cos n t  n t sin n t )i  cj  R (sin n t  n t cos n t )k
dt
and
a
dv
dt


 R n sin n t  n sin n t  n2t cos n t i


 R n cos n t  n cos n t  n2 t sin n t k
 n R[(2sin n t  n t cos n t )i  (2cos n t  n t sin n t )k ]
It then follows that the vector ( v  a) is perpendicular to the osculating plane.
i
j
k
( v  a)  n R R(cos n t  n t sin n t ) c R(sin n t  n t cos n t )
(2sin n t  n t cos n t ) 0 (2cos n t  n t sin n t )
 n R{c(2 cos n t  n t sin n t )i  R[ (sin n t  n t cos n t )(2sin n t  n t cos n t )
 (cos n t  n t sin n t )(2 cos n t  n t sin n t )]j  c(2sin n t  n t cos n t )k


 n R  c(2 cos n t  n t sin n t )i  R 2  n2 t 2 j  c(2sin n t  n t cos n t )k 


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.180* (Continued)
The angle  formed by the vector ( v  a) and the y axis is found from
cos  
Where
( v  a)  j
| ( v  a) || j |
|j| 1

( v  a)  j  n R 2 2  n2 t 2


|( v  a) |  n R c 2 (2 cos n t  n t sin n t ) 2  R 2 2  n2 t 2


2
1/ 2
 c 2 (2sin n t  n t cos n t )2 


1/ 2
2

 
 n R c 2 4  n2 t 2  R 2 2  n2t 2

Then


 R c  4   t   R  2   t  


R  2   t 

c 4   t  R 2   t 
 
 
 
n R 2 2  n2t 2
cos  
2
2 2
n
n
2
2 2
n
2
1/ 2
2 2
n
2
2 2
n
2
2 2
n
1/ 2
2
The angle  that the osculating plane forms with y axis (see the above diagram) is equal to
    90
Then
cos   cos (   90)  sin 
 sin  
Then
or
tan  


  R  2   t  
 R 2  n2t 2

c 2 4   2t 2
n


R 2  n2 t 2
2
2 2
n
2
1/ 2

c 4  n2 t 2

 R 2  n2 t 2
  tan 
 c 4   2t 2
n

1
  


PROBLEM 11.181*
Determine the direction of the binormal of the path described by the particle of Problem 11.96 when
(a) t  0, (b) t   /2 s.
SOLUTION


r  ( At cos t )i  A t 2  1 j  ( Bt sin t )k
Given:
r  ft, t  s;
A  3, B  1
First note that eb is given by
eb 
va
|va |


Now
r  (3t cos t )i  3 t 2  1 j  (t sin t )k
Then
v
dr
dt
3t
 3(cos t  t sin t )i 
t2 1

2
dv
a
 3( sin t  sin t  t cos t )i  3 t  12 t
dt
t 1
 (cos t  cos t  t sin t )k
and
 3(2sin t  t cos t )i 
(a)
j  (sin t  t cos t )k
At t  0:
t
t 2 1
j
3
j  (2 cos t  t sin t )k
(t  1)3/2
2
v  (3 ft/s)i
a  (3 ft/s2 ) j  (2 ft/s 2 )k
Then
v  a  3i  (3j  2k )
 3(2 j  3k )
and
Then
| v  a |  3 (2)2  (3)2  3 13
eb 
3(2 j  3k )
3 13
cos  x  0
or
 x  90

cos  y  
1
13
2
(2 j  3k )
13
 y  123.7
cos  z 
3
13
 z  33.7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 11.181* (Continued)
(b)
At t 
Then

2
s:

 3
  3
ft/s  i  
ft/s  j  (1 ft/s)k
v  


 2
  2 4


 
24

a  (6 ft/s 2 )i   2
ft/s 2  j   ft/s 2  k
3/ 2

 (  4)
 2
i
3
va  
2
6
j
3
2
(  4)1/2
24
2
(  4)3/2
k
1


2

 
3 2
24
3 2
 



6
i
 
2
1/ 2
4
(   4)3/ 2  
 2(  4)

 j



36
18
  2
 2
k
3/2
1/2 
(  4) 
 (  4)
 4.43984i  13.40220 j  12.99459k
and
Then
or
| v  a |  [( 4.43984) 2  ( 13.40220) 2  (12.99459) 2 ]1/ 2
 19.18829
1
(4.43984i  13.40220 j  12.99459k )
19.1829
4.43984
13.40220
12.99459
cos  x  
cos  y  
cos  z 
19.18829
19.18829
19.18829
eb 
 x  103.4
 y  134.3
 z  47.4

PROBLEM 11.182
The motion of a particle is defined by the relation x  2t 3  15t 2  24t  4, where x and t are expressed in
meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total
distance traveled when the acceleration is zero.
SOLUTION
x  2t 3  15t 2  24t  4
dx
 6t 2  30t  24
dt
dv
a
 12t  30
dt
v
so
(a)
0  6t 2  30t  24  6 (t 2  5t  4)
Times when v  0.
(t  4)(t  1)  0
(b)
t  1.00 s, t  4.00 s 
Position and distance traveled when a  0.
a  12t  30  0
t  2.5 s
x2  2(2.5)3  15(2.5)2  24(2.5)  4
so
x  1.50 m 
Final position
For
0  t  1 s, v  0.
For
1 s  t  2.5 s, v  0.
At
t  0,
At
t  1s, x1  (2)(1)3  (15)(1)2  (24)(1)  4  15 m
x0  4 m.
Distance traveled over interval: x1  x0  11 m
For
1 s  t  2.5 s,
v0
Distance traveled over interval
| x2  x1 |  |1.5  15 |  13.5 m
Total distance:
d  11  13.5
d  24.5 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.183
A drag car starts from rest and starts down the racetrack with and acceleration defined by a = 50 – 10 t ,
where a and t are in m/s2 and s, respectively. After reaching a speed of 125 m/s, a parachute is deployed to
help slow down the dragster. Knowing that this deceleration is defined by the relationship a= - 0.02v2, where
v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back
down to 10 m/s, (b) the total distance the car travels during this time.
SOLUTION
Given:
a1 2  50  10t m/s2 for 0  v  125 m/s
a23  0.02v 2 after v  125 m/s
Find v(t) as car speeds up
dv
 dv  adt
dt
a
v t 

t
dv 
0
  50  10t dt
0
v  t   50t  5t 2
Find time to reach v=125 m/s
125  50t  5t 2
Rearrange and solve for t:
t22  10t2  25  0
 t2  5 2  0
t2  5 s
dx  vdt
Find distance to reach-125 m/s:
x2
t2
 dx    50t  5t dt
2
0
0
5
x2  25t 2  t 3
3
x2  416.7 m
5
|
0
(a) Find total time to slow down 10 m/s:
a
dv
dv
 dt 
dt
a
t3
10
dv
v2
125
 dt  50 
5
t3  5 


50 10
|
v 125
t3  9.6 s 
PROBLEM 11.183 (Continued)
(b) Find total distance to slow down to 10 m/s
adx  vdv  dx 
x3
vdv
a
v3
dv
v
v2
 dx  50 
x2
10
x3  416.7  50ln v|
125
x3  543.0 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.184
A particle moves in a straight line with the acceleration
shown in the figure. Knowing that the particle starts from the
origin with v0  2 m/s, (a) construct the v  t and x t
curves for 0  t  18 s, (b) determine the position and the
velocity of the particle and the total distance traveled when
t  18 s.
SOLUTION
Compute areas under a  t curve.
A1  (0.75)(8)  6 m/s
A2  (2)(4)  8 m/s
A3  (6)(6)  36 m/s
v0  2 m/s
v8  v0  A1  8 m/s
v12  v8  A2  0
v18  v12  A3
v18  36 m/s 
Sketch v  t curve using straight line portions over the constant
acceleration periods.
Compute areas under the v  t curve.
1
(2  8)(8)  40 m
2
1
A5  (8)(4)  16 m
2
1
A6  (36)(6)  108 m
2
A4 
x0  0
x8  x0  A4  40 m
x12  x8  A5  56 m
x18  x12  A6
Total distance traveled  56  108
x18  52 m 
d  164 m 
PROBLEM 11.185
The velocities of commuter trains
A and B are as shown. Knowing
that the speed of each train is
constant and that B reaches the
crossing 10 min after A passed
through the same crossing,
determine (a) the relative velocity
of B with respect to A, (b) the
distance between the fronts of the
engines 3 min after A passed
through the crossing.
SOLUTION
(a)
v B  v A  v B/A
We have
The graphical representation of this equation is then as shown.
Then
vB2 /A  662  482  2(66)(48) cos 155
or
vB/A  111.366 km/h
and
48
111.366

sin  sin 155
  10.50
or
v B/A  111.4 km/h
(b)
10.50° 
First note that
at t  3 min, A is (66 km/h)
at t  3 min, B is (48 km/h)
Now
 603   3.3 km west of the crossing.
 607   5.6 km southwest of the crossing.
rB  rA  rB/A
Then at t  3 min, we have
rB2/A  3.32  5.62  2(3.3)(5.6) cos 25
or
rB/A  2.96 km 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.186
Knowing that slider block A starts from rest and moves to the left with a
constant acceleration of 1 ft/s2, determine (a) the relative acceleration of
block A with respect to block B, (b) the velocity of block B after 2 s.
SOLUTION
Given:
aA  1 ft/s2
From the diagram can write an expression for the length of the cable
L  x A  3 yB  constants
Differentiate:
0  x A  3 y B
0  v A  3vB
Differentiate again:
Find the acceleration of B:
Write accelerations as vectors:
0  a A  3aB
1
aB   aA
3
1 2
  ft/s
3
a A  1i ft/s2
1
a B  j ft/s2
3
(c) Find relative accelerations:
a A/ B  a A  a B
1
 1i  j ft/s2
3
a A/ B  1.054 ft/s2
(d) Find speed at t=2 s:
18.43 
vB   vB o  aB t
0

1
 2
3
2
m/s
3
vb  0.667 m/s  
PROBLEM 11.187
Collar A starts from rest at t  0 and moves downward with a constant
acceleration of 7 in./s2 . Collar B moves upward with a constant acceleration,
and its initial velocity is 8 in./s. Knowing that collar B moves through 20 in.
between t  0 and t  2 s, determine (a) the accelerations of collar B and
block C, (b) the time at which the velocity of block C is zero, (c) the distance
through which block C will have moved at that time.
SOLUTION

From the diagram

 y A  ( yC  y A )  2 yC  ( yC  yB )  constant
Then
2vA  vB  4vC  0
(1)
and
2a A  aB  4aC  0
(2)
(v A ) 0  0
Given:
(a A )  7 in./s 2
( v B )0  8 in./s


a B  constant
y  (yB )0  20 in.
At t  2 s
(a)
y B  ( y B ) 0  (vB ) 0 t 
We have
At t  2 s:
20 in.  (8 in./s)(2 s) 
aB  4 in./s2
1
aB t 2
2
1
aB (2 s) 2
2
or
a B  2 in./s2 
Then, substituting into Eq. (2)
2(7 in./s2 )  (2 in./s2 )  4aC  0
aC  3 in./s2
or
aC  3 in./s2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.187 (Continued)
(b)
Substituting into Eq. (1) at t  0
2(0)  (8 in./s)  4(vC )0  0 or (vC )0  2 in./s 
vC  (vC )0  aC t
Now
(c)
When vC  0:
0  (2 in./s)  (3 in./s2 )t
or
t
yC  ( yC )0  (vC )0 t 
We have
At t 
2
3
2
s
3
s:
t  0.667 s 
1
aC t 2
2
2  1
2 
yC  ( yC )0  (2 in./s)  s   (3 in./s 2 )  s 
3  2
3 
 0.667 in.
or
2
y C  (y C )0  0.667 in. 
PROBLEM 11.188
A golfer hits a ball with an initial velocity of
magnitude v0 at an angle  with the
horizontal. Knowing that the ball must clear
the tops of two trees and land as close as
possible to the flag, determine v0 and the
distance d when the golfer uses (a) a six-iron
with   31, (b) a five-iron with   27.
SOLUTION
The horizontal and vertical motions are
x
t cos 
1
1
y  (v0 sin  )t  gt 2  x tan   gt 2
2
2
x  (v0 cos  )t
v0 
or
(1)
2( x tan   y)
g
or
t2 
At the landing Point C:
yC  0,
And
xC  (v0 cos  )t 
(2)
t 
2v0 sin 
g
2v02 sin  cos 
g
(3)
(a)   31
To clear tree A:
xA  30 m, y A  12 m
From (2),
t A2 
From (1),
(v0 ) A 
To clear tree B:
2(30 tan 31  12)
 1.22851 s 2 ,
9.81
30
 31.58 m/s
1.1084cos 31
xB  100 m,
yB  14 m
From (2),
(t B )2 
2(100 tan 31  14)
 9.3957 s 2 ,
9.81
From (1),
(v0 ) B 
100
 38.06 m/s
3.0652 cos 31
The larger value governs,
From (3),
t A  1.1084 s
t B  3.0652 s
v0  38.06 m/s
xC 
v0  38.1 m/s 
(2)(38.06) 2 sin 31 cos 31
 130.38 m
9.81
d  xC  110
d  20.4 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.188 (Continued)
(b)
  27
By a similar calculation,
t A  0.81846 s,
tB  2.7447 s,
v0  41.138 m/s
xC  139.56 m,
(v0 ) A  41.138 m/s,
(v0 ) B  40.890 m/s,
v0  41.1 m/s 
d  29.6 m 
PROBLEM 11.189
As the truck shown begins to back up with a constant
acceleration of 4 ft/s2 , the outer section B of its boom starts to
retract with a constant acceleration of 1.6 ft/s2 relative to the
truck. Determine (a) the acceleration of section B, (b) the
velocity of section B when t  2 s.
SOLUTION
For the truck,
a A  4 ft/s2
For the boom,
a B/ A  1.6 ft/s2
50
(a) a B  a A  a B/ A
Sketch the vector addition.
By law of cosines:
aB2  a A2  aB2/ A  2a AaB/ A cos 50
 42  1.62  2(4)(1.6) cos50
aB  3.214 ft/s2
Law of sines:
sin  
aB/ A sin 50
aB
  22.4,
(b)

1.6 sin 50
 0.38131
3.214
a B  3.21 ft/s2
22.4 
v B  (vB )0  aBt  0  (3.214)(2)
v B  6.43 ft/s2
22.4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.190
A velodrome is a specially designed track used in
bicycle racing. If a rider starts from rest and accelerates
between points A and B according to the relationship
at = (11.46 – 0.01878 v2) m/s2, what is her acceleration
at point B?
SOLUTION


at  11.46 – 0.01878v 2 m/s2
Given:
Distance traveled between points A and B:
sB  18.5  r m
 18.5  28 *

2
m
 62.48 m
Find velocity at point B:
ads  vdv  ds 
62.48

0
ds 
vB
 11.46
0
vdv
a
vdv
– 0.01878v 2
vB
1
62.48 
ln 11.46  0.01878v 2 
  2  0.01878
0
2.347  ln 11.46  0.01878vB2   ln 11.46 
e0.09211  11.46  0.01878vB2
vB  23.49 m/s
Acceleration:
a B  at et 

vB2

 11.46 –
en
0.01878vB2
e
t
2
23.49 


e
28
n
a B  1.097et  19.71en m/s 2 
PROBLEM 11.191
Sand is discharged at A from a conveyor belt and falls onto the
top of a stockpile at B. Knowing that the conveyor belt forms
an angle   25 with the horizontal, determine (a) the speed
v0 of the belt, (b) the radius of curvature of the trajectory
described by the sand at Point B.
SOLUTION
The motion is projectile motion. Place the origin at Point A. Then x0  0 and y0  0.
The coordinates of Point B are xB  30 ft and yB  18 ft.
Horizontal motion:
Vertical motion:
vx  v0 cos 25
(1)
x  v0t cos 25
(2)
v y  v0 sin 25  gt
y  v0 t sin 25 
(3)
1 2
gt
2
(4)
At Point B, Eq. (2) gives
v0 t B 
xB
30

 33.101 ft
cos 25 cos 25
Substituting into Eq. (4),
1
18  (33.101)(sin 25)  (32.2)t B2
2
t B  1.40958 s
(a)
Speed of the belt.
v0 
v0t B
33.101

 23.483
1.40958
tB


v0  23.4 ft/s 
Eqs. (1) and (3) give
vx  23.483cos 25  21.283 ft/s
v y  (23.483)sin 25  (32.2)(1.40958)  35.464 ft/s
tan 
v y
vx
 1.66632
  59.03
v  41.36 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.191 (Continued)
Components of acceleration.
a  32.2 ft/s 2
at  32.2sin 
an  32.2cos   32.2cos59.03  16.57 ft/s2
(b)
Radius of curvature at B.
an 

v2

v 2 (41.36)2

an
16.57
  103.2 ft 
PROBLEM 11.192
The end Point B of a boom is originally 5 m from fixed Point A
when the driver starts to retract the boom with a constant radial
r  1.0 m/s 2 and lower it with a constant
acceleration of 
angular acceleration   0.5 rad/s2 . At t  2 s, determine
(a) the velocity of Point B, (b) the acceleration of Point B,
(c) the radius of curvature of the path.
SOLUTION
r0  5 m, r0  0, 
r  1.0 m/s2
Radial motion.
1 2

rt  5  0  0.5t 2
2
r  r0  
rt  0  1.0t
r  r0  r0t 
At t  2 s,
r  5  (0.5)(2) 2  3 m
r  ( 1.0)(2)  2 m/s
0  60 
Angular motion.

3
rad, 0  0,   0.5 rad/s 2
1

 0  0.25t 2
2
3
  0  t  0  0.5t
  0  0  t 2 


 0  (0.25)(2)2  0.047198 rad  2.70
3
  (0.5)(2)  1.0 rad/s
At t  2 s,
Unit vectors er and e .
(a)
Velocity of Point B at t  2 s.
v B  rer  re
 (2 m/s)er  (3 m)(1.0 rad/s)e


v B  (2.00 m/s)er  (3.00 m/s)e 

tan  
v 3.0

 1.5
vr 2.0
  56.31
v  vr2  v2  (2)2  (3)2  3.6055 m/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.192 (Continued)
Direction of velocity.
et 
v 2e r  3e

 0.55470er  0.83205e
3.6055
v
    2.70  56.31  59.01

(b)
v B  3.61 m/s

59.0° 
Acceleration of Point B at t  2 s.
a B  (
r  r2 )er  (r  2r)e
 [1.0  (3)(1)2 ]er  [(3)(0.5)  (2)(1.0)(0.5)]e
a B  (4.00 m/s2 )er  (2.50 m/s 2 )e 
tan  
a
2.50

 0.625
ar 4.00
  32.00
a  ar2  a2  (4) 2  (2.5) 2  4.7170 m/s 2
    2.70  32.00  29.30
a B  4.72 m/s2
Tangential component:
29.3° 
at  (a  et )et
at  (4e r  2.5e )  (0.55470er  0.83205e )et
 [(4)(0.55470)  (2.5)( 0.83205)]et
 (0.138675 m/s 2 )et  0.1389 m/s2
Normal component:
59.0°
a n  a  at
a n  4e r  2.5e  (0.138675)( 0.55470e r  0.83205e )
 (3.9231 m/s 2 )e r  (2.6154 m/s 2 )e
an  (3.9231)2  (2.6154)2  4.7149 m/s2
(c)
Radius of curvature of the path.
an 

v2

v 2 (3.6055 m/s)2

an
4.7149 m/s
  2.76 m 
PROBLEM 11.193
A telemetry system is used to quantify
kinematic values of a ski jumper
immediately before she leaves the ramp.
According to the system r  500 ft,

r  105 ft/s,
  25,
r  10 ft/s 2 ,


  0.07 rad/s,   0.06 rad/s2 . Determine
(a) the velocity of the skier immediately
before she leaves the jump, (b) the
acceleration of the skier at this instant,
(c) the distance of the jump d neglecting lift
and air resistance.
SOLUTION
(a)
Velocity of the skier.
(r  500 ft,   25°)
v  vr er  v e  rer  re
 (105 ft/s)er  (500 ft)(0.07 rad/s)e
v  (105 ft/s)er  (35 ft/s)e 
Direction of velocity:
v  (105cos 25  35cos 65)i  (35sin 65  105sin 25) j
 (109.95 ft/s)i  (12.654 ft/s) j
v y 12.654
tan  

  6.565
vx 109.95
v  (105) 2  (35) 2  110.68 ft/s
v  110.7 ft/s
6.57° 
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PROBLEM 11.193 (Continued)
(b)
Acceleration of the skier.
a  ar e r  a e  (
r  r 2 )er  (r  2r)e
ar  10  (500)(0.07) 2  12.45 ft/s 2
a  (500)(0.06)  (2)(105)(0.07)  15.30 ft/s 2
a  (12.45 ft/s2 )er  (15.30 ft/s2 )e 
a  (12.45)(i cos 25  j sin 25)  (15.30)(i cos 65  j sin 65)
 (17.750 ft/s2 )i  (8.6049 ft/s 2 ) j
ay
8.6049

tan  
  25.9
ax 17.750
a  (12.45)2  (15.30)2  19.725 ft/s 2
a  19.73 ft/s 2
(c)
25.9° 
Distance of the jump d.
Projectile motion. Place the origin of the xy-coordinate system at the end of the ramp with the
x-coordinate horizontal and positive to the left and the y-coordinate vertical and positive downward.
Horizontal motion: (Uniform motion)




Vertical motion:


x0  0
x0  109.95 ft/s


x  x0  x0t  109.95t 
(Uniformly accelerated motion)
y0  0
y0  12.654 ft/s

y  32.2 ft/s

(from Part a )
(from Part a) 
2
y  y0  y 0t 
1 2

yt  12.654t  16.1t 2 
2
At the landing point,
x  d cos 30
y  10  d sin 30 or
(1)
y  10  d sin 30
(2)
PROBLEM 11.193 (Continued)
Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract
x sin 30  ( y  10) cos 30  0
(109.95t )sin 30  (12.654t  16.1t 2  10) cos 30  0
13.943t 2  44.016t  8.6603  0
t  0.1858 s and 3.3427 s
Reject the negative root.
x  (109.95 ft/s)(3.3427 s)  367.53 ft
d
x
cos 30
d  424 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 11.CQ1
A bus travels the 100 miles between A and B at 50 mi/h and then another 100
miles between B and C at 70 mi/h. The average speed of the bus for the
entire 200-mile trip is:
(a) more than 60 mi/h
(b) equal to 60 mi/h
(c) less than 60 mi/h
SOLUTION
The time required for the bus to travel from A to B is 2 h and from B to C is 100/70  1.43 h, so the total time
is 3.43 h and the average speed is 200/3.43  58 mph.
Answer: (c) 
PROBLEM 11CQ2
Two cars A and B race each other down a straight
road. The position of each car as a function of time
is shown. Which of the following statements are
true (more than one answer can be correct)?
(a) At time t2 both cars have traveled the same
distance
(b) At time t1 both cars have the same speed
(c) Both cars have the same speed at some time t < t1
(d) Both cars have the same acceleration at some
time t < t1
(e) Both cars have the same acceleration at some
time t1 < t < t2
SOLUTION
The speed is the slope of the curve, so answer c) is true.
The acceleration is the second derivative of the position. Since A’s position increases linearly the second
derivative will always be zero. The second derivative of curve B is zero at the pont of inflection which occurs
between t1 and t2.
Answers: (c) and (e) 
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PROBLEM 11.CQ3
Two model rockets are fired simultaneously from a ledge and follow
the trajectories shown. Neglecting air resistance, which of the rockets
will hit the ground first?
(a) A
(b) B
(c) They hit at the same time.
(d ) The answer depends on h.
SOLUTION
The motion in the vertical direction depends on the initial velocity in the y-direction. Since A has a larger
initial velocity in this direction it will take longer to hit the ground.
Answer: (b) 
PROBLEM 11.CQ4
Ball A is thrown straight up. Which of the following statements about the ball are true at the
highest point in its path?
(a) The velocity and acceleration are both zero.
(b) The velocity is zero, but the acceleration is not zero.
(c) The velocity is not zero, but the acceleration is zero.
(d) Neither the velocity nor the acceleration are zero.
SOLUTION
At the highest point the velocity is zero. The acceleration is never zero.
Answer: (b) 
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PROBLEM 11.CQ5
Ball A is thrown straight up with an initial speed v0 and reaches a maximum elevation h before falling back
down. When A reaches its maximum elevation, a second ball is thrown straight upward with the same initial
speed v0. At what height, y, will the balls cross paths?
(a)
yh
(b)
y  h/2
(c)
y  h/2
(d)
y  h/2
(e)
y0
SOLUTION
When the ball is thrown up in the air it will be constantly slowing down until it reaches its apex, at which
point it will have a speed of zero. So, the time it will take to travel the last half of the distance to the apex will
be longer than the time it takes for the first half. This same argument can be made for the ball falling from the
maximum elevation. It will be speeding up, so the first half of the distance will take longer than the second
half. Therefore, the balls should cross above the half-way point.
Answer: (b) 
PROBLEM 11.CQ6
Two cars are approaching an intersection at constant speeds as shown. What velocity will
car B appear to have to an observer in car A?
(a)
(b)
(c)
(d )
(e)
SOLUTION
Since vB  vA+ vB/A we can draw the vector triangle and see
v B  v A  v B/A
Answer: (e) 
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PROBLEM 11.CQ7
Blocks A and B are released from rest in the positions shown. Neglecting friction between all surfaces, which
figure below best indicates the direction  of the acceleration of block B?
SOLUTION



Since aB  a A  aB/A we get
Answer: (d ) 
PROBLEM 11.CQ8
The Ferris wheel is rotating with a constant angular velocity . What is
the direction of the acceleration of Point A?
(a)
(b)
(c)
(d )
(e) The acceleration is zero.
SOLUTION
The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration
pointed upwards.
Answer: (b) 
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PROBLEM 11.CQ9
A racecar travels around the track shown at a constant speed. At
which point will the racecar have the largest acceleration?
(a) A
(b) B
(c) C
(d ) The acceleration will be zero at all the points.
SOLUTION
The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration. The
normal acceleration will be maximum where the radius of curvature is a minimum that is at Point A.
Answer: (a) 
PROBLEM 11.CQ10
A child walks across merry-go-round A with a constant speed u relative to
A. The merry-go-round undergoes fixed axis rotation about its center with a
constant angular velocity  counterclockwise.When the child is at the
center of A, as shown, what is the direction of his acceleration when viewed
from above.
(a)
(b)
(c)
(d )
(e) The acceleration is zero.
SOLUTION
Polar coordinates are most natural for this problem, that is,

a  (
r  r2 )er  (r  2r)e
(1)
From the information given, we know 
r  0,   0, r  0,    , r  -u. When we substitute
these values into (1), we will only have a term in the  direction.
Answer: (d ) 
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CHAPTER 12
PROBLEM 12.1
Astronauts who landed on the moon during the Apollo 15, 16 and 17 missions brought back a large collection
of rocks to the earth. Knowing the rocks weighed 139 lb when they were on the moon, determine (a) the
weight of the rocks on the earth, (b) the mass of the rocks in slugs. The acceleration due to gravity on the
moon is 5.30 ft/s2.
SOLUTION
Since the rocks weighed 139 lb on the moon, their mass is
m
(a)
Wmoon
139 lb

 26.226 lb  s2 /ft
g moon 5.30 ft/s2
On the earth, Wearth  mg earth
w  (26.226 lb  s 2 /ft)(32.2 ft/s 2 )
(b)
Since 1 slug  1 lb  s 2 /ft,
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
w  844 lb 
m  26.2 slugs 
PROBLEM 12.2
The value of g at any latitude  may be obtained from the formula
g  32.09(1  0.0053 sin 2 ) ft/s 2
which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly
spherical. Knowing that the weight of a silver bar has been officially designated as 5 lb, determine to four
significant figures (a) the mass is slugs, (b) the weight in pounds at the latitudes of 0°, 45°, and 60°.
SOLUTION
g  32.09(1  0.0053 sin 2 ) ft/s 2
  0 :
g  32.09 ft/s 2
  45:
g  32.175 ft/s 2
  90:
g  32.26 ft/s 2
m
(a) Mass at all latitudes:
(b)
Weight:
5.00 lb
32.175 ft/s2
m  0.1554 lb  s2 /ft 
W  mg
  0 :
W  (0.1554 lb  s 2 /ft)(32.09 ft/s 2 )  4.987 lb

  45:
W  (0.1554 lb  s 2 /ft)(32.175 ft/s 2 )  5.000 lb

  90:
W  (0.1554 lb  s 2 /ft)(32.26 ft/s 2 )  5.013 lb

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.3
A 400-kg satellite has been placed in a circular orbit 1500 km above the surface of the earth. The acceleration
of gravity at this elevation is 6.43 m/s2. Determine the linear momentum of the satellite, knowing that its
orbital speed is 25.6  103 km/h.
SOLUTION
Mass of satellite is independent of gravity:
m  400 kg
v  25.6  103 km/h
 1h 
3
 (25.6  106 m/h) 
  7.111  10 m/s
 3600 s 
L  mv  (400 kg)(7.111  103 m/s)
L  2.84  106 kg  m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.4
A spring scale A and a lever scale B having equal lever arms
are fastened to the roof of an elevator, and identical packages
are attached to the scales as shown. Knowing that when the
elevator moves downward with an acceleration of 1 m/s2 the
spring scale indicates a load of 60 N, determine (a) the weight
of the packages, (b) the load indicated by the spring scale and
the mass needed to balance the lever scale when the elevator
moves upward with an acceleration of 1 m/s2.
SOLUTION
Assume
g  9.81 m/s 2
m
W
g
F  ma : Fs  W  
W
a
g

a
W 1    Fs
g

or
Fs
W
1
a
g

60
1
1
9.81
W  66.8 N 
(b)
F  ma : Fs  W 
W
a
g

a
Fs  W 1  
g

1 

 66.811 

 9.81 
For the balance system B,
M 0  0: bFw  bFp  0
Fw  Fp
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Fs  73.6 N 
PROBLEM 12.4 (Continued)
But
a

Fw  Ww 1  
g

and

a
Fp  W p 1  
g

so that
Ww  W p
and
mw 
Wp
g

66.81
9.81
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
mw  6.81 kg 
PROBLEM 12.5
In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3 ft/s2 while still on a level
section of the highway. Knowing that the speed of the bus is 60 mi/h as it begins to climb the grade and that
the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus
up the grade when its speed has decreased to 50 mi/h.
SOLUTION
First consider when the bus is on the level section of the highway.
alevel  3 ft/s 2
We have
Fx  ma: P 
W
alevel
g
Now consider when the bus is on the upgrade.
We have
Substituting for P
Fx  ma: P  W sin 7 
W
a
g
W
W
alevel  W sin 7  a
g
g
a  alevel  g sin 7
or
 (3  32.2 sin 7) ft/s 2
 0.92419 ft/s 2
For the uniformly decelerated motion
2
v 2  (v0 ) upgrade
 2a ( xupgrade  0)
 5 
Noting that 60 mi/h  88 ft/s, then when v  50 mi/h   v0  , we have
 6 
2
5

2
2
 6  88 ft/s   (88 ft/s)  2(0.92419 ft/s ) xupgrade


or
or
xupgrade  1280.16 ft
xupgrade  0.242 mi 
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PROBLEM 12.6
A 0.2-lb model rocket is launched vertically from rest at time t  0 with a constant thrust
of 2 lb for one second and no thrust for t  1 s. Neglecting air resistance and the decrease
in mass of the rocket, determine (a) the maximum height h reached by the rocket, (b) the
time required to reach this maximum height.
SOLUTION
W  0.2 lbs
For 0  t  1 s Ft  2 lbs
Given:
Free Body Diagram:
For t  1 s Ft  0 lbs
For the thrust phase, 0  t  1 s :
F  ma : Ft  W  ma 
W
a
g
F

 2

a  g  t  1    32.2  
 1  289.8 ft/s 2
W
0.2




v  at   289.8 1  289.8 ft/s
At t  1 s :
y 
For free flight phase, t  1 s :
1 2 1
2
at   289.8 1  144.9 ft
2
2
a   g   32.2 ft/s
v  v1  a  t  1  289.8    32.2  t  1
(a) At maximum height:
v  0,
t 1 
289.8
 9.00 s,
32.2
t  10.00 s
v 2  v12  2a  y  y1   2 g  y  y1 
0   289.8
v 2  v12

 1304.1 ft
y  y1  
2g
 2  32.2 
2
ymax  h  1304.1  144.9
(b) Time to reach max height:
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h  1449 ft 
t  10.00 s 
PROBLEM 12.7
A tugboat pulls a small barge through a harbor. The
propeller thrust minus the drag produces a net thrust
that varies linearly with speed. Knowing that the
combined weight of the tug and barge is 3600 kN,
determine (a) the time required to increase the
speed from an initial value v1 = 1.0 m/s to a final
value v2 = 2.5 m/s, (b) the distance traveled during
this time interval.
SOLUTION
Given:
W  3600 kN
Free Body Diagram:
W
m   m  366972.5 kg
g
v1  1.0 m/s, v2  2.5 m/s
From Graph:
Fnet  81000 
From FBD:
Fx  max  Fnet  max  ax 
81000
v  81000 - 27000v N
3
81000  27000v
m
27000

 3  v  m/s2
m
Fnet
m
ax 
(a) From Kinematics:
ax 
t
dv
dv
 dt 
dt
ax
m v2 dv

27000 v1 3  v
m
2.5
t 
ln  3  v  |1
27000
 dt 
0
t  18.84 s 
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PROBLEM 12.7 (Continued)
(b) From Kinematics:
ax dx  vdv  dx 
x
m
v2
vdv
ax
vdv
 dx  27000  3  v
v1
0
x
m
2.5
3  v  3ln 3  v  |1

27000
x  36.14 m 
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PROBLEM 12.8
Determine the maximum theoretical speed that may be achieved over a distance of 60 m by a car starting from
rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent
of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume
(a) four-wheel drive, (b) front-wheel drive, (c) rear-wheel drive.
SOLUTION
(a)
Four-wheel drive
Fy  0: N1  N 2  W  0
F1  F2   N1   N 2
  ( N1  N 2 )   w
Fx  ma : F1  F2  ma
 w  ma
W
mg
  g  0.80(9.81)
m
m
a  7.848 m/s 2
a

v 2  2ax  2(7.848 m/s 2 )(60 m)  941.76 m 2 /s 2
v  30.69 m/s
(b)
v  110.5 km/h 
Front-wheel drive
F2  ma
 (0.6 W )  ma
0.6W 0.6 mg

m
m
 0.6  g  0.6(0.80)(9.81)
a
a  4.709 m/s 2
v 2  2ax  2(4.709 m/s 2 )(60 m)  565.1 m 2 /s 2
v  23.77 m/s
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v  85.6 km/h 
PROBLEM 12.8 (Continued)
(c)
Rear-wheel drive
F1  ma
 (0.4 W )  ma
0.4W 0.4 mg

m
m
 0.4  g  0.4(0.80)(9.81)
a
a  3.139 m/s 2
v 2  2ax  2(3.139 m/s 2 )(60 m)  376.7 m 2 /s 2
v  19.41 m/s
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v  69.9 km/h 
PROBLEM 12.9
If an automobile’s braking distance from 90 km/h is 45 m on level pavement, determine the automobile’s
braking distance from 90 km/h when it is (a) going up a 5° incline, (b) going down a 3-percent incline.
Assume the braking force is independent of grade.
SOLUTION
Assume uniformly decelerated motion in all cases.
For braking on the level surface,
v0  90 km/h  25 m/s, v f  0
x f  x0  45 m
v 2f  v02  2a ( x f  x0 )
a

v 2f  v02
2( x f  x0 )
0  (25) 2
(2)(45)
 6.9444 m/s 2
Braking force.
Fb  ma
W
 a
g
6.944

W
9.81
 0.70789W
(a)
Going up a 5° incline.
F  ma
W
a
g
F  W sin 5
a b
g
W
 (0.70789  sin 5)(9.81)
 Fb  W sin 5 
 7.79944 m/s 2
x f  x0 
v 2f  v02
2a
0  (25) 2

(2)(7.79944)
x f  x0  40.1 m 
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PROBLEM 12.9 (Continued)
(b)
Going down a 3 percent incline.
3
  1.71835
100
W
 Fb  W sin   a
g
a  (0.70789  sin  )(9.81)
 6.65028 m/s
0  (25)2
x f  x0 

(2)(6.65028)
tan  

x f  x0  47.0 m 
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PROBLEM 12.10
A mother and her child are skiing together, and the
mother is holding the end of a rope tied to the child’s
waist. They are moving at a speed of 7.2 km/h on a
gently sloping portion of the ski slope when the
mother observes that they are approaching a steep
descent. She pulls on the rope with an average force
of 7 N. Knowing the coefficient of friction between
the child and the ground is 0.1 and the angle of the
rope does not change, determine (a) the time required
for the child’s speed to be cut in half, (b) the distance
traveled in this time.
SOLUTION
Draw free body diagram of child.
F  ma :
x-direction:
mg sin 5  k N  T cos15  ma
y-direction:
N  mg cos 5  T sin15  0
From y-direction,
N  mg cos 5  T sin15  (20 kg)(9.81 m/s 2 ) cos 5  (7 N)sin15°  193.64 N
From x-direction,
k N
T cos15
m
(0.1)(193.64 N) (7 N) cos 15°
 (9.81 m/s 2 )sin 5 

20 kg
20 kg
a  g sin 5 
m

 0.45128 m/s 2
(in x -direction.)
v0  7.2 km/h  2 m/s
vf 
1
v0  1 m/s
2
v f  v0  at
(a)
(b)
x0  0
t
v f  v0
a

1 m/s
 2.2159 s
0.45128 m/s2
t  2.22 s 
Time elapsed.
Corresponding distance.
x  x0  v0 t 
1 2
at
2
 0  (2 m/s)(2.2159 s) 
1
(0.45128 m/s 2 )(2.2159 s) 2
2
x  3.32 m 
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PROBLEM 12.11
The coefficients of friction between the load and the flat-bed
trailer shown are s  0.40 and k  0.30. Knowing that the
speed of the rig is 72 km/h, determine the shortest distance in
which the rig can be brought to a stop if the load is not to shift.
SOLUTION
Load: We assume that sliding of load relative to trailer is impending:
F  Fm
 s N
Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration amax .
Fy  0: N  W  0
N W
Fm  s N  0.40 W
Fx  ma : Fm  mamax
0.40 W 
W
amax
g
amax  3.924 m/s 2
a max  3.92 m/s 2
Uniformly accelerated motion.
v 2  v02  2ax with v  0
v0  72 km/h  20 m/s
a  amax  3.924 m/s 2
0  (20)2  2(3.924) x
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x  51.0 m 
PROBLEM 12.12
A light train made up of two cars is traveling at 90 km/h when the
brakes are applied to both cars. Knowing that car A has a mass of 25
Mg and car B a mass of 20 Mg, and that the braking force is 30 kN
on each car, determine (a) the distance traveled by the train before it
comes to a stop, (b) the force in the coupling between the cars while
the train is showing down.
SOLUTION
v0  90 km/h  90/3.6  25 m/s
(a)
Both cars:
a  1.333 m/s2
Fx  ma: 60  103 N  (45  103 kg)a
v 2  v10  2ax: 0  (25) 2  2(1.333) x 

x  234 m 
Stopping distance:
(b)
Car A:


Fx  ma: 30  103 + P  (25  103 )a
P  (25  103 )(1.333)  30  103


Coupling force:
P  3332 N
P  3.33 kN (tension) 
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PROBLEM 12.13
The two blocks shown are originally at rest. Neglecting the
masses of the pulleys and the effect of friction in the pulleys
and between block A and the incline, determine (a) the
acceleration of each block, (b) the tension in the cable.
SOLUTION
(a)
We note that aB 
1
aA.
2
Block A
Fx  m A a A : T  (200 lb) sin 30 
200
aA
32.2
(1)
Block B
Fy  mB aB : 350 lb  2T 
(a)
350  1

aA 
32.2  2

(2)
Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:
2(200)sin 30  350  2
150 
aB 


(b)
From Eq. (1), T  (200) sin 30 
200
350  1 
aA 
aA
32.2
32.2  2 
575
aA
32.2
a A  8.40 ft/s 2
1
1
a A  (8.40 ft/s 2 ),
2
2
200
(8.40)
32.2
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30° 
a B  4.20 ft/s 2 
T  152.2 lb 
PROBLEM 12.14
Solve Problem 12.13, assuming that the coefficients of
friction between block A and the incline are s  0.25 and
k  0.20.
PROBLEM 12.13 The two blocks shown are originally at
rest. Neglecting the masses of the pulleys and the effect of
friction in the pulleys and between block A and the incline,
determine (a) the acceleration of each block, (b) the tension
in the cable.
SOLUTION
We first determine whether the blocks move by computing the friction force required to maintain block A in
equilibrium. T = 175 lb. When B in equilibrium,
Fx  0: 175  200sin 30  Freq  0
Freq  75.0 lb
Fy  0: N  200 cos 30  0 N  173.2 lb
FM  s N  0.25(173.2 lb)  43.3 lb
Since Freq  Fm , blocks will move (A up and B down).
We note that aB 
1
aA.
2
Block A
F  k N  (0.20)(173.2)  34.64 lb.
Fx  m A 0 A :  200sin 30  34.64  T 
200
aA
32.2
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(1)
PROBLEM 12.14 (Continued)
Block B
Fy  mB aB : 350lb  2T 
(a)
(2)
Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:
2(200)sin 30  2(34.64)  350  2
81.32 
aB 
(b)
350  1 
aA
32.2  2 
575
aA
32.2
200
350  1 
aA 
aA
32.2
32.2  2 
a A  4.55 ft/s 2
1
1
a A  (4.52 ft/s 2 ),
2
2
From Eq. (1), T  (200) sin 30  34.64 
30 
aB  2.28 ft/s 2 
200
(4.52)
32.2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
T  162.9 lb 
PROBLEM 12.15
Each of the systems shown is initially at rest.
Neglecting axle friction and the masses of the pulleys,
determine for each system (a) the acceleration of
block A, (b) the velocity of block A after it has
moved through 10 ft, (c) the time required for block
A to reach a velocity of 20 ft/s.
SOLUTION
Let y be positive downward for both blocks.
Constraint of cable: yA  yB  constant
aA  aB  0
For blocks A and B,
aB  aA
or
F  ma :
WA
aA
g
Block A:
WA  T 
Block B:
P  WB  T 
WB
W
aB   B a A
g
g
P  WB  WA 
WA
W
aA   B aA
g
g
Solving for aA,
or
aA 
T  WA 
WA
aA
g
WA  WB  P
g
WA  WB
(1)
v A2  (v A )02  2a A [ y A  ( y A )0 ] with (v A )0  0
v A  2a A [ y A  ( y A )0 ]
(2)
vA  (vA )0  aAt with (vA )0  0
t
vA
aA
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(3)
PROBLEM 12.15 (Continued)
(a)
Acceleration of block A.
WA  200 lb, WB  100 lb, P  0
System (1):
By formula (1),
By formula (1),
(a A )1  10.73 ft/s 2 
(a A )2 
200  100
(32.2)
200
(a A ) 2  16.10 ft/s 2 
WA  2200 lb, WB  2100 lb, P  0
System (3):
By formula (1),
(c)
200  100
(32.2)
200  100
WA  200 lb, WB  0, P  50 lb
System (2):
(b)
( a A )1 
( a A )3 
2200  2100
(32.2)
2200  2100
(a A )3  0.749 ft/s 2 
vA at y A  ( y A )0  10 ft. Use formula (2).
System (1):
(v A )1  (2)(10.73)(10)
(vA )1  14.65 ft/s 
System (2):
(v A )2  (2)(16.10)(10)
(vA )2  17.94 ft/s 
System (3):
(v A )3  (2)(0.749)(10)
(vA )3  3.87 ft/s 
Time at vA  20 ft/s. Use formula (3).
System (1):
t1 
20
10.73
t1  1.864 s 
System (2):
t2 
20
16.10
t2  1.242 s 
System (3):
t3 
20
0.749
t3  26.7 s 
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PROBLEM 12.16
Boxes A and B are at rest on a conveyor belt that is initially at
rest. The belt is suddenly started in an upward direction so
that slipping occurs between the belt and the boxes. Knowing
that the coefficients of kinetic friction between the belt and
the boxes are (k ) A  0.30 and (k ) B  0.32, determine the
initial acceleration of each box.
SOLUTION
Assume that aB  aA so that the normal force NAB between the boxes is zero.
A:
Fy  0: NA  WA cos 15  0
or
NA  WA cos 15
Slipping:
FA  ( k ) A NA
A:
 0.3WA cos 15
Fx  mAaA : FA  WA sin 15  mA aA
0.3WA cos 15  WA sin 15 
or
WA
aA
g
a A  (32.2 ft/s2 )(0.3 cos 15  sin 15)
or
 0.997 ft/s 2
B:
B:
Fy  0: N B  WB cos 15  0
or
N B  WB cos 15
Slipping:
FB  ( k ) B N B
 0.32WB cos 15
Fx  mB aB : FB  WB sin 15  mB aB
or
or
0.32WB cos 15  WB sin 15 
WB
aB
g
aB  (32.2 ft/s 2 )(0.32 cos 15  sin 15)  1.619 ft/s 2
aB  aA  assumption is correct
a A  0.997 ft/s 2
15° 
a B  1.619 ft/s 2
15° 
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PROBLEM 12.16 (Continued)
Note: If it is assumed that the boxes remain in contact ( NAB  0), then assuming NAB to be compression,
aA  aB
and find (Fx  ma) for each box.
A:
0.3WA cos 15  WA sin 15  N AB 
WA
a
g
B:
0.32WB cos 15  WB sin 15  N AB 
WB
a
g
Solving yields a  1.273 ft/s 2 and NAB  0.859 lb, which contradicts the assumption.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.17
A 5000-lb truck is being used to lift a 1000 lb boulder B that
is on a 200 lb pallet A. Knowing the acceleration of the truck
is
1 ft/s2, determine (a) the horizontal force between the tires
and the ground, (b) the force between the boulder and the
pallet.
SOLUTION
aT  1 m/s 2
Kinematics:
a A  a B  0.5 m/s 2
5000
 155.28 slugs
32.2
200
mA 
 6.211 slugs
32.2
1000
mB 
 31.056 slugs
32.2
mT 
Masses:
Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder.
Vertical components
:
2T  ( m A  mB ) g  (m A  mB ) a A
2T  (37.267)(32.2)  (37.267)(0.5)
T  609.32 lb
Apply Newton’s second law to the truck.
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PROBLEM 12.17 (Continued)
Horizontal components
(a)
: F  T  mT aT
Horizontal friction force between tires and ground.
F  T  mT aT  609.32  (155.28)(1.0)
F  765 lb 
Apply Newton’s second law to the boulder.
Vertical components  :
FAB  mB g  mB a B
FAB  mB ( g  a )  31.056(32.2  0.5)
(b)
Contact force between boulder and pallet:
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FAB  1016 lb 
PROBLEM 12.18
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
s  0.20 and k  0.15. If P  0, determine (a) the
acceleration of block B, (b) the tension in the cord.
SOLUTION
From the constraint of the cord:
2 xA  xB/A  constant
Then
2vA  vB/A  0
and
2aA  aB/A  0
Now
aB  a A  aB/A
Then
aB  aA  (2aA )
or
aB  aA
(1)
First we determine if the blocks will move for the given value of  . Thus, we seek the value of  for
which the blocks are in impending motion, with the impending motion of A down the incline.
B:
Fy  0: N AB  WB cos   0
or
N AB  mB g cos
Now
FAB  s N AB
B:
 0.2mB g cos 
Fx  0:  T  FAB  WB sin   0
or
A:
T  mB g (0.2cos  sin  )
Fy  0: N A  N AB  WA cos   0
or
N A  (mA  mB ) g cos
Now
FA  s N A
A:
 0.2(mA  mB ) g cos 
Fx  0:  T  FA  FAB  WA sin   0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.18 (Continued)
T  mA g sin   0.2(mA  mB ) g cos   0.2mB g cos 
or
 g[mA sin   0.2(mA  2mB ) cos  ]
Equating the two expressions for T
mB g (0.2cos  sin  )  g[mA sin   0.2(mA  2mB )cos ]
8(0.2  tan  )  [40 tan   0.2(40  2  8)]
or
tan   0.4
or
or   21.8 for impending motion. Since   25, the blocks will move. Now consider the
motion of the blocks.
Fy  0: N AB  WB cos 25  0
(a)
B:
or
N AB  mB g cos 25
Sliding:
FAB  k N AB  0.15mB g cos 25
Fx  mB aB :  T  FAB  WB sin 25  mB aB
or
T  mB [ g (0.15cos 25  sin 25)  aB ]
 8[9.81(0.15cos 25  sin 25)  aB ]
 8(5.47952  aB )
A:
(N)
Fy  0: N A  N AB  WA cos 25  0
or
N A  (mA  mB ) g cos 25
Sliding:
FA  k N A  0.15(mA  mB ) g cos 25
Fx  mA aA :  T  FA  FAB  WA sin 25  mA aA
Substituting and using Eq. (1)
T  m A g sin 25  0.15(m A  mB ) g cos 25
 0.15mB g cos 25  mA (aB )
 g[mA sin 25  0.15(mA  2mB ) cos 25]  m A aB
 9.81[40 sin 25  0.15(40  2  8) cos 25]  40aB
 91.15202  40aB
(N)
Equating the two expressions for T
8(5.47952  aB )  91.15202  40aB
or
aB  0.98575 m/s 2
a B  0.986 m/s 2
(b)
We have
or
25 
T  8[5.47952  (0.98575)]
T  51.7 N 
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PROBLEM 12.19
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
s  0.20 and k  0.15. If P  40 N
, determine (a) the
acceleration of block B, (b) the tension in the cord.
SOLUTION
From the constraint of the cord.
2 xA  xB/A  constant
Then
2vA  vB/A  0
and
2aA  aB/A  0
Now
aB  a A  aB/A
Then
aB  aA  (2aA )
or
aB  aA
(1)
First we determine if the blocks will move for the given value of P. Thus, we seek the value of P for which
the blocks are in impending motion, with the impending motion of a down the incline.
B:
Fy  0: N AB  WB cos 25  0
or
N AB  mB g cos 25
Now
FAB   s N AB
B:
 0.2 mB g cos 25
Fx  0:  T  FAB  WB sin 25  0
A:
T  0.2 mB g cos 25  mB g sin 25
or
 (8 kg)(9.81 m/s 2 ) (0.2 cos 25  sin 25)
 47.39249 N
A:
or
Fy  0: N A  N AB  WA cos 25  P sin 25  0
N A  (mA  mB ) g cos 25  P sin 25
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.19 (Continued)
Now
FA  s N A
or
FA  0.2[(mA  mB ) g cos 25  P sin 25]
Fx  0:  T  FA  FAB  WA sin 25  P cos 25  0
T  0.2[(mA  mB ) g cos 25  P sin 25]  0.2mB g cos 25  mA g sin 25  P cos 25  0
or
or
P(0.2 sin 25  cos 25)  T  0.2[(mA  2mB ) g cos 25]  mA g sin 25
Then
P (0.2 sin 25  cos 25)  47.39249 N  9.81 m/s 2 {0.2[(40  2  8) cos 25  40 sin 25] kg}
P  19.04 N for impending motion.
or
Since P,  40 N, the blocks will move. Now consider the motion of the blocks.
Fy  0: N AB  WB cos 25  0
(a)
or
N AB  mB g cos 25
Sliding:
FAB  k N AB
B:
 0.15 mB g cos 25
Fx  mB aB :  T  FAB  WB sin 25  mB aB
T  mB [ g (0.15 cos 25  sin 25)  aB ]
or
 8[9.81(0.15 cos 25  sin 25)  aB ]
 8(5.47952  aB )
(N)
A:
Fy  0: N A  N AB  WA cos 25  P sin 25  0
or
N A  (mA  mB ) g cos 25  P sin 25
Sliding:
FA  k N A
 0.15[(mA  mB ) g cos 25  P sin 25]
Fx  mA aA :  T  FA  FAB  WA sin 25  P cos 25  mA aA
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.19 (Continued)
Substituting and using Eq. (1)
T  mA g sin 25  0.15[(m A  mB ) g cos 25  P sin 25]
 0.15 mB g cos 25  P cos 25  mA ( aB )
 g[m A sin 25  0.15(mA  2mB ) cos 25]
 P (0.15 sin 25  cos 25)  m A aB
 9.81[40 sin 25  0.15(40  2  8) cos 25]
 40(0.15 sin 25  cos 25)  40aB
 129.94004  40aB
Equating the two expressions for T
8(5.47952  aB )  129.94004  40aB
or
aB  1.79383 m/s 2
a B  1.794 m/s 2
(b)
We have
25° 
T  8[5.47952  (1.79383)]
T  58.2 N 
or
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PROBLEM 12.20
The flat-bed trailer carries two 1500-kg beams with the upper
beam secured by a cable. The coefficients of static friction
between the two beams and between the lower beam and the
bed of the trailer are 0.25 and 0.30, respectively. Knowing that
the load does not shift, determine (a) the maximum acceleration
of the trailer and the corresponding tension in the cable, (b) the
maximum deceleration of the trailer.
SOLUTION
(a) Maximum acceleration. The cable secures the upper beam to the cab; the lower beam has impending slip.
For the upper beam, + Fy  0: N1  W  0
N1  W  mg
For the lower beam, + Fy  0: N 2  N1  W  0
or
N 2  2W  2mg
+ F  ma : 0.25 N  0.30 N   0.25  0.60  mg  ma
x
1
2
a  0.85 g   0.85  9.81 a  8.34 m/s 2

For the upper beam, + Fx  ma : T  0.25 N1  ma
T  0.25mg  ma   0.25 1500  9.81  1500  8.34   16.19  103 N
(b) Maximum deceleration of trailer.
Case 1: Assume that only the top beam has impending motion. As in Part (a) N1  mg .
+ F  ma : 0.25mg  ma
a  0.25 g  2.45 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
T  16.19 kN 
PROBLEM 12.20 (Continued)
Case 2: Assume that both beams have impending slip. As before N 2  2mg .
+ F   2m  a :
 0.30  2mg    2m  a
a  0.30 g  2.94 m/s 2
The smaller of deceleration value of Case 1 governs.
a  2.45 m/s 2  
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PROBLEM 12.21
A baggage conveyor is used to unload luggage from an airplane.
The 10-kg duffel bag A is sitting on top of the 20-kg suitcase B.
The conveyor is moving the bags down at a constant speed of
0.5 m/s when the belt suddenly stops. Knowing that the
coefficient of friction between the belt and B is 0.3 and that bag A
does not slip on suitcase B, determine the smallest allowable
coefficient of static friction between the bags.
SOLUTION
Since bag A does not slide on suitcase B, both have the same acceleration.
aa
20°
Apply Newton’s second law to the bag A – suitcase B combination treated as a single particle.
Fy  ma y :  (mB  m A ) g cos 20  N  0
N  (mA  mB ) g cos 30  (30)(9.81)cos 20  276.55 N
B N  (0.3)(276.55)  82.965 N
Fx  ma x :  B N  (m A  mB ) g sin 20  (m A  mB )a
a  g sin 20 
B N
m A mB
a  0.58972 m/s2
 9.81sin 20 
82.965
30
a  0.58972 m/s2
Apply Newton’s second law to bag A alone.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
20°
PROBLEM 12.21 (Continued)
Fy  ma y : N AB  mA g cos 20  0
N AB  ma g sin 20  (10)(9.81) cos 20  92.184 N
ZFx  max : M A g sin 20  FAB  mA a
FAB  mA ( g sin 20  a)  (10)(9.81sin 20  0.58972)
 27.655 N
Since bag A does not slide on suitcase B,
s 
FAB 27.655

 0.300
N AB 92.184
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s  0.300 
PROBLEM 12.22
To unload a bound stack of plywood from a truck, the driver first tilts the
bed of the truck and then accelerates from rest. Knowing that the
coefficients of friction between the bottom sheet of plywood and the bed
are s  0.40 and k  0.30, determine (a) the smallest acceleration of
the truck which will cause the stack of plywood to slide, (b) the
acceleration of the truck which causes corner A of the stack to reach the
end of the bed in 0.9 s.
SOLUTION
Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and aP/T be the acceleration
of the plywood relative to the truck.
(a)
Find the value of aT so that the relative motion of the plywood with respect to the truck is impending.
aP  aT and F1  s N1  0.40 N1
Fy  mP a y : N1  WP cos 20  mP aT sin 20
N1  mP ( g cos 20  aT sin 20)
Fx  max : F1  WP sin 20  mP aT cos 20
F1  mP ( g sin 20  aT cos 20)
mP ( g sin 20  aT cos 20)  0.40 mP ( g cos 20  aT sin 20)
(0.40 cos 20  sin 20)
g
cos 20  0.40sin 20
 (0.03145)(9.81)
aT 
 0.309
aT  0.309 m/s 2
(b)
xP/T  ( xP/T )o  (vP /T )t 
aP /T 
2 xP /T
t
2

1
1
aP / T t 2  0  0  aP / T t 2
2
2
(2)(2)
 4.94 m/s2
(0.9)2
a P / T  4.94 m/s 2
20
a P  aT  a P / T  (aT )  (4.94 m/s 2
20)
Fy  mP a y : N 2  WP cos 20  mP aT sin 20
N 2  mP ( g cos 20  aT sin 20)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 12.22 (Continued)
Fx  max : F2  WP sin 20  mP aT cos 20  mP aP / T
F2  mP ( g sin 20  aT cos 20  aP / T )
For sliding with friction
F2  k N2  0.30 N2
mP ( g sin 20  aT cos 20  aP /T )  0.30mP ( g cos 20  aT sin 20)
aT 
(0.30cos 20  sin 20) g  aP /T
cos 20  0.30sin 20
 (0.05767)(9.81)  (0.9594)(4.94)
 4.17
aT  4.17 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 12.23
To transport a series of bundles of shingles A
to a roof, a contractor uses a motor-driven lift
consisting of a horizontal platform BC which
rides on rails attached to the sides of a ladder.
The lift starts from rest and initially moves
with a constant acceleration a1 as shown. The
lift then decelerates at a constant rate a2 and
comes to rest at D, near the top of the ladder.
Knowing that the coefficient of static friction
between a bundle of shingles and the
horizontal platform is 0.30, determine the
largest allowable acceleration a1 and the
largest allowable deceleration a2 if the bundle
is not to slide on the platform.
SOLUTION
Acceleration a1: Impending slip.
F1  s N1  0.30 N1
N1  WA  mA a1 sin 65
Fy  mA a y :
N1  WA  m A a1 sin 65
 m A ( g  a1 sin 65)
Fx  mA ax : F1  mA a1 cos 65
F1  s N
or
mAa1 cos 65  0.30mA ( g  a1 sin 65)
a1 
0.30 g
cos 65  0.30 sin 65
 (1.990)(9.81)
 19.53 m/s 2
Deceleration a 2 : Impending slip.
a1  19.53 m/s 2
F2  s N2  0.30 N2
Fy  ma y : N1  WA   m A a2 sin 65
N1  WA  mA a2 sin 65
Fx  max :
F2  mA a2 cos 65
F2  s N2
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65 
PROBLEM 12.23 (Continued)
or
mA a2 cos 65  0.30 mA ( g  a2 cos 65)
0.30 g
cos 65  0.30 sin 65
 (0.432)(9.81)
a2 
 4.24 m/s2
a 2  4.24 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
65 
PROBLEM 12.24
An airplane has a mass of 25 Mg and its engines develop a total thrust of 40 kN during take-off. If the drag
D exerted on the plane has a magnitude D  2.25v 2 , where v is expressed in meters per second and D in
newtons, and if the plane becomes airborne at a speed of 240 km/h, determine the length of runway required
for the plane to take off.
SOLUTION
F  ma: 40  103 N  2.25v 2  (25  103 kg)a
av
Substituting

x1
0
dv
dv
: 40  103  2.25 v 2  (25  103 ) v
dx
dx
(25  103 )vdv
0 40  103  2.25v 2
25  103
x1  
[ln(40  103  2.25v 2 )]0v1
2(2.25)
dx 


v1
25  103
40  103
ln
4.5
40  103  2.25v12
For v1  240 km/h  66.67 m/s
x1 
25  103
40  103
ln
 5.556 ln1.333
4.5
40  103  2.25(66.67)2
 1.5982  103 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
x1  1.598 km 
PROBLEM 12.25
A 4-kg projectile is fired vertically with an initial velocity of 90 m/s, reaches a maximum height and falls to
the ground. The aerodynamic drag D has a magnitude D = 0.0024 v 2 where D and v are expressed in newtons
and m/s, respectively. Knowing that the direction of the drag is always opposite to the direction of the
velocity, determine (a) the maximum height of the trajectory, (b) the speed of the projectile when it reaches
the ground.
SOLUTION
Let y be the position coordinate of the projectile measured upward from
the ground. The velocity and acceleration a taken to positive upward.
D  kv 2.
Fy  ma :
(a) Upward motion.
 D  mg  ma

D
kv 2 

a    g      g 

m
m 



dv
kv 2 
k  2 mg 
v
   g 
    v 

dy
m 
m
k 

v dv
k
  dy
mg
m
v2 
k

0
v dv
k

mg
v0 2
m
v 
k

h
0
dy
0
1  2 mg 
kh
ln  v 
  
2 
k v
m
0
mg

1
1  kv 2
kh
k
ln
  ln  0  1  
2 v 2  mg
2  mg
m

0
k
  0.0024  90 2


4
m  kv02
ln 
ln 
 1 
 1
2k  mg

  2  0.0024    4  9.81
 335.36 m
h  335 m 
h 
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PROBLEM 12.25 (Continued)
F  ma :
(b) Downward motion.
D  mg  ma
D
kv 2
g 
g
m
m
dv
kv 2
k  mg

v

g   
 v2 
dy
m
m k

a 
v dv
k
  dy
mg
m
 v2
k

vf
0
v dv
k

mg
m

0
dy
h
vf
1  mg
kh

ln 
 v2 

2  k
m
0
 mg
 v 2f
1  k
 ln 
2  mg

k



kh
 
m



kv 2f 
2 kh
  
ln 1 


mg 
m

1
kv 2f
mg
vf  
 e  2 kh/m

mg
1  e 2 kh/m
k
vf 

 4  9.81 1  e 2  0.0024335.36/4 
0.0024 
 73.6 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

v f  73.6 m/s 
PROBLEM 12.26
A constant force P is applied to a piston and rod of total mass m to
make them move in a cylinder filled with oil. As the piston moves, the
oil is forced through orifices in the piston and exerts on the piston a
force of magnitude k v in a direction opposite to the motion of the
piston. Knowing that the piston starts from rest at t  0 and x  0,
show that the equation relating x, v, and t, where x is the distance
traveled by the piston and v is the speed of the piston, is linear in each
of these variables.
SOLUTION
F  ma : P  kv  ma
dv
P  kv
a
dt
m
t
v m dv
dt 
0
0 P  kv
m
v
  ln ( P  kv) 0
k
m
  [ln ( P  kv)  ln P ]
k
m P  kv
P  kv
kt
t   ln

or
ln
k
P
m
m
P  kv
P
v  (1  e  kt/m )
 e kt/m
or
m
k


Pt
x  v dt 
0
k


x
t
t
0
t
P k

   e kt/m 
k m
0
Pt P  kt/m
Pt P
 (e
 1) 
 (1  e kt/m )
k m
k m
Pt kv
 , which is linear.
k
m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 12.27
A spring AB of constant k is attached to a support at A and to a
collar of mass m. The unstretched length of the spring is .
Knowing that the collar is released from rest at x  x0 and
neglecting friction between the collar and the horizontal rod,
determine the magnitude of the velocity of the collar as it passes
through Point C.
SOLUTION
Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B
and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle
L  2  x2
The elongation of the spring is e  L  , and the magnitude of the force exerted by the spring is
Fs  ke  k (  2  x 2  )
cos  
By geometry,
x
  x2
2
Fx  max :  Fs cos  ma
 k (  2  x 2  )
a

v
0
v dv 
k
x
x
m 
2  x 2
x
  x2
2
 ma




0
 0 a dx
x
v
1 2
k
v 
2 0
m


x
x
2
x0 
  x2

0
0

k 1 2
2
2 
 dx    x     x 

m2
x

0
1 2
k
1

v    0   2  x02    2  x02 
2
m
2

k
2 2  x02  2  2  x02
v2 
m
k
   2  x02  2  2  x02   2 

m 




answer: v 
k
m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.


2  x02   
PROBLEM 12.28
Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each.
Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s,
determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect
the masses of the pulleys and axle friction.
SOLUTION
Let the position coordinate y be positive downward.
y A  yD  constant
Constraint of cord AD:
vA  vD  0,
( yB  yD )  ( yC  yD )  constant
Constraint of cord BC:
vB  vC  2vD  0,
Eliminate aD .
aA  aD  0
aB  aC  2aD  0
2aA  aB  aC  0
(1)
We have uniformly accelerated motion because all of the forces are
constant.
y B  ( y B ) 0  (vB ) 0 t 
aB 
Pulley D:
2[ yB  ( yB )0 ]
t
2

1
aB t 2 , (vB ) 0  0
2
(2)(3)
 1.5 m/s2
(2)2
Fy  0: 2TBC  TAD  0
TAD  2TBC
Block A:
or
Fy  ma y : WA  TAD  m A a A
aA 
WA  TAD WA  2TBC

mA
mA
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 12.28 (Continued)
Block C:
Fy  ma y : WC  TBC  mC aC
aC 
or
WC  TBC
mC
(3)
Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving
for TBC ,
 WC  TBC 
 W  2TBC 
2 A
0
  aB  
mA


 mC

 m g  2TBC
2 A
mA

 mC g  TBC

  aB  
mC



0

 4
1 


 TBC  3g  aB
 mA mC 
 4 1
 10  5  TBC  3(9.81)  1.5 or TBC  51.55 N


Block B:
(a)
Fy  ma y : P  WB  TBC  mB aB
Magnitude of P.
P  TBC  WB  mB aB
 51.55  5(9.81)  5(1.5)
(b)
P  10.00 N 
Tension in cord AD.
TAD  2TBC  (2)(51.55)
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TAD  103.1 N 
PROBLEM 12.29
A 40-lb sliding panel is supported by rollers at B and C. A 25-lb counterweight A is attached to a cable as
shown and, in cases a and c, is initially in contact with a vertical edge of the panel. Neglecting friction,
determine in each case shown the acceleration of the panel and the tension in the cord immediately after the
system is released from rest.
SOLUTION
(a)
F  Force exerted by counterweight
Panel:
Fx  ma :
T F 
40
a
g
(1)
Counterweight A: Its acceleration has two components
a A  a P  a A/P  a   a
Fx  max : F 
25
a
g
Fg  mag : 25  T 
(2)
25
a
g
(3)
Adding (1), (2), and (3):
40  25  25
a
g
25
25
a
g
(32.2)
90
90
T  F  F  25  T 
a  8.94 ft/s2

Substituting for a into (3):
25  T 
25  25 
g
g  90 
T  25 
625
90
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
T  18.06 lb 
PROBLEM 12.29 (Continued)
(b)
Panel:
Fy  ma :
T
40
a
g
(1)
25  T 
25
a
g
(2)
Counterweight A:
Fy  ma :
Adding (1) and (2):

40  25
a
g
25
a
g
65
T  25  T 

Substituting for a into (1):
T
(c)
a  12.38 ft/s 2
40  25  1000
g 
g  65 
65
T  15.38 lb 
Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice
versa.
Panel:
Fx  ma :
T
40
a
g
(1)
Counterweight A: Same free body as in Part (b):
Fy  ma :
25  T 
25
a
g
(2)
Since Eqs. (1) and (2) are the same as in (b), we get the same answers:
a  12.38 ft/s2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
; T  15.38 lb 
PROBLEM 12.30
An athlete pulls handle A to the left with a constant
force of P = 100 N. Knowing that after the handle
A has been pulled 30 cm its velocity is 3 m/s,
determine the mass of the weight stack B.
SOLUTION
Given:
P  100 N
x A  0.30 m
v A  3 m/s
Kinematics:
x A  4 y B  constant
v A  4v B  0
a A  4aB  0
(1)
Uniform Acceleration of handle A:

v A 2   v A o +2 a A x A   x A o
2

32  0 2  2 a A  0.3  0 
a A  15 m/s 2
From (1):
aB  3.75 m/s 2 
From FBD:
F
y
Free Body Diagram:
 mB aB
4 P  mB g  mB aB
mB 

4P
g  aB
4 100 
kg
9.81  3.75
mB  29.50 kg 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.31
A 10-lb block B rests as shown on a 20-lb bracket A. The
coefficients of friction are s  0.30 and k  0.25 between
block B and bracket A, and there is no friction in the pulley or
between the bracket and the horizontal surface. (a) Determine
the maximum weight of block C if block B is not to slide on
bracket A. (b) If the weight of block C is 10% larger than the
answer found in a determine the accelerations of A, B and C.
SOLUTION
Kinematics. Let xA and xB be horizontal coordinates of A and B measured from a fixed vertical line to the
left of A and B. Let yC be the distance that block C is below the pulley. Note that yC increases when C
moves downward. See figure.
The cable length L is fixed.
L  ( xB  xA )  ( xP  xA )  yC  constant
Differentiating and noting that xP  0,
vB  2vA  vC  0
2aA  aB  aC  0
Here, a A and aB are positive to the right, and aC is positive downward.
Kinetics. Let T be the tension in the cable and FAB be the friction force between blocks A and B. The free
body diagrams are:
Bracket A:
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(1)
PROBLEM 12.31 (Continued)
Block B:
Block C:
Bracket A:
Fx  max : 2T  FAB 
Block B:
Fx  max : FAB  T 
WA
aA
g
WB
aB
g
(2)
(3)
 Fy  ma y : N AB  WB  0
N AB  WB
or
Fy  ma y : mC  T 
Block C:
WC
aC
g
(4)
Adding Eqs. (2), (3), and (4), and transposing,
W
WA
W
a A  B aB  C aC  WC
g
g
g
(5)
Subtracting Eq. (4) from Eq. (3) and transposing,
W
WB
aB  C aC  FAB  WC
g
g
(a)
No slip between A and B.
aB  aA
From Eq. (1),
aA  aB  aC  a
From Eq. (5),
For impending slip,
a
WC g
WA  WB  WC
FAB  s N AB  sWB
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(6)
PROBLEM 12.31 (Continued)
Substituting into Eq. (6),
(WB  WC )(WC g )
 sWB  WC
WA  WB  WC
Solving for WC ,
WC 

sWB (WA  WB )
WA  2WB  sWB
(0.30)(10)(20  10)
20  (2)(10)  (0.30)(10)
WC  2.43 lbs 
(b)
WC increased by 10%.
WC  2.6757 lbs
Since slip is occurring,
FAB  k N AB  kWB
W
WB
aB  C aC  kWB  WC
g
g
Eq. (6) becomes
or
10aB  2.6757aC  [(0.25)(10)  2.6757](32.2)
(7)
With numerical data, Eq. (5) becomes
20aA  10aB  2.6757aC  (2.6757)(32.2)
(8)
Solving Eqs. (1), (7), and (8) gives
a A  3.144 ft/s 2 , aB  0.881 ft/s 2 , aC  5.407 ft/s 2
a A  3.14 ft/s 2

a B  0.881 ft/s 2

aC  5.41 ft/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 12.32
Knowing that  = 0.30, determine the acceleration of each block when
mA = mB = mC .
SOLUTION
Given:
  0.30
mA  mB  mc  m
Kinematics:
y A  2 y B  xC  constant
v A  2vB  vC  0
(1)
aC  a A  2 a B  0
Free Body Diagram of Block A:
Equation of Motion:
F
y
 ma A
mg  T  ma A
aA  g 
Free Body Diagram of Block B:
T
m
(2)
Equation of Motion:
F
y
 maB
mg  2T  maB
aB  g 
Free Body Diagram of Block B:
2T
m
(3)
Equations of Motion:
F
y
0
N  mg
F
x
 maC
0.3mg  T  maC
aC  0.3 g 
T
m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(4)
Problem 12.32 (Continued)
Substitute (2), (3) and (4)  (1)
3.3 g 
T
6T
0
m
1.1
mg
2
(5)
Substitute (5)  (2):
aA  g 
1.1
g
2
a A  0.45g  
Substitute (5)  (3):
aB  g 
2.2
g
2
aB  0.10g  
Substitute (5)  (4):
aC  0.3 g 
1.1
g
2
aC  0.25g  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.33
Knowing that  = 0.30, determine the acceleration of each block
when mA = 5 kg, mB = 30 kg, and mC = 15 kg.
SOLUTION
Given:
  0.30, g  9.81 m/s2
mA  5 kg, mB  30 kg, mC  15 kg
Kinematics:
y A  2 y B  xC  constant
v A  2vB  vC  0
a A  2aB  aC  0
Free Body Diagram of Block A:
(1)
Equation of Motion:
F
y
 mA a A
mA g  T  mA a A
aA  g 
Free Body Diagram of Block B:
T
mA
(2)
Equation of Motion:
F
y
 mB aB
mB g  2T  mB aB
aB  g 
Free Body Diagram of Block B:
2T
mB
(3)
Equations of Motion:
F
y
0
N  mC g
F
x
 mC aC
0.3mC g  T  mC aC
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Problem 12.33 (Continued)
aC  0.3 g 
Substitute (2), (3) and (4)  (1)
3.3g 
T
mC
(4)
T 4T T


0
mA mB mC
Substitute (5)  (2):
aA  g 
8.25
g
5
a A  6.377 m/s 2  
Substitute (5)  (3):
aB  g 
16.5
g
30
aB  4.415 m/s 2  
Substitute (5)  (4):
aC  0.3 g 
8.25
g
15
aC  2.453 m/s 2  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.34
A 25-kg block A rests on an inclined surface, and a 15-kg counterweight
B is attached to a cable as shown. Neglecting friction, determine the
acceleration of A and the tension in the cable immediately after the
system is released from rest.
SOLUTION
Let the positive direction of x and y be those shown in the sketch, and let
the origin lie at the cable anchor.
Constraint of cable: x A  y B / A  constant or a A  aB / A  0, where the
positive directions of a A and aB / A are respectively the x and the y
(1)
directions. Then a B / A   a A
First note that a B  a A  a B / A   a A
Block B:
+
20    aB / A
20 
Fx  mB  aB  x : mB g sin 20  N AB  mB a A
mB a A  N AB  mB g sin 20
15 a A  N AB  50.328
(2)
+ F  m a
y
B  B  y : mB g cos 20  T  mB a B / A
mB aB / A  T  mB g cos 20
15 aB / A  T  138.276
Block A:
+
(3)
Fx  m Aa A : m A g sin 20  N AB  T  m Aa A
m Aa A  N AB  T  m A g sin 20
25 aB / A  N AB  T  83.880
(4)
Eliminate aB / A using Eq. (1), then add Eq. (4) to Eq. (2) and
subtract Eq. (3).
55 a A  4.068
or
a A  0.0740 m/s 2 , a A  0.0740 m/s 2
20 
From Eq. (1), aB / A  0.0740 m/s 2
From Eq. (3), T  137.2 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
T  137.2 N 
PROBLEM 12.35
Block B of mass 10-kg rests as shown on the upper surface of a 22-kg
wedge A. Knowing that the system is released from rest and neglecting
friction, determine (a) the acceleration of B, (b) the velocity of B relative to
A at t  0.5 s.
SOLUTION
A:
Fx  mA aA : WA sin 30  N AB cos 40  mA aA
(a)
NAB 
or
22  a A  12 g 
cos 40
Now we note: a B  a A  a B /A , where aB/A is directed along the top surface of A.
B:
Fy   mB a y : NAB  WB cos 20   mB a A sin 50
NAB  10 ( g cos 20  aA sin 50)
or
Equating the two expressions for NAB
1 

22  a A  g 
2 

 10( g cos 20  a A sin 50)
cos 40
or
aA 
(9.81)(1.1  cos 20 cos 40)
 6.4061 m/s 2
2.2  cos 40 sin 50
Fx  mB ax : WB sin 20  mB aB/A  mB aA cos50
or
aB/A  g sin 20  a A cos50
 (9.81sin 20  6.4061cos50) m/s 2
 7.4730 m/s 2
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PROBLEM 12.35 (Continued)
Finally
a B  a A  a B/ A
We have
aB2  6.40612  7.47302  2(6.4061  7.4730) cos 50
or
aB  5.9447 m/s 2
and
or
7.4730 5.9447

sin 
sin 50
  74.4
a B  5.94 m/s 2
(b)
75.6 
Note: We have uniformly accelerated motion, so that
v  0  at
Now
v B/A  v B  v A  a B t  a At  aB/At
At t  0.5 s:
vB/A  7.4730 m/s 2  0.5 s
or
v B/A  3.74 m/s
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20 
PROBLEM 12.36
A 450-g tetherball A is moving along a horizontal circular path at a
constant speed of 4 m/s. Determine (a) the angle  that the cord
forms with pole BC, (b) the tension in the cord.
SOLUTION
a A  an 
First we note
v A2

  l AB sin 
where
Fy  0: TAB cos   WA  0
(a)
TAB 
or
mA g
cos 
Fx  mA a A : TAB sin   mA
v A2

Substituting for TAB and 
mA g
v A2
sin   mA
cos 
l AB sin 
1  cos 2  
or
sin 2   1  cos 2 
(4 m/s)2
cos 
1.8 m  9.81 m/s 2
cos2   0.906105cos  1  0
cos   0.64479
Solving
  49.9 
or
(b)
From above
TAB 
m A g 0.450 kg  9.81 m/s 2

cos 
0.64479
TAB  6.85 N 
or
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PROBLEM 12.37
During a hammer thrower’s practice swings, the 7.1-kg head
A of the hammer revolves at a constant speed v in a horizontal
circle as shown. If   0.93 m and   60, determine
(a) the tension in wire BC, (b) the speed of the hammer’s
head.
SOLUTION
First we note
a A  an 
v A2

Fy  0: TBC sin 60  WA  0
(a)
or
7.1 kg  9.81 m/s 2
sin 60
80.426
N

TBC 
TBC  80.4 N 
Fx  mA a A : TBC cos 60  mA
(b)
or
v A2 
v A2

(80.426 N) cos 60  0.93 m
7.1 kg
vA  2.30 m/s 
or
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PROBLEM 12.38
Human centrifuges are often used to simulate different acceleration
levels for pilots. When aerospace physiologists say that a pilot is
pulling 9g’s, they mean that the resultant normal force on the pilot from
the bottom of the seat is nine times their weight. Knowing that the
centrifuge starts from rest and has a constant angular acceleration of
1.5 RPM per second until the pilot is pulling 9g’s and then continues
with a constant angular velocity, determine (a) how long it will take for
the pilot to reach 9g’s (b) the angle  of the normal force once the pilot
reaches 9 g’s. Assume that the force parallel to the seat is zero.
SOLUTION
  1.5 RPM/s  0.157 rad/s 2
0  0
Given:
N  9 mg
R7 m
Free Body Diagram of Pilot:
Equations of Motion:
F
y
 ma y
n
N sin   mg  m  0 
N sin   mg
F
(1)
 man
N cos   mR 2
=
N cos 
mR
(2)
Substitute N=9mg into (1): 9mg sin   mg
1
 
  6.379
  sin 1  
9
Substitute N=9mg and θ into (2):
9*9.81cos 6.379
7
  3.540 rad/s
=
For constant angular acceleration:
  0   t
3.540  0  0.157 * t
(a) Solving for t:
t  22.55 s 
From earlier:
  6.379  
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PROBLEM 12.39
A single wire ACB passes through a ring at C attached to a sphere which
revolves at a constant speed v in the horizontal circle shown. Knowing that
the tension is the same in both portions of the wire, determine the speed v.
SOLUTION
Fx  ma: T (sin 30  sin 45) 
mv 2

(1)
Fy  0: T (cos 30  cos 45)  mg  0
T (cos 30  cos 45)  mg
Divide Eq. (1) by Eq. (2):
(2)
sin 30  sin 45 v 2

cos30  cos 45  g
v 2  0.76733 g  0.76733 (1.6 m)(9.81 m/s 2 )  12.044 m 2 /s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
v  3.47 m/s 
PROBLEM 12.40*
Two wires AC and BC are tied at C to a sphere which revolves at a
constant speed v in the horizontal circle shown. Determine the range of
the allowable values of v if both wires are to remain taut and if the
tension in either of the wires is not to exceed 60 N.
SOLUTION
From the solution of Problem 12.39, we find that both wires remain taut for
3.01 m/s  v  3.96 m/s 
To determine the values of v for which the tension in either wire will not exceed 60 N, we recall
Eqs. (1) and (2) from Problem 12.39:
TAC sin 30  TBC sin 45 
mv 2

(1)
TAC cos30  TBC cos 45  mg
(2)
Subtract Eq. (1) from Eq. (2). Since sin 45°  cos 45°, we obtain
TAC (cos30  sin 30)  mg 
mv 2
(3)

Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract:
TBC (sin 45 cos30  cos 45 sin 30) 
TBC sin15 
mv 2

mv 2

cos30  mg sin 30
cos30  mg sin 30
(4)
Making TAC  60 N, m  5 kg,   1.6 m, g  9.81 m/s 2 in Eq. (3), we find the value v1 of v for which
TAC  60 N:
60(cos30  sin 30)  5(9.81) 
21.962  49.05 
5v12
1.6
v12
0.32
v12  8.668,
We have TAC  60 N for v  v1 , that is, for
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
v1  2.94 m/s
v  2.94 m/s 
PROBLEM 12.40* (Continued)
Making TBC  60 N, m  5 kg,   1.6 m, g  9.81 m/s 2 in Eq. (4), we find the value v2 of v for which
TBC  60 N:
60sin 15 
5v22
cos 30  5(9.81) sin 30
1.6
15.529  2.7063v22  24.523
v22  14.80,
v2  3.85 m/s
We have TBC  60 N for v  v2 , that is, for
Combining the results obtained, we conclude that the range of allowable value is
v  3.85 m/s 
3.01 m/s  v  3.85 m/s 
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PROBLEM 12.41
A 1-kg sphere is at rest relative to a parabolic dish which rotates at a
constant rate about a vertical axis. Neglecting friction and knowing
that r  1 m, determine (a) the speed v of the sphere, (b) the
magnitude of the normal force exerted by the sphere on the inclined
surface of the dish.
SOLUTION
r 2 dy
,
 r  1  tan 
2
dx
Given:
y 
Free Body Diagram of Sphere:
Equations of Motion:
  45
or
  Fy  0: N cos   mg  0
N 
mg
cos
Fn  man : N sin   man
mg tan   m
v2
r
v 2  gr tan 
(a) v 2   9.8111.0000   9.81 m 2 /s 2
(b) N 
1 9.81
mg

cos 45
cos 45
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
v  3.13 m/s 
N  13.87 N 
PROBLEM 12.42*
As part of an outdoor display, a 12-lb model C of the earth is attached to wires AC and
BC and revolves at a constant speed v in the horizontal circle shown. Determine the
range of the allowable values of v if both wires are to remain taut and if the tension in
either of the wires is not to exceed 26 lb.
SOLUTION
aC  an 
First note
vC2

  3 ft
where
Fx  mC aC : TCA sin 40  TCB sin15 
WC vC2
g 
Fy  0: TCA cos 40  TCB cos15  WC  0
Note that Eq. (2) implies that
(a)
when
TCB  (TCB )max , TCA  (TCA )max
(b)
when
TCB  (TCB )min ,
TCA  (TCA )min
Case 1: TCA is maximum.
Let
Eq. (2)
or
TCA  26 lb
(26 lb)cos 40  TCB cos15  (12 lb)  0
TCB  8.1964 lb
(TCB )(TCA )max  26 lb
OK
[(TCB )max  8.1964 lb]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
(2)
PROBLEM 12.42* (Continued)
Eq. (1)
(vC2 )(TCA )max 
(32.2 ft/s 2 )(3 ft)
(26sin 40  8.1964 sin15) lb
12 lb
(vC )(TCA )max  12.31 ft/s
or
(cos15)(Eq. 1)  (sin15)(Eq. 2)
Now we form
TCA sin 40 cos15  TCA cos 40 sin15 
WC vC2
cos15  WC sin15
g 
TCA sin 55 
WC vC2
cos15  WC sin15
g 
or
(3)
(vc )max occurs when TCA  (TCA )max
(vC )max  12.31 ft/s
Case 2: TCA is minimum.
Because (TCA )min occurs when TCB  (TCB )min,
let TCB  0 (note that wire BC will not be taut).
TCA cos 40  (12 lb)  0
Eq. (2)
TCA  15.6649 lb, 26 lb OK
or
Note: Eq. (3) implies that when TCA  (TCA )min , vC  (vC )min . Then
(32.2 ft/s 2 )(3 ft)
(15.6649 lb)sin 40
12 lb
Eq. (1)
(vC2 )min 
or
(vC )min  9.00 ft/s
0  TCA  TCB  6 lb when
9.00 ft/s  vC  12.31 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.43*
The 1.2-lb flyballs of a centrifugal governor revolve at a constant speed v
in the horizontal circle of 6-in. radius shown. Neglecting the weights of
links AB, BC, AD, and DE and requiring that the links support only tensile
forces, determine the range of the allowable values of v so that the
magnitudes of the forces in the links do not exceed 17 lb.
SOLUTION
v2
First note
a  an 
where
  0.5 ft

W v2
g 
(1)
Fy  0: TDA cos 20  TDE cos 30  W  0
(2)
Fx  ma : TDA sin 20  TDE sin 30 
Note that Eq. (2) implies that
(a)
when
TDE  (TDE )max ,
TDA  (TDA )max
(b)
when
TDE  (TDE )min ,
TDA  (TDA )min
Case 1: TDA is maximum.
Let
Eq. (2)
TDA  17 lb
(17 lb)cos 20  TDE cos30  (1.2 lb)  0
or
TDE  17.06 lb unacceptable ( 17 lb)
Now let
TDE  17 lb
Eq. (2)
or
TDA cos 20  (17 lb)cos30  (1.2 lb)  0
TDA  16.9443 lb OK (  17 lb)
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PROBLEM 12.43* (Continued)
(TDA )max  16.9443 lb
(TDE )max  17 lb
(v 2 )(TDA )max 
Eq. (1)
(32.2 ft/s 2 )(0.5 ft)
(16.9443sin 20  17sin 30) lb
1.2 lb
v(TDA )max  13.85 ft/s
or
(cos30)  [Eq. (1)]  (sin 30)  [Eq. (2)]
Now form
TDA sin 20 cos30  TDA cos 20 sin 30 
W v2
cos30  W sin 30
g 
TDA sin 50 
W v2
cos30  W sin 30
g 
or
(3)
vmax occurs when TDA  (TDA )max
vmax  13.85 ft/s
Case 2: TDA is minimum.
Because (TDA )min occurs when TDE  (TDE )min ,
let TDE  0.
Eq. (2)
TDA cos 20  (1.2 lb)  0
or
TDA  1.27701 lb, 17 lb
OK
Note: Eq. (3) implies that when TDA  (TDA )min , v  vmin . Then
Eq. (1)
(v 2 )min 
(32.2 ft/s 2 ) (0.5 ft)
(1.27701 lb)sin 20
1.2 lb
vmin  2.42 ft/s
or
0  TAB , TBC , TAD , TDE  17 lb
when
2.42 ft/s  v  13.85 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.44
A 130-lb wrecking ball B is attached to a 45-ft-long steel cable
AB and swings in the vertical arc shown. Determine the tension
in the cable (a) at the top C of the swing, (b) at the bottom D of
the swing, where the speed of B is 13.2 ft/s.
SOLUTION
(a)
At C, the top of the swing, vB  0; thus
an 
vB2
0
LAB
Fn  0: TBA  WB cos 20  0
TBA  (130 lb)  cos 20
or
TBA  122.2 lb 
or
Fn  man : TBA  WB  mB
(b)
or
(vB ) 2D
LAB
 130 lb
TBA  (130 lb)  
2
 32.2 ft/s
2
  (13.2 ft/s)  

 
  45 ft  
TBA  145.6 lb 
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.45
During a high-speed chase, a 2400-lb sports car traveling at a
speed of 100 mi/h just loses contact with the road as it reaches
the crest A of a hill. (a) Determine the radius of curvature  of
the vertical profile of the road at A. (b) Using the value of 
found in part a, determine the force exerted on a 160-lb driver
by the seat of his 3100-lb car as the car, traveling at a constant
speed of 50 mi/h, passes through A.
SOLUTION
(a)
Note:
100 mi/h  146.667 ft/s
Fn  man : Wcar 
or
Wcar v A2
g 
(146.667 ft/s) 2
32.2 ft/s 2
 668.05 ft

  668 ft 
or
(b)
Note: v is constant  at  0; 50 mi/h  73.333 ft/s
Fn  man : W  N 
or
W v A2
g 


(73.333 ft/s) 2
N  (160 lb) 1 

2
 (32.2 ft/s )(668.05 ft) 
N  120.0 lb 
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.46
An airline pilot climbs to a new flight level along the path shown.
Knowing that the speed of the airplane decreases at a constant rate from
180 m/s at point A to 160 m/s at point C, determine the magnitude of the
abrupt change in the force exerted on a 90-kg passenger as the airplane
passes point B.
SOLUTION
Angle change over arc AB.  
8
  0.13963 rad
180
Length of arc: s AB     6000  0.13963  837.76 m
sBC  800 m, s AC  837.76  800  1637.76 m
160
v
180

1637.76
dv  0
1602 1802

 at 1637.76 
2
2
or
at ds
at  2.076 m/s 2
vB
837.76
180 v dv  0 at ds
vB2 1802

  2.076  837.76 
2
2
or
vB2  28922 m 2 /s 2
vB  170.06 m/s
Weight of passenger: mg   90  9.81  882.9 N
Just before point B.
an 
Free Body Diagram of Pilot:
+
  6000 m
v  170.06 m/s,
v2


170.06 2
6000
 4.820 m/s 2
Fn  N1  W  m  an 1 : N1  882.9   90  4.820   449.1 N
+ F  F  ma :
t
t
t
Just after point B.
+
 Ft 1
  90  2.076   186.8 N
v  170.06 m/s,
Fy  0 : N 2  W  0
  ,
an  0
N 2  W  882.9 N
+ F  ma : F  ma   90  2.076   186.8 N
x
t
t
t
Ft does not change.
N increases by 433.8 N.
magnitude of change of force  434 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.47
The roller-coaster track shown is contained in a vertical plane.
The portion of track between A and B is straight and
horizontal, while the portions to the left of A and to the right of
B have radii of curvature as indicated. A car is traveling at a
speed of 72 km/h when the brakes are suddenly applied,
causing the wheels of the car to slide on the track (k  0.20).
Determine the initial deceleration of the car if the brakes are
applied as the car (a) has almost reached A, (b) is traveling
between A and B, (c) has just passed B.
SOLUTION
Fn  man : N  mg  m
(a)
v2


v2 
N  m  g  



v2
F  k N  k m  g 





Ft  mat : F  mat
at 
Given data:

F
v2 
 k  g  
m


k  0.20, v  72 km/h  20 m/s
  30 m
g  9.81 m/s 2 ,

(20) 2 
at  0.20 9.81 

30 

(b)
an  0
at  4.63 m/s 2 
Fn  man  0: N  mg  0
N  mg
F  k N   k mg
Ft  mat : F  mat
at 
F
  k g  0.20(9.81)
m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
at  1.962 m/s 2 
PROBLEM 12.47 (Continued)
(c)
Fn  man : mg  N 
mv 2


v2 
N  m  g  



v2 
F  k N  k m  g  


Ft  mat : F  mat
at 

F
v2
 k  g 

m



(20) 2 
  0.20 9.81 

45 


at  0.1842 m/s 2 
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PROBLEM 12.48
A spherical-cap governor is fixed to a vertical shaft that rotates with angular
velocity . When the string-supported clapper of mass m touches the cap, a
cutoff switch is operated electrically to reduce the speed of the shaft.
Knowing that the radius of the clapper is small relative to the cap, determine
the minimum angular speed at which the cutoff switch operates.
SOLUTION
Given:
min is the angular velocity when the clapper hits the cap
Geometry From Figure:
Distance AB is 300 mm, the length of the Clapper Arm
Distance BC and AC is 300 mm, the radius of the Spherical Cap.
Therefore  ABC is equilateral, so   60 and   30
R  0.3*cos 
Free Body Diagram of Clapper:
Equations of Motion:
F
y
 ma y
n
FA cos   mg  m  0 
FA 
Substitute (1)into (2):
win =
mg
cos 
F
(1)
 man
2
FA sin   mRmin
win =
FA sin 
mR
(2)
m g tan 
mR
9.81tan 60
0.3cos 30
 8.087 rad/s

win =77.23 rpm 
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PROBLEM 12.49
A series of small packages, each with a mass of 0.5 kg, are discharged from a
conveyor belt as shown. Knowing that the coefficient of static friction between
each package and the conveyor belt is 0.4, determine (a) the force exerted by the
belt on a package just after it has passed Point A, (b) the angle  defining the
Point B where the packages first slip relative to the belt.
SOLUTION
Assume package does not slip.
at  0, F f   s N
On the curved portion of the belt
an 
v2


(1 m/s)2
 4 m/s 2
0.250 m
For any angle 
Fy  ma y : N  mg cos   man  
mv 2
N  mg cos  

mv 2

(1)
Fx  max : F f  mg sin   mat  0
F f  mg sin 
(a)
At Point A,
  0
N  (0.5)(9.81)(1.000)  (0.5)(4)
(b)
At Point B,
(2)
F f  s N
mg sin   s (mg cos   man )

a 
4 

sin   s  cos   n   0.40 cos  
g 
9.81 


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
N  2.905 N 
PROBLEM 12.49 (Continued)
Squaring and using trigonometic identities,
1  cos2   0.16cos2   0.130479cos  0.026601
1.16cos2   0.130479cos  0.97340  0
cos   0.97402
  13.09 
Check that package does not separate from the belt.
N
Ff
s

mg sin 
s
N  0. 
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PROBLEM 12.50
A 54-kg pilot flies a jet trainer in a half vertical loop of 1200-m
radius so that the speed of the trainer decreases at a constant rate.
Knowing that the pilot’s apparent weights at Points A and C are 1680
N and 350 N, respectively, determine the force exerted on her by the
seat of the trainer when the trainer is at Point B.
SOLUTION
First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on the
seat of the jet trainer.
At A:
Fn  man : N A  W  m
v A2

 1680 N

 9.81 m/s 2 
v A2  (1200 m) 
 54 kg

or
 25,561.3 m 2 /s 2
At C:
Fn  man : N C  W  m
or
vC2

 350 N

 9.81 m/s 2 
vC2  (1200 m) 
 54 kg

 19,549.8 m 2 /s 2
Since at  constant, we have from A to C
vC2  v A2  2at s AC
or
or
19,549.8 m 2/s 2  25,561.3 m 2 /s 2  2at (  1200 m)
at  0.79730 m/s 2
Then from A to B
vB2  v A2  2at s AB


 25,561.3 m 2/s 2  2(0.79730 m/s 2 )   1200 m 
2


 22,555 m 2/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.50 (Continued)
At B:
Fn  man : N B  m
vB2

22,555 m 2 /s 2
1200 m
or
N B  54 kg
or
N B  1014.98 N
Ft  mat : W  PB  m | at |
or
PB  (54 kg)(0.79730  9.81) m/s 2
or
PB  486.69 N
Finally,
( Fpilot ) B  N B2  PB2  (1014.98) 2  (486.69) 2
 1126 N
or
(Fpilot ) B  1126 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
25.6° 
PROBLEM 12.51
A carnival ride is designed to allow the general public to experience high acceleration motion. The ride rotates
about Point O in a horizontal circle such that the rider has a speed v0. The rider reclines on a platform A which
rides on rollers such that friction is negligible. A mechanical stop prevents the platform from rolling down the
incline. Determine (a) the speed v0 at which the platform A begins to roll upwards, (b) the normal force
experienced by an 80-kg rider at this speed.
SOLUTION
Radius of circle:
R  5  1.5cos 70  5.513 m
F  ma:
Components up the incline,
70°:
 mA g cos 20  
mv02
sin 20
R
1
(a)
Components normal to the incline,
N  mg sin 20 
(b)
1
 g R  2  (9.81 m/s) (5.513 m  2
Speed v0 : v0  
 
  12.1898 m/s
tan 20
 tan 20 


Normal force:
v0  12.19 m/s 
20°.
mv02
cos 20.
R
N  (80)(9.81)sin 20 
80(12.1898)2
cos 20  2294 N
5.513
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
N  2290 N 
PROBLEM 12.52
A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h.
(See Sample Problem 12.7 for the definition of rated speed). Knowing that a
racing car starts skidding on the curve when traveling at a speed of 180 mi/h,
determine (a) the banking angle  , (b) the coefficient of static friction between the
tires and the track under the prevailing conditions, (c) the minimum speed at
which the same car could negotiate that curve.
SOLUTION
W  mg
Weight
a
Acceleration
v2

Fx  max : F  W sin   ma cos 
F
mv 2

cos   mg sin 
(1)
Fy  ma y : N  W cos   ma sin 
N
(a)
mv 2

sin   mg cos 
(2)
Banking angle. Rated speed v  120 mi/h  176 ft/s. F  0 at rated speed.
0
mv 2

cos   mg sin 
(176)2
v2

 0.96199
 g (1000) (32.2)
  43.89
tan  
(b)
Slipping outward.
  43.9 
v  180 mi/h  264 ft/s
F  N


F v 2 cos    g sin 

N v 2 sin    g cos 
(264) 2 cos 43.89  (1000) (32.2)sin 43.89
(264)2 sin 43.89  (1000) (32.2) cos 43.89
 0.39009
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
  0.390 
PROBLEM 12.52 (Continued)
(c)
Minimum speed.
F   N
v 2 cos    g sin 
v 2 sin    g cos 
 g (sin    cos  )
v2 
cos    sin 
 

(1000) (32.2) (sin 43.89  0.39009 cos 43.89)
cos 43.89  0.39009 sin 43.89
 13.369 ft 2 /s 2
v  115.62 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
v  78.8 mi/h 
PROBLEM 12.53
Tilting trains, such as the American Flyer which will run from
Washington to New York and Boston, are designed to travel
safely at high speeds on curved sections of track which were
built for slower, conventional trains. As it enters a curve, each
car is tilted by hydraulic actuators mounted on its trucks. The
tilting feature of the cars also increases passenger comfort by
eliminating or greatly reducing the side force Fs (parallel to the
floor of the car) to which passengers feel subjected. For a train
traveling at 100 mi/h on a curved section of track banked
through an angle   6 and with a rated speed of 60 mi/h,
determine (a) the magnitude of the side force felt by a passenger
of weight W in a standard car with no tilt (  0), (b) the
required angle of tilt  if the passenger is to feel no side force.
(See Sample Problem 12.7 for the definition of rated speed.)
SOLUTION
vR  60 mi/h  88 ft/s, 100 mi/h  146.67 ft/s
Rated speed:
From Sample Problem 12.6,
vR2  g  tan 

or
vR2
(88) 2

 2288 ft
g tan  32.2 tan 6
Let the x-axis be parallel to the floor of the car.
Fx  max : Fs  W sin (   )  man cos(   )

(a)
mv 2

cos (   )
  0.
 v2

cos (   )  sin (   ) 
Fs  W 
 g

 (146.67) 2

cos 6  sin 6
W 
 (32.2)(2288)

 0.1858W
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Fs  0.1858W 
PROBLEM 12.53 (Continued)
(b)
For Fs  0,
v2
cos (   )  sin (   )  0
g
v2
(146.67) 2

 0.29199
g  (32.2)(2288)
    16.28
tan (   ) 
  16.28  6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
  10.28 
PROBLEM 12.54
Tests carried out with the tilting trains described in Problem 12.53
revealed that passengers feel queasy when they see through the car
windows that the train is rounding a curve at high speed, yet do
not feel any side force. Designers, therefore, prefer to reduce, but
not eliminate, that force. For the train of Problem 12.53, determine
the required angle of tilt  if passengers are to feel side forces
equal to 10% of their weights.
SOLUTION
vR  60 mi/h  88 ft/s, 100 mi/h  146.67 ft/s
Rated speed:
From Sample Problem 12.6,
vR2  g  tan 
or

vR2
(88) 2

 2288 ft
g tan  32.2 tan 6
Let the x-axis be parallel to the floor of the car.
Fx  max : Fs  W sin (   )  man cos(   )

Solving for Fs,
Now
So that
Let
Then
mv 2

cos (   )
 v2

cos (   )  sin (   ) 
Fs  W 
 g

v2
(146.67)2

 0.29199 and Fs  0.10W
g  (32.2)(2288)
0.10W  W [0.29199 cos (   )  sin (   )]
u  sin (   )
cos (   )  1  u 2
0.10  0.29199 1  u 2  u
or 0.29199 1  u 2  0.10  u
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PROBLEM 12.54 (Continued)
Squaring both sides,
0.08526(1  u 2 )  0.01  0.2u  u 2
or
1.08526u 2  0.2u  0.07526  0
The positive root of the quadratic equation is u  0.18685
Then,
    sin 1 u  10.77
  10.77  6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
  4.77 
PROBLEM 12.55
A 3-kg block is at rest relative to a parabolic dish which rotates at a
constant rate about a vertical axis. Knowing that the coefficient of
static friction is 0.5 and that r  2 m, determine the maximum
allowable velocity v of the block.
SOLUTION
Let  be the slope angle of the dish. tan  
At r  2 m, tan   1
or
dy
1
 r
dr
2
  45
Draw free body sketches of the sphere.
Fy  0: N cos    S N sin   mg  0
N 
mg
cos    S sin 
Fn  man: N sin   S N cos  
mv 2

mg (sin    S N cos  )
mv 2

cos    S sin 

v2   g
sin   S cos 
sin 45  0.5cos 45
 (2)(9.81)
 58.86 m2 /s2
cos   S sin 
cos 45  0.5sin 45
v  7.67 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.56
A polisher is started so that the fleece along the circumference
undergoes a constant tangential acceleration of 4 m/s 2 . Three
seconds after a polisher is started from rest, small tufts of
fleece from along the circumference of the 225-mm-diameter
polishing pad are observed to fly free of the pad. At this
instant, determine (a) the speed v of a tuft as it leaves the pad,
(b) the magnitude of the force required to free a tuft if the
average mass of a tuft is 1.6 mg.
SOLUTION
(a)
at  constant  uniformly acceleration motion
Then
v  0  at t
At t  3 s:
v  (4 m/s 2 )(3 s)
v  12.00 m/s 
or
(b)
Ft  mat : Ft  mat
or
Ft  (1.6  106 kg)(4 m/s 2 )
 6.4  106 N
Fn  man : Fn  m
At t  3 s:
Fn  (1.6  106 kg)
v2

(12 m/s)2
m
 0.225
2
 2.048  103 N
Finally,
Ftuft  Ft 2  Fn2
 (6.4  106 N) 2  (2.048  103 N)2
or
Ftuft  2.05  103 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.57
A turntable A is built into a stage for use in a theatrical production.
It is observed during a rehearsal that a trunk B starts to slide on the
turntable 10 s after the turntable begins to rotate. Knowing that the
trunk undergoes a constant tangential acceleration of 0.24 m/s 2 ,
determine the coefficient of static friction between the trunk and
the turntable.
SOLUTION
First we note that (aB )t  constant implies uniformly accelerated motion.
vB  0  (aB )t t
vB  (0.24 m/s 2 )(10 s)  2.4 m/s
At t  10 s:
In the plane of the turntable
F  mB a B : F  mB (a B )t  mB (aB )n
Then
F  mB (aB )t2  (aB ) 2n
 mB (aB )t2 
 
vB2
2

 Fy  0: N  W  0
or
N  mB g
At t  10 s:
F  s N  s mB g
Then


2
 s mB g  mB (aB )t2   vB



2
1/ 2
or
1
s 
9.81 m/s 2
2

 (2.4 m/s) 2  

2 2

(0.24
m/s
)


 
 2.5 m  

s  0.236 
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.58
The carnival ride from Prob 12.51 is modified so that
the 80 kg riders can move up and down the inclined
wall as the speed of the ride increases. Assuming
that the friction between the wall and the carriage is
negligible, determine the position h of the rider if the
speed v0 = 13 m/s.
SOLUTION
Given:
m  80 kg, v0  13 m/s
Geometry from figure:
R  5  h cos 70
Free Body Diagram of Rider:
Equations of Motion:
F
y
 ma y
n
N cos 70  mg  m  0 
N
Substitute (1)into (2):
v
mg
cos 70
F
(1)
 man
mv 2
R
N sin 70 R
v=
m
N sin 70 
(2)
m g tan 70R
m
13  9.81tan 70  5  h cos 70 
h  3.714 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.59
The carnival ride from Prob 12.51 is modified so that
the 80 kg riders can move up and down the inclined
wall as the speed of the ride increases. Knowing that
the coefficient of static friction between the wall and
the platform is 0.2, determine the range of values of
the constant speed v0 for which the platform will
remain in the position shown.
SOLUTION
Given:
m  80 kg,  s  0.2, h  1.5 m
Geometry from figure:
R  5  1.5cos 70
 5.513 m
Minimum and Maximum speeds occur if slipping is impending:
f  s N
Finding the minimum speed so that rider does not slide down the wall:
Free Body Diagram of Rider:
Equations of Motion:
F
y
 ma y
N cos 70  f sin 70  mg  m  0 
N cos 70  s N sin 70  mg  0
mg
cos 70  s sin 70
N
N  1480.9 N
F
n
 man
2
mvmin
R
2
mvmin
N sin 70   s N cos 70 
R
N sin 70  f cos 70 
vmin =
N  sin 70   s cos 70  R
m
vmin  9.430 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.59 (Continued)
Finding the maximum speed so that rider does not slide up the wall:
Free Body Diagram of Rider:
Equations of Motion:
F
y
 ma y
N cos 70  f sin 70  mg  m  0 
N cos 70  s N sin 70  mg  0
mg
cos 70  s sin 70
N
N  5093.4 N
F
n
 man
2
mvmin
R
2
mvmin
N sin 70   s N cos 70 
R
N sin 70  f cos 70 
vmin =
N  sin 70   s cos 70  R
m
vmin  18.81 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.60
A semicircular slot of 10-in. radius is cut in a flat plate which rotates about
the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is
designed to slide in the slot as the plate rotates. Knowing that the
coefficients of friction are s  0.35 and k  0.25, determine whether
the block will slide in the slot if it is released in the position corresponding to
(a)   80, (b)   40. Also determine the magnitude and the direction
of the friction force exerted on the block immediately after it is released.
SOLUTION
First note
Then
1
(26  10 sin  ) ft
12
vE  ABCD

an 
vE2

  (ABCD )2
1

  (26  10 sin  ) ft  (14 rad/s) 2
12

98
 (13  5 sin  ) ft/s 2
3
Assume that the block is at rest with respect to the plate.
Fx  max : N  W cos   m
or
vE2



v2
N  W   cos   E sin  


g


Fy  ma y : F  W sin   m
or
sin 
vE2

cos 


v2
F  W  sin   E cos  


g


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.60 (Continued)
(a)
We have
  80
Then
1
98


N  (0.8 lb)   cos80 
 (13  5sin 80) ft/s 2  sin 80
2
3
32.2 ft/s


 6.3159 lb
1
98


F  (0.8 lb) sin 80 
 (13  5sin 80) ft/s 2  cos80
2
3
32.2
ft/s


 1.92601 lb
Fmax  s N  0.35(6.3159 lb)  2.2106 lb
Now
The block does not slide in the slot, and
F  1.926 lb
(b)
We have
80° 
  40
Then
1
98


N  (0.8 lb)   cos 40 
 (13  5sin 40) ft/s 2  sin 40
2
3
32.2 ft/s


 4.4924 lb
1
98


F  (0.8 lb) sin 40 
 (13  5sin 40) ft/s 2  cos 40
2
3
32.2 ft/s


 6.5984 lb
Now
Fmax  s N ,
from which it follows that
F  Fmax
Block E will slide in the slot
and
a E  a n  a E/plate
 a n  (a E/plate )t  (a E/plate )n
At t  0, the block is at rest relative to the plate, thus (a E/plate )n  0 at t  0, so that a E/plate must be
directed tangentially to the slot.
Fx  max : N  W cos 40  m
or
vE2

sin 40


v2
N  W   cos 40  E sin 40  (as above)


g


 4.4924 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.60 (Continued)
F  k N
Sliding:
 0.25(4.4924 lb)
 1.123 lb
Noting that F and a E/plane must be directed as shown (if their directions are reversed, then Fx is
while ma x is
), we have
the block slides downward in the slot and
F  1.123 lb
40° 
Alternative solutions.
(a)
Assume that the block is at rest with respect to the plate.
F  ma: W  R  man
Then
tan (  10) 

or
and
Now
so that
W
W
g


2

man W vE  ( ABCD )2
g 
32.2 ft/s 2
98
(13  5sin 80) ft/s 2
3
(from above)
  10  6.9588
  16.9588
tan s  s
s  0.35
s  19.29
0    s  Block does not slide and R is directed as shown.
Now
F  R sin 
Then
F  (0.8 lb)
and R 
W
sin (  10)
sin16.9588
sin 6.9588
 1.926 lb
The block does not slide in the slot and
F  1.926 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
80° 
PROBLEM 12.60 (Continued)
(b)
Assume that the block is at rest with respect to the plate.
F  ma: W  R  man
From Part a (above), it then follows that
tan (  50) 
g
 (ABCD ) 2

32.2 ft/s 2
98
(13  5sin 40) ft/s 2
3
  50  5.752
or
  55.752
and
s  19.29
Now
  s
so that
The block will slide in the slot and then
  k , where
k  14.0362
or
tan k  k
k  0.25
To determine in which direction the block will slide, consider the free-body diagrams for the two
possible cases.
F  ma : W  R  ma n  ma E/plate
Now
From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding
downward. Then
Fx  max : W cos 40  R cos k  m
or

sin 40
F  R sin k
Now
Then
vE2
W cos 40 
F
W vE2

sin 40
g 
tan k


v2
F  kW   cos 40  E sin 40 


g


(see the first solution)
 1.123 lb
The block slides downward in the slot and
F  1.123 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
40° 
PROBLEM 12.61
A small block B fits inside a lot cut in arm OA which rotates in a vertical plane at a
constant rate. The block remains in contact with the end of the slot closest to A and
its speed is 1.4 m/s for 0    150. Knowing that the block begins to slide when
  150, determine the coefficient of static friction between the block and the slot.
SOLUTION
Draw the free body diagrams of the block B when the arm is at   150.
v  at  0,
g  9.81 m/s 2
Ft  mat :  mg sin 30  N  0
N  mg sin 30
Fn  man : mg cos30  F  m
F  mg cos30 
Form the ratio
s 
v2

mv 2

F
, and set it equal to s for impending slip.
N
F
g cos30  v 2 /
9.81 cos30  (1.4) 2 /0.3


9.81 sin 30
N
g sin 30
s  0.400 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.62
The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at
stations E, F, and G by picking it up at a station when   0 and depositing it at the next station when
  180. Knowing that member BC remains horizontal throughout its motion and that links AB and CD rotate
at a constant rate in a vertical plane in such a way that vB  2.2 ft/s, determine (a) the minimum value of the
coefficient of static friction between the component and BC if the component is not to slide on BC while being
transferred, (b) the values of  for which sliding is impending.
SOLUTION
Fx  max : F 
W vB2
cos 
g 
 Fy  ma y : N  W  
W vB2
sin 
g 


v2
N  W 1  B sin  


g


or


v2
Fmax   s N   s W 1  B sin  


g


Now
and for the component not to slide
F  Fmax
or
or


v2
W vB2
cos    s W 1  B sin  


g 
g


s 
cos 
g
vB2
 sin 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.62 (Continued)
We must determine the values of  which maximize the above expression. Thus
d
d




  sin 

 sin  

cos 
g
vB2

g
vB2

 sin   (cos  )(  cos  )

g
vB2
 sin 

2
or
sin  
vB2
g
Now
sin  
(2.2 ft/s) 2
 0.180373
(32.2 ft/s 2 )  10
ft 
12
s  ( s )min
  10.3915 and   169.609
or
(a)
for
0
From above,
(  s ) min 
(  s ) min 
cos 
g
vB2
 sin 
where sin  
vB2
g
cos 
cos  sin 

 tan 
 sin 
1  sin 2 
1
sin 
 tan10.3915
(s )min  0.1834 
or
(b)
We have impending motion
to the left for
  10.39 
to the right for
  169.6 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.63
Knowing that the coefficients of friction between the component I and member BC of the mechanism of
Problem 12.62 are s  0.35 and k  0.25, determine (a) the maximum allowable constant speed vB if the
component is not to slide on BC while being transferred, (b) the values of  for which sliding is impending.
SOLUTION
Fx  max : F 
W vB2
cos 
g 
 Fy  ma y : N  W  
W vB2
sin 
g 


v2
N  W 1  B sin  
g


or
Fmax   s N
Now


v2
  s W 1  B sin  


g


and for the component not to slide
F  Fmax
or
or


v2
W vB2
cos    s W 1  B sin  


g 
g


vB2  s
g
cos   s sin 
 
must be less than or equal to the minimum value
To ensure that this inequality is satisfied, vB2
max
of s g  /(cos   s sin  ), which occurs when (cos  s sin  ) is maximum. Thus
d
(cos    s sin  )   sin    s cos   0
d
or
tan    s
 s  0.35
or
  19.2900
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 12.63 (Continued)
(a)
The maximum allowed value of vB is then
v 
2
B
max
 s
g
cos    s sin 
 g
tan 
 g  sin 
cos   (tan  ) sin 
where tan    s
 10 
 (32.2 ft/s 2 )  ft  sin 19.2900
 12 
(vB )max  2.98 ft/s 
or
(b)
First note that for 90    180, Eq. (1) becomes
vB2  s
g
cos   s sin 
where   180   . It then follows that the second value of  for which motion is impending is
  180  19.2900
 160.7100
we have impending motion
to the left for
  19.29 
to the right for
  160.7 
Alternative solution.
F  ma : W  R  man
For impending motion,   s . Also, as shown above, the values of  for which motion is impending

minimize the value of vB, and thus the value of an is an 
vB2

. From the above diagram, it can be concluded
that an is minimum when man and R are perpendicular.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.63 (Continued)
Therefore, from the diagram
  s  tan 1  s
and
or
or
(as above)
man  W sin s
m
vB2

 mg sin 
vB2  g  sin 
(as above)
For 90    180, we have
from the diagram
  180  
(as above)
  s
and
or
man  W sin s
vB2  g  sin 
(as above)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.64
A small 250-g collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at a constant rate of 7.5 rad/s. Determine the three values of  for
which the collar will not slide on the rod, assuming no friction between the collar
and the rod.
SOLUTION
If the collar is not sliding, it moves at constant speed on a circle of radius   r sin  . v  
Normal acceleration. an 
v2


2 2
 (r sin  )  2

 Fy  ma y :
N cos  mg  0
N 
mg
cos
Fx  ma x :
N sin   ma
mg sin 
 (m r sin  ) 2
cos 
(1)
To satisfy (1), either:
sin   0
or
cos  
g
r 2
  0 or 180
or
cos 
9.81
 0.3488
(0.5) (7.5) 2
  0, 180, and 69.6 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.65
A small 250-g collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at a constant rate of 7.5 rad/s. Knowing that the coefficients of
friction are s = 0.25 and k = 0.20, indicate whether the collar will slide on the rod if
it is released in the position corresponding to (a)  = 75, (b)  = 40. Also,
determine the magnitude and direction of the friction force exerted on the collar
immediately after release.
SOLUTION
If the collar is not sliding, it moves at constant speed on a circle of radius   r sin  . v  
r  500 mm  0.500 m,   7.5 rad/s,
Given:
m  250 g  0.250 kg.
Normal acceleration:
an 
v2


2 2
 (r sin  ) 2

F 
ma:
F  mg sin    m (r sin  )  2 cos 
F  m ( g  r 2 cos  )sin 
F 
ma :
N  mg cos   m(r sin  ) 2 sin 
N  m( g cos   r  2 sin 2  )
(a)   75.
F  (0.25) 9.81  (0.500 cos 75)(7.5)2  sin 75
 0.61112 N
N  (0.25) 9.81cos 75  (0.500sin 2 75)(7.5) 2 
 7.1950 N
 s N  (0.25)(7.1950)  1.7987 N
Since F   s N , the collar does not slide.
F  0.611 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
75 
PROBLEM 12.65 (Continued)
F  (0.25) 9.81  (0.500 cos 40) (7.5)2  sin 40
(b)   40.
  1.8858 N
N  (0.25) 9.81cos 40  (0.500sin 2 40)(7.5)2 
 4.7839 N
 s N  (0.25) (4.7839)  1.1960 N
Since F   s N , the collar slides.
Since the collar is sliding, F   k N .
 Fn  
ma :
N  mg cos   man sin 
N  mg cos   m (r sin  ) 2 sin 
 m  g cos   (r sin 2  ) 2 
 (0.25) 9.81cos 40  (0.500sin 2 40)(7.5) 2 
 4.7839 N
F   k N  (0.20) (4.7839)  0.957 N
F  0.957 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
40 
PROBLEM 12.66
An advanced spatial disorientation trainer allows the cab to rotate around
multiple axes as well as extend inwards and outwards. It can be used to
simulate driving, fixed wing aircraft flying, and helicopter maneuvering. In
one training scenario, the trainer rotates and translates in the horizontal plane
where the location of the pilot is defined by the relationships
 
r  10  2 cos  t  and   0.1 2t 2  t  , where r,  and t are expressed
3 
in feet, radians, and seconds, respectively. Knowing that the pilot has a mass
of 175 lbs, (a) find the magnitude of the resulting force acting on the pilot at
t= 5 s (b) plot the magnitudes of the radial and transverse components of the
force exerted on the pilot from 0 to 10 seconds.
SOLUTION
 
r  10  2 cos  t  ft
3 
Given:
  0.1 2t 2  t  rad
m
175 lbs
 5.435 slugs
32.2 ft/s 2
Using Radial and Transverse Coordinates:
2
 
sin  t  ft/s
3
3 
2 2
 

r 
cos  t  ft/s 2
9
3 
r  
  0.1 4t  1 rad/s
  0.4 rad/s 2
Free Body Diagram of pilot (top and side view) showing net forces in the radial and transverse directions:
Equations of Motion:
F
r

 mar  m 
r  r 2

 2 2
2
  
  
cos  t    10  2 cos  t   0.12  4t  1 
Fr  m  
3  
 3 
 9

 F  ma  m  r  2r 


0.4
  
 
F  m 10  2 cos  t    0.4  
sin  t   4t  1 
3
 3 
3 


F
z
 0  Fz  mg
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.66 (Continued)
(a) Evaluate forces at t=5 s:
Fr  221.78 lbs
F  61.37 lbs
Fz  175 lbs
F  289.1 lb 
F  Fr2  F2  Fz2
(b) Plot of Fr and Fθ from t=0 to t=10 s:
Radial and Transverse Components of Force for t=0 to 10 seconds
100
0
-100
Force, (lbs)
-200
-300
-400
-500
-600
-700
-800
Radial Component
Transverse Component
0
1
2
3
4
5
6
Time, (sec)
7
8
9
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
10

PROBLEM 12.67
An advanced spatial disorientation trainer is programmed to only rotate
and translate in the horizontal plane. The pilot’s location is defined by the

relationships r  8 1  e
t

and
  
  2 /   sin t  , where r,  and t

2 
are expressed in feet, radians, and seconds, respectively. Determine the
radial and transverse components of the force exerted on the 175 lb pilot at
t = 3 s.
SOLUTION
r  8 1  e  t  ft
Given:
  
  2 /   sin t  rad
2


175 lbs
m
 5.435 slugs
32.2 ft/s 2
Using Radial and Transverse Coordinates:
t
r  8e ft/s

r  8et ft/s 2

  cos t rad/s
2


   sin t rad/s 2
2
2
Free Body Diagram of pilot (top view) showing net forces in the radial and transverse directions:
Equations of Motion:
F
r
 mar

Fr  m 
r  r 2

 

Fr  m  8e t  8 1  et  cos 2 t 
2 

 F  ma

F  m r  2r


 
  
F  m 8 1  et     sin t  16et cos t 
2
2 
 2

Evaluate Fr and Fθ at t=3 s:
Fr  2.165 lb 
F  64.90 lb 
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PROBLEM 12.68
The 3-kg collar B slides on the frictionless arm AA. The arm is
attached to drum D and rotates about O in a horizontal plane at the
rate   0.75t , where  and t are expressed in rad/s and seconds,
respectively. As the arm-drum assembly rotates, a mechanism
within the drum releases cord so that the collar moves outward from
O with a constant speed of 0.5 m/s. Knowing that at t  0, r  0,
determine the time at which the tension in the cord is equal to the
magnitude of the horizontal force exerted on B by arm AA.
SOLUTION
Kinematics
dr
 r  0.5 m/s
dt
We have
At t  0, r  0:

r
0
dr 

t
0
0.5 dt
or
r  (0.5t ) m
Also,

r 0
  (0.75t ) rad/s
  0.75 rad/s 2
Now
ar  
r  r 2  0  [(0.5t ) m][(0.75t ) rad/s]2  (0.28125t 3 ) m/s 2
and
a  r  2r
 [(0.5t ) m][0.75 rad/s 2 ]  2(0.5 m/s)[(0.75t ) rad/s]
 (1.125t ) m/s 2
Kinetics
Fr  mar :  T  (3 kg)(0.28125t 3 ) m/s 2
or
T  (0.84375t 3 ) N
F  mB a : Q  (3 kg)(1.125t ) m/s 2
or
Q  (3.375t ) N
Now require that
T Q
or
or
(0.84375t 3 ) N  (3.375t ) N
t 2  4.000
t  2.00 s 
or
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PROBLEM 12.69
A 0.5-kg block B slides without friction inside a slot cut in arm OA which rotates in
a vertical plane. The rod has a constant angular acceleration   10 rad/s2.
Knowing that when  º and r  0.8 m the velocity of the block is zero,
determine at this instant, (a) the force exerted on the block by the arm, (b) the
relative acceleration of the block with respect to the arm.
SOLUTION
Given:
r  0.8 m,   10 rad/s 2
m  0.5 kg, W  mg
  45,
Using Radial and Transverse Components:
vr  r  0,
FBD of Block B
v  r  0
  0
(a)

F  ma : N  W cos 45  m r  2r

N  mg cos 45  m r  2r


 (0.5)(9.81) cos 45  0.5  0.810   0 
 7.468
(b)
N  7.47 N

Fr  mar : mg sin 45  m 
r  r 2
45 

mg
sin 45  r 2
m
 g sin 45  r 2
  9.81 sin 45  0
 6.937 m/s 2

r 
a B / rod  6.94 m/s 2
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45 
PROBLEM 12.70
Pin B weighs 4 oz and is free to slide in a horizontal plane along the
rotating arm OC and along the circular slot DE of radius b  20 in.
Neglecting friction and assuming that   15 rad/s and
  250 rad/s 2 for the position   20, determine for that position
(a) the radial and transverse components of the resultant force
exerted on pin B, (b) the forces P and Q exerted on pin B,
respectively, by rod OC and the wall of slot DE.
SOLUTION
Kinematics.
From the geometry of the system, we have
r  2b cos 
Then
r   (2b sin  )

r  2b(sin    2 cos  )
and
ar  
r  r 2  2b(sin    2 cos  )  (2b cos  ) 2   2b(sin   2 2 cos  )
Now
 20 
  2  ft  [(250 rad/s2 )sin 20  2(15 rad/s)2 cos 20]  1694.56 ft/s2
 12 
a  r  2r  (2b cos  )  2(2b sin  )  2b( cos   2 2 sin  )
and
 20 
 2  ft  [(250 rad/s2 )cos 20  2(15 rad/s)2 sin 20]  270.05 ft/s2
 12 
Kinetics.
(a)
We have
Fr  mar 
and
F  ma 
1
lb
4
32.2 ft/s2
1
4
lb
32.2 ft/s 2
 (1694.56 ft/s2 )  13.1565 lb
Fr  13.16 lb 
F  2.10 lb 
 (270.05 ft/s 2 )  2.0967 lb
Fr : Fr  Q cos 20
(b)
or
Q
1
(13.1565 lb)  14.0009 lb
cos 20
F : F  P  Q sin 20
or
P  (2.0967  14.0009sin 20) lb  6.89 lb
P  6.89 lb
70 
Q  14.00 lb
40 
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PROBLEM 12.71
The two blocks are released from rest when r  0.8 m and   30.
Neglecting the mass of the pulley and the effect of friction in the
pulley and between block A and the horizontal surface, determine
(a) the initial tension in the cable, (b) the initial acceleration of block
A, (c) the initial acceleration of block B.
SOLUTION
Let r and  be polar coordinates of block A as shown, and let yB be the
position coordinate (positive downward, origin at the pulley) for the
rectilinear motion of block B.
Constraint of cable:
r  yB  constant,
r  vB  0,
For block A,
For block B,

r  aB  0

r  aB
(1)
Fx  mAaA : T cos  mAaA or T  mAaA sec
(2)
or
+ Fy  mB aB : mB g  T  mB aB
Adding Eq. (1) to Eq. (2) to eliminate T, mB g  mAaA sec  mBaB
(3)
(4)
Radial and transverse components of a A.
Use either the scalar product of vectors or the triangle construction shown,
being careful to note the positive directions of the components.

r  r 2  ar  a A  e r  a A cos 
(5)
Noting that initially   0, using Eq. (1) to eliminate r, and changing
signs gives
aB  a A cos
(6)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
aA 
mB g
(25) (9.81)

 5.48 m/s 2
m A sec   mB cos 
20sec30  25cos 30
From Eq. (6), aB  5.48cos30  4.75 m/s 2
(a)
From Eq. (2), T  (20)(5.48)sec30  126.6
(b)
Acceleration of block A.
(c)
Acceleration of block B.
T  126.6 N 
a A  5.48 m/s 2
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
a B  4.75 m/s 2 
PROBLEM 12.72
The velocity of block A is 2 m/s to the right at the instant when
r  0.8 m and   30. Neglecting the mass of the pulley and the
effect of friction in the pulley and between block A and the horizontal
surface, determine, at this instant, (a) the tension in the cable, (b) the
acceleration of block A, (c) the acceleration of block B.
SOLUTION
Let r and  be polar coordinates of block A as shown, and let yB be the
position coordinate (positive downward, origin at the pulley) for the
rectilinear motion of block B.
Radial and transverse components of v A.
Use either the scalar product of vectors or the triangle construction shown,
being careful to note the positive directions of the components.
r  vr  v A  er  vA cos30
 2cos30  1.73205 m/s
r  v  v A  e  v A sin 30
 2sin 30  1.000 m/s 2
 
v
1.000

 1.25 rad/s
r
0.8
Constraint of cable: r  yB  constant,
r  vB  0,
For block A,
For block B,

r  aB  0
or

r  aB
Fx  mAaA : T cos  mAaA or T  mAaA sec
+ Fy  mB aB : mB g  T  mB aB
Adding Eq. (1) to Eq. (2) to eliminate T, mB g  mAaA sec  mBaB
(1)
(2)
(3)
(4)
Radial and transverse components of a A.
Use a method similar to that used for the components of velocity.

r  r 2  ar  a A  e r  a A cos 
(5)
Using Eq. (1) to eliminate r and changing signs gives
aB  a A cos  r 2
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(6)
PROBLEM 12.72 (Continued)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
aA 

mB g  r2

mA sec   mB cos 

(25)[9.81  (0.8)(1.25)2 ]
 6.18 m/s 2
20sec30  25cos30
From Eq. (6), aB  6.18cos 30  (0.8)(1.25) 2  4.10 m/s 2
(a)
From Eq. (2), T  (20)(6.18)sec30  142.7
(b)
Acceleration of block A.
(c)
Acceleration of block B.
T  142.7 N 
a A  6.18 m/s 2
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
a B  4.10 m/s 2 
PROBLEM 12.73*
Slider C has a weight of 0.5 lb and may move in a slot cut in arm
AB, which rotates at the constant rate 0  10 rad/s in a horizontal
plane. The slider is attached to a spring of constant k  2.5 lb/ft,
which is unstretched when r  0. Knowing that the slider is
released from rest with no radial velocity in the position r  18 in.
and neglecting friction, determine for the position r  12 in. (a) the
radial and transverse components of the velocity of the slider,
(b) the radial and transverse components of its acceleration, (c) the
horizontal force exerted on the slider by arm AB.
SOLUTION
Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring.
Fr   k (r  l0 )
 Fr  mar :  k (r  l0 )  m(
r  r 2 )
kl
k


r    2   r  0
m
m

(1)
But
d
dr dr
dr
(r) 
 r
dt
dr dt
dr

kl 
k
   
rdr
rdr   2   r  0  dr
m
m


r
Integrate using the condition r  r0 when r  r0 .
1 2 r  1   2 k  2 kl0  r
    r 
r
r
2 r0  2 
m
m  r0


kl
1 2 1 2 1  2 k  2
r  r0      r  r02  0 (r  r0 )
2
2
2
m
m
2kl0
k

r 2  r02    2   r 2  r02 
(r  r0 )
m
m


Data:
m

W
0.5 lb

 0.01553 lb  s 2 /ft
g 32.2 ft/s 2
  10 rad/s, k  2.5 lb/ft, l0  0
r0  (vr )0  0, r0  18 in.  1.5 ft, r  12 in.  1.0 ft
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PROBLEM 12.73* (Continued)
(a)
Components of velocity when r  12 in.
2.5 

(1.02  1.52 )  0
r 2  0  102 

0.01553 

 76.223 ft 2 /s 2
vr  r  8.7306 ft/s
Since r is decreasing, vr is negative
(b)
r  8.7306 ft/s
vr  8.73 ft/s 
v  r  (1.0)(10)
v  10.00 ft/s 
Components of acceleration.
Fr  kr  kl0  (2.5)(1.0)  0  2.5 lb
ar 
Fr
2.5

0.01553
m
ar  161.0 ft/s 2 
a  r  2r  0  (2)(8.7306)(10)
a  174.6 ft/s 2 
(c)
Transverse component of force.
F  ma  (0.01553)(174.6)
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F  2.71 lb 
PROBLEM 12.74
A particle of mass m is projected from Point A with an initial velocity v0
perpendicular to line OA and moves under a central force F directed
away from the center of force O. Knowing that the particle follows a path
defined by the equation r  r0 / cos 2 and using Eq. (12.25), express
the radial and transverse components of the velocity v of the particle as
functions of  .
SOLUTION
Since the particle moves under a central force, h  constant.
Using Eq. (12.27),
h  r 2  h0  r0 v0
or
 
r0 v0
r
2

r0 v0 cos 2
r0
2

v0
cos 2
r0
Radial component of velocity.
vr  r 
 r0
r0
dr  d 


d
d  cos 2

sin 2

  r0
3/ 2
(cos
2
)


sin 2 v0
cos 2
(cos 2 )3/2 r
vr  v0
sin 2
cos 2

Transverse component of velocity.
v 
h r0 v0

cos 2
r
r0
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v  v0 cos 2 
PROBLEM 12.75
For the particle of Problem 12.74, show (a) that the velocity of the
particle and the central force F are proportional to the distance r from
the particle to the center of force O, (b) that the radius of curvature of
the path is proportional to r3.
PROBLEM 12.74 A particle of mass m is projected from Point A with
an initial velocity v0 perpendicular to line OA and moves under a central
force F directed away from the center of force O. Knowing that the
particle follows a path defined by the equation r  r0 / cos 2 and using
Eq. (12.25), express the radial and transverse components of the velocity
v of the particle as functions of .
SOLUTION
Since the particle moves under a central force, h  constant.
Using Eq. (12.27),
h  r 2  h0  r0 v0
or
 
r0 v0
r
2

r0 v0 cos 2
r0
2

v0
cos 2
r0
Differentiating the expression for r with respect to time,
r 
r0
dr  d 


d
d  cos 2

sin 2
sin 2 v0
sin 2
  r0
cos 2  v0
  r0
3/ 2
3/ 2
r
(cos 2 )
(cos 2 )
cos 2

0
Differentiating again,

r
(a)
dr  d 
sin 2  
2 cos 2 2  sin 2 2  v0 2 2 cos 2 2  sin 2 2


 v0
   v0
d
d 
r0
(cos 2 )3/ 2
cos 2 
cos 2
vr  r  v0
sin 2
cos 2

vr
v  r  0 cos 2
r0
v0 r
sin 2
r0
v  (vr )2  (v )2 
v0 r
sin 2 2  cos2 2
r0
v
v0 r

r0
v 2 2cos 2 2  sin 2 2
r0
v0 2
ar  
r  r 2  0

cos 2 2
r0
cos 2
cos 2 r0 2

v0 2 cos 2 2  sin 2 2
v0
v 2r

 02
r0
r0
r0 cos 2
cos 2
Fr  mar 
mv02 r
r02
:
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Fr 
mv02 r
r02

PROBLEM 12.75 (Continued)
Since the particle moves under a central force, a  0.
Magnitude of acceleration.
a  ar 2  a 2 
v0 2 r
r0 2
Tangential component of acceleration.
at 
v0 2 r
dv d  v0 r  v0

r
sin 2
 



dt dt  r0  r0
r0 2
Normal component of acceleration.
at  a 2  at 2 
r 
cos 2   0 
r
But
an 
Hence,
(b)
But an 
v2

or
v0 2 r
r0 2
1  sin 2 2 
v0 2 r cos 2
r0 2
2
v0 2
r

v 2 v0 2 r 2 r
 2  2
an
r0
v0
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
r3

r02
PROBLEM 12.76
A particle of mass m is projected from Point A with an initial velocity v0
perpendicular to line OA and moves under a central force F along a
semicircular path of diameter OA. Observing that r  r0 cos and using
Eq. (12.25), show that the speed of the particle is v  v0 /cos 2  .
SOLUTION
Since the particle moves under a central force, h  constant.
Using Eq. (12.27),
h  r 2  h0  r0 v0
or
 
r0 v0
r
2

r0 v0
r0 cos 
2
2

v0
r0 cos2 
Radial component of velocity.
vr  r 
d
( r0 cos  )  (r0 sin  )
dt
Transverse component of velocity.
v  r  (r0 cos  )
Speed.
v  vr 2  v 2  r0 
r0 v0
r0 cos 
2
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v
v0
cos2 

PROBLEM 12.77
For the particle of Problem 12.76, determine the tangential component Ft
of the central force F along the tangent to the path of the particle for
(a)   0, (b)   45.
PROBLEM 12.76 A particle of mass m is projected from Point A with an
initial velocity v0 perpendicular to line OA and moves under a central force
F along a semicircular path of diameter OA. Observing that r  r0 cos
and using Eq. (12.25), show that the speed of the particle is v  v0 /cos 2  .
SOLUTION
Since the particle moves under a central force, h  constant
Using Eq. (12.27),
h  r 2  h0  r0 v0
 
r0 v0
r
2

r0 v0
r0 cos 
2
2

v0
r0 cos 2 
Radial component of velocity.
vr  r 
d
( r0 cos  )  (r0 sin  )
dt
Transverse component of velocity.
v  r  (r0 cos  )
Speed.
v  vr 2  v 2  r0 
r0 v0
r0 cos 
2

v0
cos2 
Tangential component of acceleration.
at 
v0
dv
(2)( sin  ) 2v0 sin 
 v0


3
3
dt
cos 
cos  r0 cos 2 

2v0 2 sin 
r0 cos5 
Tangential component of force.
Ft  mat : Ft 
(a)
  0,
(b)
  45,
2mv0 2 sin 
r0 cos5 
Ft  0
Ft 
Ft  0 
2mv0 sin 45
cos5 45
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Ft 
8mv0 2

r0
PROBLEM 12.78
Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi
and that the moon requires 27.32 days to complete one full revolution about the earth.
SOLUTION
Mm
[Eq. (12.28)]
r2
We have
F G
and
F  Fn  man  m
Then
or
Now
G
Mm
v2

m
r
r2
M 
v
r 2
v
G
2 r

2
so that
v2
r
M
2
r  2 r 
1  2  3
r
 


G  
G   
Noting that
  27.32 days  2.3604  106 s
and
r  238,910 mi  1.26144  109 ft
we have
or
M
2


2
9
3
 (1.26144  10 ft)
9 4
4 
6

34.4  10 ft /lb  s  2.3604  10 s 
1
M  413  1021 lb  s 2 /ft 
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PROBLEM 12.79
Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration of
gravity g at the surface of the earth, and the time  required for the moon to complete one full revolution about
the earth. Compute r knowing that   27.3 days, giving the answer in both SI and U.S. customary units.
SOLUTION
Mm
r2
We have
F G
and
F  Fn  man  m
Then
G
GM
r
GM  gR 2
Now
v2 
so that
v2
r
Mm
v2

m
r
r2
v2 
or
[Eq. (12.28)]
[Eq. (12.30)]
g
gR 2
or v  R
r
r
2 r 2 r

v
R gr
For one orbit,

or
 g 2 R 2 
r  

2 
 4 
Now
  27.3 days  2.35872  106 s
1/ 3
Q.E.D.

R  3960 mi  20.9088  106 ft
1/3
 9.81 m/s 2  (2.35872  106 s) 2  (6.37  106 m)2 
r

4 2


SI:
 382.81  106 m
r  383  103 km 
or
U.S. customary units:
1/3
 32.2 ft/s 2  (2.35872  106 s) 2  (20.9088  106 ft)2 
r

4 2


 1256.52  106 ft
r  238  103 mi 
or
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PROBLEM 12.80
Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete
one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to
the ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with
which they describe their orbit. Give the answers in both SI and U.S. customary units.
SOLUTION
For gravitational force and a circular orbit,
Fr 
GMm mv 2

r
r2
or
v
GM
r
Let  be the period time to complete one orbit.
But
v  2 r
Then
GM  2
r 
4 2
3
or
GM  2
 4 2 r 2
r
 GM  2
r  
2
 4
1/3



  23.934 h  86.1624  103 s
Data:
(a)
v 2 2 
or
g  9.81 m/s 2 , R  6.37  106 m
In SI units:
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s 2
1/3
 (398.06  1012 )(86.1624  103 )2 
6
r
  42.145  10 m
2
4


h  35,800 km 
altitude h  r  R  35.775  106 m
In U.S. units:
g  32.2 ft/s 2 , R  3960 mi  20.909  106 ft
GM  gR 2  (32.2)(20.909  106 ) 2  14.077  1015 ft 3 /s 2
1/3
 (14.077  1015 )(86.1624  103 )2 
6
r
  138.334  10 ft
2
4


altitude h  r  R  117.425  106 ft
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
h  22, 200 mi 
PROBLEM 12.80 (Continued)
(b)
In SI units:
v
GM
398.06  1012

 3.07  103 m/s
r
42.145  106
v
GM
14.077  1015

 10.09  103 ft/s
6
r
138.334  10
v  3.07 km/s 
In U.S. units:
v  10.09  103 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.81
Show that the radius r of the orbit of a moon of a given planet can be determined from the radius R of the
planet, the acceleration of gravity at the surface of the planet, and the time  required by the moon to complete
one full revolution about the planet. Determine the acceleration of gravity at the surface of the planet Jupiter
knowing that R  71,492 km and that   3.551 days and r  670.9  103 km for its moon Europa.
SOLUTION
Mm
r2
We have
F G
and
F  Fn  man  m
Then
or
Now
so that
G
[Eq. (12.28)]
v2
r
Mm
v2
m
2
r
r
v2 
GM
r
GM  gR 2
v2 
[Eq. (12.30)]
gR 2
r
or
vR
g
r
2 r 2 r

v
R gr
For one orbit,

or
 g 2 R 2
r  
2
 4
Solving for g,
g  4 2
1/3




Q.E.D.
r3
 2 R2
and noting that   3.551 days  306,806 s, then
g Jupiter  4 2
 4 2
3
rEur
2
 Eur
RJup
(670.9  106 m)3
(306,806 s)2 (71.492  106 m) 2
g Jupiter  24.8 m/s 2 
or
Note:
g Jupiter  2.53 g Earth 
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PROBLEM 12.82
The orbit of the planet Venus is nearly circular with an orbital velocity of 126.5  103 km/h. Knowing that
the mean distance from the center of the sun to the center of Venus is 108  106 km and that the radius of the
sun is 695  103 km, determine (a) the mass of the sun, (b) the acceleration of gravity at the surface of the
sun.
SOLUTION
Let M be the mass of the sun and m the mass of Venus.
For the circular orbit of Venus,
GMm
mv 2
 man 
2
r
r
GM  rv 2
where r is radius of the orbit.
r  108  106 km  108  109 m
Data:
v  126.5  103 km/hr  35.139  103 m/s 
GM  (108  109 )(35.139  103 ) 2  1.3335  1020 m3 /s 2
M 
(a) Mass of sun.
(b) At the surface of the sun,
GM
1.3335  1020 m3 /s 2

G
66.73  1012
M  1.998  1030 kg 
R  695.5  103 km  695.5  106 m
GMm
 mg
R2
g 
GM
1.3335  1020

R2
(695.5  106 )2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
g  276 m/s 2 
PROBLEM 12.83
A satellite is placed into a circular orbit about the planet Saturn at an altitude of 2100 mi. The satellite
describes its orbit with a velocity of 54.7  103 mi/h. Knowing that the radius of the orbit about Saturn and the
periodic time of Atlas, one of Saturn’s moons, are 85.54  103 mi and 0.6017 days, respectively, determine
(a) the radius of Saturn, (b) the mass of Saturn. (The periodic time of a satellite is the time it requires to
complete one full revolution about the planet.)
SOLUTION
2 rA
Velocity of Atlas.
vA 
where
v A  85.54  103 mi  451.651  106 ft
and
 A  0.6017 days  51,987s
Gravitational force.
A
vA 
(2 ) (451.651  106 )
 54.587  103 ft/s
51,987
F
GMm mv 2

r
r2
from which
GM  rv2  constant
For the satellite,
rs vs2  rA v A2
rAv A2

vs 2

rs 
where
vs  54.7  103 mi/h  80.227  103 ft/s
(451.651  106 )(54.587  103 )2
 209.09  106 ft
(80.227  103 )2
rs  39,600 mi
rs 
(a)
Radius of Saturn.
R  rs  (altitude)  39,600  2100
(b)
R  37,500 mi 
Mass of Saturn.
M
rAvA2 (451.651  106 )(54.587  103 )2

G
34.4  109
M  39.1  1024 lb  s 2 /ft 
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PROBLEM 12.84
The periodic time (see Prob. 12.83) of an earth satellite in a circular polar
orbit is 120 minutes. Determine (a) the altitude h of the satellite, (b) the time
during which the satellite is above the horizon for an observer located at the
north pole.
SOLUTION
For gravitational force and a circular orbit,
Fr 
GMm
mv 2

r
r2
v 
or
GM
r
Let  be the periodic time to complete one orbit.
v  2 r
or
GM
 2 r
r

1/ 3
Solving for r,
 GM  2 
r  
2 

 4 
For earth, R  3960 mi  20.909  106 ft, g  32.2 ft/s 2

GM  gR 2   32.2  20.909  106

2
 14.077  1015 ft 3/s 2
Data:   120 min  7200 s


1/ 3
 14.077  1015  7200 2 

r 


4 2


 26.441  106 ft
(a) altitude h  r  R  5.532  106 ft
(b) cos 
h  1048 mi 
R
20.909  106

 0.79078
r
26.441  106
  37.74
t AB 
 75.48 7200   1509.6 s
2
 
360
360
t AB  25.2 min 
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PROBLEM 12.85
A 500 kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is
transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass
of the earth and that the radius of the moon is 1737 km, determine (a) the gravitational force exerted on the
spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if
the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the
surface of the moon.
SOLUTION
RE  6.37  106 m
First note that
rE  RE  hE  (6.37  106  4.5  106 ) m
Then
(a)
 10.87  106 m
F
We have
GMm
r2
[Eq. (12.28)]
GM  gR 2
and
[Eq. (12.29)]
m
R
W  
r2
r
2
Then
F  gR 2
For the earth orbit,
 6.37  106 m 
F  (500 kg)(9.81 m/s ) 

6
 10.87  10 m 
2
2
F  1684 N 
or
(b)
From the solution to Problem 12.78, we have
2
M
Then
Now

1  2  3
r
G   
2 r 3/ 2
GM
E M 
2 rE3/ 2
GM E

2 rM3/ 2
GM M
(1)
1/3
or
M 
rM   M  rE  (0.01230)1/3 (10.87  106 m)
 ME 
or
rM  2.509  106 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
rM  2510 km 
PROBLEM 12.85 (Continued)
(c)
GM  gR 2
We have
[Eq.(12.29)]
Substituting into Eq. (1)
2 rE3/ 2
RE g E

2 rM3/ 2
RM g M
2
3
2
 R  r 
 R  M
g M   E   M  gE   E   M
 RM   rE 
 RM   M E
or

 gE

using the results of Part (b). Then
2
 6370 km 
2
gM  
 (0.01230)(9.81 m/s )
1737
km


g moon  1.62 m/s 2 
or
Note:
g moon 
1
g earth
6
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PROBLEM 12.86
A space vehicle is in a circular orbit of 2200-km radius around the moon.
To transfer it to a smaller circular orbit of 2080-km radius, the vehicle is
first placed on an elliptic path AB by reducing its speed by 26.3 m/s as it
passes through A. Knowing that the mass of the moon is 73.49  1021 kg,
determine (a) the speed of the vehicle as it approaches B on the elliptic
path, (b) the amount by which its speed should be reduced as it
approaches B to insert it into the smaller circular orbit.
SOLUTION
For a circular orbit,
 Fn  man : F  m
F G
Eq. (12.28):
Then
G
or
v2
r
Mm
r2
Mm
v2
m
2
r
r
GM
v2 
r
66.73  1012 m3 /kg  s 2  73.49  1021 kg
2200  103 m
Then
2
(vA )circ

or
(vA )circ  1493.0 m/s
and
2
(vB )circ

or
(vB )circ  1535.5 m/s
(a)
We have
66.73  1012 m3 /kg  s 2  73.49  1021 kg
2080  103 m
(v A )TR  (v A )circ  v A
 (1493.0  26.3) m/s
 1466.7 m/s
Conservation of angular momentum requires that
rA m(vA )TR  rB m(vB )TR
or
2200 km
 1466.7 m/s
2080 km
 1551.3 m/s
(vB )TR 
(vB )TR  1551 m/s 
or
(b)
Now
or
(v B )circ  (vB )TR  vB
vB  (1535.5  1551.3) m/s
vB  15.8 m/s 
or
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PROBLEM 12.87
As a first approximation to the analysis of a space flight from the
earth to Mars, assume the orbits of the earth and Mars are
circular and coplanar. The mean distances from the sun to the
earth and to Mars are 149.6  106 km and 227.8  106 km,
respectively. To place the spacecraft into an elliptical transfer
orbit at point A, its speed is increased over a short interval of
time to v A which is 2.94 km/s faster than the earth’s orbital
speed. When the spacecraft reaches point B on the elliptical
transfer orbit, its speed vB is increased to the orbital speed of
Mars. Knowing that the mass of the sun is 332.8  103 times the
mass of the earth, determine the increase in speed required at B.
SOLUTION

For earth,  GM earth  gR 2   9.81 6.37  106


2
 398.06  1012 m3/s 2

For sun,  GM sun  332.8  103  GM earth  132.474  1018 m3 /s 2
For circular orbit of earth, rE  149.6  106 km  149.6  109 m/s
vE 
For transfer orbit AB,
rA  rE ,
 GM sun
rE

132.474  1018
 29.758  103 m/s
149.6  109
rB  rM  227.8  109 m
v A  vE   v  A  29.758  103  2.94  103  32.698  103 m/s
mrAv A  mrB vB



149.6  109 32.698  103
rAv A
vB 

 21.473  103 m/s
9
rB
227.8  10
For circular orbit of Mars,
vM 
 GM sun
rM

132.474  1018
 24.115  103 m/s
9
227.8  10
Speed increase at B.
 v B
 vM  vB  24.115  103  21.473  103  2.643  103 m/s
 v B
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
 2.64 km/s 
PROBLEM 12.88
To place a communications satellite into a geosynchronous orbit (see
Problem 12.80) at an altitude of 22,240 mi above the surface of the
earth, the satellite first is released from a space shuttle, which is in a
circular orbit at an altitude of 185 mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the
booster’s motor is fired to insert the satellite into an elliptic transfer
orbit. The booster is again fired at B to insert the satellite into a
geosynchronous orbit. Knowing that the second firing increases the
speed of the satellite by 4810 ft/s, determine (a) the speed of the satellite
as it approaches B on the elliptic transfer orbit, (b) the increase in speed
resulting from the first firing at A.
SOLUTION
For earth,
R  3960 mi  20.909  106 ft
GM  gR 2  (32.2)(20.909  106 ) 2  14.077  1015 ft 3 /s 2
rA  3960  185  4145 mi  21.8856  106 ft
rB  3960  22, 240  26, 200 mi  138.336  106 ft
Speed on circular orbit through A.
(v A )circ 

GM
rA
14.077  1015
21.8856  106
 25.362  103 ft/s
Speed on circular orbit through B.
(vB )circ 

GM
rB
14.077  1015
138.336  106
 10.088  103 ft/s
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PROBLEM 12.88 (Continued)
(a)
Speed on transfer trajectory at B.
(vB ) tr  10.088  103  4810
 5.278  103
Conservation of angular momentum for transfer trajectory.
5280 ft/s 
rA (v A ) tr  rB (vB ) tr
(v A ) tr 
rB (vB ) tr
rA
(138.336  106 )(5278)
21.8856  106
 33.362  103 ft/s

(b)
Change in speed at A.
v A  (v A ) tr  (v A )circ
 33.362  103  25.362  103
 8.000  103
vA  8000 ft/s 
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PROBLEM 12.89
A space vehicle is in a circular orbit of 1400-mi radius around the moon.
To transfer to a smaller orbit of 1300-mi radius, the vehicle is first placed
in an elliptic path AB by reducing its speed by 86 ft/s as it passes through
A. Knowing that the mass of the moon is 5.03  1021 lb  s 2 /ft, determine
(a) the speed of the vehicle as it approaches B on the elliptic path, (b) the
amount by which its speed should be reduced as it approaches B to insert
it into the smaller circular orbit.
SOLUTION
Circular orbits: v 
GM
r
rA  1400 mi  7.392  106 ft
 vA 1 
 34.4  10  5.03  10 
9
21
7.392  106
 4.8382  103 ft/s
rB  1300 mi  6.864  106 ft
 vB  2
 34.4  10  5.03  10 
9

21
6.864  106
 5.0208  103 ft/s
(a) Transfer orbit AB.
 vA 2   vA 1   v  A
 4.8382  103  86  4.7522  103 ft/s
mrA  v A 2  mrB  vB 1
 vB 1 
rA  v A 2
rB
 7.392  10  4.7522  10   5.1178  10

6
3
3
6.864  103
ft/s
 vB 1  5.12  103 ft/s 
(b) Speed change at B.
 vB    vB 2   vB 1  5.0208  103 ft/s  5.1178  103 ft/s  97.0 ft/s
Speed reduction at B.
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vB  97.0 ft/s 
PROBLEM 12.90
A 1 kg collar can slide on a horizontal rod, which is free to
rotate about a vertical shaft. The collar is initially held at A by a
cord attached to the shaft. A spring of constant 30 N/m is
attached to the collar and to the shaft and is undeformed when
the collar is at A. As the rod rotates at the rate   16 rad/s, the
cord is cut and the collar moves out along the rod. Neglecting
friction and the mass of the rod, determine (a) the radial and
transverse components of the acceleration of the collar at A,
(b) the acceleration of the collar relative to the rod at A, (c) the
transverse component of the velocity of the collar at B.
SOLUTION
Fsp  k (r  rA )
First note
(a)
F  0 and at A,
Fr   Fsp  0
(a A ) r  0 
(aA )  0 
Fr  mar :
(b)
Noting that
Fsp  m (
r  r 2 )
acollar/rod  
r , we have at A
0  m[acollar/rod  (150 mm)(16 rad/s)2 ]
acollar/rod  38400 mm/s 2
(acollar/rod ) A  38.4 m/s 2 
or
(c)
After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular
momentum about the shaft is conserved.
rA m (vA )  rB m (vB )
Then
or
(vB ) 
where (v A )  rA 0
150 mm
[(150 mm)(16 rad/s)]  800 mm/s
450 mm
(vB )  0.800 m/s 
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PROBLEM 12.91
A 1-lb ball A and a 2-lb ball B are mounted on a horizontal rod
which rotates freely about a vertical shaft. The balls are held in
the positions shown by pins. The pin holding B is suddenly
removed and the ball moves to position C as the rod rotates.
Neglecting friction and the mass of the rod and knowing that
the initial speed of A is vA  8 ft/s, determine (a) the radial and
transverse components of the acceleration of ball B
immediately after the pin is removed, (b) the acceleration of
ball B relative to the rod at that instant, (c) the speed of ball A
after ball B has reached the stop at C.
SOLUTION
Let r and  be polar coordinates with the origin lying at the shaft.
Constraint of rod:  B   A   radians; B  A  ; B  A  .
(a) Components of acceleration
Sketch the free body diagrams of the balls showing the radial and
transverse components of the forces acting on them. Owing to
frictionless sliding of B along the rod, (FB )r  0.
Radial component of acceleration of B.
Fr  mB (aB )r :
(aB )r  0 
Transverse components of acceleration.
(a A )  rA  2rA  ra
(aB )  rB  2rB
(1)
Since the rod is massless, it must be in equilibrium. Draw its free body
diagram, applying Newton’s 3rd Law.
M 0  0: rA (FA )  rB (FB )  rAmA (aA )  rBmB (aB )  0
rAmArA  rB mB (rB  2rB)  0
 
At t  0,
2mB rB
m ArA2  mB rB 2
rB  0 so that   0.
From Eq. (1),
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(aB )  0 
PROBLEM 12.91 (Continued)
(b)
Acceleration of B relative to the rod.
(v )
96
At t  0, (v A )  8 ft/s  96 in./s,   A  
 9.6 rad/s
rA
10

rB  rB 2  (aB ) r  0

rB  rB2  (8) (9.6) 2  737.28 in./s 2

rB  61.4 ft/s 2 
(c)
Speed of A.
Substituting
d
(mr 2) for rF in each term of the moment equation
dt
gives




d
d
m ArA2 
mB rB 2  0
dt
dt
Integrating with respect to time,

mArA2  mB rB 2  mArA2
   m r  
2
0
B B
0
Applying to the final state with ball B moved to the stop at C,
 WA 2 WB 2  
W

W
rA 
rC   f   A rA2  B (rB )02  0

g
g
 g

 g

 f 
WArA2  WB (rB )02 
(1)(10)2  (2)(8)2
0 
(9.6)  3.5765 rad/s
2
2
WArA  WB rC
(1)(10) 2  (2)(16) 2
(v A ) f  rA f  (10)(3.5765)  35.765 in./s
(v A ) f  2.98 ft/s 
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PROBLEM 12.92
Two 2.6-lb collars A and B can slide without friction on a frame, consisting
of the horizontal rod OE and the vertical rod CD, which is free to rotate
about CD. The two collars are connected by a cord running over a pulley
that is attached to the frame at O and a stop prevents collar B from moving.
The frame is rotating at the rate   12 rad/s and r  0.6 ft when the stop
is removed allowing collar A to move out along rod OE. Neglecting
friction and the mass of the frame, determine, for the position r  1.2 ft,
(a) the transverse component of the velocity of collar A, (b) the tension in
the cord and the acceleration of collar A relative to the rod OE.
SOLUTION
Masses: m A  mB 
2.6
 0.08075 lb  s 2 /ft
32.2
(a) Conservation of angular momentum of collar A: ( H 0 )2  (H 0 )1
mAr1(v )1  mAr2 (v )2
(v )2 
r1(v )1 r121 (0.6)2 (12)


 3.6
1.2
r2
r2
(v )2  3.60 ft/s 
2 
(v ) 2
3.6

 3.00 rad/s
1.2
rA
(b) Let y be the position coordinate of B, positive upward with origin at O.
Constraint of the cord: r  y  constant
or

y  
r
Kinematics:
(aB ) y  
y  
r
Collar B:
Collar A:
and
(a A ) r  
r  r 2
Fy  mB aB : T  WB  mB 
y  mB 
r
(1)
Fr  m A (a A )r :  T  m A (
r  r 2 )
(2)
Adding (1) and (2) to eliminate T,
WB  (m A  mB )
r  m Ar 2
a A/rod  
r
mAr 2  WB (0.08075)(1.2)(3.00)2  (2.6)

 10.70 ft/s 2
0.08075  0.08075
mA  mB
T  mB (
r  g )  (0.08075)(10.70  32.2)
T  1.736 lb 
a A / rod  10.70 ft/s 2 radially inward. 
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PROBLEM 12.93
A small ball swings in a horizontal circle at the end of a cord of length l1 ,
which forms an angle 1 with the vertical. The cord is then slowly drawn
through the support at O until the length of the free end is l2 . (a) Derive a
relation among l1 , l2 , 1 , and 2 . (b) If the ball is set in motion so that
initially l1  0.8 m and 1  35, determine the angle 2 when l2  0.6 m.
SOLUTION
(a)
For state 1 or 2, neglecting the vertical component of acceleration,
Fy  0: T cos   W  0
T  W cos 
Fx  man : T sin   W sin  cos  
But    sin 
mv 2

so that
v2 
W
m
sin 2  cos    g sin  tan 
v1  1 g sin 1 tan 1
v2   2 g sin  2 tan  2
and
M y  0: H y  constant
r1mv1  r2 mv2
or
v11 sin 1  v2 2 sin 2
3/2
3/2
1 g sin 1 sin 1 tan 1   2 sin  2 sin  2 tan  2
31 sin 3 1 tan 1  32 sin 3  2 tan  2 
(b)
With 1  35,
1  0.8 m, and  2  0.6 m
(0.8)3 sin 3 35 tan 35  (0.6)3 sin 3  2 tan  2
sin 3  2 tan  2  0.31320  0
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2  43.6 
PROBLEM 12.94
A particle of mass m is projected from Point A with an initial velocity
v0 perpendicular to OA and moves under a central force F along an
elliptic path defined by the equation r  r0 /(2  cos ). Using Eq.
(12.35), show that F is inversely proportional to the square of the
distance r from the particle to the center of force O.
SOLUTION
u
1 2  cos 

,
r
r0
d 2u
2
F
u  
2
r0 mh 2 u 2
d
Solving for F,
F
du sin 

,
d
r0
d 2 u cos 

r0
d 2
by Eq. (12.37).
2mh 2u 2 2mh 2

r0
r0 r 2
Since m, h, and r0 are constants, F is proportional to
1
r2
2
, or inversely proportional to r . 
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PROBLEM 12.95
A particle of mass m describes the logarithmic spiral r  r0 eb under a central force F directed toward the
center of force O. Using Eq. (12.35) show that F is inversely proportional to the cube of the distance r from
the particle to O.
SOLUTION
u
1 1 b
 e
r r0
du
b
  eb
d
r0
d 2u b 2 b
e

r0
d 2
d 2u
b 2  1 b
F

u

e

r0
d 2
mh 2u 2
F 

(b 2  1)mh 2u 2 b
e
r0
(b 2  1)mh 2u 2 (b 2  1)mh 2

r
r3
Since b, m, and h are constants, F is proportional to
1
r3
, or inversely proportional to r 3.
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PROBLEM 12.96
A particle of mass m describes the path defined by the equation
r  r0 (6 cos   5) under a central force F directed away from the
center of force O. Using Eq. (12.35), show that F is inversely
proportional to the square of the distance r from the particle to O.
SOLUTION
1 6 cos   5

r
r0
6sin 
du

d
r0
2
d u
6cos 

d
r0
u 
d 2u
5
F
u   
r0
d 2
mh 2u 2
F 
Solving for F,
Substitute u 
by Eq. (12.35).
5mh 2u 2
r0
1
r
F 
5mh 2

r0r 2
1
, or inversely proportional to r 2 . The minus sign
r2
indicates that the force is repulsive, as shown in Fig. P12.96.
Since m, h, and r0 are constants, F is proportional to
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PROBLEM 12.97
A particle of mass m describes the parabola y  x 2 4r0 under a central
force F directed toward the center of force C. Using Eq. (12.35) and
Eq. 12.37 ' with   1, show that F is inversely proportional to the square
of the distance r from the particle to the center of force and that the angular
momentum per unit mass h  2GMr0 .
SOLUTION
From Fig. P12.97,
But
y
x2
4r0
x  r sin ,
y  r0  r cos
or r0  r cos 

r 2 1  cos2 
r 2 sin 2 

4r0
4r0
 1  cos 2 

4r0


 2
 r   cos  r  r0  0

Solving the quadratic equation for r,
r 
2r0
1  cos 2 

2

  cos  cos 2   4  1  cos 

4r0





  r0  



2r 1  cos 
2r0
 cos  1  0
2 
1  cos 
1  cos 2 
since r  0. Simplifying gives

r 
2r0
1  cos
1 1  cos

u
r
2r0
or
du
sin 

d
2r0
and
(1)
d 2u
cos

2r0
d 2
d 2u
1
F
u 

2
2r0
d
mh 2u 2
Solving for F,
mh 2u 2
2r0
F 
Since m, h, and r0 are constants, F is proportional to
F 
mh 2

2r0r 2
1
, or inversely proportional to r 2.
r2
1 GM
 2 1   cos   u
r
h
GM
1
Comparing with (1) shows that   1 and 2 
2r0
h
By Eq. 12.37 ' ,
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h
2GMr0 
PROBLEM 12.98
It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 14.1 km/s as
it reached its minimum altitude of 303 km above the surface of the earth. Determine the eccentricity of the
trajectory of the spacecraft during this portion of its flight.
SOLUTION
For earth, R  6.37  106 m
r0  6.37  106  303.  103  6.673  106 m
h  r0v0  (6.673  106 )(14.1  103 )  94.09  109 m 2 /s
GM  gR 2  (9.81)(6.37  106 ) 2  398.06  1012 m 3 /s 2
1
GM
 2 (1   )
r0
h
1 
h2
(94.09  109 ) 2

 3.33
r0GM
(6.673  106 )(398.06  1012 )
  3.33  1
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  2.33 
PROBLEM 12.99
It was observed that during the Galileo spacecraft’s first flyby of the earth, its maximum altitude was 600 mi
above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the
maximum velocity of Galileo during its first flyby of the earth.
SOLUTION
For the earth: R  3960 mi  20.909  106 ft
GM  gR 2  (32.2) (20.909  106 ) 2  14.077  1015 ft 3 /s 2
For a parabolic trajectory,   1.
Eq. (12.39) :
At   0,
1 GM
 2 (1  cos  )
r
h
1
2GM
2GM

 2 2
2
r0
h
r0 v0
or
v0 
2GM
r0
At r0  3960  600  4560 mi  24.077  106 ft,
v0 
(2)(14.077  1015 )
 34.196  103 ft/s
24.077  106
v0  6.48 mi/s 
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PROBLEM 12.100
As a space probe approaching the planet Venus on a parabolic trajectory reaches
Point A closest to the planet, its velocity is decreased to insert it into a circular
orbit. Knowing that the mass and the radius of Venus are 4.87  1024 kg and 6052
km, respectively, determine (a) the velocity of the probe as it approaches A,
(b) the decrease in velocity required to insert it into the circular orbit.
SOLUTION
First note
(a)
rA  (6052  280) km  6332 km
From the textbook, the velocity at the point of closest approach on a parabolic trajectory is given by
v0 
2GM
r0
1/ 2
Thus,
(v A ) par
 2  66.73  1012 m3 /kg  s 2  4.87  1024 kg 


6332  103 m


 10,131.4 m/s
(v A ) par  10.13 km/s 
or
(b)
We have
(v A )circ  (v A ) par  v A
Now
(v A )circ 

Then
v A 
GM
r0
1
2
1
2
Eq. (12.44)
(v A )par
(v A ) par  (v A )par
 1


 1 (10.1314 km/s)
 2

 2.97 km/s
| vA |  2.97 km/s 
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PROBLEM 12.101
It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn,
it was at a distance of 185  103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing
that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295  103 km at a speed of 11.35 km/s,
determine the eccentricity of the trajectory of Voyager I on its approach to Saturn.
SOLUTION
For a circular orbit,
v
Eq. (12.44)
GM
r
For the orbit of Tethys,
GM  rT vT2
For Voyager’s trajectory, we have
1 GM
 2 (1   cos  )
r
h
where h  r0v0
At O,
r  r0 ,   0
Then
1
GM

(1   )
r0 (r0 v0 )2
or

r0 v02
r v2
 1  0 02  1
GM
rT vT
2

185  103 km  21.0 km/s 

 1
295  103 km  11.35 km/s 
  1.147 
or
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PROBLEM 12.102
A satellite describes an elliptic orbit about a planet of mass M.
Denoting by r0 and r1, respectively, the minimum and maximum
values of the distance r from the satellite to the center of the planet,
derive the relation
1 1 2GM
  2
r0 r1
h
where h is the angular momentum per unit mass of the satellite.
SOLUTION
Using Eq. (12.39),
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B .
rB
h
But
 B   A  180,
so that
Adding,
cos  A   cos  B .
1 1 1 1 2GM
    2 
rA rB r0 r1
h
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PROBLEM 12.103
A space probe is describing a circular orbit about a planet of radius R. The altitude of the probe above the
surface of the planet is  R and its speed is v0. To place the probe in an elliptic orbit which will bring it closer
to the planet, its speed is reduced from v0 to  v0 , where   1, by firing its engine for a short interval of
time. Determine the smallest permissible value of  if the probe is not to crash on the surface of the planet.
SOLUTION
GM
rA
For the circular orbit,
v0 
where
rA  R   R  R(1   )
Eq. (12.44),
GM  v02 R (1   )
Then
From the solution to Problem 12.102, we have for the elliptic orbit,
1
1 2GM
  2
rA rB
h
h  hA  rA (v A ) AB
Now
 [ R(1   )](  v0 )
Then
2v02 R (1   )
1
1
 
R (1   ) rB [ R (1   )  v0 ]2

2
 R(1   )
2
Now min corresponds to rB  R.
Then
1
1
2
  2
R(1   ) R  min
R(1   )
 min 
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
2

2 
PROBLEM 12.104
A satellite describes a circular orbit at an altitude of 19 110 km above
the surface of the earth. Determine (a) the increase in speed required at
point A for the satellite to achieve the escape velocity and enter a
parabolic orbit, (b) the decrease in speed required at point A for the
satellite to enter an elliptic orbit of minimum altitude 6370 km, (c) the
eccentricity  of the elliptic orbit.
SOLUTION


For earth, GM  gR 2   9.81 6.37  106  398.06  1012 m3/s 2
rA  6370  19110  25480 km  25.48  106 m
vcirc 
GM

rA
vesc 
398.06  1012
 3.9525  103 m/s
6
25.48  10
2GM

rA
2 vcirc  5.5897  103 m/s
(a) Increase in speed at A.
v  vesc  vcirc  1.637  103 m/s
v  1.637  103 m/s 
Elliptical orbit with rB  6370  6370  12740 km  12.74  106 m.
Using Eq. (12.39),
But
and
1
GM
 2  C cos B .
rB
h
 B   A  180, so that cos A   cos B
Adding,
h
1
GM
 2  C cos A
rA
h
1
1
r  rB
2GM

 A

rA rB
rArB
h2
2GMrArB

rA  rB
 2   398.06  1012  25.48  106 12.74  106 
38.22  106
 82.230  109 m 2 /s
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PROBLEM 12.104 (Continued)
vA 
h
82.230  109

 3.2272  103 m/s
6
rA
25.48  10
v  725 m/s 
(b) Decrease in speed. v  vcirc  v A  725 m/s
(c)
1
1
r  rB

 A
 C cos B  C cos A  2C
rB rA
rArB
C 
By Eq. (12.40),
rA  rB
12.74  106

 19.623  109 m 1
6
6
2rArB
 2  25.48  10 12.74  10




19.623  109 82.230  109
Ch 2

 
GM
398.06  1012


2
  0.333 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.105
A space probe is to be placed in a circular orbit of 5600 mi radius
about the planet Venus in a specified plane. As the probe reaches
A, the point of its original trajectory closest to Venus, it is inserted
in a first elliptic transfer orbit by reducing its speed by ΔvA . This
orbit brings it to Point B with a much reduced velocity. There the
probe is inserted in a second transfer orbit located in the specified
plane by changing the direction of its velocity and further reducing
its speed by ΔvB . Finally, as the probe reaches Point C, it is
inserted in the desired circular orbit by reducing its speed by ΔvC .
Knowing that the mass of Venus is 0.82 times the mass of the earth,
that rA  9.3  103 mi and rB  190  103 mi, and that the probe
approaches A on a parabolic trajectory, determine by how much the
velocity of the probe should be reduced (a) at A, (b) at B, (c) at C.
SOLUTION
R  3690 mi  20.9088  106 ft, g  32.2 ft/s 2
For Earth,
GM earth  gR 2  (32.2)(20.9088  106 )2  14.077  1015 ft 3 /s2
GM  0.82GM earth  11.543  1015 ft 3 /s 2
For Venus,
For a parabolic trajectory with
rA  9.3  103 mi  49.104  106 ft
(v A )1  vesc 
First transfer orbit AB.
2GM
(2)(11.543  1015 )

 21.683  103 ft/s
rA
49.104  106
rB  190  103 mi  1003.2  106 ft
At Point A, where   180
1 GM
GM
 2  C cos 180  2  C
rA
hAB
hAB
(1)
1 GM
GM
 2  C cos 0  2  C
rB
hAB
hAB
(2)
At Point B, where   0
Adding,
1
1 rB  rA 2GM


 2
rA rB
rA rB
hAB
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PROBLEM 12.105 (Continued)
Solving for hAB ,
hAB 
2GMrA rB
(2)(11.543  1015 )(49.104  106 )(1003.2  106 )

 1.039575  1012 ft 2 /s
6
rB  rA
1052.3  10
(v A ) 2 
hAB 1.039575  1012

 21.174  103 ft/s
rA
49.104  106
(vB )1 
hAB 1.039575  1012

 1.03626  103 ft/s
6
rB
1003.2  10
rC  5600 mi  29.568  106 ft
Second transfer orbit BC.
At Point B, where   0
1 GM
GM
 2  C cos 0  2  C
rB
hBC
hBC
At Point C, where   180
1 GM
GM
 2  C cos 180  2  C
rC
hBC
hBC
Adding,
1
1 rB  rC 2GM


 2
rB rC
rB rC
hBC
hBC 
2GMrB rC
(2)(11.543  1015 )(1003.2  106 )(29.568  106 )

 814.278  109 ft 2 /s
rB  rC
1032.768  106
( vB ) 2 
hBC 814.278  109

 811.69 ft/s
rB
1003.2  106
(vC )1 
hBC 814.278  109

 27.539  103 ft/s
rC
29.568  106
Final circular orbit.
rC  29.568  106 ft
(vC )2 
GM
11.543  1015

 19.758  103 ft/s
rC
29.568  106
Speed reductions.
(a)
At A:
(v A )1  (v A ) 2  21.683  103  21.174  103
vA  509 ft/s 
(b)
At B:
(vB )1  (vB ) 2  1.036  103  811.69
vB  224 ft/s 
(c)
At C:
(vC )1  (vC ) 2  27.539  103  19.758  103 
vC  7.78  103 ft/s 
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PROBLEM 12.106
For the space probe of Problem 12.105, it is known that rA  9.3  103 mi and that the velocity of the probe is
reduced to 20,000 ft/s as it passes through A. Determine (a) the distance from the center of Venus to Point B,
(b) the amounts by which the velocity of the probe should be reduced at B and C, respectively.
SOLUTION
Data from Problem 12.105:
rC  29.568  106 ft, M  0.82 M earth
For Earth,
R  3960 mi  20.9088  106 ft, g  32.2 ft/s 2
GM earth  gR 2  (32.2)(20.9088  106 ) 2  14.077  1015 m 3 /s 2
GM  0.82GM earth  11.543  1015 ft 3 /s 2
For Venus,
v A  20, 000 ft/s, rA  9.3  103 mi  49.104  106 ft
Transfer orbit AB:
hAB  rA v A  (49.104  106 )(20, 000)  982.08  109 ft 2 /s
At Point A, where   180
1 GM
GM
 2  C cos 180  2  C
rA
hAB
hAB
At Point B, where   0
1 GM
GM
 2  C cos 0  2  C
rB
hAB
hAB
Adding,
1
1 2GM

 2
rA rB
hAB
1 GM 1
 2 
rB
rA
hAB

(2)(11.543  1015 )
1

9 2
(982.08  10 )
49.104  106
 3.57125  109 ft 1
(a)
Radial coordinate rB .
rB  280.01  106 ft
(vB )1 
rB  53.0  103 mi 
hAB 982.08  109

 3.5073  103 ft/s
6
rB
280.01  10
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PROBLEM 12.106 (Continued)
rC  5600 mi  29.568  106 ft
Second transfer orbit BC.
At Point B, where   0
1 GM
GM
 2  C cos 0  2  C
rB
hBC
hBC
At Point C, where   180
1 GM
GM
 2  C cos 180  2  C
rC
hBC
hBC
r  rC 2GM
1
1

 B
 2
rB rC
rB rC
hBC
Adding,
hBC 
2GMrB rC

rB  rC
(2)(11.543  1015 )(280.01  106 )(29.568  106 )
309.578  106
 785.755  109 ft 2 /s
( vB ) 2 
hBC 785.755  109

 2.8062  103 ft/s
rB
280.01  106
(vC )1 
hBC 785.755  109

 26.575  103 ft/s
rC
29.568  106
rC  29.568  106 ft
Circular orbit with
(vC )2 
(b)
GM
11.543  1015

 19.758  103 ft/s
rC
29.568  106
Speed reductions at B and C.
At B:
(vB )1  (vB ) 2  3.5073  103  2.8062  103
vB  701 ft/s 

At C:
(vC )1  (vC ) 2  26.575  103  19.758  103
vC  6.82  103 ft/s 
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PROBLEM 12.107
As it describes an elliptic orbit about the sun, a spacecraft
reaches a maximum distance of 202  106 mi from the center
of the sun at Point A (called the aphelion) and a minimum
distance of 92  106 mi at Point B (called the perihelion). To
place the spacecraft in a smaller elliptic orbit with aphelion at
A and perihelion at B , where A and B are located
164.5  106 mi and 85.5  106 mi, respectively, from the center
of the sun, the speed of the spacecraft is first reduced as it
passes through A and then is further reduced as it passes
through B . Knowing that the mass of the sun is 332.8  103
times the mass of the earth, determine (a) the speed of the
spacecraft at A, (b) the amounts by which the speed of the
spacecraft should be reduced at A and B to insert it into the
desired elliptic orbit.
SOLUTION
First note
Rearth  3960 mi  20.9088  106 ft
rA  202  106 mi  1066.56  109 ft
rB  92  106 mi  485.76  109 ft
From the solution to Problem 12.102, we have for any elliptic orbit about the sun
1 1 2GM sun
 
r1 r2
h2
(a)
For the elliptic orbit AB, we have
r1  rA , r2  rB , h  hA  rAvA
Also,
GM sun  G[(332.8  103 ) M earth ]
2
 gRearth
(332.8  103 )
Then
using Eq. (12.30).
2
(332.8  103 )
1
1 2 gRearth
 
rA rB
(rA v A ) 2
1/ 2
or
R
v A  earth
rA
 665.6 g  103 


1
1



rA
rB


3960 mi  665.6  103  32.2 ft/s 2

202  106 mi  1066.561 109 ft  485.761 109 ft

 52, 431 ft/s
or
1/ 2




v A  52.4  103 ft/s 
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PROBLEM 12.107 (Continued)
(b)
From Part (a), we have
1 1
2GM sun  (rA v A ) 2   
 rA rB 
Then, for any other elliptic orbit about the sun, we have

2 1
1 1 (rA v A ) rA 
 
r1 r2
h2
1
rB

For the elliptic transfer orbit AB , we have
r1  rA , r2  rB , h  htr  rA (vA )tr
Then
or

2 1
1
1
1 (rA v A ) rA  rB


rA rB
[rA (v A ) tr ]2

(v A ) tr  v A 


1
rA

1
rB
1
rA

1
rB
1/ 2





 1  rA
rB
 vA 
 1  rA
rB

1/ 2




1/2
 1  202

92
 (52, 431 ft/s) 
 1  202 
85.5 

 51,113 ft/s
htr  (hA )tr  (hB )tr : rA (vA )tr  rB (vB )tr
Now
Then
(vB ) tr 
202  106 mi
 51,113 ft/s  120,758 ft/s
85.5  106 mi
For the elliptic orbit AB, we have
r1  rA , r2  rB , h  rBvB
Then
or

2 1
1
1 (rA v A ) rA 


rA rB
(rB vB ) 2
r 
vB   v A A 
rB 

1
rB
1
rA

1
rB
1
rA

1
rB

1/ 2




1
1
202  106 mi  202  106  92  106
 (52, 431 ft/s)
85.5  106 mi  164.51 106  85.51 106

 116,862 ft/s
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1/ 2




PROBLEM 12.107 (Continued)
Finally,
or
(vA )tr  vA  vA
vA  (51,113  52, 431) ft/s
|vA |  1318 ft/s 
or
and
or
vB  (vB )tr  vB
vB  (116,862  120, 758) ft/s
 3896 ft/s
|vB |  3900 ft/s 
or
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PROBLEM 12.108
Halley’s comet travels in an elongated elliptic orbit for which the minimum distance from the sun is
approximately 12 rE , where rE  150  106 km is the mean distance from the sun to the earth. Knowing that the
periodic time of Halley’s comet is about 76 years, determine the maximum distance from the sun reached by
the comet.
SOLUTION
We apply Kepler’s Third Law to the orbits and periodic times of earth and Halley’s comet:
2
H 
 aH 

 

 E 
 aE 
Thus
3
 
aH  a E  H 
 E 
2/3
 76 years 
 rE 

 1 year 
 17.94rE
But
2/3
1
(rmin  rmax )
2
11

17.94rE   rE  rmax 
22

aH 
1
rE
2
 (35.88  0.5)rE
rmax  2(17.94 rE ) 
 35.38 rE
rmax  (35.38)(150  106 km)
rmax  5.31  109 km 
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PROBLEM 12.109
Based on observations made during the 1996 sighting of comet Hyakutake, it was concluded that the
trajectory of the comet is a highly elongated ellipse for which the eccentricity is approximately   0.999887.
Knowing that for the 1996 sighting the minimum distance between the comet and the sun was 0.230RE ,
where RE is the mean distance from the sun to the earth, determine the periodic time of the comet.
SOLUTION
For Earth’s orbit about the sun,
v0 
2 RE 2 RE 3/ 2
GM
, 0 

RE
v0
GM
or
GM 
2 RE3/ 2
0
(1)
For the comet Hyakutake,
1 GM
 2  (1   ),
r0
h
a
1 GM
1 
r0
 2 (1   ), r1 
r1
1 
h
r
1
1 
(r0  r1 )  0 , b  r0 r1 
r0
2
1 
1 
h  GMr0 (1   )


2 r02 (1   )1/ 2
2 ab

h
(1   )3/ 2 GMr0 (1   )
2 r03/2
GM (1   )3/ 2
 r 
 0 
 RE 
3/2
1
(1   )3/ 2
 (0.230)3/ 2
Since

2 r03/2 0
2 RE3 (1   )3/ 2
0
1
 0  91.8  103 0
3/ 2
(1  0.999887)
 0  1 yr,   (91.8  103 )(1.000)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
  91.8  103 yr 
PROBLEM 12.110
A space probe is to be placed in a circular orbit of radius 4000
km about the planet Mars. As the probe reaches A, the point of
its original trajectory closest to Mars, it is inserted into a first
elliptic transfer orbit by reducing its speed. This orbit brings it
to Point B with a much reduced velocity. There the probe is
inserted into a second transfer orbit by further reducing its
speed. Knowing that the mass of Mars is 0.1074 times the
mass of the earth, that rA  9000 km and rB  180,000 km, and
that the probe approaches A on a parabolic trajectory,
determine the time needed for the space probe to travel from A
to B on its first transfer orbit.
SOLUTION
For earth, R  6373 km  6.373  106 m
GM  gR 2  (9.81) (6.373  106 )2  398.43  1012 m3 /s 2
For Mars, GM  (0.1074)(398.43  1012 )  42.792  1012 m3 /s 2
rA  9000 km  9.0  106 m
rB  180000 km  180  106 m
For the parabolic approach trajectory at A,
(v A )1 
2GM

rA
(2)(42.792  1012 )
 3.0837  103 m/s
9.0  106
First elliptic transfer orbit AB.
Using Eq. (12.39),
But
1
GM
 2  C cos A
rA
hAB
 B   A  180,
Adding,
so that
and
1
GM
 2  C cos B .
rB
hAB
cos A   cos B.
1
1
r  rB
2GM

 A
 2
rA rB
rArB
hAB
hAB 
2GMrArB

rA  rB
(2)(42.792  1012 )(9.0  106 )(180  106 )
189.0  106
hAB  27.085  109 m 2 /s
a 
b
1
1
(rA  rB )  (9.0  106  180  106 )  94.5  106 m
2
2
rArB 
(9.0  106 )(180  106 )  40.249  106 m
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PROBLEM 12.110 (Continued)
Periodic time for full ellipse:
For half ellipse AB,
 AB 
 AB 
 
2 ab
h
1
 ab
 
2
h
 (94.5  106 )(40.249  106 )
27.085  109
 444.81  103 s
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 AB  122.6 h 
PROBLEM 12.111
A spacecraft and a satellite are at diametrically opposite positions
in the same circular orbit of altitude 500 km above the earth. As it
passes through point A, the spacecraft fires its engine for a short
interval of time to increase its speed and enter an elliptic orbit.
Knowing that the spacecraft returns to A at the same time the
satellite reaches A after completing one and a half orbits,
determine (a) the increase in speed required, (b) the periodic time
for the elliptic orbit.
SOLUTION
g  9.81 m/s 2
For earth, R  6.37  106 m ,

GM  gR 2   9.81 6.37  106

2
 398.06  1012 m3/s 2
For circular orbit of satellite,
r0  6370  500  6870 km  6.87  106 m/s
v0 
0 
398.06  1012
 7.6119  103 m/s
6.87  106
GM

r0


 2  6.87  106
2 r0

 5.6708  103 s
v0
7.6119  103
For elliptic orbit of spacecraft it is given that
3
 0  8.5062  103 s
2
 
a 
Using Eq. (12.39),
But
1
GM
 2  C cos A
rA
h
 B   A  180,
Adding,
1
 rA  rB  ,
2
so that

rArB
1
GM
 2  C cos B .
rB
h
cos  A   cos  B .
1
1
r  rB
2a
2GM

 A
 2 
rA rB
rArB
b
h2
Periodic time:  
or
h
GMb 2
a
2 ab 2 ab a
2 a3 / 2


h
GM
GMb2

398.06  1012 8.5062  103
GM  2
a 

4 2
4 2
3
and
b

2
 729.558  1018 m3
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PROBLEM 12.111 (Continued)
a  9.0023  106 m,
rA  r0  6.87  106 m
rB  2a  rA  11.1346  106 m,
h
2 ab



b

rArB  8.7461  106 m
2 9.0023  106 8.7461  106
vA 
3
8.5062  10
  58.159  10
9
m 2 /s
h
58.159  109

 8.4656  103 m/s
6
rA
6.87  10
(a) Increase in speed at A.
v A  v A  v0  8.4656  103  7.6119  103
v A  854 m/s 
(b) Periodic time for elliptic orbit.
As calculated above   8.5062 s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
  141.8 min 
PROBLEM 12.112
The Clementine spacecraft described an elliptic orbit of
minimum altitude hA  400 km and a maximum altitude of
hB  2940 km above the surface of the moon. Knowing that
the radius of the moon is 1737 km and that the mass of the
moon is 0.01230 times the mass of the earth, determine the
periodic time of the spacecraft.
SOLUTION
For earth,
R  6370 km  6.370  106 m
GM  gR 2  (9.81)(6.370  106 )2  398.06  1012 m3 /s 2
For moon,
GM  (0.01230)(398.06  1012 )  4.896  1012 m3 /s 2
rA  1737  400  2137 km  2.137  106 m
rB  1737  2940  4677 km  4.677  106 m
Using Eq. (12.39),
1 GM
 2  C cos  A
rA
h
But
 B   A  180, so that cos  A   cos  B .
Adding,
and
1 GM
 2  C cos  B .
rB
h
1 1 rA  rB 2GM
 
 2
rA rB
rA rB
hAB
hAB 
2GMrA rB
(2)(4.896  1012 )(2.137  106 )(4.677  106 )

rA  rB
6.814  106
 3.78983  109 m 2 /s
1
(rA  rB )  3.402  106 m
2
b  rA rB  3.16145  106 m
a
Periodic time.

2 ab 2 (3.402  106 )(3.16145  106 )

 17.831  103 s
9
hAB
3.78983  10
  4.95 h 

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PROBLEM 12.113
Determine the time needed for the space probe of Problem 12.100 to travel from B to C.
SOLUTION
From the solution to Problem 12.100, we have
(v A ) par  10,131.4 m/s
(v A )circ 
and
1
2
(v A ) par  7164.0 m/s
rA  (6052  280) km  6332 km
Also,
For the parabolic trajectory BA, we have
1 GM v
 2 (1   cos )
r
hBA
[Eq. (12.39)]
where   1. Now
at A,   0:
1 GM v
 2 (1  1)
rA
hBA
or
rA 
at B,   90:
1 GM v
 2 (1  0)
rB
hBA
or
rB 
2
hBA
2GM v
2
hBA
GM v
rB  2rA
As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and
Point B. Thus,
(Area swept out) BA  ABA 
Now
2
4
rA rB  rA2
3
3
dA 1
 h
dt 2
where h  constant
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PROBLEM 12.113 (Continued)
Then
A
2A
1
ht or t BA  BA
2
hBA
t BA 

hBA  rA v A
2  43 rA2

rA v A
8 rA
3 vA
8 6332  103 m
3 10,131.4 m/s
 1666.63 s
For the circular trajectory AC,
t AC 
Finally,
 r
2 A
(v A )circ

 6332  103 m
2 7164.0 m/s
 1388.37 s
t BC  t BA  t AC
 (1666.63  1388.37) s
=3055.0 s
or
tBC  50 min 55 s 
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PROBLEM 12.114
A space probe is describing a circular orbit of radius nR with a velocity v0
about a planet of radius R and center O. As the probe passes through Point
A, its velocity is reduced from v0 to  v0, where   1, to place the probe on a
crash trajectory. Express in terms of n and  the angle AOB, where B denotes
the point of impact of the probe on the planet.
SOLUTION
r0  rA  nR
For the circular orbit,
v0 
GM
GM

r0
nR
The crash trajectory is elliptic.
v A   v0 
 2GM
nR
h  rA v A  nRv A   2 nGMR
GM
1
 2
2
 nR
h
1 GM
1   cos 
 2 (1   cos  ) 
r
h
 2 nR
At Point A,   180
1
1
1 

 2
rA nR  nR
or  2  1  
or   1   2
At impact Point B,     
1 1

rB R
1 1   cos (   ) 1   cos 


R
 2 nR
 2 nR
 cos   1  n 2 or cos  
1  n 2


1  n 2
1  2
  cos 1[(1  n 2 ) /(1   2 )] 
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PROBLEM 12.115
A long-range ballistic trajectory between Points A and B on the earth’s
surface consists of a portion of an ellipse with the apogee at Point C.
Knowing that Point C is 1500 km above the surface of the earth and
the range R of the trajectory is 6000 km, determine (a) the velocity
of the projectile at C, (b) the eccentricity  of the trajectory.
SOLUTION
For earth, R  6370 km  6.37  106 m
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s 2
For the trajectory, rC  6370  1500  7870 km  7.87  106 m
rA  rB  R  6.37  106 m,
rC
7870

 1.23548
rA
6370
Range A to B: s AB  6000 km  6.00  106 m
s AB
6.00  106

 0.94192 rad  53.968
R
6.37  106
 
For an elliptic trajectory,
At A,
  180 
At C,
  180 ,

2
1 GM
 2 (1   cos  )
r
h
1
GM
 2 (1   cos153.016)
rA
h
 153.016,
1
GM
 2 (1   )
rC
h
Dividing Eq. (1) by Eq. (2),
rC
1   cos153.016

 1.23548
1
rA
 
1.23548  1
 0.68384
1.23548  cos153.016
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(1)
(2)
PROBLEM 12.115 (Continued)
From Eq. (2), h 
h
GM (1   )rC
(398.06  1012 )(0.31616)(7.87  106 )  31.471  109 m 2 /s
vC 
(a) Velocity at C.
(b)
h
31.471  109

 4.00  103 m/s
6
rC
7.87  10
Eccentricity of trajectory.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
vC  4 km/s 
  0.684 
PROBLEM 12.116
A space shuttle is describing a circular orbit at an altitude of 563 km above
the surface of the earth. As it passes through Point A, it fires its engine for
a short interval of time to reduce its speed by 152 m/s and begin its descent
toward the earth. Determine the angle AOB so that the altitude of the
shuttle at Point B is 121 km. (Hint: Point A is the apogee of the elliptic
descent trajectory.)
SOLUTION
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s 2
rA  6370  563  6933 km  6.933  106 m
rB  6370  121  6491 km  6.491  106 m
For the circular orbit through Point A,
vcirc 
GM

rA
398.06  1012
 7.5773  103 m/s
6.933  106
For the descent trajectory,
v A  vcirc  v  7.5773  103  152  7.4253  103 m/s
h  rAvA  (6.933  106 )(7.4253  103 )  51.4795  109 m2 /s
1 GM
 2 (1   cos  )
r
h
At Point A,
  180,
r  rA
1
GM
 2 (1   )
rA
h
1 
h2
(51.4795  109 )2

 0.96028
GM rA
(398.06  1012 )(6.933  106 )
  0.03972
1
GM
 2 (1   cos  B )
rB
h
1   cos  B 
h2
(51.4795  109 )2

 1.02567
GM rB
(398.06  1012 )(6.491  106 )
cos B 
 B  49.7
1.02567  1

 0.6463
AOB  180   B  130.3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
AOB  130.3 
PROBLEM 12.117
As a spacecraft approaches the planet Jupiter, it releases a probe which is to
enter the planet’s atmosphere at Point B at an altitude of 280 mi above the
surface of the planet. The trajectory of the probe is a hyperbola of eccentricity
  1.031. Knowing that the radius and the mass of Jupiter are 44423 mi and
1.30  1026 slug, respectively, and that the velocity vB of the probe at B forms
an angle of 82.9° with the direction of OA, determine (a) the angle AOB,
(b) the speed vB of the probe at B.
SOLUTION
rB  (44.423  103  280) mi  44.703  103 mi
First we note
(a)
1 GM j
 2 (1   cos  )
r
h
We have
1 GM j
 2 (1   )
rA
h
At A,   0:
h2
 rA (1   )
GM j
or
At B,    B   AOB :
or
Then
or
[Eq. (12.39)]
1 GM j
 2 (1   cos  B )
rB
h
h2
 rB (1   cos  B )
GM j
rA (1   )  rB (1   cos  B )
cos  B 


1  rA
 (1   )  1
  rB


1  44.0  103 mi
(1  1.031)  1

3
1.031  44.703  10 mi

 0.96902
or
 B  14.2988
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 AOB  14.30 
PROBLEM 12.117 (Continued)
(b)
From above
where
Then
h 2  GM j rB (1   cos  B )
1
| rB  mv B |  rB vB sin 
m
  ( B  82.9)  97.1988
h
( rB vB sin  ) 2  GM j rB (1   cos  B )
or
1/ 2

1  GM j
(1   cos  B ) 
vB 

sin   rB

1/ 2

 34.4  109 ft 4 /lb  s 4  (1.30  1026 slug)

1
 [1  (1.031)(0.96902)]

6
sin 97.1988 
236.03  10 ft

vB  196.2 ft/s 
or
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PROBLEM 12.118
A satellite describes an elliptic orbit about a planet. Denoting by r0
and r1 the distances corresponding, respectively, to the perigee and
apogee of the orbit, show that the curvature of the orbit at each of
these two points can be expressed as
1 1 1 
   
 2  r0 r1 
1
SOLUTION
Using Eq. (12.39),
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B .
rB
h
But
 B   A  180,
so that
Adding,
cos  A   cos  B
1 1 2GM
  2
rA rB
h
At Points A and B, the radial direction is normal to the path.
an 
But
Fn 
1


v2


h2
r 2
GMm
mh 2
ma


n
r2
r 2
GM 1  1
1
   
2
2  rA rB 
h
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1


1 1 1 
  
2  r0 r1 
PROBLEM 12.119
(a) Express the eccentricity  of the elliptic orbit described by a
satellite about a planet in terms of the distances r0 and r1
corresponding, respectively, to the perigee and apogee of the orbit.
(b) Use the result obtained in Part a and the data given in Problem
12.109, where RE  149.6  106 km, to determine the approximate
maximum distance from the sun reached by comet Hyakutake.
SOLUTION
(a)
We have
1 GM
 2 (1   cos  )
r
h
At A,   0:
1 GM
 2 (1   )
r0
h
Eq. (12.39)
h2
 r0 (1   )
GM
or
At B,   180:
1 GM
 2 (1   )
r1
h
h2
 r1 (1   )
GM
or
r0 (1   )  r1 (1   )
Then

or
(b)
r1  r0

r1  r0
1 
r0
1 
From above,
r1 
where
r0  0.230 RE
Then
r1 
1  0.999887
 0.230(149.6  109 m)
1  0.999887
r1  609  1012 m 
or
Note: r1  4070 RE
or r1  0.064 lightyears.
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PROBLEM 12.120
Derive Kepler’s third law of planetary motion from Eqs. (12.37) and (12.43).
SOLUTION
For an ellipse,
2a  rA  rB
Using Eq. (12.37),
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B .
rB
h
But
 B   A  180,
so that
Adding,
and b  rA rB
cos  A   cos  B .
1
1 r r
2a 2GM
  A B  2  2
rA rB
rA rB
b
h
hb
GM
a
By Eq. (12.43),

2 
2 ab 2 ab a 2 a3/ 2


h
b GM
GM
4 2 a 3
GM
For Orbits 1 and 2 about the same large mass,
and
12 
4 2 a13
GM
 22 
4 2 a23
GM
2
3
 1   a1 
    
  2   a2 
Forming the ratio,
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PROBLEM 12.121
Show that the angular momentum per unit mass h of a satellite describing an elliptic orbit of semimajor axis a
and eccentricity  about a planet of mass M can be expressed as
h  GMa (1   2 )
SOLUTION
By Eq. (12.39),
1 GM
 2 (1   cos  )
r
h
At A,   0 :
1 GM
 2  (1   )
rA
h
or
rA 
h2
GM (1   )
At B,   180 :
1 GM
 2  (1   )
rB
h
or
rB 
h2
GM (1   )
h2
1
 1


GM  1   1  
Adding,
rA  rB 
But for an ellipse,
rA  rB  2a
2a 
2h 2
GM (1   2 )
2
2h


2
 GM (1   )
h  GMa (1   2 ) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.122
In the braking test of a sports car its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with
slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static
friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if
the car skids. Ignore air resistance and rolling resistance.
SOLUTION
(a)
Coefficient of static friction.
Fy  0:
N W  0
N W
v0  70 mi/h  102.667 ft/s
v 2 v02

 at (s  s0 )
2
2
at 
v 2  v02
0  (102.667)2

  31.001 ft/s2
2(s  s0 )
(2)(170)
For braking without skidding   s , so that s N  m | at |
Ft  mat :  s N  mat
s  
(b)
mat
a
31.001
  t 
W
g
32.2
s  0.963 
Stopping distance with skidding.
Use   k  (0.80)(0.963)  0.770
F  mat : k N  mat
at  
k N
m
   k g   24.801 ft/s 2
Since acceleration is constant,
(s  s0 ) 
v 2  v02
0  (102.667) 2

2at
(2)( 24.801)
s  s0  212 ft 
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PROBLEM 12.123
A bucket is attached to a rope of length L  1.2 m and is made to revolve in a
horizontal circle. Drops of water leaking from the bucket fall and strike the floor
along the perimeter of a circle of radius a. Determine the radius a when   30.
SOLUTION
Initial velocity of drop  velocity of bucket
Fy  0:
T cos30  mg
Fx  max :
Divide (2) by (1):
(1)
T sin 30  man
tan 30 
(2)
an
v2

g
pg
Thus
v 2   g tan 30
But
  L sin 30  (1.2 m) sin 30°  0.6 m
Thus
v 2  0.6(9.81) tan 30  3.398 m 2 /s 2
v  1.843 m/s
Assuming the bucket to rotate clockwise (when viewed from
above), and using the axes shown, we find that the components
of the initial velocity of the drop are
(v0 ) x  0, (v0 ) y  0, (v0 )2  1.843 m/s
Free fall of drop
y  y0  (v0 ) y t 
1 2
gt
2
y  y0 
1 2
gt
2
When drop strikes floor:
y 0
y0 
1 2
gt  0
2
But
y0  2L  L cos30  2(1.2)  1.2cos30  1.361 m
Thus
1.361 
1
(9.81) t 2  0
2
t  0.5275
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PROBLEM 12.123 (Continued)
Projection on horizontal floor (uniform motion)
x  x0  (v0 ) z t  L sin 30  0,
x  0.6 m
z  z0  (v0 ) z t  0  1.843(0.527)  0.971 m
Radius of circle: a 
a
x2  z 2
(0.6)2  (0.971)2
a  1.141 m 
Note: The drop travels in a vertical plane parallel to the yz plane.
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PROBLEM 12.124
A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A.
Neglecting friction, determine immediately after the system is released from
rest (a) the acceleration of A, (b) the acceleration of B relative to A.
SOLUTION
Acceleration vectors:
a A  aA
30, a B/A  aB/A
a B  a A  a B/ A
Block B:
Fx  max : mB aB /A  mB a A cos30  0
aB/A  a A cos30
(1)
Fy  ma y : N AB  WB  mB a A sin 30
N AB  WB  (WB sin 30)
Block A:
aA
g
(2)
F  ma : WA sin 30  N AB sin 30  WA
WA sin 30  WB sin 30  (WB sin 2 30)
aA 
(WA  WB )sin 30
2
WA  WB sin 30
g
aA
g
aA
a
 WA A
g
g
(30  12)sin 30
(32.2)  20.49 ft/s2
30  12sin 2 30
a A  20.49 ft/s 2
(a)
30 
aB/ A  (20.49) cos 30  17.75 ft/s 2
(b)
a B/A  17.75 ft/s2
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
PROBLEM 12.125
A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A
which rides on an inclined I-beam as shown. Knowing that at the instant
shown the trolley has an acceleration of 1.2 ft/s2 up and to the right,
determine (a) the acceleration of B relative to A, (b) the tension in cable CD.
SOLUTION
(a)
First we note: a B  a A  a B/A , where a B/A is directed perpendicular to cable AB.
Fx  mB ax : 0  mB ax  mB a A cos 25
B:
aB/A  (1.2 ft/s2 ) cos 25
or
a B/A  1.088 ft/s2
or
(b)

For crate B
Fy  mB a y : TAB  WB 
or
WB
a A sin 25
g
A:
 (1.2 ft/s 2 )sin 25 
TAB  (500 lb) 1 

32.2 ft/s 2


 507.87 lb
For trolley A
Fx  mA a A : TCD  TAB sin 25  WA sin 25 
or
WA
aA
g

1.2 ft/s 2
TCD  (507.87 lb)sin 25  (40 lb)  sin 25 
32.2 ft/s 2




TCD  233 lb 
or
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PROBLEM 12.126
The roller-coaster track shown is contained in a vertical plane.
The portion of track between A and B is straight and
horizontal, while the portions to the left of A and to the right
of B have radii of curvature as indicated. A car is traveling at a
speed of 72 km/h when the brakes are suddenly applied,
causing the wheels of the car to slide on the track (k  0.25).
Determine the initial deceleration of the car if the brakes are
applied as the car (a) has almost reached A, (b) is traveling
between A and B, (c) has just passed B.
SOLUTION
v  72 km/h  20 m/s
(a)
Almost reached Point A.
  30 m
an 
v2


(20)2
 13.333 m/s 2
30
Fy  ma y : N R  N F  mg  man
N R  N F  m( g  an )
F  k ( N R  N F )  k m( g  an )
Fx  max :  F  mat
at  
F
   k ( g  an )
m
| at |  k ( g  an )  0.25(9.81  13.33)
(b)
Between A and B.
| at |  5.79 m/s2 
 
an  0
| at |  k g  (0.25)(9.81)
(c)
Just passed Point B.
  45 m
an 
v2


(20)2
 8.8889 m/s 2
45
Fy  ma y : N R  N F  mg  man
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| at |  2.45 m/s 2 
PROBLEM 12.126 (Continued)
or
N R  N F  m( g  an )
F  k ( N R  N F )  k m( g  an )
Fx  max :  F  mat
at  
F
   k ( g  an )
m
| at |  k ( g  an )  (0.25)(9.81  8.8889)
| at |  0.230 m/s 2 
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PROBLEM 12.127
The parasailing system shown uses a
winch to pull the rider in towards the
boat, which is travelling with a
constant velocity. During the interval
when  is between 20º and 40º, (where t
= 0 at  = 20º) the angle increases at the
constant rate of 2 º/s. During this time,
the length of the rope is defined by the
relationship r  125 
1 3/ 2
t , where r
3
and t are expressed in meters and
seconds, respectively. At the instant
when the rope makes a 30 degree angle
with the water, the tension in the rope is
18 kN. At this instant, what is the
magnitude and direction of the force of
the parasail on the 75 kg parasailor?
SOLUTION
1 3
r  125  t 2 m
3
0  20,   30,   2 / s  0.0349 rad/s, =0
Given:
T  18 kN, m  75 kg
  0  t
Kinematics:
aP  aB  aP/ B
a P  0  ar er  a e
30  20  2t  t  5 s
Using Radial and Transverse Coordinates:
Free Body Diagram of parasailor (side view):
1 12
r   t m/s
2
1  12

r   t m/s2
4
Equations of Motion:
F
r
 mar

FP cos   T  mg sin   m 
r  r 2
FP cos   T  m  g sin   
r  r 2 
 F  ma


FP sin   mg cos   m r  2r
(1)
FP sin   m  g cos   r  2r 
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
(2)
PROBLEM 12.127 (Continued)
r  121.27 m, r  1.118 m/s, 
r  0.1118 m/s2
Evaluate at t=5 s:
(1)
2
FP cos   18000  75 9.81sin 30  0.1118  121.27  0.0349    FP cos   18348.4 N


(2)
FP sin   75 9.81cos 30  0  2  1.118  0.0349    FP sin   631.33 N
FP sin 
FP cos 
FP cos1.97  18348.4 N

631.33
 tan   0.03441    1.97
18348.4
FP  18.4 kN
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31.97 
PROBLEM 12.128
A small 200-g collar C can slide on a semicircular rod which is made to rotate about
the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value
of the coefficient of static friction between the collar and the rod if the collar is not to
slide when (a)   90, (b)   75, (c)   45. Indicate in each case the direction
of the impending motion.
SOLUTION
vC  (r sin  )AB
First note
 (0.6 m)(6 rad/s)sin 
 (3.6 m/s)sin 
(a)
With   90,
vC  3.6 m/s
Fy  0: F  WC  0
or
F  mC g
Now
F  s N
or
N
1
s
Fn  mC an : N  mC
or
or
1
s
mC g  mC
s 
mC g
vC2
r
vC2
r
gr (9.81 m/s 2 )(0.6 m)

vC2
(3.6 m/s)2
( s )min  0.454 
or
The direction of the impending motion is downward. 
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PROBLEM 12.128 (Continued)
(b) and (c)
First observe that for an arbitrary value of , it is not known whether the impending motion will be upward or
downward. To consider both possibilities for each value of , let Fdown correspond to impending motion
downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown,
we have
Fy  0: N cos   F sin   WC  0
F  s N
Now
N cos   s N sin   mC g  0
Then
or
N
mC g
cos    s sin 
and
F
s mC g
cos    s sin 
Fn  mC an : N sin   F cos   mC
vC2

  r sin 
Substituting for N and F
mC g
 s mC g
v2
sin  
cos   mC C
r sin 
cos    s sin 
cos    s sin 
s
vC2
tan 


1   s tan  1   s tan  gr sin 
or
s  
or
1
vC2
gr sin 
vC2
gr sin 
tan 
vC2
[(3.6 m/s)sin  ]2

 2.2018sin 
gr sin  (9.81 m/s 2 )(0.6 m)sin 
Now
Then
(b)
tan  
s  
tan   2.2018 sin 
1  2.2018sin  tan 
s  
tan 75  2.2018sin 75
  0.1796
1  2.2018sin 75 tan 75
  75
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.128 (Continued)
Then
downward:
s   0.1796
upward:
s  0
not possible
( s )min  0.1796 
The direction of the impending motion is downward. 
(c)
  45
s  
tan 45  2.2018sin 45
  (  0.218)
1  2.2018sin 45 tan 45
Then
downward:
s  0
upward:
s  0.218
not possible
( s )min  0.218 
The direction of the impending motion is upward. 
Note: When
or
tan   2.2018sin   0
  62.988,
s  0. Thus, for this value of  , friction is not necessary to prevent the collar from sliding on the rod.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 12.129
Telemetry technology is used to quantify kinematic values of a
200-kg roller coaster cart as it passes overhead. According to
r  2 m/s2 ,   90,
the system, r  25 m, r  10 m/s, 
  0.4 rad/s,   0.32 rad/s2. At this instant, determine
(a) the normal force between the cart and the track, (b) the
radius of curvature of the track.
SOLUTION
Find the acceleration and velocity using polar coordinates.
vr  r  10 m/s
v  r  (25 m)( 0.4 rad/s)  10 m/s
So the tangential direction is
45° and v  10 2 m/s.
ar  
r  r 2  2 m/s  (25 m)( 0.4 rad/s)2
 6 m/s 2
a  r  2r
 (25 m)(0.32) rad/s 2 | 2(10 m/s)(  0.4 rad/s)
0
So the acceleration is vertical and downward.
(a)
To find the normal force use Newton’s second law.
y-direction
N  mg sin 45  ma cos 45
N  m( g sin 45  a cos 45)
 (200 kg)(9.81) m/s 2  6 m/s 2 )(0.70711)
 538.815 N
N  539 N 
(b)
Radius l curvature of the track.
an 
 
v2

(10 2)2
v2

6cos 45
an
  47.1 m 
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PROBLEM 12.130
The radius of the orbit of a moon of a given planet is equal to twice the radius of that planet. Denoting
by  the mean density of the planet, show that the time required by the moon to complete one full revolution
about the planet is (24  /G  )1/2 , where G is the constant of gravitation.
SOLUTION
For gravitational force and a circular orbit,
Fr 
GMm mv 2

r
r2
or
v
GM
r
Let  be the periodic time to complete one orbit.
v  2 r

Solving for ,
But

GM
 2 r
r
hence,
GM  2
2 r 3/2
GM
4
M   R3  ,
3
Then

or
3  r 
G   R 

3
G  R3/2
3/2
Using r  2R as a given leads to
  23/2
3

G
24
G
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  (24 /G  )1/2 
PROBLEM 12.131
At engine burnout on a mission, a shuttle had reached Point A at an
altitude of 40 mi above the surface of the earth and had a horizontal
velocity v0. Knowing that its first orbit was elliptic and that the shuttle
was transferred to a circular orbit as it passed through Point B at an
altitude of 170 mi, determine (a) the time needed for the shuttle to travel
from A to B on its original elliptic orbit, (b) the periodic time of the
shuttle on its final circular orbit.
SOLUTION
R  3960 mi  20.909  106 ft, g  32.2 ft/s 2
For Earth,
GM  gR 2  (32.2)(20.909  106 )2  14.077  1015 ft 3 /s2
(a)
For the elliptic orbit,
rA  3960  40  4000 mi  21.12  106 ft
rB  3960  170  4130 mi  21.8064  106 ft
1
( rA  rB )  21.5032  106 ft
2
b  rA rB  21.4605  106 ft
a
Using Eq. 12.39,
1 GM
 2  C cos  A
rA
h
and
1 GM
 2  C cos  B
rB
rB
But  B   A  180, so that cos  A   cos  B
1
1 r r
2a 2GM
  A B  2  2
rA rB
rA rB
b
h
Adding,
or
h
Periodic time.


GMb 2
a
2 ab 2 ab a 2 a 3/ 2


h
GM
GMb 2
2 (21.5032  106 )3/ 2
14.077  1015
 5280.6 s  1.4668 h
The time to travel from A to B is one half the periodic time
 AB  0.7334 h
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 AB  44.0 min 
PROBLEM 12.131 (Continued)
(b)
For the circular orbit,
a  b  rB  21.8064  106 ft
 circ 
2 a 3/ 2
GM

2 (21.8064  106 )3/ 2
14.077  1015
 5393 s
 circ  1.498 h
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 circ  89.9 min 
PROBLEM 12.132
A space probe in a low earth orbit is inserted into an elliptic
transfer orbit to the planet Venus. Knowing that the mass of the
3
sun is 332.8  10 times the mass of the earth and assuming that
the probe is subjected only to the gravitational attraction of the
sun, determine the value of  , which defines the relative position
of Venus with respect to the earth at the time the probe is inserted
into the transfer orbit.
SOLUTION
First determine the time tprobe for the probe to travel from the earth to Venus:
1
tprobe   tr
2
where  tr is the periodic time of the elliptic transfer orbit. Applying Kepler’s Third Law to the orbits about
the sun of the earth and the probe, we obtain
 tr2
2
 earth
where

atr3
3
aearth
1
(rE  rv )
2
1
 (93  106  67.2  106 ) mi
2
 80.1  106 mi
atr 
and
aearth  rE
Then
tprobe 

( Note:  earth  0.0167)
1  atr 
 
2  rE 
3/ 2
 earth
1  80.1  106 mi 


2  93.0  106 mi 
3/ 2
(365.25 days)
 145.977 days
 12.6124  106 s
In time tprobe , Venus travels through the angle  v given by
v  v tprobe 
vv
tprobe
rv
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PROBLEM 12.132 (Continued)
Assuming that the orbit of Venus is circular (note:  venus  0.0068), then, for a circular orbit
vv 
Now
GM sun
rv
[Eq. (12.44)]
GM sun  G (332.8  103 M earth )

2
 332.8  103 gRearth
Then

using Eq. (12.30)
3
2
tprobe  332.8  10 gRearth

v 
rv 
rv

 tprobe Rearth
where

 
1/ 2


(332.8 g  103 )1/ 2
rv3/ 2
Rearth  3960 mi  20.9088  106 ft
and
rv  67.2  106 mi  354.816  109 ft
Then
v  (12.6124  106 s)(20.9088  106 ft)
(332.8  103  32.2 ft/s 2 )1/ 2
(354.816  109 ft)1/2
 4.0845 rad
 234.02
Finally,
   v  180  234.02  180
  54.0 
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PROBLEM 12.133*
Disk A rotates in a horizontal plane about a vertical axis at the
constant rate 0  10 rad/s. Slider B has mass 1 kg and moves in a
frictionless slot cut in the disk. The slider is attached to a spring of
constant k, which is undeformed when r  0. Knowing that the slider
is released with no radial velocity in the position r  500 mm,
determine the position of the slider and the horizontal force exerted
on it by the disk at t  0.1 s for (a) k  100 N/m, (b) k  200 N/m.
SOLUTION
First we note
r  0, xsp  0  Fsp  kr
when
r0  500 mm  0.5 m
and
  0  12 rad/s
then
  0
Fr  mB ar :  Fsp  mB (
r  r02 )
 k


r 
 02  r  0
 mB

F  mB a : FA  mB (0  2r0 )
(a)
k  100 N/m
Substituting the given values into Eq. (1)
100 N/m


r
 (10 rad/s)2  r  0
 1 kg


r 0
Then
dr
 
r  0 and at t  0, r  0 :
dt

r
0
dr 

0.1
0
(0) dt
r  0
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(1)
(2)
PROBLEM 12.133* (Continued)
dr
 r  0 and at t  0, r0  0.5 m
dt
and

r
r0
dr 

0.1
0
(0) dt
r  r0
r  0.5 m 
Note: r  0 implies that the slider remains at its initial radial position.
With r  0, Eq. (2) implies
FH  0 
(b)
k  200 N/m
Substituting the given values into Eq. (1)
 200


r
 (10 rad/s) 2  r  0
1 kg


r  100 r  0
d
(r)
dt
Now

r
Then

r  vr
so that
d dr d
d

 vr
dt dt dr
dr
dvr
dr
dvr
 100r  0
dr
vr

At t  0, vr  0, r  r0 :
r  vr
vr
0
vr d vr  100

r
r dr
r0
vr2  100(r 2  r02 )
vr  10 r02  r 2
vr 
Now
At t  0, r  r0 :

r
r0
dr
r02  r 2

dr
 10 r02  r 2
dt
t
 10 dt  10t
0
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PROBLEM 12.133* (Continued)
r  r0 sin  ,
Let
sin 1 ( r/r0 )

Then
/2
dr  r0 cos  d
r0 cos  d
r02  r02 sin 2 
sin 1 ( r/r0 )

/2
r
sin 1 
 r0
 10t
d  10t
 
   10 t
 2


r  r0 sin 10 t    r0 cos10 t  (0.5 ft) cos10 t
2

r  (5 m/s)sin10 t
Then
Finally,
at t  0.1 s:
r  (0.5 ft) cos (10  0.1)
r  0.270 m 
Eq. (2)
FH  1 kg  2  [(5 ft/s)sin (10  0.1)] (10 rad/s)
FH  84.1 N 
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PROBLEM 12.CQ1
A 1000 lb boulder B is resting on a 200 lb platform A when truck
C accelerates to the left with a constant acceleration. Which of
the following statements are true (more than one may be true)?
(a) The tension in the cord connected to the truck is 200 lb
(b) The tension in the cord connected to the truck is 1200 lb
(c) The tension in the cord connected to the truck is greater
than 1200 lb
(d ) The normal force between A and B is 1000 lb
(e) The normal force between A and B is 1200 lb
( f ) None of the above
SOLUTION
The boulder and platform are accelerating upwards; therefore it will require a force larger than their weight to
achieve this acceleration:
Answers: (c) The tension will be greater than 1200 lb and (d) the normal force will be greater than 1000 lb.
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PROBLEM 12.CQ2
Marble A is placed in a hollow tube, and the tube is swung in a horizontal
plane causing the marble to be thrown out. As viewed from the top, which
of the following choices best describes the path of the marble after leaving
the tube?
(a) 1
(b) 2
(c) 3
(d ) 4
(e) 5
SOLUTION
Answer: (d ) The particle will have velocity components along the tube and perpendicular to the tube when it
leaves. After it leaves, it will not accelerate in the horizontal plane since there are no forces in that plane;
therefore, it will travel in a straight line.
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PROBLEM 12.CQ3
The two systems shown start from rest. On
the left, two 40 lb weights are connected
by an inextensible cord, and on the right,
a constant 40 lb force pulls on the cord.
Neglecting all frictional forces, which of
the following statements is true?
(a) Blocks A and C will have the same
acceleration
(b) Block C will have a
acceleration than block A
larger
(c) Block A will have a
acceleration than block C
larger
(d ) Block A will not move
(e) None of the above
SOLUTION
Answer: (b) If you draw a FBD of B, you will see that since it is accelerating downward, the tension in the
cable will be less than 40 lb, so the acceleration of A will be less than the acceleration of C. Also, the system
on the left has more inertia, so it will require more force to have the same acceleration as the system on the
right. 
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PROBLEM 12.CQ4
The system shown is released from rest in the position shown. Neglecting
friction, the normal force between block A and the ground is
(a) less than the weight of A plus the weight of B
(b) equal to the weight of A plus the weight of B
(c) greater than the weight of A plus the weight of B
SOLUTION
Answer: (a) Since B has an acceleration component downward the normal force between A and the ground
will be less than the sum of the weights. To see this, you can draw an FBD of both block A and B. From
block A you will find that the normal force on the ground is the sum of the weight of block A plus the vertical
component of the normal force of block B on A since block A does not accelerate in the vertical direction.
From the FBD of block B and using F=ma, you will see that the vertical component of the normal force of B
on A is less than the weight of block B since it is accelerating downwards.
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PROBLEM 12.CQ5
People sit on a Ferris wheel at Points A, B, C and D. The Ferris wheel
travels at a constant angular velocity. At the instant shown, which
person experiences the largest force from his or her chair (back and
seat)? Assume you can neglect the size of the chairs, that is, the people
are located the same distance from the axis of rotation.
(a) A
(b) B
(c) C
(d ) D
(e) The force is the same for all the passengers.
SOLUTION
Answer: (c) Draw a FBD and KD at each location and it will be clear that the maximum force will be
experienced by the person at Point C. At this point the normal force isdirectly opposite the force of gravity
and the acceleration is in the same direction as the normal force making the normal force the sum of mg and
ma.
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PROBLEM 12.CQ6
A uniform crate C with mass mC is being
transported to the left by a forklift with a
constant speed v1. What is the magnitude
of the angular momentum of the crate
about Point D, that is, the upper left
corner of the crate?
(a) 0
(b) mv1a
(c) mv1b
(d ) mv1 a 2  b 2
SOLUTION
Answers: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum,
mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point D.
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PROBLEM 12.CQ7
A uniform crate C with mass mC is being
transported to the left by a forklift with
a constant speed v1. What is the
magnitude of the angular momentum of
the crate about Point A, that is, the point
of contact between the front tire of the
forklift and the ground?
(a) 0
(b) mv1d
(c) 3mv1
(d ) mv1 32 + d 2
SOLUTION
Answer: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum,
mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point A.
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PROBLEM 12.F1
Crate A is gently placed with zero initial velocity onto a moving
conveyor belt. The coefficient of kinetic friction between the crate and
the belt is k. Draw the FBD and KD for A immediately after it
contacts the belt.
SOLUTION
Answer:
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PROBLEM 12.F2
Two blocks weighing WA and WB are at rest on a conveyor that is initially at
rest. The belt is suddenly started in an upward direction so that slipping
occurs between the belt and the boxes. Assuming the coefficient of friction
between the boxes and the belt is k, draw the FBDs and KDs for blocks
A and B. How would you determine if A and B remain in contact?
SOLUTION
Answer:
Block A
Block B
To see if they remain in contact assume aA  aB and then check to see if NAB is greater than zero.
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PROBLEM 12.F3
Objects A, B, and C have masses mA, mB, and mC respectively.
The coefficient of kinetic friction between A and B is k, and
the friction between A and the ground is negligible and the
pulleys are massless and frictionless. Assuming B slides on A
draw the FBD and KD for each of the three masses A, B and C.
SOLUTION
Answer:
Block A
Block B
Block C
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PROBLEM 12.F4
Blocks A and B have masses mA and mB respectively. Neglecting friction
between all surfaces, draw the FBD and KD for each mass.
SOLUTION
Block A
Block B
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PROBLEM 12.F5
Blocks A and B have masses mA and mB respectively.
Neglecting friction between all surfaces, draw the FBD
and KD for the two systems shown.
SOLUTION
System 1
System 2
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PROBLEM 12.F6
A pilot of mass m flies a jet in a half vertical loop of radius R so that
the speed of the jet, v, remains constant. Draw a FBD and KD of the
pilot at Points A, B and C.
SOLUTION
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PROBLEM 12.F7
Wires AC and BC are attached to a sphere which revolves at a constant speed v in
the horizontal circle of radius r as shown. Draw a FBD and KD of C.
SOLUTION
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PROBLEM 12.F8
A collar of mass m is attached to a spring and slides without
friction along a circular rod in a vertical plane. The spring has
an undeformed length of 5 in. and a constant k. Knowing that
the collar has a speed v at Point C, draw the FBD and KD of
the collar at this point.
SOLUTION
where x  2/12 ft and r  5/12 ft.
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PROBLEM 12.F9
Four pins slide in four separate slots cut in a horizontal circular plate as
shown. When the plate is at rest, each pin has a velocity directed as shown
and of the same constant magnitude u. Each pin has a mass m and
maintains the same velocity relative to the plate when the plate rotates
about O with a constant counterclockwise angular velocity . Draw the
FBDs and KDs to determine the forces on pins P1 and P2.
SOLUTION
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PROBLEM 12.F10
At the instant shown, the length of the boom AB is being decreased at the
constant rate of 0.2 m/s, and the boom is being lowered at the constant rate of
0.08 rad/s. If the mass of the men and lift connected to the boom at Point B is m,
draw the FBD and KD that could be used to determine the horizontal and
vertical forces at B.
SOLUTION
r  0,   0.08 rad/s,   0
Where r  6 m, r  0.2 m/s, 
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PROBLEM 12.F11
Disk A rotates in a horizontal plane about a vertical axis at the constant
rate 0 . Slider B has a mass m and moves in a frictionless slot cut in
the disk. The slider is attached to a spring of constant k, which is
undeformed when r  0. Knowing that the slider is released with no
radial velocity in the position r  r0, draw a FBD and KD at an arbitrary
distance r from O.
SOLUTION
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PROBLEM 12.F12
Pin B has a mass m and slides along the slot in the rotating arm
OC and along the slot DE which is cut in a fixed horizontal
plate. Neglecting friction and knowing that rod OC rotates at
the constant rate 0 , draw a FBD and KD that can be used to
determine the forces P and Q exerted on pin B by rod OC and
the wall of slot DE, respectively.
SOLUTION
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CHAPTER 13
PROBLEM 13.1
A 400-kg satellite is placed in a circular orbit 6394 km above the surface of the earth. At this elevation the
acceleration of gravity is 4.09 m/s 2. Knowing that its orbital speed is 20 000 km/h, determine the kinetic
energy of the satellite.
SOLUTION
m  400 kg
Given: Mass of satellite,
v  20.0  103 km/h
Speed of satellite,
Find: Kinetic energy, T


 1h 
v  20.0  103 km/h 
 1000 m/km 
 3600 s 
 5555 m/s
T 

1 2 1
mv   400 kg  5.555  103 m/s
2
2

2
T  6.17  109 N  m
Note: Acceleration of gravity has no effect on the mass of the satellite.
T  6.17 GJ 
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PROBLEM 13.2
A 1-lb stone is dropped down the “bottomless pit” at
Carlsbad Caverns and strikes the ground with a speed
of 95 ft/s. Neglecting air resistance, determine (a) the
kinetic energy of the stone as it strikes the ground and
the height h from which it was dropped, (b) Solve Part
a assuming that the same stone is dropped down a hole
on the moon. (Acceleration of gravity on the moon 
5.31 ft/s2.)
SOLUTION
W lb
1 lb

 0.031056 lb  s 2 /ft
2
g
32.2 ft/s
Mass of stone:
m
Initial kinetic energy:
T1  0
(a)
(rest)
Kinetic energy at ground strike:
T2 
1 2 1
mv2  (0.031056)(95)2  140.14 ft  lb
2
2
T2  140.1 ft  lb 
Use work and energy:
where
T1  U12  T2
U12  wh  mgh
0  mgh 
h
(b)
On the moon:
1 2
mv2
2
v22
(95)2

2 g (2)(32.2)
h  140.1 ft 
g  5.31 ft/s2
T2  140.1 ft  lb 
T1 and T2 will be the same, hence
h
v22
(95) 2

2 g (2)(5.31)
h  850 ft 
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PROBLEM 13.3
A baseball player hits a 5.1-oz baseball with an initial velocity of 140 ft/s at
an angle of 40° with the horizontal as shown. Determine (a) the kinetic
energy of the ball immediately after it is hit, (b) the kinetic energy of the ball
when it reaches its maximum height, (c) the maximum height above the
ground reached by the ball.
SOLUTION
 1 lb 
W  (5.1 oz) 
  0.31875 lb
 16 oz 
Mass of baseball:
m
(a)
W 0.31875 lb

 0.009899 lb  s 2 /ft
g
32.2 ft/s 2
Kinetic energy immediately after hit.
v  v0  140 ft/s
T1 
(b)
1 2 1
mv  (0.009899)(140)2
2
2
T1  97.0 ft  lb 
Kinetic energy at maximum height:
v  v0 cos 40  140cos 40  107.246 ft/s
T2 
Principle of work and energy:
1 2 1
mv  (0.009899)(107.246)2
2
2
T2  56.9 ft  lb 
T1  U12  T2
U12  T2  T1  40.082 ft  lb
Work of weight:
U12  Wd
Maximum height above impact point.
d
(c)
T2  T1 40.082 ft  lb

 125.7 ft
W
0.31875 lb
125.7 ft 
Maximum height above ground:
h  125.7 ft  2 ft
h  127.7 ft 
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PROBLEM 13.4
A 500-kg communications satellite is in a circular geosynchronous orbit and completes one revolution about
the earth in 23 h and 56 min at an altitude of 35800 km above the surface of the earth. Knowing that the radius
of the earth is 6370 km, determine the kinetic energy of the satellite.
SOLUTION
Radius of earth:
R  6370 km
Radius of orbit:
r  R  h  6370  35800  42170 km  42.170  106 m
Time one revolution:
t  23 h  56 min
t  (23 h)(3600 s/h)  (56 min)(60 s/min)  86.160  103 s
Speed:
v
2 r 2 (42.170  106 )

 3075.2 m/s
t
86.160  103
Kinetic energy:
T
1 2
mv
2
T
1
(500 kg)(3075.2 m/s)2  2.3643  109 J
2
T  2.36 GJ 
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PROBLEM 13.5
In an ore-mixing operation, a bucket full of ore is suspended
from a traveling crane which moves along a stationary bridge.
The bucket is to swing no more than 10 ft horizontally when
the crane is brought to a sudden stop. Determine the maximum
allowable speed v of the crane.
SOLUTION
Let position  be the position with bucket B directly below A, and position  be that of maximum swing
where d  10 ft. Let L be the length AB.
T1 
Kinetic energies:
1 2
mv , T2  0
2
U12  Wh  mgh
Work of the weight:
where h is the vertical projection of position  above position 
From geometry (see figure),
y  L2  d 2
hL y
 L  L2  d 2
 30  (30)2  (10) 2
 1.7157 ft
Principle of work and energy:
T1  U12  T2
1 2
mv  mgh  0
2
v 2  2 gh  (2)(32.2 ft/s2 )(1.7157 ft)  110.49 ft 2 /s2
v  10.51 ft/s 
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PROBLEM 13.6
In an ore-mixing operation, a bucket full of ore is suspended
from a traveling crane which moves along a stationary bridge.
The crane is traveling at a speed of 10 ft/s when it is brought to
a sudden stop. Determine the maximum horizontal distance
through which the bucket will swing.
SOLUTION
Let position  be the position with bucket B directly below A, and position  be that of maximum swing
where the horizontal distance is d. Let L be the length AB.
T1 
Kinetic energies:
1 2
mv , T2  0
2
U12  Wh  mgh
Work of the weight:
where h is the vertical projection of position  above position .
Principle of work and energy:
T1  U12  T2
1 2
mv  mgh  0
2
v2
(10 ft/s) 2
h

 1.5528 ft
2 g (2)(32.2 ft/s 2 )
From geometry (see figure),
d  L2  y 2
 L2  ( L  h)2
 (30)2  (30  1.5528)2
 9.53 ft
d  9.53 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.7
Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting
from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75,
and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels.
Assume (a) front-wheel drive, (b) rear-wheel drive.
SOLUTION
Let W be the weight and m the mass.
W  mg
(a)
N  0.60W  0.60 mg
s  0.75
Front wheel drive:
Maximum friction force without slipping:
F   s N  (0.75)(0.60W )  0.45mg
U12  Fd  0.45 mgd
T1  0,
Principle of work and energy:
T2 
1 2
mv2
2
T1  U12  T2
1 2
mv2
2
v22  (2)(0.45 gd )  (2)(0.45)(9.81 m/s2 )(110 m)  971.19 m 2 /s 2
0  0.45mgd 
v2  31.164 m/s
(b)
v2  112.2 km/h 
N  0.40W  0.40mg
s  0.75
Rear wheel drive:
Maximum friction force without slipping:
F   s N  (0.75)(0.40W )  0.30mg
U12  Fd  0.30mgd
T1  0,
Principle of work and energy:
T2 
1 2
mv2
2
T1  U12  T2
1
mv22
2
v22  (2)(0.30) gd  (2)(0.30)(9.81 m/s 2 )(110 m)  647.46 m 2 /s 2
0  0.30 mgd 
v2  25.445 m/s
v2  91.6 km/h 
Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for
static and dynamic conditions. Compare with sample Problem 16.1 where the vehicle is treated as a rigid
body.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.8
A 2000-kg automobile starts from rest at point A on a 6o
incline and coasts through a distance of 150 m to point B. The
brakes are then applied, causing the automobile to come to a
stop at point C, 20 m from B. Knowing that slipping is
impending during the braking period and neglecting air
resistance and rolling resistance, determine (a) the speed of the
automobile at point B, (b) the coefficient of static friction
between the tires and the road.
SOLUTION
Given: Automobile Weight W = mg = (2000 kg) (9.81)
W  19, 620 N
Initial Velocity A,
v A  0 m/s
Incline Angle,
  6
Vehicle brakes at impending slip for 20 m from B to C
vC  0
Find; speed of automobile at point B, vB
Coefficient of static friction, 
(a)
U A  B  WhA  B  (19620 N) (150 m)sin 6
 307.63  103 N  m
U A  B  TB  TA 
307.63  103 N  m 
1 2
mv  0
2
1
(2000 kg) vB2  0
2
vB  17.54 m/s 
(b)
U A C  WhA C  Fd B C  TC  TA  0
d B C  20 m
F  N
Where   coefficient of static friction
U AC  (19620 N)(sin 6) (170 m)  F (20 m)
F   (19620 N) cos 6
(19620 N) (sin 6) (170 m)   (19620 N) (cos 6) (20 m)  0

170
tan 6 0.893
20
  0.893 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.9
A package is projected up a 15 incline at A with an initial
velocity of 8 m/s. Knowing that the coefficient of kinetic
friction between the package and the incline is 0.12, determine
(a) the maximum distance d that the package will move up the
incline, (b) the velocity of the package as it returns to its
original position.
SOLUTION
(a)
Up the plane from A to B:
1 2 1W
W
mv A 
(8 m/s)2  32
TB  0
2
2 g
g
F  k N  0.12 N
 (W sin15  F )d
TA 
U AB
F  0 N  W cos15  0 N  W cos15
U AB  W (sin15  0.12cos15)d  Wd (0.3747)
TA  U AB  TB : 32
W
 Wd (0.3743)  0
g
d
(b)
32
(9.81)(0.3747)
d  8.70 m 
Down the plane from B to A: (F reverses direction)
1W 2
vA
TB  0
2 g
 (W sin15  F )d
TA 
U BA
d  8.71 m/s
 W (sin15  0.12 cos15)(8.70 m/s)
U BA  1.245W
TB  U BA  TA
0  1.245W 
1W 2
vA
2 g
v A2  (2)(9.81)(1.245)
 24.43
v A  4.94 m/s
vA  4.94 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
15 
PROBLEM 13.10
A 1.4 kg model rocket is launched vertically from rest with a constant thrust of 25 N until the rocket reaches
an altitude of 15 m and the thrust ends. Neglecting air resistance, determine (a) the speed of the rocket when
the thrust ends, (b) the maximum height reached by the rocket, (c) the speed of the rocket when it returns to
the ground.
SOLUTION
Weight:
W  mg  (1.4)(9.81)  13.734 N
(a)
T1  0
First stage:
U12  (25  13.734)(15)  169.0 N  m
T1  U12  T2
1
T2  mv 2  U12  169.0 N  m
2
2U1 2
(2) (169.0)

m
1.4
v2 
(b)
v2  15.54 m/s 
Unpowered flight to maximum height h:
T2  169.0 N  m
T3  0
U 23  W (h  15)
T2  U 23  T3
W (h  15)  T2
h  15 
(c)
T2 169.0


W 13.734
h  27.3 m 
Falling from maximum height:
T3  0
T4 
1 2
mv4
2
U 34  Wh  mgh
T3  U 34  T4 :
0  mgh 
1 2
mv4
2
v42  2 gh  (2)(9.81 m/s2 )(27.3 m)  535.6 m2 /s2
v4  23.1 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.11
Packages are thrown down an incline at A with a velocity of 1
m/s. The packages slide along the surface ABC to a conveyor
belt which moves with a velocity of 2 m/s. Knowing that
k  0.25 between the packages and the surface ABC,
determine the distance d if the packages are to arrive at C with
a velocity of 2 m/s.
SOLUTION
N AB  mg cos 30
On incline AB:
FAB  k N AB  0.25 mg cos 30
U A B  mgd sin 30  FAB d
 mgd (sin 30  k cos 30)
N BC  mg
On level surface BC:
xBC  7 m
FBC   k mg
U B C    k mg xBC
At A,
TA 
1 2
mvA
2
and vA  1 m/s
At C,
TC 
1 2
mvC
2
and vC  2 m/s
Assume that no energy is lost at the corner B.
TA  U A B  U BC  TC
Work and energy.
1 2
1
mvA  mgd (sin 30  k cos30)  k mg xBC  mv02
2
2
Dividing by m and solving for d,
vC2 /2 g   k xBC  v A2 /2 g 

d
(sin 30   k cos 30)

(2) 2/(2)(9.81)  (0.25)(7)  (1) 2/(2)(9.81)
sin 30  0.25cos 30
d  6.71 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.12
Packages are thrown down an incline at A with a velocity of
1 m/s. The packages slide along the surface ABC to a conveyor
belt which moves with a velocity of 2 m/s. Knowing that
d  7.5 m and k  0.25 between the packages and all
surfaces, determine (a) the speed of the package at C, (b) the
distance a package will slide on the conveyor belt before it
comes to rest relative to the belt.
SOLUTION
(a)
On incline AB:
N AB  mg cos 30
FAB  k N AB  0.25 mg cos 30
UA B  mgd sin 30  FAB d
 mgd (sin 30  k cos 30)
On level surface BC:
N BC  mg
xBC  7 m
FBC   k mg
U B C    k mg xBC
At A,
TA 
1 2
mvA
2
and vA  1 m/s
At C,
TC 
1 2
mvC
2
and vC  2 m/s
Assume that no energy is lost at the corner B.
Work and energy.
TA  U A B  U BC  TC
1 2
1
mvA  mgd (sin 30  k cos30)  k mg xBC  mv02
2
2
Solving for vC2 ,
vC2  vA2  2 gd (sin 30  k cos30)  2 k g xBC
 (1)2  (2)(9.81)(7.5)(sin 30  0.25cos30)  (2)(0.25)(9.81)(7)
 8.3811 m2/s2
(b)
vC  2.90 m/s 
Box on belt: Let xbelt be the distance moves by a package as it slides on the belt.
Fy  ma y
N  mg  0
N  mg
Fx   k N   k mg
At the end of sliding,
v  vbelt  2 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.12 (Continued)
Principle of work and energy:
1 2
1 2
mvC   k mg xbelt  mvbelt
2
2
2
v 2  vbelt
xbelt  C
2 k g

8.3811  (2) 2
(2)(0.25)(9.81)
xbelt  0.893 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.13
Boxes are transported by a conveyor belt with a velocity v0 to
a fixed incline at A where they slide and eventually fall off at
B. Knowing that k  0.40, determine the velocity of the
conveyor belt if the boxes leave the incline at B with a velocity
of 8 ft/s.
SOLUTION
Forces when box is on AB.
Fy  0: N  W cos15  0
N  W cos15
Box is sliding on AB.
F f  k N  kW cos15
Distance
AB  d  20 ft
Work of gravity force:
(U A B ) g  Wd sin15
 Ff d   kWd cos15
Work of friction force:
Total work
U A B  Wd (sin15  k sin15)
Kinetic energy:
TA 
1W 2
v0
2 g
1W 2
TB 
vB
2 g
Principle of work and energy:
TA  U A B  TB
1W 2
1W 2
v0  Wd (sin15  k cos15) 
vB
2 g
2 g
v02  vB2  2 gd (sin15  k cos15)
 (8) 2  (2)(32.2)(20)[sin15  (0.40)(cos15)]
 228.29 ft 2 /s 2
v0  15.11 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
15 
PROBLEM 13.14
Boxes are transported by a conveyor belt with a velocity v0 to
a fixed incline at A where they slide and eventually fall off at
B. Knowing that k  0.40, determine the velocity of the
conveyor belt if the boxes are to have zero velocity at B.
SOLUTION
Forces when box is on AB.
Fy  0: N  W cos15  0
N  W cos15
Box is sliding on AB.
F f  k N  kW cos15
Distance
AB  d  20 ft
Work of gravity force:
(U A B ) g  Wd sin15
 Ff d   kWd cos15
Work of friction force:
Total work
U A B  Wd (sin15  k cos15)
Kinetic energy:
TA 
1W 2
v0
2 g
1W 2
TB 
vB
2 g
Principle of work and energy:
TA  U A B  TB
1W 2
1W 2
v0  Wd (sin15  k cos15) 
vB
2 g
2 g
v02  vB2  2 gd (sin15   k cos15)
 0  (2)(32.2)(20)[sin15  (0.40)(cos15)]
 164.29 ft 2 /s 2
v0  12.81 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
15 
PROBLEM 13.15
A 1200-kg trailer is hitched to a 1400-kg car. The car and trailer are
traveling at 72 km/h when the driver applies the brakes on both the car
and the trailer. Knowing that the braking forces exerted on the car and
the trailer are 5000 N and 4000 N, respectively, determine (a) the
distance traveled by the car and trailer before they come to a stop,
(b) the horizontal component of the force exerted by the trailer hitch
on the car.
SOLUTION
Let position 1 be the initial state at velocity v1  72 km/h  20 m/s and position 2 be at the end of braking
(v2  0). The braking forces and FC  5000 N for the car and 4000 N for the trailer.
(a)
Car and trailer system.
(d  braking distance)
1
(mC  mT )v12
2
 ( FC  FT )d
T1 
U1 2
T2  0
T1  U1 2  T2
1
(mC  mT )v12  ( FC  FT )d  0
2
d
(b)
(mC  mT )v12 (2600)(20)2

 57.778
2( FC  FT )
(2)(9000)
d  57.8 m 
Car considered separately.
Let H be the horizontal pushing force that the trailer exerts on the car through the hitch.
1
mC v12
2
 ( H  FC )d
T1 
U1 2
T2  0
T1  U1 2  T2
1
mC v12  ( H  FC )d  0
2
H  FC 
Trailer hitch force on car:
mC v12
(1400)(20) 2
 5000 
2d
(2)(57.778)
H  154 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.16
A trailer truck enters a 2 percent uphill grade traveling at 72 km/h and reaches a speed of 108 km/h in 300 m.
The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average force at the wheels of the
cab, (b) the average force in the coupling between the cab and the trailer.
SOLUTION
Initial speed:
v1  72 km/h  20 m/s
Final speed:
v2  108 km/h  30 m/s
Vertical rise:
h  (0.02)(300)  6.00 m
Distance traveled:
d  300 m
(a)
Traction force. Use cab and trailer as a free body.
m  1800  5400  7200 kg
T1  U12  T2
Work and energy:
Ft 
W  mg  (7200)(9.81)  70.632  103 N
1 2
1
mv1  Wh  Ft d  mv22
2
2
1 1 2
1 1

1

mv1  Wh  mv12  
(7200)(30)2  (70.632  103 )(6.00)  (7200)(20)2 


d 2
2

 300  2
 7.4126  103 N
(b)
Ft  7.41 kN 
Coupling force Fc . Use the trailer alone as a free body.
W  mg  (5400)(9.81)  52.974  103 N
m  5400 kg
Assume that the tangential force at the trailer wheels is zero.
Work and energy:
T1  U12  T2
1 2
1
mv1  Wh  Fc d  mv22
2
2
The plus sign before Fc means that we have assumed that the coupling is in tension.
Fc 
1 1 2
1
1 1
1


mv2  Wh  mv12  
(5400)(30) 2  (52.974  103 )(6.00)  (5400)(20)2 

d  2
2
300
2
2


 5.5595  103 N
Fc  5.56 kN (tension) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.17
The subway train shown is traveling at a speed of 30 mi/h when
the brakes are fully applied on the wheels of cars B and C,
causing them to slide on the track, but are not applied on the
wheels of car A. Knowing that the coefficient of kinetic friction
is 0.35 between the wheels and the track, determine (a) the
distance required to bring the train to a stop, (b) the force in
each coupling.
SOLUTION
k  0.35 FB  (0.35)(100 kips)  35 kips
FC  (0.35)(80 kips)  28 kips
v1  30 mi/h  44 ft/s
(a)
Entire train:
v2  0 T2  0
T1  U1 2  T2
1 (80 kips  100 kips  80 kips)
(44 ft/s) 2  (28 kips 35 kips) x  0
2
32.2 ft/s 2
x  124.07 ft 


(b)
Force in each coupling: Recall that x  124.07 ft
x  124.1 ft 
Car A: Assume FAB to be in tension
T1  V1 2  T2
1 80 kips
(44) 2  FAB (124.07 ft)  0
2 32.2
FAB  19.38 kips
FAB  19.38 kips (tension) 
Car C:
T1  U1 2  T2
1 80 kips
(44) 2  ( FBC  28 kips)(124.07 ft)  0
2 32.2
FBC  28 kips  19.38 kips
FBC  8.62 kips
FBC  8.62 kips (tension) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.18
The subway train shown is traveling at a speed of 30 mi/h when
the brakes are fully applied on the wheels of cars A, causing it
to slide on the track, but are not applied on the wheels of cars A
or B. Knowing that the coefficient of kinetic friction is 0.35
between the wheels and the track, determine (a) the distance
required to bring the train to a stop, (b) the force in each
coupling.
SOLUTION
(a)
Entire train:
FA   N A  (0.35)(80 kips)  28 kips
v2  0 T2  0
v1  30 mi/h  44 ft/s 
T1  V1 2  T2
1 (80 kips  100 kips  80 kips)
(44 ft/s)2  (28 kips) x  0
2
32.2 ft/s 2
x  279.1 ft
(b)
x  279 ft 
Force in each coupling:
Car A: Assume FAB to be in tension
T1  V1 2  T2
1 80 kips
(44 ft/s)2  (28 kips  FAB )(279.1 ft)  0
2 32.2 ft/s 2
28 kips  FAB  8.62 kips
FAB  19.38 kips
Car C:
FAB  19.38 kips (compression) 
T1  V1 2  T2
1 80 kips
(44 ft/s) 2  FBC (279.1 ft)  0
2 32.2 ft/s 2
FBC  8.617 kips
FBC  8.62 kips (compression) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.19
Blocks A and B weigh 25 lbs and 10 lbs, respectively, and they are both at a
height 6 ft above the ground when the system is released from rest. Just before
hitting the ground block A is moving at a speed of 9 ft/s. Determine (a) the
amount of energy dissipated in friction by the pulley, (b) the tension in each
portion of the cord during the motion.
SOLUTION
By constraint of the cable block B moves up a distance h when block A moves down a distance h. (h  6 ft)
Their speeds are equal.
Let FA and FB be the tensions on the A and B sides, respectively, of the pulley.
Masses:
WA
25

 0.7764 lb  s 2 /ft
32.2
g
WB
10
MB 

 0.31056 lb  s 2 /ft
32.2
g
MA 
Let position 1 be the initial position with both blocks a distance h above the ground and position 2 be just
before block A hits the ground.
Kinetic energies:
(T1 ) A  0,
(T1 ) B  0
1
1
m A v 2  (0.7764)(9) 2  31.444 ft  lb
2
2
1
1
2
(T2 ) B  mB v  (0.31056)(9)2  12.578 ft  lb
2
2
(T2 ) A 
Principle of work and energy:
T1  U12  T2
Block A:
U12  (WA  FA )h
0  (25  FA )(6)  31.444
Block B:
FA  19.759 lb
U12  ( FB  WB )h
0  ( FB  10)(6)  12.578
FB  12.096 lb
At the pulley FA moves a distance h down, and FB moves a distance h up. The work done is
U12  ( FA  FB )h  (19.759  12.096)(6)  46.0 ft  lb
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PROBLEM 13.19 (Continued)
Since the pulley is assumed to be massless, it cannot acquire kinetic energy; hence,
(a)
Energy dissipated by the pulley:
(b)
Tension in each portion of the cord:
E p  46.0 ft  lb 
A :19.76 lb 
B :12.10 lb 
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PROBLEM 13.20
The system shown is at rest when a constant 30 lb force is
applied to collar B. (a) If the force acts through the entire
motion, determine the speed of collar B as it strikes the
support at C. (b) After what distance d should the 30 lb
force be removed if the collar is to reach support C with
zero velocity?
SOLUTION
Let F be the cable tension and vB be the velocity of collar B when it strikes the support. Consider the collar B.
Its movement is horizontal so only horizontal forces acting on B do work. Let d be the distance through which
the 30 lb applied force moves.
(T1 ) B  (U1 2 ) B  (T2 ) B
0  30d  (2 F )(2) 
1 18 2
vB
2 32.2
30d  4F  0.27950vB2
(1)
Now consider the weight A. When the collar moves 2 ft to the left, the weight moves 4 ft up, since the cable
length is constant. Also, v A  2vB .
(T1 ) A  (U1 2 ) A  (T2 ) B
1 WA 2
vA
2 g
1 6
4 F  (6)(4) 
(2vB )2
2 32.2
0  ( F  WA )(4) 
4 F  24  0.37267 vB2
(2)
30d  24  0.65217vB2
(3)
Add Eqs. (1) and (2) to eliminate F.
(a)
Case a:
d  2 ft, vB  ?
(30)(2)  (24)  0.65217vB2
vB2  55.2 ft 2 /s2
(b)
Case b:
vB  7.43 ft/s 
d  ?, vB  0.
30d  24  0
d  0.800 ft 
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PROBLEM 13.21
Car B is towing car A at a constant speed of 10 m/s on an uphill
grade when the brakes of car A are fully applied causing all four
wheels to skid. The driver of car B does not change the throttle
setting or change gears. The masses of the cars A and B are
1400 kg and 1200 kg, respectively, and the coefficient of
kinetic friction is 0.8. Neglecting air resistance and rolling
resistance, determine (a) the distance traveled by the cars
before they come to a stop, (b) the tension in the cable.
SOLUTION
Given: Car B tows car A at 10 m/s uphill.
k  0.8
Car A brakes so 4 wheels skid.
Car B continues in same gear and throttle setting.
Find: (a) Distance d, traveled to stop
(b) Tension in cable
(a)
F1  traction force (from equilibrium)
F1  (1400 g )sin 5  (1200 g )sin 5
F  0.8N A
 2600(9.81)sin 5
For system: A  B
U1 2  [( F1  1400g sin 5  1200 g sin 5)  F ]d
 T2  T1  0 
Since
1
1
mA  Bv 2   (2600)(10)2
2
2
( F1  1400g sin 5  1200 g sin 5)  0
 Fd  0.8[1400(9.81) cos 5]d  130, 000 N  m
d  11.88 m 
(b)
Cable tension, T
U1 2  [T  0.8N A ](11.88)  T2  T1
(T  0.8(1400)(9.81)cos5)11.88  
1400
(10)2
2
(T  10945)  5892
 5.053 kN
T  5.05 kN 
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PROBLEM 13.22
The system shown is at rest when a constant 250-N force is applied to
block A. Neglecting the masses of the pulleys and the effect of friction
in the pulleys and between block A and the horizontal surface,
determine (a) the velocity of block B after block A has moved 2 m, (b)
the tension in the cable.
SOLUTION
Constraint of cable:
x A  3 yB  constant
x A  3yB  0
v A  3vB  0
Let F be the tension in the cable.
Block A:
mA  30 kg, P  250 N, (T1 ) A  0
(T1 ) A  (U12 ) A  (T2 ) A
1
mA v A2
2
1
0  (250  F )(2)  (30)(3vB )2
2
0  ( P  F )(x A ) 
500  2 F  135vB2
Block B:
(1)
mB  25 kg, WB  mB g  245.25 N
(T1 ) B  (U1 2 ) B  (T2 ) B
1
mB vB2
2
2 1
(3F )  245.25)    (25) vB2
3 2
0  (3F  WB )(yB ) 
2 F  163.5  12.5 vB2
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(2)
PROBLEM 13.22 (Continued)
Add Eqs. (1) and (2) to eliminate F.
500  163.5  147.5vB2
vB2  2.2814 m 2 /s 2
(a)
Velocity of B.
(b)
Tension in the cable.
From Eq. (2),
v B  1.510 m/s
2 F  163.5  (12.5)(2.2814)

F  96.0 N 
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PROBLEM 13.23
The system shown is at rest when a constant 250-N force is applied to
block A. Neglecting the masses of the pulleys and the effect of friction
in the pulleys and assuming that the coefficients of friction between
block A and the horizontal surface are  s  0.25 and k  0.20,
determine (a) the velocity of block B after block A has moved 2 m, (b)
the tension in the cable.
SOLUTION
Check the equilibrium position to see if the blocks move. Let F be the tension in the cable.
Block B:
3 F  mB g  0
F
Block A:
mB g (25)(9.81)

 81.75 N
3
3
Fy  0:
N A  mA g  0
N A  mA g  (30)(9.81)  294.3 N
Fx  0: 250  FA  F  0
FA  250  81.75  168.25 N
s N A  (0.25)(294.3)  73.57 N
Available static friction force:
Since FA   s N A , the blocks move.
The friction force, FA, during sliding is
FA  k N A  (0.20)(294.3)  58.86 N
Constraint of cable:
x A  3 yB  constant
 x A  3yB  0
v A  3vB  0
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PROBLEM 13.23 (Continued)
m A  30 kg, P  250 N, (T1 ) A  0.
Block A:
(T1 ) A  (U12 ) A  (T2 ) A
1
mA v A2
2
1
0  (250  58.86  F )(2)  (30)(3vB ) 2
2
0  ( P  FA  F )(x A ) 
382.28  2 F  135vB2
(1)
M B  25 kg, WB  mB g  245.25 N
Block B:
(T1 ) B  (U12 ) B  (T2 ) B
1
mB vB2
2
2 1
(3F  245.25)    (25)vB2
3 2
0  (3F  WB )(yB ) 
2 F  163.5  12.5vB2
(2)
Add Eqs. (1) and (2) to eliminate F.
382.28  163.5  147.5vB2
vB2  1.48325 m 2 /s 2
v B  1.218 m/s
(a)
Velocity of B:
(b)
Tension in the cable:
From Eq. (2),
2 F  163.5  (12.5)(1.48325)

F  91.0 N 
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PROBLEM 13.24
Two blocks A and B, of mass 4 kg and 5 kg, respectively, are connected
by a cord which passes over pulleys as shown. A 3 kg collar C is placed
on block A and the system is released from rest. After the blocks have
moved 0.9 m, collar C is removed and blocks A and B continue to move.
Determine the speed of block A just before it strikes the ground.
SOLUTION
Position  to Position .
v1  0
T1  0
At  before C is removed from the system
1
1
(m A  mB  mC )v22  (12 kg)v22  6v22
2
2
 (m A  mC  mB ) g (0.9 m)
T2 
U1 2
U1 2  (4  3  5)( g )(0.9 m)  (2 kg)(9.81 m/s 2 )(0.9 m)
U1 2  17.658 J
T1  U1 2  T2 :
0  17.658  6v22
v22  2.943
At Position , collar C is removed from the system.
Position  to Position .
T2 
1
9 
(mA  mB )v22   kg  (2.943)  13.244 J
2
2 
1
9
T3  (mA  mB )(v3 )2  v32
2
2
U 23  (m A  mB )( g )(0.7 m)  (1 kg)(9.81 m/s 2 )(0.7 m)  6.867 J
T2  U 2 3  T3
13.244  6.867  4.5v32
v32  1.417
v A  v3  1.190 m/s
vA  1.190 m/s 
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PROBLEM 13.25
Four 3-kg packages are held in place by friction on a conveyor
which is disengaged from its drive motor. When the system is
released from rest, package 1 leaves the belt at A just as
package 4 comes onto the inclined portion of the belt at B.
Determine (a) the velocity of package 2 as it leaves the belt at
A, (b) the velocity of package 3 as it leaves the belt at A.
Neglect the mass of the belt and rollers.
SOLUTION
Given: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass of
belt and rollers. Package 1 leaves the belt as package 4 comes onto the belt.
Find: (a) Velocity of package 2 as it leaves the belt at A.
(b) Velocity of package 3 as it leaves the belt at A.
(a) Package 1 falls off the belt, and 2, 3, 4 move down.
2.4
 0.8 m
3
1

T2  3  mv22 
2

T2 
3
 3 kg  v22
2
T2  4.5v22
U1 2   3 W  0.8    3  3 kg   9.81 m/s 2  0.8 
U1 2  70.632 J
T1  U1 2  T2
0  70.632  4.5v22
v22  15.696
v2  3.9618
v2  3.96 m/s 
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PROBLEM 13.25 (Continued)
(b) Package 2 falls off the belt and its energy is lost to the system, and 3 and 4 move down 2 ft.
1

T2   2   mv22 
2

T2   3 kg 15.696 
T2  47.088 J 
 
1

T3   2   mv32    3 kg  v32
2

T3  3v32


U 2  3  2 W  0.8    2  3 kg  9.81 m/s 2  0.8 m 
U 2  3  47.088 J
T2  U 2  3  T3  47.088  47.088  3v32
v32  31.392
v3  5.6029
v3  5.60 m/s 
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PROBLEM 13.26
A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of
constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum
speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.
SOLUTION
Call blocks A and B.
(a)
mA  2 kg, mB  3 kg
Position 1: Block B has just been removed.
Spring force:
FS  (mA  mB ) g  k x
Spring stretch:
x1  
(mA  mB ) g
(5 kg)(9.81 m/s 2 )

 1.22625 m
k
40 N/m
Let position 2 be a later position while the spring still contacts block A.
Work of the force exerted by the spring:
(U1 2 )e  

x2
x1
k x dx
1
  k x2
2

Work of the gravitational force:
x2
x1

1
1
k x12  k x22
2
2
1
1
(40)(1.22625) 2  (40) x22  30.074  20 x22
2
2
(U1 2 ) g   m A g ( x2  x1 )
 (2)(9.81)( x2  1.22625)  19.62 x2  24.059
Total work:
Kinetic energies:
U12  20 x22  19.62 x2  6.015
T1  0
T2 
Principle of work and energy:
1
1
m A v22  (2)v22  v22
2
2
T1  U12  T2
0  20 x22  19.62 x2  6.015  v22
Speed squared:
v22  20 x22  19.62 x2  6.015
At maximum speed,
dv2
0
dx2
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(1)
PROBLEM 13.26 (Continued)
Differentiating Eq. (1), and setting equal to zero,
2v2
Substituting into Eq. (1),
dv2
 40 x2  19.62  0
dx
19.62
x2  
 0.4905 m
40
v22  (20)(0.4905)2  (19.62)(0.4905)  6.015  10.827 m2 /s2
v 2  3.29 m/s 
Maximum speed:
(b)
Position 3: Block A reaches maximum height. Assume that the block has separated from the spring.
Spring force is zero at separation.
Work of the force exerted by the spring:
(U13 )e  

0
x1
kxdx 
1 2 1
kx1  (40)(1.22625)2  30.074 J
2
2
Work of the gravitational force:
(U13 ) g  mA gh  (2)(9.81)h  19.62 h
Total work:
U13  30.074  19.62 h
At maximum height,
v3  0, T3  0
Principle of work and energy:
T1  U13  T3
0  30.074  19.62 h  0
Maximum height:
h  1.533 m 
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PROBLEM 13.27
Solve Problem 13.26, assuming that the 2-kg block is attached to the spring.
PROBLEM 13.26 A 3-kg block rests on top of a 2-kg block supported by, but not
attached to, a spring of constant 40 N/m. The upper block is suddenly removed.
Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height
reached by the 2-kg block.
SOLUTION
Call blocks A and B.
(a)
mA  2 kg, mB  3 kg
Position 1: Block B has just been removed.
Spring force:
FS  (mA  mB ) g  kx1
Spring stretch:
x1  
( m A  mB ) g
(5 kg)(9.81 m/s 2 )

 1.22625 m
k
40 N/m
Let position 2 be a later position. Note that the spring remains attached to block A.
Work of the force exerted by the spring:
(U1 2 )e  

x2
x1
kxdx
1
  kx 2
2
x2
x1

1 2 1 2
kx1  kx2
2
2
1
1
 (40)(1.22625) 2  (40) x22  30.074  20 x22
2
2
Work of the gravitational force:
(U1 2 ) g   m A g ( x2  x1 )
 (2)(9.81)( x2  1.22625)  19.62 x2  24.059
Total work:
Kinetic energies:
U12  20 x22  19.62 x2  6.015
T1  0
T2 
Principle of work and energy:
1
1
m A v22  (2)v22  v22
2
2
T1  U12  T2
0  20 x22  19.62 x2  6.015  v22
Speed squared:
v22  20 x22  19.62 x2  6.015
At maximum speed,
dv2
0
dx2
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(1)
PROBLEM 13.27 (Continued)
Differentiating Eq. (1) and setting equal to zero,
2v2
dv2
 40 x2  19.62  0
dx2
x2  
Substituting into Eq. (1),
19.62
 0.4905 m
40
v22  (20)(0.4905)2  (19.62)(0.4905)  6.015  10.827 m1 /s2
v2  3.29 m/s 
Maximum speed:
(b)
Maximum height occurs when v2  0.
Substituting into Eq. (1),
0  20 x22  19.62 x2  6.015
Solving the quadratic equation
x2  1.22625 m and 0.24525 m
Using the larger value,
x2  0.24525 m
Maximum height:
h  x2  x1  0.24525  1.22625
h  1.472 m 
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PROBLEM 13.28
People with mobility impairments can gain great health
and social benefits from participating in different
recreational activities. You are tasked with designing an
adaptive spring-powered shuffleboard attachment that
can be utilized by people who use wheelchairs.
Knowing that the coefficient of kinetic friction between
the 15 ounce puck A and the wooden surface is 0.3, the
maximum spring displacement you desire is 6 inches,
and that you want the puck to travel at least 30 ft/s,
determine (a) the spring constant k, (b) how far the
athlete should pull back the spring to make the puck
come to rest after 34 ft.
SOLUTION
Given:
k  0.3
Sketch:
15 oz
 32.2ft/s 2  0.02911 slugs
oz
16
lbs
6 in
 0.5 ft
s1,max 
in
12
ft
v2,max  30 ft/s
m
d3  34 ft
(a) Find spring constant k, for maximum conditions:
T1  U12  T2
where:
T1  0 and T2 
U12 


s1,max
0
1 2
mv2,max
2
kxdx 

s1,max
0
k mgdx
1 2
ks1,max  k mgs1,max
2
1
1
2
2
0  k  0.5  0.3  0.02911 32.2  0.5   0.02911 30 
2
2
k  105.9 lb/ft
k  8.828 lb/in 
(b) Find pull back distance, s1 so that puck stops after d=34 ft.
T1  0 and T3  0
T1  U13  T3
where:
U13 


s1
0
kxdx 

s1  d
0
k mgdx
1 2
ks1,max  k mg  s1  d 
2
1
105.9 s12  0.3 0.02911 32.2 s1  34  0
2
52.95s12  0.28125s1  9.5625  0
0
s1  0.4276 ft or s1  0.4221 ft
s1 =5.13 in 
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PROBLEM 13.29
A 7.5-lb collar is released from rest in the position shown, slides
down the inclined rod, and compresses the spring. The direction of
motion is reversed and the collar slides up the rod. Knowing that
the maximum deflection of the spring is 5 in., determine (a) the
coefficient of kinetic friction between the collar and the rod, (b) the
maximum speed of the collar.
SOLUTION
Position 1, initial condition
Position 2, spring deflected 5 inches
Position 3, initial contact of spring with collar
2
 18  5  1
 5 
 18  5 
U1  2   F 
   60    + 7.5 
 sin 30
 12  2
 12 
 12 
(Friction)
(Spring)
(Gravity)
T1  T2  0,  U1  2  0
2
 23  1
 5
 23 
0     7.5  0.866      60     7.5    0.5
 12  2
 12 
 12 
  0.1590 
(a)
(b)
Max speed occurs just before contact with the spring
1  7.5  2
 18 
 18 
U1 3     7.5  0.866    7.5    0.5  T3  
 vmax
 12 
 12 
2  32.2 
vmax  5.92 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.30
A 10-kg block is attached to spring A and connected to spring B by a cord and
pulley. The block is held in the position shown with both springs unstretched when
the support is removed and the block is released with no initial velocity. Knowing
that the constant of each spring is 2 kN/m, determine (a) the velocity of the block
after it has moved down 50 mm, (b) the maximum velocity achieved by the block.
SOLUTION
(a)
W  weight of the block  10 (9.81)  98.1 N
xB 
1
xA
2
1
1
k A ( xA )2  kB ( xB )2
2
2
(Gravity) (Spring A) (Spring B)
U1 2  W ( xA ) 
U1  2  (98.1 N)(0.05 m) 

1
(2000 N/m)(0.05 m)2
2
1
(2000 N/m)(0.025 m)2
2
U1 2 
1
1
(m)v 2  (10 kg)v 2
2
2
4.905  2.5  0.625 
1
(10)v 2
2
v  0.597 m/s 
(b)
Let x  distance moved down by the 10 kg block
2
U1 2  W ( x) 
1
1 x
1
k A ( x ) 2  k B    ( m )v 2
2
2 2
2
d 1
k

(m)v 2   0  W  k A ( x)  B (2 x)

dx  2
8

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.30 (Continued)
0  98.1  2000 ( x) 
2000
(2 x)  98.1  (2000  250) x
8
x  0.0436 m (43.6 mm)
For
x  0.0436, U  4.2772  1.9010  0.4752 
1
(10)v 2
2
vmax  0.6166 m/s
vmax  0.617 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.31
A 5-kg collar A is at rest on top of, but not attached to, a spring
with stiffness k1  400 N/m; when a constant 150-N force is
applied to the cable. Knowing A has a speed of 1 m/s when the
upper spring is compressed 75 mm, determine the spring
stiffness k2. Ignore friction and the mass of the pulley.
SOLUTION
Use the method of work and energy applied to the collar A.
T1  U12  T2
Since collar is initially at rest,
T1  0.
In position 2, where the upper spring is compressed 75 mm and v2  1.00 m/s, the kinetic energy is
T2 
1 2 1
mv2  (5 kg)(1.00 m/s)2  2.5 J
2
2
As the collar is raised from level A to level B, the work of the weight force is
(U12 ) g  mgh
where m  5 kg, g  9.81 m/s 2 and h  450 mm  0.450 m
Thus, (U12 ) g  (5)(9.81)(0.450)  22.0725 J
In position 1, the force exerted by the lower spring is equal to the weight of collar A.
F1  mg  (5 kg)(9.81 m/s)  49.05 N
As the collar moves up a distance x1, the spring force is
F  F1  k1 x2
until the collar separates from the spring at
xf 
F1 49.05 N

 0.122625 m  122.625 mm
k1 400 N/m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.31 (Continued)
Work of the force exerted by the lower spring:
(U12 )1 

xf
0
( F1  k1 x)dx
 F1 x f 

1 2
1
1
kx f  k1 x 2f  k1 x 2f  k1 x 2f
2
2
2
1
(400 N/m)(0.122625) 2  3.0074 J
2
In position 2, the upper spring is compressed by y  75 mm  0.075 m. The work of the force exerted
by this spring is
1
1
(U12 )2   k2 y 2   k2 (0.075)2  0.0028125 k 2
2
2
Finally, we must calculate the work of the 150 N force applied to the cable. In position 1, the length AB is
(l AB )1  (450) 2  (400) 2  602.08 mm
In position 2, the length AB is (l AB )2  400 mm.
The displacement d of the 150 N force is
d  (l AB )1  (l AB )2  202.08 mm  0.20208 m
The work of the 150 N force P is
(U12 ) P  Pd  (150 N)(0.20208 m)  30.312 J
Total work:
U1 2  22.0725  3.0074  0.0028125k2  30.312
 11.247  0.0028125k2
Principle of work and energy:
T1  U12  T2
0  11.247  0.0028125k2  2.5
k2  3110 N/m
k2  3110 N/m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.32
A piston of mass m and cross-sectional area A is equilibrium
under the pressure p at the center of a cylinder closed at both
ends. Assuming that the piston is moved to the left a distance
a/2 and released, and knowing that the pressure on each side of
the piston varies inversely with the volume, determine the
velocity of the piston as it again reaches the center of the
cylinder. Neglect friction between the piston and the cylinder
and express your answer in terms of m, a, p, and A.
SOLUTION
Pressures vary inversely as the volume
pL Aa

P
Ax
pR
Aa

P
A(2a  x)
Initially at ,
v0
x
pa
x
pa
pR 
(2a  x)
pL 
a
2
T1  0
At ,
x  a, T2 
U1 2 

a
a/2
1 2
mv
2
( pL  pR ) Adx 

a
a/2
1 
1
paA  
 dx
 x 2a  x 
U1 2  paA[ln x  ln (2a  x)]aa/2

a
 3a  
U1 2  paA ln a  ln a  ln    ln   
2
 2 


3a 2 
4
U1 2  paA ln a 2  ln
  paA ln  
4 
3

4 1
T1  U1 2  T2 0  paA ln    mv 2
3 2
v2 
2 paA ln  43 
m
 0.5754
paA
m
v  0.759
paA

m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.33
An uncontrolled automobile traveling at 65 mph strikes squarely a highway crash cushion of the type shown
in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F of the force
required to crush the barrels is shown as a function of the distance x the automobile has moved into the
cushion. Knowing that the weight of the automobile is 2250 lb and neglecting the effect of friction, determine
(a) the distance the automobile will move into the cushion before it comes to rest, (b) the maximum
deceleration of the automobile.
SOLUTION
(a)
65 mi/h  95.3 ft/s
Assume auto stops in 5  d  14 ft.
v1  95.33 ft/s
1 2 1  2250 lb
mv1 
2
2  32.2 ft/s 2
T1  317,530 lb  ft
T1 

2
 (95.3 ft/s)

 317.63 k  ft
v2  0
T2  0
U1 2  (18 k)(5 ft)  (27 k)(d  5)
 90  27 d  135
 27 d  45 k  ft
T1  U1 2  T2
317.53  27 d  45
d  13.43 ft 
Assumption that d  14 ft is ok.
(b)
Maximum deceleration occurs when F is largest. For d  13.43 ft, F  27 k. Thus, F  maD
 2250 lb
(27, 000 lb)  
2
 32.2 ft/s

 ( aD )

aD  386 ft/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.34
Two types of energy-absorbing fenders designed to be used on a
pier are statically loaded. The force-deflection curve for each type
of fender is given in the graph. Determine the maximum
deflection of each fender when a 90-ton ship moving at 1 mi/h
strikes the fender and is brought to rest.
SOLUTION
W1  (90 ton)(2000 lb/ton)  180  103 lb
Weight:
W 180  103

 5590 lb  s 2 /ft
g
32.2
Mass:
m
Speed:
v1  1 mi/h 
5280 ft
 1.4667 ft/s
3600 s
1 2 1
mv1  (5590)(1.4667)2
2
2
 6012 ft  lb
T2  0
(rest)
T1 
Kinetic energy:
Principle of work and energy:
T1  U12  T2
6012  U1 2  0
U1 2  6012 ft  lb  72.15 kip  in.
The area under the force-deflection curve up to the maximum deflection is equal to 72.15 kip  in.
Fender A: From the force-deflection curve F  kx
Area 

x
0
fdx 
k

x
0
Fmax 60

 5 kip/in.
xmax 12
kx dx 
1 2
kx
2
1
(5) x2  72.51
2
x 2  28.86 in.2
Fender B: We divide area under curve B into trapezoids
Partial area
1
(2 in.)(4 kips)  4 kip  in.
From x  0
to x  2 in.:
2
x  5.37 in. 
Total Area
4 kip  in.
From x  2 in.
to x  4 in.:
1
(2 in.)(4  10)  14 kip  in.
2
18 kip  in.
From x  4 in.
to x  6 in.:
1
(2 in.)(10  18)  28 kip  in.
2
46 kip  in.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.34 (Continued)
We still need U  72.15  46  26.15 kip  in.
Equation of straight line approximating curve B from x  6 in. to x  8 in. is
x F  18

F  18  6x
2 30  18
1
U  18x  (6x)x  26.15 kip  in.
2
2
(x)  6x  8.716  0
x  1.209 in.
Thus:
x  6 in.  1.209 in.  7.209 in.
x  7.21 in. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.35
Nonlinear springs are classified as hard or soft, depending upon the curvature of their force-deflection curve
(see figure). If a delicate instrument having a mass of 5 kg is placed on a spring of length l so that its base is
just touching the undeformed spring and then inadvertently released from that position, determine the
maximum deflection xm of the spring and the maximum force Fm exerted by the spring, assuming (a) a linear
spring of constant k  3 kN/m, (b) a hard, nonlinear spring, for which F  (3 kN/m)( x  160 x 2 ) .
SOLUTION
W  mg  (5 kg) g
W  49.05 N
T1  T2  0, T1  U1 2  T2 yields U12  0
Since
U1 2  Wxm 
(a)

xm
0
Fdx  49.05 xm 

xm
0
Fdx  0
For F  kx  (300 N/m) x
Eq. (1):
49.05 xm 

xm
0
3000 x dx  0
xm  32.7  103 m  32.7 mm 
49.05 xm  1500 xm2  0
Fm  3000 xm  3000(32.7  103 )
(b)
(1)
For
Eq. (1)
Fm  98.1 N 
F  (3000 N/m) x(1  160 x 2 )
49.05 xm 

xm
0
3000( x  160 x3 )dx  0
1

49.05 xm  3000  xm2  40 xm4   0
2

Solve by trial:
(2)
xm  30.44  103 m
xm  30.4 mm 
Fm  (3000)(30.44  103 )[1  160(30.44  103 )2 ]
Fm  104.9 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.36
A meteor starts from rest at a very great distance from the earth. Knowing that the radius of the earth is 6370
km and neglecting all forces except the gravitational attraction of the earth, determine the speed of the meteor
(a) when it enters the ionosphere at an altitude of 1000 km, (b) when it enters the stratosphere at an altitude of
50 km, (c) when it strikes the earth’s surface.
SOLUTION
R  6370 km
Given:
U1 2  T2  T1

GMm GMm

r2
r1
Since r1 is very large,
GMm
0
r1
thus
T1  0
GMm
1
mv 2 
r2
2
v2
R2
 g
r2
2
v2 

(a) For
2 gR 2
r2


2 9.81 m/s 2  6,370, 000 m 
2
r2
r2  6,370,000 m  1,000,000 m  7,370,000 m
v  10.3933 km/s
(b) For
r2  6,370,000 m  50,000 m  6, 420,000 m
v  11.1358 km/s
(c) For
v  10.39 km/s 
v  11.14 km/s 
r2  6,370,000 m
v  11.1794 km/s
v  11.18 km/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.37
Express the acceleration of gravity gh at an altitude h above the surface of the earth in terms of the
acceleration of gravity g0 at the surface of the earth, the altitude h, and the radius R of the earth. Determine
the percent error if the weight that an object has on the surface of the earth is used as its weight at an altitude
of (a) 0.625 mi, (b) 625 mi.
SOLUTION
GM E m
F 
h  R
2



1
h
R
2
 mg h
GM E m
 mg0
R2
At earths’ surface  h  0 
GM E
GM E
 g0
R2
gh 
Thus
GM E m / R 2
g0
 Rh  1
gh 
R2

h
R

1
2
2
R  3960 mi
F  mgh  WT
At altitude h, “true” weight
W0  mg0
Assume weight
Error  E 
gh 
(a)
(b)
g0
 Rh  1
W0  WT
mg0  mg h
g  gh

 0
W0
mg0
g0
g0 
2
E 
g0
1 Rh 
g0

1
h  0.625 mi: P  100 E  100 1 

1  0.625

3960


1
h  625 mi: P  100 E  100 1 

625
1  3960


2



2




2


1
 1 

1  Rh





2

P  0.0316% 
P  25.4% 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.38
A golf ball struck on earth rises to a maximum height of 60 m
and hits the ground 230 m away. How high will the same golf
ball travel on the moon if the magnitude and direction of its
velocity are the same as they were on earth immediately after
the ball was hit? Assume that the ball is hit and lands at the
same elevation in both cases and that the effect of the
atmosphere on the earth is neglected, so that the trajectory in
both cases is a parabola. The acceleration of gravity on the
moon is 0.165 times that on earth.
SOLUTION
Solve for hm .
At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the
same in both cases.
1 2
mv
2
 mge he
T1 
U1 2
T2 
Earth
U1 2  mg m hm
Earth
1 2
1
mv  mge he  mvH2
2
2
Moon
1 2
1
mv  mg m hm  mvH2
2
2
 ge he  g m hm  0
Subtracting
1 2
mvH
2
Moon
hm ge

he g m
 ge 
hm  (60 m) 

 0.165 ge 
hm  364 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.39
The sphere at A is given a downward velocity v0 of magnitude
5 m/s and swings in a vertical plane at the end of a rope of
length l  2 m attached to a support at O. Determine the angle 
at which the rope will break, knowing that it can withstand a
maximum tension equal to twice the weight of the sphere.
SOLUTION
1 2 1
mv0  m (5) 2
2
2
T1  12.5 m
T1 
1 2
mv
2
 mg (l ) sin 
T2 
U1 2
T1  U1 2  T2
12.5m  2mg sin  
1 2
mv
2
25  4 g sin   v 2
(1)
Newton’s law at .
v2
v2
m

2
2
v  4 g  2 g sin 
2mg  mg sin   m
(2)
Substitute for v2 from Eq. (2) into Eq. (1)
25  4 g sin   4 g  2 g sin 
(4)(9.81)  25
sin  
 0.2419
(6)(9.81)
  14.00 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.40
The sphere at A is given a downward velocity v0 and swings
in a vertical circle of radius l and center O. Determine the
smallest velocity v0 for which the sphere will reach Point B as
it swings about Point O (a) if AO is a rope, (b) if AO is a
slender rod of negligible mass.
SOLUTION
1 2
mv0
2
1
T2  mv 2
2
U1 2  mgl
T1 
1 2
1
mv0  mgl  mv 2
2
2
2
2
v0  v  2 gl
T1  U1 2  T2
Newton’s law at 
(a)
For minimum v, tension in the cord must be zero.
Thus, v 2  gl
v02  v 2  2 gl  3gl
(b)
v0  3 gl 
Force in the rod can support the weight so that v can be zero.
Thus,
v02  0  2 gl
v0  2 gl 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.41
A bag is gently pushed off the top of a wall at A and swings in a vertical
plane at the end of a rope of length l Determine the angle  for which the
rope will break, knowing that it can withstand a maximum tension equal to
twice the weight of the bag.
SOLUTION
Use work - energy: position 1 is at A, position 2 is at B.
T1  U1 2  T2
(1)
Where
T1  0; U1 2  mg l sin  ; T2 
1 2
mvB
2
Substitute
0  mg l sin  
1 2
mvB
2
vB2  2 g l sin 
(2)
For T = 2 W use Newton’s 2nd law.
Fn  man  2W  W sin  
mvB2
l
(3)
Substitute (2) into (3)
2 mg  mg sin   2 mg
l sin 
l
2  3sin 
or sin 
2
   41.81
3
  41.8 
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PROBLEM 13.42
A roller coaster starts from rest at A, rolls down the
track to B, describes a circular loop of 40-ft diameter,
and moves up and down past Point E. Knowing that h =
60 ft and assuming no energy loss due to friction,
determine (a) the force exerted by his seat on a 160-lb
rider at B and D, (b) the minimum value of the radius of
curvature at E if the roller coaster is not to leave the
track at that point.
SOLUTION
Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position
2 be at P. Apply the principle of work and energy.
1
T2  mvP2
2
T1  0
U12  mgyP
T1  U1 2  T2 : 0  mgyP 
1 2
mvP
2
vP2  2 gyP
Magnitude of normal acceleration at P:
( aP ) n 
(a)
vP2
P

2 gyP
P
Rider at Point B.
yB  h  60 ft
 B  r  20 ft
an 
(2 g )(60)
 6g
20
F  ma:
N B  mg  m(6 g )
N B  7 mg  7W  (7)(160 lb)
N B  1120 lb 
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PROBLEM 13.42 (Continued)
Rider at Point D.
yD  h  2r  20 ft
 D  20 ft
an 
(2 g )(20)
 2g
20
F  ma:
N D  mg  m(2 g )
N D  mg  W  160 lb
N D  160 lb 
(b)
Car at Point E.
yE  h  r  40 ft
NE  0
F  man :
mg  m 
 E  2 yE
2 gyE
E
  80.0 ft 
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PROBLEM 13.43
In Problem 13.42, determine the range of values of h
for which the roller coaster will not leave the track at D
or E, knowing that the radius of curvature at E is
  75 ft. Assume no energy loss due to friction.
PROBLEM 13.42 A roller coaster starts from rest at A,
rolls down the track to B, describes a circular loop of
40-ft diameter, and moves up and down past Point E.
Knowing that h  60 ft and assuming no energy loss
due to friction, determine (a) the force exerted by his
seat on a 160-lb rider at B and D, (b) the minimum
value of the radius of curvature at E if the roller coaster
is not to leave the track at that point.
SOLUTION
Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position
2 be at P. Apply the principle of work and energy.
1
T2  mvP2
2
T1  0
U12  mg yP
T1  U1 2  T2 : 0  mgy p 
1 2
mvP
2
vP2  2 g yP
Magnitude of normal acceleration of P:
( aP ) n 
vP2
P

2 g yP
P
The condition of loss of contact with the track at P is that the curvature of the path is equal to p and the
normal contact force N P  0.
Car at Point D.
 D  r  20 ft
yD  h  2r
( aD ) n 
2 g (h  2 r )
r
F  ma
2 g ( h  2r )
r
2 h  5r
N D  mg
r
N D  mg  m
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PROBLEM 13.43 (Continued)
For N D  0
2h  5r  0
h
5
r  50 ft
2
Car at Point E.
 E    75 ft
yE  h  r  h  20 ft
( aE ) n 
2 g (h  20)
75
F  ma
2mg (h  20)
75
115  2h
N E  mg
75
N E  mg  
N E  0,
For
115  2h  0
h  57.5 ft
Range of values for h:
50.0 ft  h  57.5 ft 
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PROBLEM 13.44
A small block slides at a speed v on a horizontal surface. Knowing that
h  0.9 m, determine the required speed of the block if it is to leave the
cylindrical surface BCD when   30°.
SOLUTION
At Point C where the block leaves the surface BCD the contact force is reduced to zero. Apply Newton’s
second law at Point C.
n-direction:
N  mg cos   man  
mvC2
h
vc2  gh cos 
With N  0, we get
Apply the work-energy principle to the block sliding over the path BC. Let position 1 correspond to Point B
and position 2 to C.
T1 
1 2
mvB
2
T2 
1
1
mvC2  mgh cos 
2
2
U12  weight  change in vertical distance
 mgh (1  cos  )
1 2
1
mvB  mg (1  cos  )  mgh cos 
2
2
2
vB  gh cos   2 gh(1  cos  )  gh(3cos   2)
T1  U1 2  T2 :
Data:
g  9.81 m/s 2 , h  0.9 m,   30.
vB2  (9.81)(0.9)(3cos 30  2)  5.2804 m 2 /s 2
vB  2.30 m/s 
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PROBLEM 13.45
A small block slides at a speed v  8 ft/s on a horizontal surface at a
height h  3 ft above the ground. Determine (a) the angle  at which it
will leave the cylindrical surface BCD, (b) the distance x at which it
will hit the ground. Neglect friction and air resistance.
SOLUTION
Block leaves surface at C when the normal force N  0.
mg cos  man
g cos 
vC2
h
vC2  gh cos   gy
(1)
Work-energy principle.
TB 
(a)
1 2 1
mv  m(8)2  32m
2
2
1 2
mvC
2
 TC
TC 
TB  U B C
Use Eq. (1)
32m  mg (h  y ) 
1 2
mvC
2
32  g (h  yC ) 
1
gyC
2
U B C  W (h  g )  mg (h  yC )
(2)
3
g yC
2
(32  gh)
yC 
 32 g 
32  gh 
yC 
(32  (32.2)(3))
3
(32.2)
2
yC  2.6625 ft
yC  h cos 
cos  
yC 2.6625

 0.8875
h
3
(3)
  27.4 

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PROBLEM 13.45 (Continued)
(b)
From (1) and (3)
vC  gy
vC  (32.2)(2.6625)
vC  9.259 ft/s
At C:
(vC ) x  vC cos   (9.259)(cos 27.4)  8.220 ft/s
(vC ) y  vC sin   (9.259)(sin 27.4)  4.261 ft/s
y  yC  (vC ) y t 
At E:
yE  0:
1 2
gt  2.6625  4.261t  16.1t 2
2
t 2  0.2647t  0.1654  0
t  0.2953 s
At E:
x  h(sin  )  (vC ) x t  (3)(sin 27.4)  (8.220)(0.2953)
x  1.381  2.427  3.808 ft
x  3.81 ft 
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PROBLEM 13.46
A chair-lift is designed to transport 1000 skiers per hour from the base A to
the summit B. The average mass of a skier is 70 kg and the average speed of
the lift is 75 m/min. Determine (a) the average power required, (b) the
required capacity of the motor if the mechanical efficiency is 85 percent and
if a 300 percent overload is to be allowed.
SOLUTION
Note: Solution is independent of speed.
(a)
Average power 
U (1000)(70 kg)(9.81 m/s 2 )(300 m)
Nm

 57, 225
t
3600 s
s
Average power  57.2 kW 
(b)
Maximum power required with 300% over load

100  300
(57.225 kW)  229 kW
100
Required motor capacity (85% efficient)
Motor capacity 
229 kW
 269 kW 
0.85
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PROBLEM 13.47
It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic
car-lift platform to a height of 2.8 m. Determine (a) the average output
power delivered by the hydraulic pump to lift the system, (b) the average
power electric required, knowing that the overall conversion efficiency
from electric to mechanical power for the system is 82 percent.
SOLUTION
(a)
( PP ) A  ( F )(vA )  (mC  mL )( g )(vA )
v A  s/t  (2.8 m)/(15 s)  0.18667 m/s
( PP ) A  [(1200 kg)  (300 kg)](9.81 m/s2 )(0.18667 m/s)3
(b)
( PP ) A  2.747 kJ/s
( PP ) A  2.75 kW 
( PE ) A  ( PP )/  (2.75 kW)/(0.82)
( PE ) A  3.35 kW 
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PROBLEM 13.48
The velocity of the lift of Problem 13.47 increases uniformly from zero to
its maximum value at mid-height 7.5 s and then decreases uniformly to
zero in 7.5 s. Knowing that the peak power output of the hydraulic pump is
6 kW when the velocity is maximum, determine the maximum life force
provided by the pump.
PROBLEM 13.47 It takes 15 s to raise a 1200-kg car and the supporting
300-kg hydraulic car-lift platform to a height of 2.8 m. Determine
(a) the average output power delivered by the hydraulic pump to lift the
system, (b) the average power electric required, knowing that the overall
conversion efficiency from electric to mechanical power for the system is
82 percent.
SOLUTION
Newton’s law
Mg  ( M C  M L ) g  (1200  300) g
Mg  1500 g
F  F  1500 g  1500a
(1)
Since motion is uniformly accelerated, a  constant
Thus, from (1), F is constant and peak power occurs when the velocity is a maximum at 7.5 s.
a
vmax
7.5 s
P  (6000 W)  ( F )(vmax )
vmax  (6000)/F
a  (6000)/(7.5)( F )
Thus,
(2)
Substitute (2) into (1)
F  1500 g  (1500)(6000)/(7.5)( F )
F 2  (1500 kg)(9.81 m/s 2 ) F 
(1500 kg)(6000 N  m/s)
0
(7.5 s)
F 2  14, 715F  1.2  106  0
F  14,800 N
F  14.8 kN 
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PROBLEM 13.49
(a) A 120-lb woman rides a 15-lb bicycle up a 3-percent slope at a constant speed of 5 ft/s. How much power
must be developed by the woman? (b) A 180-lb man on an 18-lb bicycle starts down the same slope and
maintains a constant speed of 20 ft/s by braking. How much power is dissipated by the brakes? Ignore air
resistance and rolling resistance.
SOLUTION
tan  
3
100
  1.718
W  WB  WW  15  120
(a)
W  135 lb
PW  W  v  (W sin  ) (v)
PW  (135)(sin 1.718)(5)
(a)
PW  20.24 ft  lb/s
PW  20.2 ft  lb/s 
W  WB  Wm  18  180
(b)
W  198 lb
Brakes must dissipate the power generated by the bike and the man going down the slope at 20 ft/s.
PB  W  v  (W sin  )(v)
PB  (198)(sin 1.718)(20)
PB  118.7 ft  lb/s 




(b)
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PROBLEM 13.50
A power specification formula is to be
derived for electric motors which drive
conveyor belts moving solid material at
different rates to different heights and
distances. Denoting the efficiency of the
motors by  and neglecting the power
needed to drive the belt itself, derive a
formula (a) in the SI system of units for
the power P in kW, in terms of the mass
flow rate m in kg/h, the height b and
horizontal distance l in meters, and (b) in
U.S. customary units, for the power in hp,
in terms of the material flow rate w in
tons/h, and the height b and horizontal
distance l in feet.
SOLUTION
(a)
Material is lifted to a height b at a rate, (m kg/h)( g m/s2 )  [mg (N/h)]
Thus,
U [mg (N/h)][b(m)]  mgb 


 N  m/s
(3600 s/h)
t
 3600 
1000 N  m/s  1 kW
Thus, including motor efficiency, 
P (kw) 
mgb (N  m/s)
 1000 N  m/s 
(3600) 
 ( )
kW


P(kW)  0.278  106
mgb


U [W (tons/h)(2000 lb/ton)][b(ft)]

t
3600 s/h
(b)

With  ,
Wb
ft  lb/s; 1hp  550 ft  lb/s
1.8
 Wb
  1 hp   1 
hp  
(ft  lb/s)  
 
 1.8
  550 ft  lb/s   
hp 
1.010  103Wb

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
PROBLEM 13.51
A 1400-kg automobile starts from rest and travels 400 m during a
performance test. The motion of the automobile is defined by the relation
x  4000 ln(cosh 0.03t ), where x and t are expressed in meters and seconds,
respectively. The magnitude of the aerodynamic drag is D  0.35v 2 , where
D and v are expressed in newtons and m/s, respectively. Determine the
power dissipated by the aerodynamic drag when (a) t  10 s, (b) t  15 s.
SOLUTION
Motion is determined as a function of time as
x  4000ln  cosh 0.03t 
Velocity
v
1
dx


 4000 
  sinh 0.03t  0.03
dt
 cosh 0.03t 
v
Power dissipated
120sinh 0.03t
cosh 0.03t


P  Dv  0.35v 2 v  0.35v 3
3  sinh 0.03t 
P  0.35 120  

 cosh 0.03t 
3 e
0.03t
 604.8 10   0.03t
e
3
 e0.03t 

 e0.03t 
3
(a) t  10 s,
P  14.95 10  W  14.95 kW 
(b) t  15 s,
P  45.4 10  W  45.4 kW 
3
3
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PROBLEM 13.52
A 1400-kg automobile starts from rest and travels 400 m during a
performance test. The motion of the automobile is defined by the relation
2
a  3.6e0.0005 x , where a and x are expressed in m/s and meters,
respectively. The magnitude of the aerodynamic drag is D  0.35v 2 ,
where D and v are expressed in newtons and m/s, respectively. Determine
the power dissipated by the aerodynamic drag when (a) x  200 m,
(b) x  400 m.
SOLUTION
Motion is defined by the following function:
a  3.6e0.0005x
v
dv
dx
v
x 0.0005 x
dx
0 vdv  3.60 e

3.6 0.0005 x u
e du

0.0005 0
v2
 7200 e 0.0005 x  1
2




v 2  14400 1  e 0.0005 x

v  120 1  e 0.0005 x

1
2
Power dissipated  P  Dv  0.35v3
3
P  604.8 10  1  e0.0005 x  2
3
x  200 m,
P  17.75 10  W  17.75 kW 
(b) x  400 m,
P  46.7 10  W  46.7 kW 
(a)
3
3
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PROBLEM 13.53
The fluid transmission of a 15-Mg truck allows the engine to deliver an essentially constant power of 50 kW
to the driving wheels. Determine the time required and the distance traveled as the speed of the truck is
increased (a) from 36 km/h to 54 km/h, (b) from 54 km/h to 72 km/h.
SOLUTION
For constant power, P:
P  Fv
 mav
m
dv
v
dt
Separate variables
m dv
m  v2
v2 
t
v1
1
0
 0 dt   0 P v  t  P  2  2 


(1)
Distance
P  mv
dv dx
dv
 mv 2
dx dt
dx
Separate variables
m
m  v3
v3 
x
v1 2
1
0
 0 dx  P  v0 v dv  x  P  3  3 


(2)
with numbers
(a) v0  36 km/h  10 m/s; v1  54 km/h  15 m/s, so,
t

x
15  103 kg
15 m/s 2  10 m/s 2 

2 50 × 10 W 

3
15  103

153  103   237.5 m 

3 

3
50
10



 t  18.75 s 
 x  238 m 
(b) v0  54 km/h  15 m/s; v1  72 km/h  20 m/s
t

x
15  103
 2  50  10
15  103
3

 20 2  152   26.25 s


 t  26.2 s 
 203  153   462.5 m 

 x  462 m 
 3  50  103  
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PROBLEM 13.54
The elevator E has a weight of 6600 lbs when fully loaded and is connected as
shown to a counterweight W of weight of 2200 lb. Determine the power in hp
delivered by the motor (a) when the elevator is moving down at a constant speed
of 1 ft/s, (b) when it has an upward velocity of 1 ft/s and a deceleration of
0.18 ft/s2 .
SOLUTION
(a) Acceleration  0
Counterweight
Elevator
Motor
F  0: 2TC  TW  6600  0
Fy  0: TW  WW  0
TW  2200 lb
Kinematics:
TC  2200 lb
2 xE  xC , 2 xE  xC , vC  2vE  2 ft/s
P  TC  vC  (2200 lb)(2 ft/s)  4400 lb  ft/s  8.00 hp
P  8.00 hp 
aE  0.18 ft/s2 , vE  1 ft/s
(b)
Counterweight
Elevator
Counterweight:
F  Ma : TW  W 
W
(aW )
g
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PROBLEM 13.54 (Continued)
TW  (2200 lb) 
(2200 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
TW  2212 lb
Elevator
F  ma
2TC  TW  WE 
2TC  (2212 lb)  (6600 lb) 
WE
(a E )
g
(6600 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
2TC  4351 lb
TC  2175.6 lb
vC  2 ft/s (see part(a))
P  TC  vC  (2175.6 lb)(2 ft/s)  4351.2 lb  ft/s
 7.911 hp



P  7.91 hp 
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PROBLEM 13.55
A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0 . In each of the
two cases shown, derive an expression for the constant ke , in terms of k1 and k2 , of the single spring
equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when
subjected to the same force P.
SOLUTION
System is in equilibrium in deflected x0 position.
Case (a)
Force in both springs is the same  P
x0  x1  x2
Thus,
x0 
P
ke
x1 
P
k1
x2 =
P
k2
P P P
 
ke k1 k2
1
1 1
 
ke k1 k2
Case (b)
ke 
k1k2

k1  k2
Deflection in both springs is the same  x0
P  k1 x0  k2 x0
P  (k1  k2 ) x0
P  ke x0
Equating the two expressions for
P  (k1  k2 ) x0  ke x0
ke  k1  k2 
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PROBLEM 13.56
A loaded railroad car of mass m is rolling at a
constant velocity v0 when it couples with a
massless bumper system. Determine the
maximum deflection of the bumper assuming the
two springs are (a) in series (as shown), (b) in
parallel.
SOLUTION
Let position A be at the beginning of contact and position B be at maximum deflection.
1 2
mv0
2
VA  0
(zero force in springs)
TB  0
(v  0 at maximum deflection)
TA 
1 2 1
k1 x1  k2 x22
2
2
where x1 is deflection of spring k1 and x2 is that of spring k2.
VB 
Conservation of energy:
TA  VA  TB  VB
1 2
1
1
mv0  0  0  k1 x12  k2 x22
2
2
2
k1 x12  k2 x22  mv02
(a)
(1)
Springs are in series.
Let F be the force carried by the two springs.
F
k1
Then,
x1 
Eq. (1) becomes
1
1 
F 2     mv02
 k1 k2 
so that
The maximum deflection is
and x2 
F
k2
1
1 
F  v0 m /   
 k1 k 2 
1
1 
 F
 k1 k2 
  x1  x2  
1
1
1 
1 
    v0 m /   
 k1 k2 
 k1 k2 
1
1 
 v0 m   
 k1 k 2 
  v0 m ( k1  k 2 )/k1k 2 
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PROBLEM 13.56 (Continued)
(b)
Springs are in parallel.
x1  x2  
Eq. (1) becomes
(k1  k2 ) 2  mv02
  v0
m

k1  k2
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PROBLEM 13.57
A 750-g collar can slide along the horizontal rod shown. It is
attached to an elastic cord with an undeformed length of 300 mm
and a spring constant of 150 N/m. Knowing that the collar is
released from rest at A and neglecting friction, determine the
speed of the collar (a) at B, (b) at E.
SOLUTION
T1  V1  T2  V2
Use conservation of energy
(1)
(a) Position 1 is at A and position 2 is at B
T1  0;
V1 
1 2
kx1
2
where
x1     0
1
  500 2  4002  3502  2  729.726 mm
So    0  429.726 mm
V1 
1
1500 N/m  0.429726 m 2  13.8498 J
2
1
At B
  3502  4002  2  531.507 mm     0  231.507 mm
1 2 1
mv2   0.75  v22  0.375 v22
2
2
1
2
V2  150  0.231507   4.01966 J
2
T2 
Substituting into (1)
0  13.8498  0.375 v22  4.01966
v2  vB  5.12 m/s 
(b) At E
T1  0; V1  13.8498  same as before 
1
At E
Substituting into (1)
  3502  500 2  2  610.328 mm     0  310.328 mm
1
T3  mvE2  0.375 vE2
2
1
1
2
vE  kx 2  150  0.310328  7.223 J
2
2
0  13.8498  0.375 vE2  7.223  vE  4.20 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.58
A 4-lb collar can slide without friction along a horizontal rod and is
in equilibrium at A when it is pushed 1 in. to the right and released
from rest. The springs are undeformed when the collar is at A and
the constant of each spring is 2800 lb/in. Determine the maximum
velocity of the collar.
SOLUTION
1  62  92  10.817 in.
0 
 6 2   8 2
 10 in.  0.8333 ft
Stretch  10.817  10  0.817 in.
S1  0.06805 ft
2 
 7 2   6 2
 9.215 in.
Stretch  9.2195  10   0.7805 in.
S2  0.06504 ft
T1  0, V2  0
1 2
mv2
2
1 4  2
 
 v2
2  32.2 
T2 
V1 

1
 33,600 lb/ft  S12  S22
2

V1  16,800  0.008861  148.86 ft  lb
T1  V1  T2  V2
148.86 
1 4  2

 v2
2  32.2 
v22  2396.7
v2  49.0 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.59
A 4-lb collar can slide without friction along a horizontal rod and is
released from rest at A. The undeformed lengths of springs BA and
CA are 10 in. and 9 in., respectively, and the constant of each spring
is 2800 lb/in. Determine the velocity of the collar when it has moved
1 in. to the right.
SOLUTION
k  2800 lb/in.  33, 600 lb/ft
T1  0
V1 
1
1
2
 1 
k  l1    33, 600   
2
2
 12 
2
 116.667 ft  lb
1  62  92  10.817 in.  0.9014 ft
S1  Stretch  10.817  10  0.817 in.  0.06808 ft
 2  62  7 2  9.2195 in.
S2  Stretch  9.2195  9  0.2195 in.  0.018295 ft
T2 
1 2 1 4  2
mv2  
v2  0.0621 v22
2
2  32.2 
V2 
1
 33,600  S12  S22
2


2
2
 16800  0.06808    0.018295  


 83.489 ft  lb
T1  V1  T2  V2
116.667  0.06211 v22  83.489
v2  23.1 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.60
A 500-g collar can slide without friction on the curved rod BC in a horizontal
plane. Knowing that the undeformed length of the spring is 80 mm and that
k  400 kN/m, determine (a) the velocity that the collar should be given at A
to reach B with zero velocity, (b) the velocity of the collar when it eventually
reaches C.
SOLUTION
(a)
Velocity at A:
TA 
1 2  0.5

mv A  
kg  v A2
2
2


TA  (0.25)v A2
LA  0.150 m  0.080 m
LA  0.070 m
1
k (LA )2
2
1
VA  (400  103 N/m)(0.070 m) 2
2
VA  980 J
VA 
vB  0
TB  0
LB  0.200 m  0.080 m  0.120 m
1
1
k (LB )2  (400  103 N/m)(0.120 m) 2
2
2
VB  2880 J
VB 
Substitute into conservation of energy.
TA  VA  TB  VB
v A2 
0.25v A2  980  0  2880
(2880  980)
(0.25)
v A2  7600 m 2 /s 2
vA  87.2 m/s 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.60 (Continued)
(b)
Velocity at C:
Since slope at B is positive, the component of the spring force FP , parallel to the rod, causes the block
to move back toward A.
TB  0, VB  2880 J [from part (a )]
1 2 (0.5 kg) 2
mvC 
vC  0.25vC2
2
2
LC  0.100 m  0.080 m  0.020 m
TC 
VC 
1
1
k ( LC )2  (400  103 N/m)(0.020 m)2  80.0 J
2
2
Substitute into conservation of energy.
TB  VB  TC  VC
0  2880  0.25vC2  80.0
vC2  11, 200 m 2 /s2
vC  105.8 m/s 
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PROBLEM 13.61
For the adapted shuffleboard device in Prob 13.28, you decide to utilize an elastic
cord instead of a compression spring to propel the puck forward. When the cord is
stretched directly between points A and B, the tension is 20 N. The 425 gram puck
is placed in the center and pulled back through a distance of 400 mm; a force of
100 N is required to hold it at this location. Knowing that the coefficient of friction
is 0.3, determine how far the puck will travel.
SOLUTION
Given:
k  0.3
Sketch:
m  0.425 kg
FC,2  20 N, Force in cord at positon 2
li = length of cord at positon i
si = stretch of cord in position i
lO  unstretched length of cord
At Position 1:


l1 = 2 0.32  0.42  l1  1 m
 40 
  53.13
 30 
 =tan 1 
Find Force in cord:
FBD:
For Equilibrium:
F
x
0
2 FC ,1 sin   100  0
FC ,1  62.5 N
At Position 2:
l2 = 0.6 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.61 (Continued)
Find Stiffness of cord, k:
FC ,1  ks1 , FC ,2  k s2
FC ,1  FC ,2  k (s1  s2 )
62.5  20  k [(1.0  l0 )  (0.6  l0 )]
k  106.25 N/m
Find stretch in cord at positions 1, 2 and 3:
s1 
FC ,1
k
62.5 N

106.25 N/m
 0.5882 m
s2  s3 
FC ,2
k
20 N

106.25 N/m
 0.1882 m
Work Energy from position 1 to 3:
T1  Vg1  Ve1  U1NC
3  T3  Vg3  Ve3
where:
Ve1 
U 1NC

3
Ve3 
Therefore:
1 2
ks1
2

x1  d
0
k mgdx
1 2
ks3
2
1 2
1
ks1  k mg  x1  d   ks32
2
2
Substitute known values:
1
1
106.25 0.58822  0.3 0.425 9.81 0.4  d   106.25 0.18822
2
2
Solve for d:
d  12.79 m 
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PROBLEM 13.62
An elastic cable is to be designed for bungee jumping from a
tower 130 ft high. The specifications call for the cable to be 85
ft long when unstretched, and to stretch to a total length of 100
ft when a 600-lb weight is attached to it and dropped from the
tower. Determine (a) the required spring constant k of the
cable, (b) how close to the ground a 186-lb man will come if
he uses this cable to jump from the tower.
SOLUTION
(a)
Conservation of energy:
V1  0 T1  0 V1  100 W
Datum at :
V1  (100 ft)(600 lb)
 6  10 4 ft  lb
V2  0 T2  0
V2  Vg  Ve  0 
1
k (15 ft) 2
2
T1  V1  T2  V2
0  6  104  0  (112.5)k
k  533 lb/ft 

(b)
From (a),
k  533 lb/ft
T1  0
W  186 lb
V1  (186)(130  d )
T2  0
Datum:
1
(533)(130  85  d )2
2
V2  (266.67)(45  d ) 2
V2  Vg  Ve  0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.62 (Continued)
d  distance from the ground
T1  V1  T2  V2
0  (186)(130  d )  0  (266.67)(45  d )2
266.7d 2  23815d  515827  0
d
23815  (23815)2  4(266.7)(515827) 36.99 ft

52.3 ft
(2)(266.7)
Discard 52.3 ft (since the cord acts in compression when rebound occurs).
d  37.0 ft 
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PROBLEM 13.63
It is shown in mechanics of materials that the stiffness of an elastic cable is k  AE/L
where A is the cross sectional area of the cable, E is the modulus of elasticity and L is
the length of the cable. A winch is lowering a 4000-lb piece of machinery using at a
constant speed of 3ft/s when the winch suddenly stops. Knowing that the steel cable
has a diameter of 0.4 in., E  29  106 lb/in2, and when the winch stops L  30 ft,
determine the maximum downward displacement of the piece of machinery from the
point it was when the winch stopped.
SOLUTION
m
Mass of machinery:
W 4000

 124.22 lb  s 2 /ft
g
32.2
Let position 1 be the state just before the winch stops and the gravitational potential Vg be equal to zero at this
state.
For the cable,
A

4
(diameter)2 

4
(0.4 in.)2  0.12566 in 2
AE  (0.12556 in.2 )(29  106 lb/in 2 )  3.6442  106 lb
k
AE 3.6442  106 lb

 121.47  103 lb/ft
L
30 ft
For
L  30 ft,
Initial force in cable (equilibrium):
F1  W  4000 lb.
Elongation in position 1:
x1 
F1
4000

 0.03293 ft
k 121.47  103
Potential energy:
V1 
1 2 F12
kx1 
2
2k
V1 
(4000 lb)2
 65.860 ft  lb
(2)(121.47  103 lb/ft)
T1 
1 2
mv1
2
Kinetic energy:
1
T1  (124.22 lb  s2 /ft)(3 ft/s2 )  558.99 ft  lb
2
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PROBLEM 13.63 (Continued)
Let position 2 be the position of maximum downward displacement. Let x2 be the elongation in this position.
Potential energy:
V2 
1 2
kx2  W ( x2  x1 )
2
1
(121.47  103 ) x22  (4000)( x2  0.03293)
2
 60.735  103 x 2  4000 x2  131.72
V2 
Kinetic energy:
T2  0
(since v2  0)
Principle of work and energy:
T1  V1  T2  V2
558.99  65.860  60.735  103 x22  4000 x2  131.72
60.735  103 x22  4000 x2  493.13  0
x2  0.12887 ft
Maximum displacement:
  x2  x1  0.09594 ft
  1.151 in.
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PROBLEM 13.64
A 2-kg collar is attached to a spring and slides without friction in
a vertical plane along the curved rod ABC. The spring is
undeformed when the collar is at C and its constant is 600 N/m. If
the collar is released at A with no initial velocity, determine its
velocity (a) as it passes through B, (b) as it reaches C.
SOLUTION
Spring elongations:
At A,
x A  250 mm  150 mm  100 mm  0.100 m
At B,
xB  200 mm  150 mm  50 mm  0.050 m
At C ,
xC  0
Potential energies for springs.
1 2 1
kx A  (600)(0.100)2  3.00 J
2
2
1 2 1
(VB )e  kxB  (600)(0.050)2  0.75 J
2
2
(VC )e  0
(VA )e 
Gravitational potential energies: Choose the datum at level AOC.
(VA ) g  (VC ) g  0
(VB ) g  mg y  (2)(9.81)(0.200)  3.924 J
Kinetic energies:
TA  0
1 2
mvB  1.00 vB2
2
1
TC  mvC2  1.00 vC2
2
TB 
(a)
Velocity as the collar passes through B.
Conservation of energy:
TA  VA  TB  VB
0  3.00  0  1.00 vB2  0.75  3.924
vB2  6.174 m2 /s2
v B  2.48 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.64 (Continued)
(b)
Velocity as the collar reaches C.
Conservation of energy:
TA  VA  TC  VC
0  3.00  0  1.00vC2  0  0
vC2  3.00 m2 /s2
vC  1.732 m/s 
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PROBLEM 13.65
A 500-g collar can slide without friction along the semicircular rod BCD.
The spring is of constant 320 N/m and its undeformed length is 200 mm.
Knowing that the collar is released form rest at B, determine (a) the speed
of the collar as it passes through C, (b) the force exerted by the rod on the
collar at C.
SOLUTION
Given:
m  500 g
Sketch:
k  320 N/m
Undeformed length of spring  200 mm
vB  0
Finding Speed at C:
 3002  150 2   752
LAB 
 343.69318 mm
k  320 N/m
At B
TB  0
VB  VB e  VB  g
LAB  343.69318 mm  200 mm
LAB  143.69318 mm  0.14369318 m
1
1
2
2
k  LAB    320 N/m  0.1436932 m 
2
2
VB e

VB e
 3.303637 J
VB  g
 Wr   0.5 kg  9.81 m/s 2  0.15 m   0.73575 J


VB  VB e  VB  g  3.303637 J  0.73575 J  4.03939 J
At C
TC 
 
1 2
1
mvC   0.5 kg  vC2
2
2
TC  0.25vC2
VC e

1
2
k  LAC 
2
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PROBLEM 13.65 (Continued)
LAC  309.23 mm  200 mm  109.23 mm  0.10923 m
VC e

1
 320 N/m  0.10923 m 2  1.90909 J
2
TB  VB  TC  VC
0  4.0394  0.25vC2  1.90909
vC2 
(a)
Find force of rod on collar AC
4.0394  1.90909
 8.5212 m2 /s2
0.25
Free Body Diagram
vC  2.92 m/s 
Fz  0 (no friction)
F  Fxi  Fy j
  tan 1
75
 14.04
300
Fe   k LAC   cos i  sin  k 
Fe   320  0.10923 cos14.04i  sin14.04k 
Fe  33.909i  8.4797k (N)


F   Fx  33.909  i  Fy  4.905 j  8.4797k 
Fx  33.909 N  0
Fy  4.905 N   0.5 
mv 2
j  mgk
r
8.5212 m /s 
2 2
0.15 m
Fx  33.909 N
Fy  33.309 N
(b)
F  33.9 N i  33.3 N j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.66
A thin circular rod is supported in a vertical plane by a bracket at A.
Attached to the bracket and loosely wound around the rod is a spring
of constant k  3 lb/ft and undeformed length equal to the arc of
circle AB. An 8-oz collar C, not attached to the spring, can slide
without friction along the rod. Knowing that the collar is released
from rest at an angle  with the vertical, determine (a) the smallest
value of  for which the collar will pass through D and reach Point A,
(b) the velocity of the collar as it reaches Point A.
SOLUTION
(a)
Smallest angle occurs when the velocity at D is close to zero.
vC  0
vD  0
TC  0
TD  0
V  Ve  Vg
Point C:
 LBC  (1 ft)( )   ft
1
k (LBC ) 2
2
3
(VC )e   2
2
(VC )e 
R  12 in.  1 ft
(VC ) g  WR (1  cos  )
 8 oz 
(VC ) g  
 (1 ft)(1  cos )
 16 oz/lb 
(VC ) g 
1
(1  cos  )
2
3
1
VC  (VC )e  (VC ) g   2  (1  cos  )
2
2
Point D:
(VD )e  0 (spring is unattached)
(VD ) g  W (2 R)  (2)(0.5 lb)(1 ft)  1 lb  ft
TC  VC  TD  VD
By trial,
3
1
0   2  (1  cos  )  1
2
2
(1.5) 2  (0.5)cos   0.5
  0.7592 rad
  43.5 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.66 (Continued)
(b)
Velocity at A:
Point D:
VD  0 TD  0 VD  1 lb  ft[see Part (a)] 
Point A:
TA 
1 2 1 (0.5 lb) 2
mv A 
vA
2
2 (32.2 ft/s 2 )
TA  0.0077640v A2
VA  (VA ) g  W ( R)  (0.5 lb)(1 ft)  0.5 lb  ft
TA  VA  TD  VD
0.0077640v A2  0.5  0  1
v A2  64.4 ft 2 /s 2
vA  8.02 ft/s  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.67
Cornhole is a game that requires you to toss beanbags
through a hole in a wooden board. People with limited
arm mobility often have difficulty enjoying this favorite
tailgating activity. An adapted launching device attaches
to a wheelchair so that points O and A are fixed. The
device mimics an underhand throw by utilizing an elastic
band to power the arm OC, which rotates about pin O.
The elastic cord has an unstretched length of 1 ft and is
attached to fixed point A and to point B on the arm. The
combined weight of the beanbag and holder at C is 4 lbs,
and you can neglect the weight of the rod OB. Knowing
that the starting position is 30 degrees from the
horizontal as shown in the figure, determine the spring
constant if the velocity of the bean bag is 31 ft/s when
the bag is released at an angle of = 45 degrees.
SOLUTION
Given:
lO  1 ft
Sketch:
4 lbs
=0.1242 slugs
32.2 ft/s 2
v2  31 ft/s
m
 2  45
li = length of cord at positon i
si = stretch of cord in position i
hi = height of beanbag at position i
At Position 1:
From law of cosines:
l12 =22  22  2  2  2  cos30
l1  3.864 ft
Stretch in cord:
s1  l1  l0
s1  2.864 ft
Height of beanbag:
h1  3sin 30
h1  1.5 ft
At Position 2:
From law of cosines:
l22 =22  22  2  2  2  cos 45
l2  1.531 ft
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.67 (Continued)
s2  l2  l0
Stretch in cord:
s2  0.5307 ft
h2  3sin 45
Height of beanbag:
h2  2.1213 ft
Work Energy from position 1 to 2:
T 1  Vg1  Ve1  U1NC
2  T2  Vg 2  Ve2
Vg1  mgh1 , Ve1 
where:
1 2
ks1
2
1 2
mv2 , Vg2  mgh2
2
1
Ve2  ks22
2
T2 
Therefore:
1
1
1
mgh1  ks12  mv22  mgh2  ks22
2
2
2
Substitute known values:
 0.1242  32.2 1.5 
Solve for k:
1
1
1
2
2
k  2.864    0.1242  312   0.1242  32.2  2.1213  k  0.5307 
2
2
2
k  14.447 lb/ft
k  1.204 lb/in 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.68
A spring is used to stop a 50-kg package which is moving down a 20º
incline. The spring has a constant k  30 kN/m and is held by cables so
that it is initially compressed 50 mm. Knowing that the velocity of the
package is 2 m/s when it is 8 m from the spring and neglecting friction,
determine the maximum additional deformation of the spring in
bringing the package to rest.
SOLUTION
Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the
cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the
spring. Use the principle of conservation of energy. T1  V1  T2  V2 .
Position 1:
1 2 1
mv1  (50)(2)2  100 J
2
2
V1g  mgh1  (50)(9.81)(8 sin 20)  1342.09 J
T1 
V1e 
1 2 1
ke1  (30  103 )(0.050) 2  37.5 J
2
2
1 2
mv2  0 since v2  0.
2
 mgh2  (50)(9.81)( x sin 20)  167.76 x
T2 
Position 2:
V2 g
V2e 
1 2 1
ke2  (30  103 )(0.05  x)2  37.5  1500 x  15000 x 2
2
2
Principle of conservation of energy:
100  1342.09  37.5  167.61x  37.5  1500 x  15000 x 2
15,000 x2  1332.24 x  1442.09  0
Solving for x,
x  0.26882 and 0.357 64
x  0.269 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.69
Solve Problem 13.68 assuming the kinetic coefficient of friction between
the package and the incline is 0.2.
PROBLEM 13.68 A spring is used to stop a 50-kg package which is
moving down a 20 incline. The spring has a constant k  30 kN/m
and is held by cables so that it is initially compressed 50 mm. Knowing
that the velocity of the package is 2 m/s when it is 8 m from the spring
and neglecting friction, determine the maximum additional deformation
of the spring in bringing the package to rest.
SOLUTION
Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the
cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the
spring. Use the principle of work and energy. T1  V1  U12  T2  V2
Position 1.
1 2 1
mv1  (50)(2)2  100 J
2
2
V1g  mgh1  (50)(9.81)(8sin 20)  1342.09 J
T1 
V1e 
Position 2.
1 2 1
ke1  (30  103 )(0.05)2  37.5 J
2
2
1 2
mv2  0 since v2  0.
2
 mgh2  (50)(9.81)( x sin 20)  167.76 x
T2 
V2 g
V2e 
1 2 1
ke2  (30  103 )(0.05  x)2  37.5  1500 x  15,000 x 2
2
2
Work of the friction force.
Fn  0
N  mg cos 20  0
N  mg cos 20
 (50)(9.81) cos 20
 460.92 N
F f  k N
 (0.2)(460.92)
 92.184
U1 2   F f d
 92.184(8  x)
 737.47  92.184 x
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.69 (Continued)
Principle of work and energy:
T1  V1  U12  T2  V2
100         x
 167.76 x  37.5  1500 x  15, 000 x 2
15,000 x 2  1424.42 x  704.62  0
Solving for x,
x  0.17440 and  0.26936
x  0.1744 m 
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PROBLEM 13.70
A section of track for a roller coaster consists of two circular
arcs AB and CD joined by a straight portion BC. The radius
of AB is 27 m and the radius of CD is 72 m. The car and its
occupants, of total mass 250 kg, reach Point A with
practically no velocity and then drop freely along the track.
Determine the normal force exerted by the track on the car as
the car reaches point B. Ignore air resistance and rolling
resistance.
SOLUTION
Calculate the speed of the car as it reaches Point B using the principle of conservation of energy as the car
travels from position A to position B.
Position A:
v A  0,
TA 
Position B:
VB  mgh
1 2
mvA  0, VA  0 (datum)
2
where h is the decrease in elevation between A and B.
TB 
Conservation of energy:
1 2
mvB
2
TA  VA  TB  VB :
1 2
mvB  mgh
2
vB2  2 gh
00
 (2)(9.81 m/s 2 )(27 m)(1  cos 40)
 123.94 m 2 /s 2
Normal acceleration at B:
( aB ) n 
vB2


123.94 m 2 /s 2
 4.59 m/s 2
27 m
(a B )n  4.59 m/s2
50°
Apply Newton’s second law to the car at B.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.70 (Continued)
50Fn  man : N  mg cos 40  man
N  mg cos 40  man  m( g cos 40  an )
 (250 kg)[(9.81 m/s2 ) cos 40  4.59 m/s2 ]
 1878.7  1147.5
N  731 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.71
A section of track for a roller coaster consists of two circular
arcs AB and CD joined by a straight portion BC. The radius of
AB is 27 m and the radius of CD is 72 m. The car and its
occupants, of total mass 250 kg, reach Point A with
practically no velocity and then drop freely along the track.
Determine the maximum and minimum values of the normal
force exerted by the track on the car as the car travels from A
to D. Ignore air resistance and rolling resistance.
SOLUTION
Calculate the speed of the car as it reaches Point P, any point on the roller coaster track. Apply the principle of
conservation of energy.
Position A:
v A  0,
TA 
Position P:
VP  mgh
1 2
mvA  0, VA  0 (datum)
2
where h is the decrease in elevation along the track.
TP 
Conservation of energy:
1
mv 2
2
TA  VA  TP  VP
00
1 2
mv  mgh
2
v 2  2 gh
(1)
Calculate the normal force using Newton’s second law. Let  be the slope angle of the track.
Fn  man : N  mg cos   man
N  mg cos   man
Over portion AB of the track,
and
h   (1  cos  )
an  
mv 2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.71 (Continued)
where  is the radius of curvature. (   27 m)
N  mg cos  
2mg  (1  cos  )

 mg (3cos   2)
At Point A (  0)
N A  mg  (250)(9.81)  2452.5 N
At Point B (  40)
N B  (2452.5)(3cos 40  2)
N B  731 N
Over portion BC,
  40,
an  0
(straight track)
N BC  mg cos 40  2452.5cos 40
N BC  1879 N
Over portion CD,
h  hmax  r (1  cos )
an 
and
mv 2
r
where r is the radius of curvature. ( r  72 m)
2mgh
r
h

 mg cos   2mg  max  1  cos  
r


N  mg cos  
2h 

 mg  3cos   2  max 
r 

which is maximum at Point D, where
 2h 
N D  mg 1  max 
r 

hmax  27  18  45 m,
Data:
r  72 m
 (2) (45) 
N D  (2452.5) 1 
 5520 N
72 

Summary:

minimum (just above B ):
maximum (at D ):
731 N 
5520 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.72
A 1-lb collar is attached to a spring and slides without friction
along a circular rod in a vertical plane. The spring has an
undeformed length of 5 in. and a constant k  10 lb/ft.
Knowing that the collar is released from being held at A
determine the speed of the collar and the normal force between
the collar and the rod as the collar passes through B.
SOLUTION
W
1

 0.031056 lb  s 2 /ft
g 32.2
For the collar,
m
For the spring,
k  10 lb/ft l0  5 in.
 A  7  5  5  17 in.
At A:
    0  12 in.  1 ft
 B  (7  5) 2  52  13 in.
At B:
 B   0  1.8 in. 
2
ft
3
Velocity of the collar at B.
Use the principle of conservation of energy.
TA  VA  TB  VB
Where
TA 
1 2
mvA  0
2
1
k ( A   0 )2  W (0)
2
1
 (10)(1) 2  0  5 ft  lb
2
1 2 1
TB  mvB  (0.031056)vB2  0.015528vB2
2
2
1
VB  k ( B   0 )2  Wh
2
VA 
2
1
2
 5
(10)    (1)   
2
3
 12 
 1.80556 ft  lb

0  5  0.015528vB2  1.80556
vB2  205.72 ft 2 /s 2
vB  14.34 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.72 (Continued)
Forces at B.
2
Fs  k ( B   0 )  (10)    6.6667 lb.
3
5
sin 
13
5
  5 in.  ft
12
mvB2
man 

(0.031056)(205.72)
5/12
 15.3332 lb

Fy  ma y : Fs sin   W  N  man
N  man  W  Fs sin 
5
 15.3332  1  (6.6667)  
 13 
N  13.769 lb
N  13.77 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.73
A 10-lb collar is attached to a spring and slides without
friction along a fixed rod in a vertical plane. The spring
has an undeformed length of 14 in. and a constant k  4
lb/in. Knowing that the collar is released from rest in the
position shown, determine the force exerted by the rod on
the collar at (a) Point A, (b) Point B. Both these points are
on the curved portion of the rod.
SOLUTION
Mass of collar:
m
W
10

 0.31056 lb  s 2 /ft
g 32.2
Let position 1 be the initial position shown, and calculate the potential energies of the spring for positions 1,
A, and B. l0  14 in.
l1  (14  14) 2  (14)2  31.305 in.
x1  l1  l0  31.305  14  17.305 in.
(V1 )e 
1 2 1
kx1  (4)(17.305) 2  598.92 in  lb  49.910 ft  lb
2
2
l A  (14) 2  (14)2  19.799 in.
x A  l A  l0  19.799  14  5.799 in.
(V A ) e 
1 2 1
kx A  (4)(5.799) 2  67.257 in  lb  5.605 ft  lb
2
2
lB  14  14  28 in.
xB  lB  l0  28  14  14 in.
(VB )e 
Gravitational potential energies:
1 2 1
kxB  (4)(14)2  392 in  lb  32.667 ft  lb
2
2
Datum at level A.
(V1 ) g  0 (VA ) g  0
(VB ) g  Wy  (10 lb)(14 in.)  140 in  lb  11.667 ft  lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.73 (Continued)
Total potential energies:
V  Ve  Vg
V1  49.910 ft  lb, VA  5.605 ft  lb, VB  21.0 ft  lb
Kinetic energies:
T1  0
1 2 1
mv A  (0.31056)v A2  0.15528v A2
2
2
1 2 1
TB  mvB  (0.31056)vB2  0.15528vB2
2
2
TA 
Conservation of energy:
T1  V1  TA  VA :
0  49.910  0.15528vA2  5.605
Conservation of energy:
T1  V1  TB  VB
0  49.910  0.15528vB2  21.0
Normal accelerations at A and B.
vA2  285.32 ft 2 /s2
vB2  186.18 ft 2 /s2
an  v2 /
  14 in.  1.16667 ft
Spring forces at A and B:
(a A )n 
285.32 ft 2 /s 2
1.16667 ft
(a A ) n  244.56 ft/s 2
( aB ) n 
189.10 ft 2 /s 2
1.16667 ft
(a B ) n  159.58 ft/s 2
F  kx
FA  (4 lb/in.)(5.799 in.)
FB  (4 lb/in.)(14 in.)
FA  23.196 lb
45
FB  56.0 lb
To determine the forces (N A and N B ) exerted by the rod on the collar, apply Newton’s second law.
(a)
At Point A:
F  m(a A )n :
W  N A  FA sin 45  m(a A ) n
10  N A  23.196sin 45  (0.31056)(244.56)
N A  82.4 lb 
(b) At Point B:
F  m(aB )n :
N B  (0.31056)(159.58)
N B  49.6 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.74
An 8-oz package is projected upward with a
velocity v0 by a spring at A; it moves around a
frictionless loop and is deposited at C. For each of
the two loops shown, determine (a) the smallest
velocity v0 for which the package will reach C, (b)
the corresponding force exerted by the package on
the loop just before the package leaves the loop at
C.
SOLUTION
Loop 1
(a) The smallest velocity at B will occur when the force exerted by the
tube on the package is zero.
F  0  mg 
mvB2
r
vB2  rg  1.5 ft(32.2 ft/s2 )
vB2  48.30
TA 
At A
1 2
mv0
2
0.5


VA  0  8 oz  0.5 lb  
 0.01553 
32.2


At B
TB 
1 2
1
mvB  m (48.30)  24.15 m
2
2
VB  mg (7.5  1.5)  9 mg  9(0.5)  4.5 lb  ft
TA  VA  TB  VB :
1
(0.01553)v02  24.15(0.01553)  4.5
2
v02  627.82

v0  25.056 
v0  25.1 ft/s 
At C
TC 
1 2
mvC  0.007765vC2
2
VC  7.5 mg  7.5(0.5)  3.75
TA  VA  TC  VC : 0.007765v02  0.007765vC2  3.75
0.007765(25.056)2  3.75  0.007765vC2
vC2  144.87
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.74 (Continued)
(b)
Loop 2
F  man : N  0.01553
(144.87)
1.5
N  1.49989
{Package in tube} NC  1.500 lb

(a) At B, tube supports the package so,
vB  0
vB  0, TB  0
VB  mg (7.5  1.5)
 4.5 lb  ft
TA  VA  TB  VB
1
(0.01553)v A2  4.5  v A  24.073
2
v A  24.1 ft/s 
(b) At C
TC  0.007765vC2 , VC  7.5 mg  3.75
TA  VA  TC  VC : 0.007765(24.073)2  0.007765vC2  3.75
vC2  96.573
 96.573 
NC  0.01553 
  0.99985
 1.5 
{Package on tube} NC  1.000 lb 
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PROBLEM 13.75
If the package of Problem 13.74 is not to hit the
horizontal surface at C with a speed greater than
10 ft/s, (a) show that this requirement can be
satisfied only by the second loop, (b) determine
the largest allowable initial velocity v0 when the
second loop is used.
SOLUTION
(a)
Loop 1
From Problem 13.74, at B
vB2  gr  48.3 ft 2 /s 2  vB  6.9498 ft/s
TB 
1 2
1
mvB  (0.01553)(48.3)  0.37505
2
2
VB  mg (7.5  1.5)  (0.5)(9)  4.5 lb  ft
TC 
1 2
1
mvC  (0.01553)vC2  0.007765vC2
2
2
VC  7.5(0.5)  3.75 lb  ft
TB  VB  TC  VC : 0.37505  4.5  0.007765vC2  3.75
vC2  144.887  vC  12.039 ft/s
12.04 ft/s  10 ft/s  Loop (1) does not work 
1
TA  mv02  0.007765v02
2
VA  0
(b) Loop 2 at A
vC  10 ft/s
At C assume
TC 
1 2
mvC  0.007765(10)2  0.7765
2
vC  7.5(0.5)  3.75
TA  VA  TC  VC : 0.007765v02  0.7765  3.75

 v0  24.144 
v0  24.1 ft/s 
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PROBLEM 13.76
A small package of weight W is projected into a
vertical return loop at A with a velocity v0. The
package travels without friction along a circle
of radius r and is deposited on a horizontal
surface at C. For each of the two loops shown,
determine (a) the smallest velocity v0 for which
the package will reach the horizontal surface at
C, (b) the corresponding force exerted by the
loop on the package as it passes Point B.
SOLUTION
Loop 1:
(a)
Newton’s second law at position C:
F  ma:
mg  m
vc2
r
vc2  gr
Conservation of energy between position A and B.
1 2
mv0
2
VA  0
TA 
1 2
1
mvC  mgr
2
2
VC  mg (2r )  2mgr
TC 
TA  VA  TC  VC :
1 2
1
mv0  0  mgr  2mgr
2
2
v02  5gr
v0 
Smallest velocity v 0:
5 gr

(b) Conservation of energy between positions A and B.
(b) TB 
1 2
mvB ;
2
VB  mg (r )
TA  VA  TB  VB:
1 2
1
mvA  0  mvB2  mgr
2
2
1
1
m(5 gr )  0  mvB2  mgr 
2
2
vB2  3gr

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PROBLEM 13.76 (Continued)
Newton’s second law at position B.
man  m
vB2
3gr
m
 3 mg
r
r
F  Feff :
N B  3 mg
N B  3W
Force exerted by the loop:

Loop 2:
(a)
At point C, vc  0
Conservation of energy between positions A and C.
1 2
mvC  0
2
VC  mg (2r )  2mgr
TC 
TA  VA  TC  VC :
1 2
mv0  0  0  2mgr
2
v02  4 gr
v0 
Smallest velocity v 0:
(b)
4 gr

Conservation of energy between positions A and B.
TA  VA  TB  VB:
1 2
1
mv0  0  mvB2  mgr
2
2
1
1
m(4 gr )  mvB2  mgr  vB2  2 gr
2
2
Newton’s second law at position B.
man  m
vB2
2 gr
m
 2 mg
r
r
F  eff : N  2 mg
Force exerted by loop:
N  2W
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
PROBLEM 13.77
The 1 kg ball at A is suspended by an inextensible cord and given an initial
horizontal velocity of 5 m/s. If l  0.6 m and xB  0, determine yB so that the
ball will enter the basket.
SOLUTION
v1  v0
Let position 1 be at A.
Let position 2 be the point described by the angle where the path of the ball changes from circular to
parabolic. At position 2, the tension Q in the cord is zero.
Relationship between v2 and  based on Q  0. Draw the free body diagram.
F  0: Q  mg sin   man 
With
Q  0, v22  gl sin 
or
v2 
mv22
l
gl sin 
(1)
Relationship among v0 , v2 and  based on conservation of energy.
T1  V1  T2  V2
1 2
1
mv0  mgl  mv22  mgl sin 
2
2
v02  v22  2 gl (1  sin  )
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(2)
PROBLEM 13.77 (Continued)
Eliminating v2 from Eqs. (1) and (2),
v02  gl sin   2 gl (1  sin  )
 1  (5) 2

1  v02
 2   0.74912
  2  
3  gl
 3  (9.81)(0.6)

  48.514
sin  
From Eq. (1),
v22  (9.81)(0.6) sin 48.514  4.4093 m 2/s 2
v2  2.0998 m/s
x and y coordinates at position 2.
x2  l cos   0.6cos 48.514  0.39746 m
y2  l sin   0.6sin 48.514  0.44947 m
Let t2 be the time when the ball is a position 2.
Motion on the parabolic path. The horizontal motion is
x  v2 sin   2.0998 sin 48.514
 1.5730 m/s
x  x2  1.5730(t  t2 )
At Point B,
xB  0
0  0.39746  1.5730(tB  t2 ) t B  t2  0.25267 s
The vertical motion is
y  y2  v2 cos  (t  t2 ) 
At Point B,
1
g (t  t2 )2
2
yB  y2  v2 cos  (tB  t2 ) 
1
g (tB  t2 )2
2
yB  0.44947  (2.0998 cos 48.514)(0.25267)
1
 (9.81)(0.25267) 2
2
 0.48779 m
yB  0.448 m 
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PROBLEM 13.78
The pendulum shown is released from rest at A and swings through 90°
before the cord touches the fixed peg B. Determine the smallest value of
a for which the pendulum bob will describe a circle about the peg.
SOLUTION
Use conservation of energy from the point of release (A) and the top of the circle.
T1  V1  T2  V2
(1) (datum at lowest point)
where
T1  0;
V1  mg 
1
mv 2 ; V2  mgz  mg  2    a 
2
At 2
T2 
Substituting into (1)
0  mg  
1
m v 2  2mg    a 
2
(2)
We need another equation – use Newton’s 2nd law at the top.  Tension, T0  0 at top
 Fn  man  m g 
m v2

v2  g   g   a 
Substituting into (2)
mg  
1
mg    a   2 mg    a 
2
2    a  4  4a
5 a  3
a
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3

5
PROBLEM 13.79*
Prove that a force F(x, y, z) is conservative if, and only if, the following relations are satisfied:
 Fx  Fy

y
x
 Fy  Fz

z
y
 Fz  Fx

x
z
SOLUTION
For a conservative force, Equation (13.22) must be satisfied.
Fx  
We now write
Since
V
x
 Fx
 2V

y
 x y
Fy  
V
y
Fz  
V
z
 Fy
 2V

x
 y x
 Fx  Fy


y
x
 2V
 2V

:
 x y  y x
We obtain in a similar way
 Fy  Fz

z
y
 Fz  Fx


x
z
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PROBLEM 13.80
The force F  ( yzi  zxj  xyk )/xyz acts on the particle P ( x, y , z ) which moves in space. (a) Using the
relation derived in Problem 13.79, show that this force is a conservative force. (b) Determine the potential
function associated with F.
SOLUTION
Fx 
(a)
yz
xyz
Fy 
zx
xyz
Fz 
xy
xyz
 
 Fy  1y
 Fx   1x 

0

0
y
y
x
x
Thus,
 Fx  Fy

y
x
The other two equations derived in Problem 13.79 are checked in a similar way.
(b)
Recall that
Fx  
V
,
x
Fy  
V
,
y
Fz  
V
z
Fx 
1
V

x
x
V   ln x  f ( y, z )
(1)
Fy 
1
V

y
y
V   ln y  g ( z , x)
(2)
Fz 
1
V

z
z
V   ln z  h( x, y )
(3)
Equating (1) and (2)
 ln x  f ( y, z )   ln y  g ( z , x)
Thus,
f ( y , z )   ln y  k ( z )
(4)
g ( z , x)   ln x  k ( z )
(5)
Equating (2) and (3)
 ln z  h( x, y )   ln y  g ( z , x)
g ( z , x)   ln z  l ( x)
From (5),
g ( z , x)   ln x  k ( z )
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PROBLEM 13.80 (Continued)
Thus,
k ( z )   ln z
l ( x)   ln x
From (4),
f ( y, z )   ln y  ln z
Substitute for f ( y , z ) in (1)
V   ln x  ln y  ln z

V   ln xyz 
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PROBLEM 13.81*
A force F acts on a particle P(x, y) which moves in the xy plane.
Determine whether F is a conservative force and compute the work of
F when P describes the path ABCA knowing that
(a) F   kx  y  i   kx  y  j, (b) F   kx  y  i   x  ky  j.
SOLUTION
(a)
a
U AB   0 kxdx  k
a2
2
Fx  Fy ,  F is normal to BC, U BC  0
U CA   0   a  u du 
a
U ABCA   k  1
(b) From Problem 13.79,
a 2
2
a2
, not conservative 
2
Fy
Fx
1
y
x
Conservative, U ABCA  0 
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PROBLEM 13.82*
The potential function associated with a force P in space is known
to be V ( x, y, z )  ( x 2  y 2  z 2 )1/2. (a) Determine the x, y, and z
components of P. (b) Calculate the work done by P from O to D
by integrating along the path OABD, and show that it is equal to
the negative of the change in potential from O to D.
SOLUTION
(a)
(b)
Px  
V
 [ ( x 2  y 2  z 2 )1/ 2 ]

 x( x 2  y 2  z 2 ) 1/ 2
x
x

Py  
V
 [ ( x 2  y 2  z 2 )1/ 2 ]

 y ( x 2  y 2  z 2 )1/ 2
y
y

Pz  
V
 [( x 2  y 2  z 2 )1/ 2 ]

 z ( x 2  y 2  z 2 )1/ 2
z
z

U OABD  U OA  U AB  U BD
O–A: Py and Px are perpendicular to O–A and do no work.
x  y  0 and Pz  1
Also, on O–A
UO A 
Thus,

a
0
Pz dz 

a
0
dz  a
A–B: Pz and Py are perpendicular to A–B and do no work.
y  0, z  a and Px 
Also, on A–B
U A B 
Thus,

a
0
x
(x  a 2 )1/ 2
2
xdx
( x  a 2 )1/ 2
2
 a ( 2  1)
B–D: Px and Pz are perpendicular to B–D and do no work.
On
BD,
k a
za
Py 
y
( y  2a 2 )1/ 2
2
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PROBLEM 13.82* (Continued)
Thus,
U BD 

a
0
y
dy  ( y 2  2a 2 )1/ 2
( y  2a 2 )1/ 2
2
a
0
U BD  ( a 2  2a 2 )1/ 2  (2a 2 )1/ 2  a ( 3  2 )
U OABD  U O  A  U A B  U B  D
 a  a ( 2  1)  a ( 3  2)
U OABD  a 3 
VOD  V (a, a, a )  V (0, 0, 0)
 (a 2  a 2  a 2 )1/ 2  0
Thus,
VOD   a 3 
U OABD  VOD
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PROBLEM 13.83*
(a) Calculate the work done from D to O by the force P of
Problem 13.82 by integrating along the diagonal of the cube.
(b) Using the result obtained and the answer to part b of Problem
13.82, verify that the work done by a conservative force around
the closed path OABDO is zero.
PROBLEM 13.82 The potential function associated with a force
P in space is known to be V(x, y, z)  ( x 2  y 2  z 2 )1/ 2.
(a) Determine the x, y, and z components of P. (b) Calculate the
work done by P from O to D by integrating along the path OABD,
and show that it is equal to the negative of the change in potential
from O to D.
SOLUTION
From solution to (a) of Problem 13.82
P
(a)
U OD 

D
O
xi  yj  zk
( x  y 2  z 2 )1/ 2
2
P  dr
r  xi  yj  zk
dr  dxi  dyj  dzk
xi  yj  zk
P 2
( x  y 2  z 2 )1/ 2
Along the diagonal.
Thus,
x yz
P  dr 
3x
 3
(3x 2 )1/ 2
UO D 

a
0
3 dx  3a
U OD  3a 
U OABDO  U OABD  U DO
(b)
From Problem 13.82
U OABD  3a
at left
The work done from D to O along the diagonal is the negative of the work done from O to D.
U DO  U OD   3a
[see part (a)]
Thus,
U OABDO  3a  3a  0
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
PROBLEM 13.84*
The force F  ( xi  yj  zk )/(x 2  y 2  z 2 )3/2 acts on the particle P( x, y, z ) which moves in space. (a) Using
the relations derived in Problem 13.79, prove that F is a conservative force. (b) Determine the potential
function V(x, y, z) associated with F.
SOLUTION
Fx 
(a)
x
2
2
( x  y  z 2 )3/2
y
Fy  2
2
( x  y  z 2 )1/2
x   2  (2 y )
 Fx
 2
 y ( x  y 2  z 2 )5/2
3
y   32  2 y
 Fy

 x ( x 2  y 2  z 2 )5/2
Thus,
 Fx  Fy

y
x
The other two equations derived in Problem 13.79 are checked in a similar fashion.
(b)
Recalling that
V
V
V
, Fy  
, Fz  
x
y
z
x
V
Fx  
V 
dx
x
( x 2  y 2  z 2 )3/2
Fx  

V  ( x 2  y 2  z 2 ) 1/2  f ( y, z )
Similarly integrating  V/ y and  V/ z shows that the unknown function f ( x, y ) is a constant.

V
1
2
2
( x  y  z 2 )1/2
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
PROBLEM 13.85
(a) Determine the kinetic energy per unit mass which a missile must have after being fired from the surface of
the earth if it is to reach an infinite distance from the earth. (b) What is the initial velocity of the missile
(called the escape velocity)? Give your answers in SI units and show that the answer to part b is independent
of the firing angle.
SOLUTION
g  9.81 m/s2
At the surface of the earth,
r1  R  6370 km  6.37  106 m
Centric force at the surface of the earth,
GMm
R2
2
GM  gR  (9.81)(6.37  106 )2  398.06  1012 m3 /s2
F  mg 
Let position 1 be on the surface of the earth (r1  R ) and position 2 be at r2  OD. Apply the conservation of
energy principle.
T1  V1  T2  V2
1 2 GMm 1 2 GMm
mv1 
 mv2 
2
2
r1
r2
GMm GMm

R

T1 T2 GM T2
 
  gR
m m
R
m
T1  T2 
For the escape condition set
T2
0
m
T1
 gR  (9.81 m/s 2 )(6.37  106 m)  62.49  106 m2 /s 2
m
T1
 62.5 MJ/kg 
m
(a)
1 2
mvesc  mgr
2
vesc  2 gR
(b)
vesc  (2)(9.81)(6.37  106 )  11.18  103 m/s
vesc  11.18 km/s 
Note that the escape condition depends only on the speed in position 1 and is independent of the direction of
the velocity (firing angle).
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PROBLEM 13.86
A satellite describes an elliptic orbit of minimum altitude 606 km
above the surface of the earth. The semimajor and semiminor axes
are 17,440 km and 13,950 km, respectively. Knowing that the speed
of the satellite at Point C is 4.78 km/s, determine (a) the speed at
Point A, the perigee, (b) the speed at Point B, the apogee.
SOLUTION
rA  6370  606  6976 km  6.976  106 m
rC  (17440  6976) 2  (13950) 2  17438.4 km  17.4384  106 m
rB  (2)(17440)  6976  27904 km  27.904  106 m
For earth,
R  6370 km  6.37  106 m
GM  gR 2  (9.81 m/s 2 )(6.370  106 )2  398.06  1012 m3 /s2
vC  4.78 km/s  4780 m/s
(a)
Speed at Point A: Use conservation of energy.
TA  VA  TC  VC
1 2 GMm 1 2 GMm
mv A 
 mvC 
rA
rC
2
2
1
1 
v A2  vC2  2GM   
 rA rC 
1
1


 (4780) 2  (2)(398.06  1012 ) 

6
6


6.976
10
17.4384
10


 91.318  106 m2 /s 2
VA  9.556  103 m/s
vA  9.56 km/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.86 (Continued)
(b)
Speed at Point B: Use conservation of energy.
TB  VB  TC  VC
1 2 GMm 1 2 GMm
mvB 
 mvC 
rB
rC
2
2
 1
1 
vB2  vC2  2GM   
 rB rC 
1
1


 (4780)2  (2)(398.06  1012 ) 

6
6
 27.904  10 17.4384  10 
 5.7258  106 m 2 /s 2
vB  2.39  103 m/s
vB  2.39 km/s 
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PROBLEM 13.87
While describing a circular orbit 200 mi above the earth a space vehicle launches a 6000-lb communications
satellite. Determine (a) the additional energy required to place the satellite in a geosynchronous orbit at an
altitude of 22,000 mi above the surface of the earth, (b) the energy required to place the satellite in the same
orbit by launching it from the surface of the earth, excluding the energy needed to overcome air resistance.
(A geosynchronous orbit is a circular orbit in which the satellite appears stationary with respect to the ground).
SOLUTION
Geosynchronous orbit
r1  3960  200  4160 mi  21.965  106 ft
r2  3960  22,000  25,960 mi  137.07  106 ft
E  T V 
Total energy
1 2 GMm
mv 
2
r
M  mass of earth
m  mass of satellite
Newton’s second law
T 
F  man :
1 2
GM
mv  m
2
2r
E  T V 
V 
GMm
r
1 GMm GMm
1 GMm


2 r
r
2 r
GM  gRE2
E 
GMm
mv 2
GM

 v2 
2
r
r
r
E 
1 gRE2 m
1 RE2W

where (W  mg )
2 r
2 r
1 (6000)(20.9088  106 ft) 2
1.3115  1018
ft  lb

2
r
r
Geosynchronous orbit at r2  137.07  106 ft
EGs 
1.3115  1018
 9.5681  109 ft  lb
6
137.07  10
(a) At 200 mi,
E200  
r1  21.965  106 ft
1.3115  1018
 5.9709  1010
21.965  106
E300  EGs  E200  5.0141  1010
 E300  50.1  109 ft  lb 
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PROBLEM 13.87 (Continued)
(b)
Launch from earth
At launch pad
EE  
 gRE2 m
GMm

 WRE
RE
RE
EE  6000(3960  5280)  1.25453  1011
 EE  EGs  EE  9.5681  109  125.453  109
 E E  115.9  109 ft  lb 
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PROBLEM 13.88
How much energy per pound should be imparted to a satellite in order to place it in a circular orbit at an
altitude of (a) 400 mi, (b) 4000 mi?
SOLUTION
r1  R  3960 mi: v1  0, T  0
Before launching:
E1  
GMm
gR 2m

 mgR
R
R
In circular orbit of radius r2  cf . Equation 12.30
Newton’s second law
GMm mv22

r2
r22
F  man
E2 
E  E2  E1
E  
(a)

1 gmR 2
R 
  mgR   mgR  1 

2 r2
2r2 

W  mg

E
R 
 R 1 

W
2r2 

r2  3960  400  4360 mi
E
3960 

 3960 1 
  5280 ft/mi
W
4360 

(b)
GM
gR 2

r2
r2
1 2 GMm 1
R 2 gR 2m
1 gRm
mv2 
 mg


r2
r2
r2
2
2
2 r2
Energy imparted is
Unit weight
v22 
E
 1.918  106 ft  lb/lb 
W
r2  3960  4000  7960 mi
E
3960 

 3960 1 
  5280 ft/mi
W
7960 

E
 10.51  106 ft  lb/lb 
W
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.89
Knowing that the velocity of an experimental space probe fired
from the earth has a magnitude v A  32.5 Mm/h at Point A,
determine the speed of the probe as it passes through Point B.
SOLUTION
rA  R  hA  6370  4300  10740 km  10.670  106 m
rB  6370  12700  19070 km  19.070  106 m
GM  gR 2  (9.81)(6.370  106 )2  398.06  1012 m3 /s 2
v A  32.5 Mm/h  9.0278  103 m/s
Use conservation of energy.
TB  VB  TA  VA
1 2 GMm 1 2 GMm
mvB 
 mv A 
2
rB
2
rA
1 1
vB2  v A2  2GM   
 rB rA 
1
1


 (9.0278  103 )2  (2)(398.06  1012 ) 

6
6
19.070  10 10.670  10 
 48.635  106 m 2 /s 2
vB  6.97  103 m/s
vB  25.1 Mm/h 
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PROBLEM 13.90
A spacecraft is describing a circular orbit at an altitude of 1500 km above
the surface of the earth. As it passes through Point A, its speed is reduced by
40 percent and it enters an elliptic crash trajectory with the apogee at Point
A. Neglecting air resistance, determine the speed of the spacecraft when it
reaches the earth’s surface at Point B.
SOLUTION
Circular orbit velocity
vC2
GM
 2 , GM  gR 2
r
r
vC2 
GM
gR 2
(9.81 m/s 2 )(6.370  106 m)2


r
r
(6.370  106 m  1.500  106 m)
vC2  50.579  106 m2 /s2
vC  7112 m/s
Velocity reduced to 60% of vC gives vA  4267 m/s.
Conservation of energy:
TA  VA  TB  VB
1
GM m
1
GM m
 m vB2 
m v A2 
rA
rB
2
2
1
9.81(6.370  106 ) 2
vB2 9.81(6.370  106 ) 2
(4.267  103 )2 


2
2
(7.870  106 )
(6.370  106 )
vB  6.48  103 m/s
vB  6.48 km/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.91
Observations show that a celestial body traveling at 1.2  106 mi/h appears to be describing about Point B a
circle of radius equal to 60 light years. Point B is suspected of being a very dense concentration of mass called
a black hole. Determine the ratio MB/MS of the mass at B to the mass of the sun. (The mass of the sun is
330,000 times the mass of the earth, and a light year is the distance traveled by light in one year at a velocity
of 186,300 mi/s.)
SOLUTION
One light year is the distance traveled by light in one year.
Speed of light  186,300 mi/s
r  (60 yr)(186,300 mi/s)(5280 ft/mi)(365 days/yr)(24 h/day)(3600 s/h)
r  1.8612  1018 ft
Newton’s second law
F
r
2
m
v2
r
2
rv
G
2
 gRearth
MB 
GM earth
GM B m
 (32.2 ft/s 2 )(3960 mi  5280 ft/mi)2
 14.077  1015 (ft 3 /s 2 )
M sun  330,000M E : GM sun  330,000 GM earth
GM sun  (330, 000)(14.077  1015 )
 4.645  1021 ft 3 /s 2
G
MB 
4.645  1021
M sun
rv 2 M sun
rv 2

G
4.645  1021
MB
(1.8612  1018 )(1.76  106 )2

M sun
4.645  1021
MB
 1.241  109 
M sun
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PROBLEM 13.92
(a) Show that, by setting r  R  y in the right-hand member of Eq. (13.17) and expanding that member in a
power series in y/R, the expression in Eq. (13.16) for the potential energy Vg due to gravity is a first-order
approximation for the expression given in Eq. (13.17). (b) Using the same expansion, derive a second-order
approximation for Vg.
SOLUTION
Vg  
WR 2
WR 2
WR
setting r  R  y : Vg  

r
R y
1  Ry
y

Vg  WR 1  
R

1
 (1) y (1)(2)  y 2

 WR 1 

 


1 R
1 2  R 


We add the constant WR, which is equivalent to changing the datum from r   to r  R :
 y  y 2

Vg  WR      
 R  R 

(a)
First order approximation:
 y
Vg  WR    Wy 
R
[Equation 13.16]
(b)
Second order approximation:
 y  y 2 
Vg  WR     
 R  R  
Vg  Wy 
Wy 2

R
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.93
Collar A has a mass of 3 kg and is attached to a spring of constant
1200 N/m and of undeformed length equal to 0.5 m. The system is set
in motion with r  0.3 m, v  2 m/s, and vr  0. Neglecting the mass
of the rod and the effect of friction, determine the radial and
transverse components of the velocity of the collar when r  0.6 m.
SOLUTION
Let position 1 be the initial position.
r1  0.3 m
(vr )1  0,
(v )1  2 m/s,
v1  2 m/s
x1  r1  l0  (0.3  0.5)  0.2 m
Let position 2 be when r  0.6 m.
r2  0.6 m
(vr )2  ?,
(v )2  ?,
v2  ?
x2  r2  l0  (0.6  0.5)  0.1 m
Conservation of angular momentum:
r1m(v )1  v2 m(v ) 2
(v ) 2 
Conservation of energy:
r1 (v )1 (0.3)(2)

 1.000 m/s
r2
0.6
T1  V1  T2  V2
1 2 1 2 1 2 1 2
mv1  kx1  mv2  kx2
2
2
2
2
k
v22  v12 
x12  x22
m
1200
(0.2)2  (0.1)2   16 m 2 /s 2
 (2)2 

3 
(vr )22  v22  (v )2  16  1  15 m 2 /s 2


vr  3.87 m/s
vr  3.87 m/s 
v  1.000 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.94
Collar A has a mass of 3 kg and is attached to a spring of constant
1200 N/m and of undeformed length equal to 0.5 m. The system is set
in motion with r  0.3 m, v  2 m/s, and vr  0. Neglecting the mass
of the rod and the effect of friction, determine (a) the maximum
distance between the origin and the collar, (b) the corresponding
speed. (Hint: Solve the equation obtained for r by trial and error.)
SOLUTION
Let position 1 be the initial position.
r1  0.3 m
(vr )1  0,
(v )1  2 m/s,
v1  2 m/s
x1  r1  l0  0.3  0.5  0.2 m
1 2 1
mv  (3)(2) 2  6 J
2
2
1 2 1
V1  kx1  (1200)(0.2) 2  24 J
2
2
T1 
Let position 2 be when r is maximum. (vr )2  0
r2  rm
x2  (rm  0.5)
1 2 1
mv2  (3)(v ) 22  1.5(v )22
2
2
1
1
V2  kx22  (1200)(rm  0.5) 2
2
2
T2 
 600(rm  0.5)2
Conservation of angular momentum:
r1m(v )1  v2 m(v ) 2
(v )2 
Conservation of energy:
r1
(0.3)
0.6
(v ), 
(2) 
r2
rm
rm
T1  V1  T2  V2
6  24  1.5(v )22  600(rm  0.5) 2
2
 0.6 
2
30  (1.5) 
  600(rm  0.5)
 rm 
0.54
f (rm )  2  600(rm  0.5)2  30  0
rm
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PROBLEM 13.94 (Continued)
Solve for rm by trial and error.
rm (m)
0.5
1.0
0.8
0.7
0.72
0.71
f (rm )
–27.8
120.5
24.8
–4.9
0.080
–2.469
rm  0.72 
(a)
Maximum distance.
(b)
Corresponding speed.
(0.01)(0.08)
 0.7197 m
2.467  0.08
rm  0.720 m 
(v )2 
0.6
 0.8337 m/s
0.7197
(vr )2  0
v2  0.834 m/s 
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PROBLEM 13.95
A governor is designed so that the valve of negligible mass at D will
open once a vertical force greater than 20 lbs is exerted on it. In
initial testing of the device, the two 1 lb masses are at x = 1 in., and
are prevented from sliding along the rod by stops. Each mass is
connected to the valve by a 10 lb/in spring, which are both
unstretched at x = 1 in. The governor is rotating so that v1 = 30 ft/s
when the stops are removed. When the valve opens, determine the
position and velocity of the masses.
SOLUTION
Given:
1
ft
12
1 lbs
m
=0.03106 slugs
32.2 ft/s 2
v ,1  30 ft/s, vr,1  0
x1  1 in 
Sketch: Pos. 1 with stops, Pos. 2 when valve opens.
li = length of spring at positon i
si = stretch of spring in position i
Fy ,2  20 lbs
k  10 lb/in  120 lb/ft
42  x12
At Position 1:
s1  0
l1 
At Position 2:
FBD of valve where: Fs  ks2
Equilibrium in y-direction:
12
17
l1 
ft
12
F
y
 20 lbs
2 Fs cos  2  20 lbs
 4 / 12 
2ks2 
  20
 l2 
l
s2  30 2
k
Note s2  l2  l1 so:
l2
k
l1k
l2 
 l2  0.4581 ft
k  30
s2  0.1145 ft
l2  l1  30
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PROBLEM 13.95 (Continued)
 4 
Location of mass, x2  l22   
 12 
2

x2  3.771 in 
x2  0.3143 ft
Conservation of angular momentum for each mass:
mx1v ,1  mx2 v ,2
v ,2 
x1
v ,1
x2
v ,2  7.955 ft/s
Work Energy from position 1 to 2:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where:
1

T2  2  mv12 
2

1

T2  2  mv22 
2

1

Ve2  2  ks22 
2

Therefore:
1

1

1

2  mv12   2  mv22   2  ks22 
2

2

2

v2  v12 
k 2
s2
m
Substitute known values:
v2  302 
120
2
 0.1145
0.03106
v2  29.14 ft/s
Solve for vr,2:
vr ,2  v22  v2,2
vr ,2  28.04 ft/s
v 2   28.04 ft/s  e r   7.96 ft/s  e r 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.96
A 1.5-lb ball that can slide on a horizontal frictionless surface is attached to a fixed
Point O by means of an elastic cord of constant k  1 lb/in. and undeformed length 2
ft. The ball is placed at Point A, 3 ft from O, and given an initial velocity v0
perpendicular to OA. Determine (a) the smallest allowable value of the initial speed v0
if the cord is not to become slack, (b) the closest distance d that the ball will come to
Point O if it is given half the initial speed found in part a.
SOLUTION
Let L1 be the initial stretched length of the cord and L2 the length of the closest approach to Point O if the cord
does not become slack. Let position 1 be the initial state and position 2 be that of closest approach to Point O.
The only horizontal force acting on the ball is the conservative central force due to the elastic cord. At the point
of closest approach the velocity of the ball is perpendicular to the cord.
Conservation of angular momentum:
r1m v1  r2m v2
L1m v0  L2m v2 or v2 
L1v0
L2
T1  V1  T2  V2
Conservation of energy:
1 2 1
1
1
mv1  k ( L1  L0 )2  mv22  k ( L2  L0 ) 2
2
2
2
2
k
v12  v22   [( L1  L0 )2  ( L2  L0 )2 ]
m
v02 
L12 2
k
v   [( L1  L0 )2  ( L2  L0 )2 ]
2 0
m
L2
L0  2 ft, L1  3 ft
Data:
L2  L0  2 ft for zero tension in the cord at the point of closest approach.
k  1 lb/in.  12 lb/ft
m  W /g  1.5/32.2  0.04658 lb  s 2 /ft
v02 
(3)2
12
v 
[(3  2)2  (2  2) 2 ]
2 0
0.04658
(2)
1.25 v02  257.6
(a)
v02  206.1 ft 2 /s2
v0  14.36 ft/s 
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PROBLEM 13.96 (Continued)
(b)
Let v0 
1
(14.36
2
ft/s)  7.18 ft/s2 so that the cord is slack in the position of closest approach to Point O.
Let position 1 be the initial position and position 2 be position of closest approach with the cord being
slack.
Conservation of energy:
T1  V1  T2  V2
1 2 1
1
mv0  k ( L1  L0 )2  mv22
2
2
2
k
( L  L0 ) 2
m
12
 (7.18)2 
(3  2)2  309.17 ft 2 /s 2
0.04658
v2  17.583 ft/s
v22  v02 
Conservation of angular momentum:
r1m v1  r2m v2 sin 
r2 sin   d 
d 
r1v1
Lv
 10
v2
v2
(3)(7.18)
17.583
d  1.225 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.97
A 1.5-lb ball that can slide on a horizontal frictionless surface is attached to a fixed
Point O by means of an elastic cord of constant k  1 lb/in. and undeformed length 2
ft. The ball is placed at Point A, 3 ft from O, and given an initial velocity v 0
perpendicular to OA, allowing the ball to come within a distance d  9 in. of Point O
after the cord has become slack. Determine (a) the initial speed v0 of the ball, (b) its
maximum speed.
SOLUTION
Let L1 be the initial stretched length of the cord. Let position 1 be the initial position. Let position 2 be the
position of closest approach to point after the cord has become slack. While the cord is slack there are no
horizontal forces acting on the ball, so the velocity remains constant. While the cord is stretched, the only
horizontal force acting on the ball is the conservative central force due to the elastic cord. At the point of
closest approach the velocity of the ball is perpendicular to the radius vector.
Conservation of angular momentum:
r1m v1  r2 mv2
L1v0  d v2
Conservation of energy:
v2 
or
L1
v0
d
T1  V1  T2  V2
1 2 1
1
mv0  k ( L1  L0 )2  mv22  0
2
2
2
k
2
2
v0  v2   ( L1  L0 )2
m
2
v02
k
L

  12 v0    ( L1  L0 )
m
d

L0  2 ft, L1  3 ft, d  9 in.  0.75 ft
Data:
k  1 lb/in.  12 lb/ft
m  W /g  1.5/32.2  0.04658 lb  s 2 /ft
2
v02
12
3v 
 0   
(3  2)2
0.04658
 0.75 
 15 v02  257.6
v02  17.17 ft 2 /s2
(a)
(b)
Maximum speed.
vm  v2 
3v0
0.75
v0  4.14 ft/s 
vm  16.58 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.98
Using the principles of conservation of energy and conservation of angular momentum, solve part a of Sample
Problem 12.14.
SOLUTION
R  6370 km
r0  500 km  6370 km
r0  6870 km  6.87  106 m
v0  36,900 km/h

36.9  106 m
3.6  103 s
 10.25  103 m/s
Conservation of angular momentum:
r0 mv0  r1mv A,
r
v A   0
 r1
v A 
r0  rmin , r1  rmax
 6.870  106

v

 0 
r1



3
 (10.25  10 )

70.418  109
r1
(1)
Conservation of energy:
Point A:
v0  10.25  103 m/s
1 2 1
mv0  m(10.25  103 )2
2
2
TA  (m)(52.53  106 )(J)
TA 
VA  
GMm
r0
GM  gR 2  (9.81 m/s 2 )(6.37  106 m)2
GM  398  1012 m3 /s 2
r0  6.87  106 m
VA  
(398  1012 m3 /s 2 )m
(6.87  106 m)
 57.93  106 m (J)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.98 (Continued)
Point A
1 2
mv A
2
GMm
VA  
r1
TA 

398  1012 m
(J)
r1
TA  VA  TA  VA
52.53  106 m  57.93  106 m 
1
398  1012 m
m v A2  
2
r1
Substituting for vA from (1)
5.402  106 
(70.418  109 ) 2 398  1012

r1
(2)(r1 ) 2
5.402  106 
(2.4793  1021 ) 398  1012

r1
r12
(5.402  106 )r12  (398  1012 )r1  2.4793  1021  0
r1  66.7  106 m, 6.87  106 m
rmax  66,700 km 
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PROBLEM 13.99
Solve sample Problem 13.11, assuming that the elastic cord is
replaced by a central force F of magnitude (80/r2) N directed
toward O.
SAMPLE PROBLEM 13.11 A sphere of mass m  6 kg is
attached to an elastic cord of constant k  100 N/m, which is
undeformed when the sphere is located at the origin O. Knowing
that the sphere may slide without friction on the horizontal surface
and that in the position shown its velocity vA has a magnitude of
20 m/s, determine (a) the maximum and minimum distances from
the sphere to the origin O, (b) the corresponding values of its
speed.
SOLUTION
(a)
The force exerted on the sphere passes through O. Angular momentum about O is conserved.
Minimum velocity is at B, where the distance from O is maximum.
Maximum velocity is at C, where distance from O is minimum.
rA mv A sin 60  rm mvm
(0.5 m)(0.6 kg)(20 m/s)sin 60  rm (0.6 kg)vm
vm 
8.66
rm
(1)
Conservation of energy:
At Point A,
1 2 1
mv A  (0.6 kg)(20 m/s)2  120 J
2
2
80
80
,
V  Fdr  2 dr 
r
r
80
VA 
 160 J
0.5
TA 


At Point B,
1
1
TB  mvm2  (0.6 kg)vm2  0.3vm2
2
2
and Point C :
VB 
80
rm
TA  VA  TB  VB
120  160  0.3vm2 
80
rm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.99 (Continued)
Substitute (1) into (2)
2
 8.66  80
40  (0.3) 
 
rm
 rm 
rm2  2 rm  0.5625  0
rm  0.339 m and rm  1.661 m
rmax  1.661 m 
rmin  0.339 m 
(b)
Substitute rm and rm from results of part (a) into (1) to get corresponding maximum and minimum
values of the speed.
vm 
8.66
 25.6 m/s
0.339
vmax  25.6 m/s 
vm 
8.66
 5.21 m/s
1.661
vmin  5.21 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.100
A spacecraft is describing an elliptic orbit of minimum altitude
hA  2400 km and maximum altitude hB  9600 km above
the surface of the earth. Determine the speed of the spacecraft
at A.
SOLUTION
rA  6370 km  2400 km
rA  8770 km
rB  6370 km  9600 km
 15,970 km
rA mvA  rB mvB
Conservation of momentum:
vB 
Conservation of energy:
TA 
1 2
mv A
2
rA
8770
vA 
v A  0.5492v A
15,970
rB
VA 
GMm
rA
TB 
1 2
mvB
2
(1)
VB 
GMm
rB
GM  gR 2  (9.81 m/s 2 )(6370  103 m) 2  398.1  1012 m 3/s 2
(398.1  1012 ) m
 45.39  106 m
8770  103
(398.1  1012 ) m
VB 
 24.93 m
(15,970  103 )
VA 
TA  VA  TB  VB :
1
1
m v A2  45.39  106 m  m vB2  24.93  106 m
2
2
(2)
Substituting for vB in (2) from (1)
v A2 [1  (0.5492) 2 ]  40.92  106
v A2  58.59  106 m 2/s 2
vA  7.65  103 m/s
v A  27.6  103 km/h 
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PROBLEM 13.101
While describing a circular orbit, 185 mi above the surface of
the earth, a space shuttle ejects at Point A an inertial upper stage
(IUS) carrying a communication satellite to be placed in a
geosynchronous orbit (see Problem 13.87) at an altitude of
22,230 mi above the surface of the earth. Determine (a) the
velocity of the IUS relative to the shuttle after its engine has
been fired at A, (b) the increase in velocity required at B to
place the satellite in its final orbit.
SOLUTION
For earth,
R  3960 mi  20.909  106 ft
g  32.2 ft 2
GM  gR 2  (32.2)(20.909  106 )2  14.077  1015 ft 3 /s 2
Speed on a circular orbits of radius r, rA, and rB.
F  man
GMm mv 2

r
r2
GM
v2 
r
v
GM
r
rA  3960  185  4145 mi  21.886  106 ft
(v A )circ 
14.077  1015
 25.362  103 ft/s
21.886  106
rB  3960  22230  26190 mi  138.283  106 ft
(vB )circ 
14.077  1015
 10.089  103 ft/s
138.283  106
Calculate speeds at A and B for path AB.
Conservation of angular momentum: mrAvA sin  A  mrB vB sin  A
vB 
rA v A sin 90 21.886  106 v A

 0.15816 v A
rB sin 90
138.283  106
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PROBLEM 13.101 (Continued)
TA  VA  TB  VB
Conservation of energy:
1 2 GMm 1 2 GMm
mv A 
 mvB 
2
rA
2
rB
1
1  2GM (rB  rA )
v A2  vB2  2GM    
rA rB
 rA rB 
v A2  (0.15816v A ) 2 
(2)(14.077  1015 )(116.397  106 )
(21.886  106 )(138.283  106 )
0.97499v A2  1.082796  109
v A  33.325  103 ft/s
vB  (0.15816)(33.325  106 )  5.271  103 ft/s
(a)
(b)
Increase in speed at A:
vA  33.325  103  25.362  103  7.963  103 ft/s
vA  7960 ft/s 
vB  10.089  103  5.271  103  4.818  103 ft/s
vB  4820 ft/s 
Increase in speed at B:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.102
A spacecraft approaching the planet Saturn reaches Point A with a
velocity v A of magnitude 68.8  103 ft/s. It is to be placed in an
elliptic orbit about Saturn so that it will be able to periodically
examine Tethys, one of Saturn’s moons. Tethys is in a circular
orbit of radius 183  103 mi about the center of Saturn, traveling at
a speed of 37.2  103 ft/s. Determine (a) the decrease in speed
required by the spacecraft at A to achieve the desired orbit, (b) the
speed of the spacecraft when it reaches the orbit of Tethys at B.
SOLUTION

(a)
rA  607.2  106 ft
rB  966.2  106 ft
vA  speed of spacecraft in the elliptical orbit after its speed has been
decreased.
Elliptical orbit between A and B.
Conservation of energy
Point A:



1
mvA2
2
GM sat m
VA 
rA
TA 
M sa  Mass of Saturn, determine GM sa from the speed of Tethys in
its circular orbit.
(Eq. 12.44)
vcirc 
GM sat
r
2
GM sat  rB vcirc
GM sat  (966.2  106 ft 2 )(37.2  103 ft/s) 2
 1.337  1018 ft 3/s 2
VA  
(1.337  1018 ft 3/s 2 ) m
(607.2  106 ft)
 2.202  109 m

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PROBLEM 13.102 (Continued)
TB 
Point B:
1 2
mvB
2
VB 
GM sat m
(1.337  1018 ft 3 /s 2 ) m

rB
(966.2  106 ft)
VB  1.384  109
TA  VA  TB  VB ;
1
1
m vA2  2.202  109 m  m vB2  1.384  109 m
2
2
vA2  vB2  1.636  109
Conservation of angular momentum:
rA mvA  rB mvB
vB 
rA
607.2  106
vA 
vA  0.6284vA
rB
966.2  106
vA2 [1  (0.6284)2 ]  1.636  109
vA  52, 005 ft/s
(a)
v A  v A  vA  68,800  52,005
(b)
vB 
rA
vA  (0.6284)(52,005)
rB
v A  16,795 ft/s 
vB  32,700 ft/s 
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PROBLEM 13.103
A spacecraft traveling along a parabolic path toward the planet Jupiter is
expected to reach Point A with a velocity vA of magnitude 26.9 km/s. Its
engines will then be fired to slow it down, placing it into an elliptic orbit
which will bring it to within 100  103 km of Jupiter. Determine the
decrease in speed v at Point A which will place the spacecraft into the
required orbit. The mass of Jupiter is 319 times the mass of the earth.
SOLUTION
Conservation of energy.
Point A:
1
m(v A  v A )2
2
GM J m
VA 
rA
TA 
GM J  319GM E  319 gRE2
RE  6.37  106 m
GM J  (319)(9.81 m/s 2 )(6.37  106 m)2
GM J  126.98  1015 m3 /s 2
rA  350  106 m
VA 
(126.98  1015 m3 /s 2 )m
(350  106 m)
VA  (362.8  106 )m
Point B:
1 2
mvB
2
GM J m (126.98  1015 m3 /s 2 )m
VB 

rB
(100  106 m)
TB 
VB  (1269.8  106 )m
TA  VA  TB  VB
1
1
m(v A  v A )2  362.8  106 m  mvB2  1269.8  106 m
2
2
(vA  vA )2  vB2  1814  106
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 13.103 (Continued)
Conservation of angular momentum.
rA  350  106 m
rB  100  106 m
rA m(v A  v A )  rB mvB
r 
vB   A  (v A  v A )
 rB 
 350 

 (v A  v A )
 100 
(2)
Substitute vB in (2) into (1)
(v A  v A ) 2 [1  (3.5)2 ]  1814  106
(v A  v A ) 2  161.24  106
(v A  v A )  12.698  103 m/s
(Take positive root; negative root reverses flight direction.)
vA  26.9  103 m/s
(given)
vA  (26.9  103 m/s  12.698  103 m/s)
v A  14.20 km/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.104
As a first approximation to the analysis of a space flight from the
earth to Mars, it is assumed that the orbits of the earth and Mars are
circular and coplanar. The mean distances from the sun to the
earth and to Mars are 149.6  106 km and 227.8  106 km,
respectively. To place the spacecraft into an elliptical transfer orbit
at Point A, its speed is increased over a short interval of time to vA
which is faster than the earth’s orbital speed. When the spacecraft
reaches Point B on the elliptical transfer orbit, its speed vB is
increased to the orbital speed of Mars. Knowing that the mass of
the sun is 332.8  103 times the mass of the earth, determine the
increase in velocity required (a) at A, (b) at B.
SOLUTION
M  mass of the sun
GM  332.8(10)3 (9.81 m/s2 )(6.37  106 m)2  1.3247(10)20 m3 /s2
Circular orbits
Earth vE 
GM
 29.758 m/s
149.6(10)9
Mars vM 
GM
 24.115 m/s
227.8(10)9
Conservation of angular momentum
Elliptical orbit
v A (149.6)  vB (227.8)
Conservation of energy
1 2
GM
1
GM
vA 
 vB2 
9
2
2
149.6(10)
227.8(10)9
v A  vB
(227.8)
 1.52273 vB
(149.6)
1
1.3247(10) 20
1 2 1.3247(10)20
(1.52273) 2 vB2 

vB 
2
2
149.6(10)9
227.8(10)9
0.65935vB2  3.0398(10)8
vB2  4.6102(10)8
vB  21, 471 m/s, v A  32,695 m/s
(a)
Increase at A,
(b)
Increase at B,
vA  vE  32.695  29.758  2.94 km/s 
vB  vM  24.115  21.471  2.64 km/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.105
The optimal way of transferring a space vehicle from an inner circular
orbit to an outer coplanar circular orbit is to fire its engines as it passes
through A to increase its speed and place it in an elliptic transfer orbit.
Another increase in speed as it passes through B will place it in the
desired circular orbit. For a vehicle in a circular orbit about the earth at
an altitude h1  200 mi, which is to be transferred to a circular orbit at
an altitude h2  500 mi, determine (a) the required increases in speed
at A and at B, (b) the total energy per unit mass required to execute the
transfer.
SOLUTION
Elliptical orbit between A and B
Conservation of angular momentum
mrAvA  mrBvB
vA 
rB
7.170
vB 
vB
rA
6.690
rA  6370 km  320 km  6690 km,
rA  6.690  106 m
vA  1.0718vB
rB  6370 km  800 km  7170 km,
(1)
rB  7.170  106 m
R  (6370 km)  6.37  106 m
Conservation of energy
GM  gR 2  (9.81 m/s2 )(6.37  106 m)2  398.060  1012 m3 /s 2
Point A:
TA 
1 2
mv A
2
VA  
GMm
(398.060  1012 )m

rA
(6.690  106 )
VA  59.501  106 m
Point B:
TB 
1 2
mvB
2
VB  
GMm
(398.060  1012 )m

rB
(7.170  106 )
VB  55.5  106 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.105 (Continued)
TA  VA  TB  VB
1 2
1
mvA  59.501  106 m  mvB2  55.5  106 m
2
2
v A2  vB2  8.002  106
From (1)
vB2 [(1.0718)2  1]  8.002  106
vA  1.0718vB
vB2  53.79  106 m2 /s2 ,
vB  7334 m/s
v A  (1.0718)(7334 m/s)  7861 m/s
Circular orbit at A and B
(Equation 12.44)
(v A )C 
GM

rA
398.060  1012
 7714 m/s
6.690  106
(vB )C 
GM

rB
398.060  1012
 7451 m/s
7.170  106
(a) Increases in speed at A and B
v A  v A  (vA )C  7861  7714  147 m/s 
vB  (vB )C  vB  7451  7334  117 m/s 
(b) Total energy per unit mass
E/m 
E/m 
1
[(v A )2  (v A )C2  (vB )C2  (vB )2 ]
2
1
[(7861)2  (7714)2  (7451)2  (7334)2 ]
2
E/m  2.01  106 J/kg 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.106
During a flyby of the earth, the velocity of a spacecraft is 10.4 km/s
as it reaches its minimum altitude of 990 km above the surface at
Point A. At Point B the spacecraft is observed to have an altitude of
8350 km. Determine (a) the magnitude of the velocity at Point B,
(b) the angle B .
SOLUTION
At A:
hA  vr  [1.04(10)4 m/s][6.37(10)6 m  0.990(10)6 m]
hA  76.544(10)9 m2 /s
1
1
GM
(TA  VA )  v 2 
m
2
r

1
(9.81)[6.37(10)6 ]2
0
[1.04(10) 4 ]2 
2
[6.37(10)6  0.990(10)6 ]
(Parabolic orbit)
At B:
1
1
GM
(TB  VB )  vB2 
0
m
2
rB
1 2
(9.81)[6.37(106 )]2
vB 
2
[6.37(10)6  8.35(10)6 ]
vB2  54.084(10)6
vB  7.35 km/s 
(a)
hB  vB sin B rB  76.544(10)9
sin B 
76.544(10)9
7.35(10 )[6.37(10)6  8.35(10)6 ]
6
 0.707483
(b)
B  45.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.107
A space platform is in a circular orbit about the earth at an
altitude of 300 km. As the platform passes through A, a rocket
carrying a communications satellite is launched from the
platform with a relative velocity of magnitude 3.44 km/s in a
direction tangent to the orbit of the platform. This was intended
to place the rocket in an elliptic transfer orbit bringing it to
Point B, where the rocket would again be fired to place the
satellite in a geosynchronous orbit of radius 42,140 km. After
launching, it was discovered that the relative velocity imparted
to the rocket was too large. Determine the angle  at which the
rocket will cross the intended orbit at Point C.
SOLUTION
R  6370 km
rA  6370 km  300 km
rA  6.67  106 m
rC  42.14  106 m
GM  gR 2
GM  (9.81 m/s)(6.37  106 m) 2
GM  398.1  1012 m3 /s 2
For any circular orbit:
Fn  man 
2
mv circ
r
2
mv
GMm
 m circ
2
r
r
GM

r
Fn 
vcirc
Velocity at A:
(v A )circ 
GM
(398.1  1012 m3/s3 )

 7.726  103 m/s
rA
(6.67  106 m)
v A  (v A )circ  (v A ) R  7.726  103  3.44  103  11.165  103 m/s
PROBLEM 13.107 (Continued)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
Velocity at C:
Conservation of energy:
TA  VA  TC  VC
1
GM m 1
GM m
m v A2 
 m vC2 
rA
rC
2
2
 1 1
vC2  v A2  2GM   
 rC rA 
1
1


 (11.165  103 ) 2  2(398.1  1012 ) 


6
6.67  106 
 42.14  10
vC2  124.67  106  100.48  106
 24.19  106 m 2 /s 2
vC  4.919  103 m/s
Conservation of angular momentum:
rA mv A  rC mvC cos 
cos  
rAv A
rC vC
(6.67  106 )(11.165  103 )
(42.14  106 )(4.919  103 )
cos   0.35926

  68.9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.108
A satellite is projected into space with a velocity v0 at a distance r0 from the
center of the earth by the last stage of its launching rocket. The velocity v0
was designed to send the satellite into a circular orbit of radius r0.
However, owing to a malfunction of control, the satellite is not projected
horizontally but at an angle  with the horizontal and, as a result, is
propelled into an elliptic orbit. Determine the maximum and minimum
values of the distance from the center of the earth to the satellite.
SOLUTION
For circular orbit of radius r0
F  man
v02 
v02
GMm

m
r0
r02
GM
r0
But v0 forms an angle  with the intended circular path.
For elliptic orbit.
Conservation of angular momentum:
r0 mv0 cos   rA mvA
r

v A   0 cos   v0
 rA

Conservation of energy:
1 2 GMm 1 2 GMm
mv0 
 mv A 
2
r0
2
rA
r0 
2GM 
v02  v A2 
1  
r0  rA 
Substitute for vA from (1)
v02
  r 2
 2GM 
r0 
1   0  cos 2   
1  
r0 
rA 
  rA 



v02
But
GM

,
r0
2
thus
r 

r 
1   0  cos 2   2 1  0 
 rA 
 rA 
2
r 
r 
cos   0   2  0   1  0
 rA 
 rA 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 13.108 (Continued)
Solving for
r0
rA
r0  2  4  4cos 2 
1  sin 


2
rA
2cos 
1  sin 2 
(1  sin  )(1  sin  )
rA 
r0  (1  sin  )r0
1  sin 
also valid for Point A
Thus,
rmax  (1  sin  )r0
rmin  (1  sin  )r0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.109
A space vehicle is to rendezvous with an orbiting laboratory which circles
the earth at a constant altitude of 360 km. The shuttle has reached an
altitude of 60 km when its engine is shut off, and its velocity v0 forms an
angle 0 = 50° with the vertical OB at that time. What magnitude should v0
have if the shuttle’s trajectory is to be tangent at A to the orbit of the
laboratory?
SOLUTION
Conservation of energy
1 2 GMm 1 2 GMm
mv0 
 mv A 
2
rB
2
rA
v A2  v02 
So
2GM
rB

rB 
1  
rA 

(1)
Given
R  6370 km = 6.37  106 m

GM  gR 2   9.81 6.37  106

2
 398  1012
rA  6370  360  6730 km  6.73  106 m
rB  6370  60  6430 km  6.43  106 m
Substitute into (1)
vA2

v02


2 398  1012 
6.43  106 
1




6.43  106 
6.73  106 
v A2  v02  5.518  106
(2)
We need another equation  conservation of angular momentum
rB mv sin   rA mvA
rB v0 sin   6.43  106 
 
6
 v0 sin 50 
rA
 6.73  10 
vA  0.7319 v0
vA 
Substitute into (2)
 0.7319 v0 2  v02  5.518  106
0.46433 v02  5.518  106
v0  3477 m/s
v0  3450 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.110
A space vehicle is in a circular orbit at an altitude of 225 mi above the
earth. To return to earth, it decreases its speed as it passes through A by
firing its engine for a short interval of time in a direction opposite to the
direction of its motion. Knowing that the velocity of the space vehicle
should form an angle B  60° with the vertical as it reaches Point B at an
altitude of 40 mi, determine (a) the required speed of the vehicle as it
leaves its circular orbit at A, (b) its speed at Point B.
SOLUTION
(a)
rA  3960 mi  225 mi  4185 mi
rA  4185 mi  5280 ft/mi  22, 097  103 ft
rB  3960 mi  40 mi  4000 mi
rB  4000  5280  21,120  103 ft
R  3960 mi  20,909  103 ft
GM  gR 2  (32.2 ft/s2 )(20,909  103 ft)2
GM  14.077  1015 ft 3 /s 2
Conservation of energy:
1 2
mv A
2
GMm
VA 
rA
TA 

14.077  1015 m
22,097  103
 637.1  106 m
1 2
mvB
2
GMm
VB 
rB
TB 

14.077  1015 m
21,120  103
 666.5  106 m
TA  VA  TB  VB
1
1
m v A2  637.1  106 m  m vB2  666.5  106 m
2
2
2
v A  vB2  58.94  106
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 13.110 (Continued)
Conservation of angular momentum:
rA mv A  rB mvB sin B
vB 
(rA )v A
4185  1 
vA

(rB )(sin B ) 4000  sin 60 
vB  1.208 v A
(2)
Substitute vB from (2) in (1)
v A2  (1.208v A ) 2  58.94  106
v A2 [(1.208) 2  1]  58.94  106
v A2  128.27  106 ft 2 /s 2
vA  11.32  103 ft/s 
(a)
(b)
From (2)
vB  1.208v A
 1.208(11.32  106 )
 13.68  103 ft/s
vB  13.68  103 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.111*
In Problem 13.110, the speed of the space vehicle was decreased as it passed through A by firing its engine in
a direction opposite to the direction of motion. An alternative strategy for taking the space vehicle out of its
circular orbit would be to turn it around so that its engine would point away from the earth and then give it an
incremental velocity v A toward the center O of the earth. This would likely require a smaller expenditure of
energy when firing the engine at A, but might result in too fast a descent at B. Assuming this strategy is used
with only 50 percent of the energy expenditure used in Problem 13.109, determine the resulting values of B
and vB .
SOLUTION
rA  3960 mi  225 mi
rA  4185 mi  22.097  106 ft
rB  3960 mi  40 mi  4000 mi
rB  21.120  106 ft
GM  gR 2  (32.2 ft/s 2 )[(3960)(5280) ft 2 ]
GM  14.077  1015 ft 3 /s 2
Velocity in circular orbit at 225 m altitude:
Newton’s second law
F  man :
(v A )circ 
2
GMm m(v A )circ

rA
rA2
GM
14.077  1015

rA
22.097  106
 25.24  103 ft/s
Energy expenditure:
From Problem 13.110,
Energy,
v A  11.32  103 ft/s
1
1
2
m(v A )circ
 mv A2
2
2
1
1
 m(25.24  103 ) 2  m(11.32  103 )2
2
2
6
 254.46  10 m ft  lb
E109 
E109
E109
E110  (0.50)E109 
(254.46  106 m)
ft  lb
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.111* (Continued)
Thus, additional kinetic energy at A is
1
(254.46  106 m)
m(v A ) 2  E110 
ft  lb
2
2
(1)
Conservation of energy between A and B:
TA 
1
2
 (v A )2 ]
m[(v A )circ
2
TB 
1 2
mvB
2
VB 
VA 
GMm
rA
GMm
rA
TA  VA  TB  VB
1
254.46  106 m 14.077  1015 m 1 2 14.077  1015 m
m(25.24  103 )2 

 mvB 
2
2
2
22.097  106
21.120  106
vB2  637.06  106  254.46  106  1274.1  106  1333  106
vB2  950.4  103
vB  30.88  103 ft/s 
Conservation of angular momentum between A and B:
rA m(vA )circ  rB mvB sin B
 r  (v )
(4185) (25.24  103 )
 0.8565
sin B   A  A circ 
(4000) (30.88  103 )
 rB  (vB )
B  58.9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.112
Show that the values vA and vP of the speed of an earth satellite at the
apogee A and the perigee P of an elliptic orbit are defined by the
relations
v A2 
2GM rP
rA  rP rA
vP2 
2GM rA
rA  rP rP
where M is the mass of the earth, and rA and rP represent,
respectively, the maximum and minimum distances of the orbit to the
center of the earth.
SOLUTION
Conservation of angular momentum:
rA mv A  rP mvP
vA 
rP
vP
rA
(1)
Conservation of energy:
1 2 GMm 1 2 GMm
mvP 
 mv A 
2
rP
2
rA
(2)
Substituting for vA from (1) into (2)
2
vP2
2GM  rP  2 2GM

   vP 
rP
rA
 rA 
  r 2 


1   P   vP2  2GM  1  1 
  rA  
 rP rA 


rA2  rP2
rA2
vP2  2GM
rA  rP
rA rP
rA2  rP2  (rA  rP )(rA  rP )
with
vP2 
2GM
rA  rP
 rA 
  (3)
 rP 

Exchanging subscripts P and A
v A2 
2GM  rP 
 
rA  rP  rA 
Q.E.D.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.113
Show that the total energy E of an earth satellite of mass m
describing an elliptic orbit is E  GMm /(rA  rP ), where M is the
mass of the earth, and rA and rP represent, respectively, the
maximum and minimum distances of the orbit to the center of the
earth. (Recall that the gravitational potential energy of a satellite was
defined as being zero at an infinite distance from the earth.)
SOLUTION
See solution to Problem 13.112 (above) for derivation of Equation (3).
vP2 
2GM rA
(rA  rP ) rP
Total energy at Point P is
E  TP  VP 
1 2 GMm
mvP 
2
rP
1  2GMm rA  GMm


2  (rA  r0 ) rP 
rP

rA
1
 GMm 
 
 rP (rA  rP ) rP 
(r  r  r )
 GMm A A P
rP (rA  rP )

E
GMm

rA  rP
Note: Recall that gravitational potential of a satellite is defined as being zero at an infinite distance from the
earth.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.114*
A space probe describes a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O.
Show that (a) in order for the probe to leave its orbit and hit the planet at an angle  with the vertical, its
velocity must be reduced to  v0, where
  sin 
2(n  1)
n  sin 2 
(b) the probe will not hit the planet if  is larger than
2
2 /(1  n).
SOLUTION
(a)
Conservation of energy:
1
m ( v0 )2
2
GMm
VA  
nR
TA 
At A:
1
mv 2
2
GMm
VB  
R
TB 
At B:
M  mass of planet
m  mass of probe
TA  VA  TB  VB
1
GMm 1
GMm
m ( v0 )2 
 mv 2 
2
nR
2
R
(1)
Conservation of angular momentum:
nR m v0  Rmv sin 
v
n v0
sin 
(2)
Replacing v in (1) by (2)
2
( v0 ) 2 
2GM  n v0 
2GM

  R
sin
nR



Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(3)
PROBLEM 13.114* (Continued)
For any circular orbit.
an 
v2
r
Newton’s second law
2
GMm m(v ) circ

r
r2
GM
vcirc 
r
v0  vcirc 
For r  nR,
GM
nR
Substitute for v0 in (3)
GM 2GM
n 2 2  GM  2GM




nR
nR
R
sin 2   nR 
2

n 
 2 1  2   2(1  n)
 sin  
2
2 
2(1  n)(sin 2  ) 2(n  1)sin 2 
 2
(sin 2   n 2 )
(n  sin 2  )
  sin 
(b)
2(n  1)
n  sin 2 
2
Q.E.D.

Probe will just miss the planet if   90,
  sin 90
Note:
2(n  1)

n  sin 2 90
2
2
n 1
n2  1  (n  1)(n  1) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.115
A missile is fired from the ground with an initial velocity v0 forming an angle  0 with the vertical. If the
missile is to reach a maximum altitude equal to  R , where R is the radius of the earth, (a) show that the
required angle  0 is defined by the relation
sin 0  (1   ) 1 
  vesc 


1    v0 
2
where vesc is the escape velocity, (b) determine the range of allowable values of v0 .
SOLUTION
rA  R
(a)
Conservation of angular momentum:
Rmv0 sin 0  rB mvB
rB  R   R  (1   ) R
vB 
Rv0 sin 0 v0 sin 0

(1   ) R
(1   )
(1)
Conservation of energy:
1 2 GMm 1 2
GMm
 mvB 
mv0 
2
R
2
(1   ) R
TA  VA  TB  VB
v02  vB2 
2 GMm 
1  2 GMm   

1

R  1   
R  1   
Substitute for vB from (1)

sin 2 0
v02 1 
 (1   )2

From Equation (12.43):
 2 GMm   
 
R  1   

2
vesc


sin 2 0
v02 1 
 (1   ) 2

2GM
R
 2   
  vesc 

1 

2
v  
 1   esc 
2
(1   )
 v0  1  
sin 2 0
sin 0  (1   ) 1 
  vesc 


1    v0 
(2)
2
Q.E.D.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.115 (Continued)
(b)
Allowable values of v0 (for which maximum altitude   R)
0  sin 2 0  1
For sin 0  0, from (2)
2
v  
0  1   esc 
 v0  1  
v0  vesc

1
For sin 0  1, from (2)
2
v  
1
 1   esc 
2
(1   )
 v0  1  
2
 vesc 
1
1

  1   

1
 v0 
v0  vesc
2
 1  2    1 2  



 (1   )
1

1
2 
vesc

1
 v0  vesc

1 
2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.116
A spacecraft of mass m describes a circular orbit of radius r1 around
the earth. (a) Show that the additional energy E which must be
imparted to the spacecraft to transfer it to a circular orbit of larger
radius r2 is
E 
GMm(r2  r1 )
2r1r2
where M is the mass of the earth. (b) Further show that if the transfer
from one circular orbit to the other is executed by placing the
spacecraft on a transitional semielliptic path AB, the amounts of
energy  EA and  EB which must be imparted at A and B are,
respectively, proportional to r2 and r1 :
 EA 
r2
r1  r2
E
 EB 
r1
r1  r2
E
SOLUTION
(a)
For a circular orbit of radius r
F  man :
GMm
v2

m
r
r2
GM
v2 
r
E  T V 
1 2 GMm
1 GMm
mv 

2
r
2 r
(1)
Thus E required to pass from circular orbit of radius r1 to circular orbit of radius r2 is
1 GMm 1 GMm

2 r1
2 r2
GMm(r2  r1 )
Q.E.D.
E 
2r1r2
 E  E1  E2  
(b)
For an elliptic orbit, we recall Equation (3) derived in
Problem 13.113 (with vP  v1 )
v12 
2Gm r2
(r1  r2 ) r1
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.116 (Continued)
At Point A: Initially spacecraft is in a circular orbit of radius r1.
GM
2
vcirc

r1
1 2
1 GM
Tcirc  mvcirc  m
r1
2
2
After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall
and
v12 
2GM r2

(r1  r2 ) r1
T1 
2GMr2
1 2 1
mv1  m
2
2 r1 (r1  r2 )
At Point A, the increase in energy is
E A  T1  Tcirc 
Recall Equation (2):
2GMr2
1
1 GM
m
 m
r1
2 r1 (r1  r2 ) 2
E A 
GMm(2r2  r1  r2 ) GMm(r2  r1 )

2r1 (r1  r2 )
2r1 (r1  r2 )
E A 
r2
r1  r2
E A 
r2
E
(r1  r2 )
Q.E.D.
r1
E
(r1  r2 )
Q.E.D.
 GMm(r2  r1 ) 


2r1r2


A similar derivation at Point B yields,
EB 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.117*
Using the answers obtained in Problem 13.108, show that the intended circular orbit and the resulting elliptic
orbit intersect at the ends of the minor axis of the elliptic orbit.
PROBLEM 13.108 A satellite is projected into space with a velocity v0 at a distance r0 from the center of the
earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular
orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at
an angle  with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and
minimum values of the distance from the center of the earth to the satellite.
SOLUTION
If the point of intersection P0 of the circular and elliptic orbits is at an end of
the minor axis, then v0 is parallel to the major axis. This will be the case
only if   90  0 , that is if cos 0   sin  . We must therefore prove that
cos0   sin 
(1)
We recall from Equation (12.39):
1 GM
 2  C cos 
r
h
When   0,
r  rmin
(2)
and rmin  r0 (1  sin  )
1
GM
 2 C
r0 (1  sin  )
h

For   180,
(3)
r  rmax  r0 (1  sin  )
1
GM
 2 C
r0 (1  sin  )
h
(4)
Adding (3) and (4) and dividing by 2:
GM
1 
1
1




2
2r0  1  sin  1  sin  
h
1

r0 cos 2 
Subtracting (4) from (3) and dividing by 2:
1 
1
1


2r0  1  sin  1  sin 
sin 
C
r0 cos 2 
C
  1

  2r0
 2sin 

2
 1  sin  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.117* (Continued)
Substituting for
GM
h2
and C into Equation (2)
1
1
(1  sin  cos  )

r r0 cos 2 
Letting r  r0 and   0 in Equation (5), we have
cos 2   1  sin  cos  0
cos 2   1
sin 
sin 2 

sin 
  sin 
cos  0 
This proves the validity of Equation (1) and thus P0 is an end of the minor axis of the elliptic orbit.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(5)
PROBLEM 13.118*
(a) Express in terms of rmin and vmax the angular momentum per unit mass, h, and the total energy per unit
mass, E/m, of a space vehicle moving under the gravitational attraction of a planet of mass M (Figure 13.15).
(b) Eliminating vmax between the equations obtained, derive the formula
1
rmin

2
GM 
2 E  h  
1
1


m  GM  
h2 


(c) Show that the eccentricity  of the trajectory of the vehicle can be expressed as
  1
2E  h 
m  GM 
2
(d ) Further show that the trajectory of the vehicle is a hyperbola, an ellipse, or a parabola, depending on
whether E is positive, negative, or zero.
SOLUTION
(a)
Point A:
Angular momentum per unit mass.
H0
m
rmin mvmax

m
h
h  rmin vmax
(1) 
Energy per unit mass
E 1
 (T  V )
m m
E 1 1 2
GMm  1 2
GM
  mvmax 
  vmax 
m m2
rmin  2
rmin
(b)
From Eq. (1): vmax  h/rmin substituting into (2)
E 1 h 2 GM


2
m 2 rmin
rmin
E
2
2 
 1 
2GM 1
m
  2  0

  2 
rmin
h
h
 rmin 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2) 
PROBLEM 13.118* (Continued)
1
Solving the quadratic:
rmin
2
2  mE 
GM
 GM 


 2 
h2
h2
 h 

Rearranging
1
rmin
(c)
GM
 2
h
2

1  1  2 E  h 

m  GM 

(3) 
Eccentricity of the trajectory:
Eq. (12.39)
1 GM
 2 (1   cos  )
r
h
When   0,
cos   1 and r  rmin
Thus,
1
rmin
GM
(1   )
h2
  1
Comparing (3) and (4),
(d )

(4)
2E  h 
m  GM 
2
(5)
Recalling discussion in section 12.12 and in view of Eq. (5)

1. Hyperbola if   1, that is, if E  0

2. Parabola if   1, that is, if E  0

3. Ellipse if   1, that is, if E  0

Note: For circular orbit   0 and
2
1
2E  h 
0
m  GM 
GM
r
2
or
 GM  m
,
E  

 h  2
but for circular orbit
v2 
thus
1 (GM )2
1 GMm

E m

2
2 r
GMr
and
h2  v 2 r 2  GMr ,
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.119
A 35,000 Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water,
determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of
150 kN.
SOLUTION
m  35,000 Mg  35  106 kg
F  150  103 N
v1  4 km/hr  1.1111 m/s
mv1  Ft  0
(35  106 kg)(1.1111 m/s)  (150  103 N)t  0
t  259.26 s
t  4 min19 s 
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PROBLEM 13.120
A 2500-lb automobile is moving at a speed of 60 mi/h when the brakes are fully applied, causing all four
wheels to skid. Determine the time required to stop the automobile (a) on dry pavement (  k  0.75), (b) on an
icy road (  k  0.10).
SOLUTION
v1  60 mph  88 ft/s
mv1  kWt  0
t
(a)
(b)
mv1
k W

mv1
k mg

v1
k g
For k  0.75
t
88 ft/s
(0.75)(32.2 ft/s2 )
t  3.64 s 
t
88 ft/s
(0.10)(32.2 ft/s2 )
t  27.3 s 
For k  0.10
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.121
A sailboat weighing 980 lb with its occupants is running down
wind at 8 mi/h when its spinnaker is raised to increase its
speed. Determine the net force provided by the spinnaker over
the 10-s interval that it takes for the boat to reach a speed of
12 mi/h.
SOLUTION
v1  8 mi/h  11.73 ft/s t1 2  10 sec
v2  12 mi/h  17.60 ft/s
m  v1  imp1 2  mv2
m(11.73 ft/s)  Fn (10 s)  m(17.60 ft/s)
Fn 
(980 lb)(17.60 ft/s  11.73 ft/s)
(32.2 ft/s 2 )(10 s)
Fn  17.86 lb 
Note: Fn is the net force provided by the sails. The force on the sails is actually greater and includes the force
needed to overcome the water resistance on the hull.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.122
A truck is hauling a 300-kg log out of a ditch
using a winch attached to the back of the truck.
Knowing the winch applies a constant force of
2500 N and the coefficient of kinetic friction
between the ground and the log is 0.45,
determine the time for the log to reach a speed
of 0.5m/s.
SOLUTION
Apply the principle of impulse and momentum to the log.
mv1  Imp1  2  mv 2
Components in y-direction:
0  Nt  mgt cos 20  0
N  mg cos 20
Components in x-direction:
0  Tt  mgt sin 20  k Nt  mv2
(T  mg sin 20  k mg cos 20)t  mv2
[T  mg (sin 20  k cos 20)]t  mv2
Data:
T  2500 N, m  300 kg,
g  9.81 m/s 2 ,
 k  0.45,
v2  0.5 m/s
[2500  (300)(9.81)(sin 20  0.45cos 20)] t  (300)(0.5)
248.95 t  150
t  0.603 s 
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PROBLEM 13.123
The coefficients of friction between the load and the flatbed
trailer shown are  s  0.40 and k  0.35. Knowing that the
speed of the rig is 55 mi/h, determine the shortest time in
which the rig can be brought to a stop if the load is not to shift.
SOLUTION
Apply the principle impulse-momentum to the crate, knowing that, if the crate does not shift, the velocity of
the crate matches that of the truck. For impending slip the friction and normal components of the contact
force between the crate and the flatbed trailer satisfy the following equation:
Ff  s N
v1  55mi h  80.667 ft s
Given:
s  0.40, k  0.35
Impulse Momentum Diagram:
Using impulse momentum in the x-Direction
m v1  μs m gt  0
t
v1
80.667 ft s

μs g  0.4  32.2 ft s 2


t  6.26 s 
Since this is the shortest time the load can be brought to rest and the load does not slide it is also the shortest
time the rig can be brought to rest.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.124
Steep safety ramps are built beside mountain highways to
enable vehicles with defective brakes to stop. A 10-ton truck
enters a 15 ramp at a high speed v0  108 ft/s and travels for 6
s before its speed is reduced to 36 ft/s. Assuming constant
deceleration, determine (a) the magnitude of the braking force,
(b) the additional time required for the truck to stop. Neglect air
resistance and rolling resistance.
SOLUTION
W  20, 000 lb
m
20,000
 621.118 lb  s 2 /ft
32.2
Momentum in the x direction
x : mv0  ( F  mg sin15)t  mv1
621.118(108)  ( F  mg sin15)6  (621.118)(36)
F  mg sin15  7453.4
(a)
F  7453.4  20, 000 sin15  2277 lb
F  2280 lb 
(b)
mv0  ( F  mg sin15)t  0
t  total time
621.118(108)  7453.4 t  0;
t  9.00 s
Additional time  9 – 6
t  3.00 s 
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PROBLEM 13.125
Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by
friction. The train is travelling down a 5 percent grade when it decreases its speed at a constant rate from
120 mi/h to 60 mi/h in a time interval of 12 s. Determine the smallest allowable value of the coefficient of
static friction between a trunk and the floor of the baggage car if the trunk is not to slide.
SOLUTION
v1  120 mi/h  176 ft/s
v2  60 mi/h  88 ft/s
t12  12 s
Nt1 2  Wt1 2 cos 
m v1   s m gt1 2 cos   m gt1 2 sin   m v2
(176 ft/s)  s (32.2 ft/s 2 )(12 s)(cos 2.86)  (32.2 ft/s2 )(12 s)(sin 2.86)  88 ft/s
s 
176  88  (32.2)(12)(sin 2.86)
(32.2)(12)(cos 2.86)
s  0.278 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.126
The 18000-kg F-35B uses thrust vectoring to allow it
to take off vertically. In one maneuver, the pilot
reaches the top of her static hover at 200 m. The
combined thrust and lift force on the airplane applied
at the end of the static hover can be expressed as F=
(44t + 2500t2)i + (250t2 + t + 176580)j, where F and
t are expressed in newtons and seconds, respectively.
Determine (a) how long it will take the airplane to
reach a cruising speed of 1000 km/hr (cruising speed
is defined to be in the x-direction only), (b) the
altitude of the plane at this time.
SOLUTION
Given:
y1  200 m
Impulse-Momentum Diagram:
m  18000 kg
Fx  44t  2500t 2
Fy  250t 2  t  176580
vx,2  1000 km/hr  277.78 m/s

Impulse-Momentum in the x-direction:
0  Fx dt  mvx ,2
  44t  2500t dt  18000 * 277.78
t
2
0
2500 3
t  22t 2  5, 000, 000
3
33 2
t3 
t  6000  0
1250
t  18.162 s 
(a) Solve for t:
Impulse-Momentum in the y-direction:

0  Fy dt  mvy,2
  250t
t
2
0

 t  176580 dt  mvy,2
250 3 1 2
t  t  176580t  mvy,2
3
2
Solve for velocity as function of time:
vy,2 
1  250 3 1 2

t  t  176580t 

m 3
2

From kinematics:
vy,2 
dy
dt
 vy,2 dt  dy
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.126 (Continued)
Integrating:

t
0
vy,2 dt 

y2
200
y2
250 3 1 2

t  t  176580t  dt 
dy

0
200
2
 3

1  125 4 1 3

t  t  88290t 2   y2  200
18000  6
6

1
m
(b) Find y2 for t=18.162 s:

18.162 
dy
y2  1944 m

y2  1.944 km 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.127
A truck is traveling down a road with a 4-percent grade at a speed of 60 mi/h when its brakes are applied to
slow it down to 20 mi/h. An antiskid braking system limits the braking force to a value at which the wheels of
the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels
is 0.60, determine the shortest time needed for the truck to slow down.
SOLUTION
  tan 1
4
 2.29
100
mv1   imp1 2  mv 2
mv1  Wt sin   Ft  mv2
v1  60 mi/h  88 ft/s
N  W cos  W  mg
v2  20 mi/h  29.33 ft/s
F   s N   sW cos 
( m )(88 ft/s)  ( m )(32.2 ft/s2 )(t )(sin 2.29)  (0.60)( m )(32.2 ft/s 2 )(cos 2.29)(t )  ( m )(29.33 ft/s)
t
88  29.33
32.2[(0.60) cos 2.29  sin 2.29]
t  3.26 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.128
In anticipation of a long 6 upgrade, a bus driver accelerates at a constant rate from 80 km/h to 100 km/h in
8 s while still on a level section of the highway. Knowing that the speed of the bus is 100 km/h as it begins to
climb the grade at time t  0 and that the driver does not change the setting of the throttle or shift gears,
determine (a) the speed of the bus when t  10 s, (b) the time when the speed is 60 km/h.
SOLUTION
Given:
v2  100 km h  27.778 m s , v1  80 km h  22.222 m s , t  8 s
Impulse Momentum Diagram for acceleration on level ground:
Impulse-Momentum in the x-direction while on level ground:
mv1 

8
0
Fdt  mv2
F  8   m  v2  v1   F  0.6944m
Impulse Momentum Diagram for acceleration on incline:
Impulse-Momentum in the x-direction while on incline:
mv2 

10
0
 F  mg sin 6  dt  mv3
m  0.6944   9.81 sin 6  10   m  v3  27.778 
v3  24.47 m/s
(a) Solve for v3:
mv2 
v3  88.08 km h 
t
  F  mg sin 6 dt  mv
0
3
m  0.6944   9.81 sin 6   t   m 16.667  27.778 
(b)
Solve for t:
t  33.57 s 
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PROBLEM 13.129
A light train made of two cars travels at 45 mi/h. Car A weighs
18 tons, and car B weighs 13 tons. When the brakes are applied,
a constant braking force of 4300 lb is applied to each car.
Determine (a) the time required for the train to stop after the
brakes are applied, (b) the force in the coupling between the
cars while the train is slowing down.
SOLUTION
(a)
Entire train:
v1  45 mi/h  66 ft/s
WA  WB  18  13  31 tons  62,000 lb
0  (4300  4300)t1 2 
t1 2 
(b)
Car A:
62, 000
(66)
g
(62,000 lb)(66 ft/s)
(32.2 ft/s2 )(8600 lb)
t1 2  14.78 s 
WA  18 tons  36,000 lb, t12  14.78 s
0  [(4300 lb)  FC ][14.78 s] 
(36,000 lb)
(66 ft/s)
(32.2 ft/s2 )
FC  692.5 lb
FC  693 lb (tension) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.130
Solve Problem 13.129, assuming that a constant braking force
of 4300 lb is applied to car B, but the brakes on car A are not
applied.
PROBLEM 13.129 A light train made of two cars travels at
45 mi/h. Car A weighs 18 tons, and car B weighs 13 tons. When
the brakes are applied, a constant braking force of 4300 lb is
applied to each car. Determine (a) the time required for the train
to stop after the brakes are applied, (b) the force in the coupling
between the cars while the train is slowing down.
SOLUTION
(a)
Entire train:
v1  45 mi/h  66 ft/s
WA  WB  18  13  31 tons  62,000 lb
0  (4300 lb)t1 2 
(62, 000 lb)
(66 ft/s)
(32.2 ft/s 2 )
t1 2  29.55 s
(b)
t1 2  29.6 s 
Car A:
0   FC (t1 2 )  mAv1 t1 2  29.55 s
FC 
(36,000 lb)(66 ft/s)
 2497 lb
(32.2 ft/s 2 )(29.55 s)
FC  2500 lb (tension) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.131
A tractor-trailer rig with a 2000-kg tractor, a 4500-kg trailer,
and a 3600-kg trailer is traveling on a level road at 90 km/h.
The brakes on the rear trailer fail and the antiskid system of the
tractor and front trailer provide the largest possible force which
will not cause the wheels to slide. Knowing that the coefficient
of static friction is 0.75, determine (a) the shortest time for the
rig to a come to a stop, (b) the force in the coupling between
the two trailers during that time. Assume that the force exerted
by the coupling on each of the two trailers is horizontal.
SOLUTION
v1  90 km h  25 m s , v2  0
Given:
W1   6500  9.81  63765 N; W2   3600  9.81  35316 N
Impulse Momentum diagram for combined cab and trailer:
From Equilibrium:
N1  W1;
Impulse-Momentum in the x-direction:
mv1 
N 2  W2
F  0.75 N1
t
 0.75 N dt  mv
0
1
2
0.75 N1t  m  v2  v1 
t
t
(a)

m v1  v2

0.75N1
10,100  25 
 5.2798 s
 0.75 63765
t  5.28 s 
Impulse Momentum diagram for the trailer alone:
Impulse-Momentum in the x-direction:
mv1 

5.2798
0
Cdt  mv2
5.2798C  m  v1  0 
C
(b)

3600 25  v2
5.2798
  C  17046 N
C  17.05 kN Compression
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.132
The system shown is at rest when a constant 150-N force
is applied to collar B. Neglecting the effect of friction,
determine (a) the time at which the velocity of collar B
will be 2.5 m/s to the left, (b) the corresponding tension
in the cable.
SOLUTION
Constraint of cord. When the collar B moves 1 unit to the left, the weight A moves up 2 units. Thus
v A  2vB
Masses and weights:
vB 
mA  3 kg
1
vA
2
WA  29.43 N
mB  8 kg
Let T be the tension in the cable.
Principle of impulse and momentum applied to collar B.
: 0  150t  2Tt  mB (vB )2
For (vB )2  2.5 m/s
150t  2Tt  (8 kg)(2.5 m/s)
150t  2Tt  20
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 13.132 (Continued)
Principle of impulse and momentum applied to weight A.
: 0  Tt  WAt  mA (v A )2
Tt  WAt  mA (2VB 2 )
Tt  29.43t  (3 kg)(2)(2.5 m/s)
Tt  29.43t  15
(2)
To eliminate T multiply Eq. (2) by 2 and add to Eq. (1).
(a)
(b)
Time:
91.14t  50
From Eq. (2),
T
Tension in the cable.
t  0.549 s 
15
 29.43
t
T  56.8 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.133
An 8-kg cylinder C rests on a 4-kg platform A supported by a cord which
passes over the pulleys D and E and is attached to a 4-kg block B. Knowing
that the system is released from rest, determine (a) the velocity of block B
after 0.8 s, (b) the force exerted by the cylinder on the platform.


SOLUTION
(a)
Blocks A and C:
[(mA  mC )v]1  T (t1 2 )  (mA  mC ) gt1 2  [(mA  mC )v]2
0  (12 g  T )(0.8)  12v
(1)
Block B:
[mB v]1  (T )t1 2  mB gt1 2  (mB v)2

0  (T  4 g )(0.8)  4v

(2)
Adding (1) and (2), (eliminating T)
(12 g  4 g )(0.8)  (12  4)

v
(b)
(8 kg)(9.81 m/s2 )(0.8 s)
16 kg
v  3.92 m/s 
Collar A:
(mAv)1  0 0  ( FC  mA g )
(3)
From Eq. (2) with v  3.92 m/s
4v
T
 4g
0.8
(4 kg)(3.92 m/s)
T
 (4 kg)(9.81 m/s 2 )
(0.8 s)
T  58.84 N


Solving for FC in (3)
FC 

(4 kg)(3.92 m/s)
 (4 kg)(9.81 m/s 2 )  58.84 N
(0.8 s)
FC  39.2 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.134
An estimate of the expected load on overthe-shoulder seat belts is to be made before
designing prototype belts that will be
evaluated in automobile crash tests.
Assuming that an automobile traveling at 45
mi/h is brought to a stop in 110 ms,
determine (a) the average impulsive force
exerted by a 200-lb man on the belt, (b) the
maximum force Fm exerted on the belt if
the force-time diagram has the shape shown.
SOLUTION
(a)
Force on the belt is opposite to the direction shown.
v1  45 mi/h  66 ft/s,
W  200 lb

mv1  Fdt  mv 2
 Fdt  F
ave t
t  0.110 s
(200 lb)(66 ft/s)
 Fave (0.110 s)  0
(32.2 ft/s 2 )
Fave 
(b)
(200)(66)
 3727 lb
(32.2)(0.110)
Fave  3730 lb 
1
Fm (0.110 s)
2
From (a), impulse  Fave t  (3727 lb)(0.110 s)
Impulse  area under F t diagram 
1
Fm (0.110)  (3727)(0.110)
2
Fm  7450 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.135
A 60-g model rocket is fired vertically. The engine
applies a thrust P which varies in magnitude as shown.
Neglecting air resistance and the change in mass of the
rocket, determine (a) the maximum speed of the rocket
as it goes up, (b) the time for the rocket to reach its
maximum elevation.
SOLUTION
m  0.060 kg
Mass:
mg  (0.060)(9.81)  0.5886 N
Weight:
Forces acting on the model rocket:
Thrust:
P (t )(given function of t )
Weight:
W (constant)
Support:
S (acts until P  W )
Over
0  t  0.2 s:
13
t  65t
0.2
W  0.5886 N
P
S  W  P  0.5886  65t
Before the rocket lifts off,
S become zero when t  t1.
0  0.5886  65t1
t1  0.009055 s.
Impulse due to S: (t  t1 )

t
0
Sdt 

t1
0
Sdt
1
mgt1
2
 (0.5)(0.5886)(0.009055)
 0.00266 N  s

The maximum speed occurs when
dv
 a  0.
dt
At this time, W  P  0, which occurs at t2  0.8 s.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.135 (Continued)
(a)
Maximum speed (upward motion):
Apply the principle of impulse-momentum to the rocket over 0  t  t2 .

0.8
0
Pdt  area under the given thrust-time plot.
1
1
(0.2)(13)  (0.1)(13  5)  (0.8  0.3)(5)
2
2
 4.7 N  s


0.8
0
m1v1 

Wdt  (0.5886)(0.8)  0.47088 N  S
0.8
0
Pdt 

0.8
0
Sdt 

0.8
0
Wdt  mv2
0  4.7  0.00266  0.47088  0.060 v2
v2  70.5 m/s 
(b)
Time t3 to reach maximum height: (v3  0)
mv1 

t3
0
Pdt 

t3
0
Sdt  Wt3  mv3
0  4.7  0.00266  0.5886t3  0
t3  7.99 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.136
A simplified model consisting of a single straight line is to be obtained for the
variation of pressure inside the 10-mm-diameter barrel of a rifle as a 20-g bullet
is fired. Knowing that it takes 1.6 ms for the bullet to travel the length of the
barrel and that the velocity of the bullet upon exit is 700 m/s, determine the value
of p0.
SOLUTION
At
p  p0  c1  c2 t
t  0,
c1  p0
At
t  1.6  103 s,
p0
0  c1  c2 (1.6  103 s)
c2 
p0
1.6  103 s
m  20  103 kg
0 A

1.6  103 s
0
pdt  mv2
 (10  103 ) 2
A
4
A  78.54  106 m 2
0 A

1.6 103 s
0
(c1  c2 t ) dt 
20  103
g

(c )(1.6  103 s)2 
3
(78.54  106 m 2 ) (c1 )(1.6  103 s)  2
  (20  10 kg)(700 m/s)
2


1.6  103 c1  1.280  106 c2  178.25  103
(1.6  103 m 2  s) p0 
(1.280  106 m 2  s 2 )
p0  178.25  103 kg  m/s
(1.6  103 s)
p0  222.8  106 N/m 2
p0  223 MPa  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.137
A crash test is performed between an SUV A and a 2500 lb compact car B.
The compact car is stationary before the impact and has its brakes applied. A
transducer measures the force during the impact, and the force P varies as
shown. Knowing that the coefficients of friction between the tires and road
are s= 0.9 and k= 0.7, determine (a) the time at which the compact car will
start moving, (b) the maximum speed of the car, (c) the time at which the car
will come to a stop.
SOLUTION
2500 lb
 77.64 slugs
32.2 ft/s 2
s  0.9, k  0.7
mB 
Given:
Impulse-Momentum Diagram:
0  t  0.1  P  300000t lb
0.1  t  0.2  P  300000  0.2  t  lb
vB,1  0
  P  F  dt  m v
Impulse-Momentum in the x-direction:
0
From Equilibrium:
N  mB g
(a) Compact car starts moving when:
Ff  s N  s mB g
f
(1)
B B,2
P   m
 dt  0
or P   s mB g  0
Substitute into (1):
0
Solving for P:
P  2250 lbs when car starts to move
For 0  t  0.1, P  300000t
2250  300000t
s
Bg
ts  0.0075 s 
P  Ff  k N  k mB g
(b) Maximum Velocity will occur near the end of the impulse when:
For
0.1  t  0.2, P  300000  0.2  t 
0.7  2500   300000  0.2  t 
tm  0.19417 s
Use (1) to find maximum velocity:
0

0.1
ts
300000tdt 
vB ,max 
1
mB

tm
0.1

2
150000t
300000  0.2  t  dt 
|
0.1
ts


tm
ts
k mB gdt  mB vB,max
 60000t  150000t 2
|
tm

0.1

|
 k g
tm
ts
vB ,max  34.26 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.137 (Continued)
(c) Use impulse momentum from tm until car stops at tstp:
mB vB ,max 
  P  F  dt  0
mB vB,max 

tstp
tm
0.2
tm

f
300000  0.2  t  dt 
mB vB,max  60000t  150000t 2
|
0.2
tm

tstp
tm
k mB gdt  0
t stp
|
  k mB g
tm
0
tstp  1.717 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.138
A crash test is performed between a 4500 lb SUV A and a compact
car B. A transducer measures the force during the impact, and the
force P varies as shown. Knowing that the SUV is travelling 30
mph when it hits the car, determine the speed of the SUV
immediately after the impact.
SOLUTION
Given:
4500 lb
 139.75 slugs
32.2 ft/s 2
0  t  0.1  P  300000t lb
m
Impulse-Momentum Diagram of SUV:
0.1  t  0.2  P  300000  0.2  t  lb
v1  30 mph  44.0 ft/s

Impulse-Momentum in the x-direction:
mv1  Pdt  mv2
Put in load function P(t):
mv1 
Integrate:
mv1  150000t 2
Solve for v2:
v2  v1 

0.1
0
300000t 
0.1
|
0

0.2
0.1

300000  0.2  t   mv2
 60000t  150000t 2
|
0.2
0.1
 mv2
1
3000
m
v2  22.533 ft/s
v2  15.36 mi/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.139
A baseball player catching a ball can soften the impact by
pulling his hand back. Assuming that a 5-oz ball reaches his
glove at 90 mi/h and that the player pulls his hand back during
the impact at an average speed of 30 ft/s over a distance of 6
in., bringing the ball to a stop, determine the average impulsive
force exerted on the player’s hand.
SOLUTION
v  90 mi/h  132 ft/s
5
m  /g
16
 
d
 12  1 s
60
vav
30
6
t
0  Fav t  mv Fav 
 
Wv
gt
Fav 

mv
t
 165 lb  (132 ft/s)
1
(32.2 ft/s 2 )  60
s
Fav  76.9 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.140
A 1.62 ounce golf ball is hit with a golf club and leaves it with a velocity of 100 mi/h. We assume that for
0  t  t0, where t0 is the duration of the impact, the magnitude F of the force exerted on the ball can be
expressed as F  Fm sin ( t/t0 ). Knowing that t0  0.5 ms, determine the maximum value Fm of the force
exerted on the ball.
SOLUTION
W  1.62 ounces  0.10125 lb m  3.1444  103 slug
t  0.5 ms  0.5  103 s
v  100 mi/h  146.67 ft/s
The impulse applied to the ball is

t0
0
Fdt 

t0
0

Fm sin
Fm t0

t
t0
dt  
Fm t0

(cos   cos 0) 
cos
t
t0
t0
0
2 Fm t0

Principle of impulse and momentum.
mv1 

t0
0
F dt  mv 2
with v1  0,
0
Solving for Fm,
Fm 
2 Fmt0

 mv2
2t0
 mv2

 (3.1444  103 )(146.67)
(2)(0.5  103 )
 1.4488  103 lb
Fm  1.45 kip  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.141
The triple jump is a track-and-field event in which an athlete
gets a running start and tries to leap as far as he can with a hop,
step, and jump. Shown in the figure is the initial hop of the
athlete. Assuming that he approaches the takeoff line from the
left with a horizontal velocity of 10 m/s, remains in contact
with the ground for 0.18 s, and takes off at a 50° angle with a
velocity of 12 m/s, determine the vertical component of the
average impulsive force exerted by the ground on his foot. Give
your answer in terms of the weight W of the athlete.
SOLUTION
mv1  (P  W)t  mv 2
t  0.18 s
Vertical components
W
(12)(sin 50)
g
(12)(sin 50)
Pv  W 
W
(9.81)(0.18)
0  ( Pv  W )(0.18) 
Pv  6.21W 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.142
The last segment of the triple jump track-and-field event is the
jump, in which the athlete makes a final leap, landing in a sandfilled pit. Assuming that the velocity of a 80-kg athlete just
before landing is 9 m/s at an angle of 35° with the horizontal
and that the athlete comes to a complete stop in 0.22 s after
landing, determine the horizontal component of the average
impulsive force exerted on his feet during landing.
SOLUTION
m  80 kg
t  0.22 s
mv1  (P  W)t  mv 2
Horizontal components
m(9)(cos 35)  PH (0.22)  0
PH 
(80 kg)(9 m/s)(cos 35)
 2.6809 kN
(0.22 s)
PH  2.68 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.143
The design for a new cementless hip implant is to be studied using an
instrumented implant and a fixed simulated femur. Assuming the punch applies
an average force of 2 kN over a time of 2 ms to the 200 g implant determine
(a) the velocity of the implant immediately after impact, (b) the average
resistance of the implant to penetration if the implant moves 1 mm before
coming to rest.
SOLUTION
m  200 g  0.200 kg
Fave  2 kN  2000 N
t  2 ms  0.002 s
(a)
Velocity immediately after impact:
Use principle of impulse and momentum:
v1  0
v2  ?
Imp12  Fave (t )
mv1  Imp12  mv 2
0  Fave (t )  mv2
v2 
(b)
Fave (t ) (2000)(0.002)

m
0.200
v2  20.0 m/s 
Average resistance to penetration:
x  1 mm  0.001 m
v2  20.0 ft/s
v3  0
Use principle of work and energy.
T2  U 23  T3
Rave 
or
1 2
mv2  Rave (x)  0
2
mv22
(0.200)(20.0)2

 40  103 N
2(x)
(2)(0.001)
Rave  40.0 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.144
A 28-g steel-jacketed bullet is fired with a velocity of 650 m/s toward
a steel plate and ricochets along path CD with a velocity 500 m/s.
Knowing that the bullet leaves a 50-mm scratch on the surface of the
plate and assuming that it has an average speed of 600 m/s while in
contact with the plate, determine the magnitude and direction of the
impulsive force exerted by the plate on the bullet.
SOLUTION
Given:
v1  650 m/s,
m  0.028 kg,
v2  500 m/s
Impulse-momentum diagram of the bullet:
Impulse momentum in the x-dir
mv1 cos 20 Fx t  mv2 cos10
So,
Fx t  mv1 cos 20 mv2 cos10
 0.028  650 cos 20 0.028  500 cos10 3.3151 N  s
y -dir
 mv1 sin 20 Fy t  mv2 sin10
So,
Fy t  mv2 sin10 mv1 sin 20
 0.028  500  sin10 0.028  650  sin 20 8.6558 N  s
We need t. The average velocity is 600 m/s
x  vave t; t 
x
0.05 m

 83.33  106 s
vave 600 m/s
So
3.3151

 39.78 kN 
6
83.33  10


8.6558


Fy 
103.87
kN

83.33  106
Fx 
F  111.2 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.145
A 25-ton railroad car moving at
2.5 mi/h is to be coupled to a 50
ton car which is at rest with
locked wheels ( k  0.30).
Determine (a) the velocity of
both cars after the coupling is
completed, (b) the time it takes
for both cars to come to rest.
SOLUTION
Weight and mass: (Label cars A and B.)
Car A : WA  50 tons  100, 000 lb, mA  3106 lb  s 2 /ft
Car B: WB  25 tons  50,000 lb, mB  1553 lb  s 2 /ft
Initital velocities:
vA  0
vB  2.5 mi/h  3.6667 ft/s
(a)
v B  3.6667 ft/s
The momentum of the system consisting of the two cars is conserved immediately before and after
coupling.
Let v be the common velocity of that cars immediately after coupling. Apply conservation of
momentum.
: mB vB  mAv  mB v
v 
(b)
mB vB
(3106)(3.6667)

 2.444 ft/s
4569
m A  mB
v  1.667 mi/h
After coupling: The friction force acts only on car A.
 F  0 A : N A  WA  0
N A  WA
FA  k N A  kWA
(sliding)
FB  0 (Car B is rolling.)
Apply impuslse-momentum to the coupled cars.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.145 (Continued)
: (mA  mB )v  FAt  0
t
(mA  mB )v11 mB vB

FA
 k WA
t
(1553)(3.6667)
 0.1898
(0.30)(100, 000)
t  0.190 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.146
At an intersection car B was traveling south and car A was
traveling 30° north of east when they slammed into each other.
Upon investigation it was found that after the crash the two cars
got stuck and skidded off at an angle of 10° north of east. Each
driver claimed that he was going at the speed limit of 50 km/h and
that he tried to slow down but couldn’t avoid the crash because the
other driver was going a lot faster. Knowing that the masses of
cars A and B were 1500 kg and 1200 kg, respectively, determine
(a) which car was going faster, (b) the speed of the faster of the
two cars if the slower car was traveling at the speed limit.
SOLUTION
(a)
Total momentum of the two cars is conserved.
mv, x :
mAvA cos 30  (mA  mB )v cos 10
(1)
mv, y :
mAvA sin 30  mB vB  (mA  mB )v sin 10
(2)
Dividing (1) into (2),
mB v B
sin 30
sin 10


cos 30 m A v A cos 30 cos 10
vB (tan 30  tan 10)(m A cos 30)

vA
mB
vB
(1500)
cos 30
 (0.4010)
vA
(1200)
vB
 0.434
vA
v A  2.30 vB
A was going faster. 
Thus,
(b)
Since vB was the slower car.
vB  50 km/h
vA  (2.30)(50)
v A  115.2 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.147
The 650-kg hammer of a drop-hammer pile driver falls from a height of
1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground.
Assuming perfectly plastic impact (e  0), determine the average
resistance of the ground to penetration.
SOLUTION
Velocity of the hammer at impact:
Conservation of energy.
T1  0
VH  mg (1.2 m)
VH  (650 kg)(9.81 m/s 2 )(1.2 m)
V1  7652 J
1
m
2
650 2
VH2 
v  325 vH2
2
V2  0
T2 
T1  V1  T2  V2
0  7652  325 v 2
v 2  23.54 m 2 /s2
v  4.852 m/s
Velocity of pile after impact:
Since the impact is plastic (e  0), the velocity of the pile and hammer are the same after impact.
Conservation of momentum:
The ground reaction and the weights are non-impulsive.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.147 (Continued)
mH v H  ( mH  m p ) v 
Thus,
v 
Work and energy:
mH v H
(650)

(4.852 m/s)  3.992 m/s
(mH  m p ) (650  140)
d  0.110 m
T2  U 23  T3
1
(mH  mH )(v) 2
2
T3  0
T2 
1
(650  140)(3.992) 2
2
T2  6.295  103 J
T2 
U 2 3  (mH  m p ) gd  FAV d
 (650  140)(9.81)(0.110)  FAV (0.110)
U 2 3  852.49  (0.110) FAV
T2  U 23  T3
6.295  103  852.49  (0.110) FAV  0
FAV  (7147.5)/(0.110)  64.98  103 N
FAV  65.0 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.148
A small rivet connecting two pieces of sheet metal is being clinched
by hammering. Determine the impulse exerted on the rivet and the
energy absorbed by the rivet under each blow, knowing that the head
of the hammer has a weight of 1.5 lbs and that it strikes the rivet with
a velocity of 20 ft/s. Assume that the hammer does not rebound and
that the anvil is supported by springs and (a) has an infinite mass
(rigid support), (b) has a weight of 9 lb.
SOLUTION
Weight and mass:
Hammer: WH  1.5 lb
mH  0.04658 lb  s 2 /ft
Anvil: Part a: WA  
mA  
Part b: WA  9 lb
m A  0.2795 lb  s 2 /ft
Kinetic energy before impact:
1
1
mH vH2  (0.04658)(20)2  9.316 ft  lb
2
2
Let v2 be the velocity common to the hammer and anvil immediately after impact. Apply the principle of
conservation of momentum to the hammer and anvil over the duration of the impact.
T1 
: mv1   mv2
mH vH  (mH  mA )v2
v2 
mH vH
mH  mA
TA 
1
1 mH2 vH2
(mH  mA )v22 
2
2 mH  mA
T2 
mH
T1
mH  m A
(1)
Kinetic energy after impact:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.148 (Continued)
Impulse exerted on the hammer:
: mH vH  F (t )  mH v2
F t  mH (vH  v2 )
(a)
(3)
WA   :
By Eq. (1),
v2  0
By Eq. (2),
T2  0
T1  T2  9.32 ft  lb 
Energy absorbed:
By Eq. (3),
F (t )  (0.04658)(20  0)  0.932 lb  s
The impulse exerted on the rivet the same magnitude but opposite to direction.
F t  0.932 lb  s 
(b)
WA  9 lb:
By Eq. (1),
v2 
(0.04658)(20)
 2.857 ft/s
0.32608
By Eq. (2),
T2 
(0.04658)(9.316)
 1.331 ft  lb
0.32608
T1  T2  7.99 ft  lb 
Energy absorbed:
By Eq. (3),
F (t )  (0.04658)(20  2.857)
F (t )  0.799 lb  s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.149
Bullet B weighs 0.5 oz and blocks A and C both weigh
3 lb. The coefficient of friction between the blocks and
the plane is k  0.25. Initially the bullet is moving at
v0 and blocks A and C are at rest (Figure 1). After the
bullet passes through A it becomes embedded in block
C and all three objects come to stop in the positions
shown (Figure 2). Determine the initial speed of the
bullet v0.
SOLUTION
Masses:
0.5
 970.5  106 lb  s 2 /ft
(16)(32.2)
Bullet:
mB 
Blocks A and C:
mA  mC 
Block C  bullet:
3
 93.168  103 lb  s 2 /ft
32.2
mC  mB  94.138  103 lb  s 2 /ft
Normal forces for sliding blocks from N  mg  0
Block A:
N A  mA g  3.00 lb.
Block C  bullet:
NC  (mC  mB ) g  3.03125 lb.
Let v0 be the initial speed of the bullet;
v1 be the speed of the bullet after it passes through block A;
vA be the speed of block A immediately after the bullet passes through it;
vC be the speed block C immediately after the bullet becomes embedded in it.
Four separate processes and their governing equations are described below.
1.
The bullet hits block A and passes through it. Use the principle of conservation of momentum.
(v A ) 0  0
mB v0  mA (v A )0  mB v1  mAv A
v0  v1 
2.
mAv A
mB
(1)
The bullet hits block C and becomes embedded in it. Use the principle of conservation of momentum.
(vC )0  0
mB v1  mC (vC )0  (mB  mC )vC
v1 
(mB  mC )vC
mB
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.149 (Continued)
3.
Block A slides on the plane. Use principle of work and energy.
T1  U12  T2
2k N A d A
1
mAvA2  k N A d A  0 or vA 
mA
2
4.
(3)
Block C with embedded bullet slides on the plane. Use principle of work and energy.
dC  4 in.  0.33333 ft
T1  U1 2  T2
1
(mC  mB )vC2  k N C dC  0 or vC 
2
2 k N C dC
mC  mB
(4)
Applying the numerical data:
(2)(0.25)(3.03125)(0.33333)
94.138  103
 2.3166 ft/s
From Eq. (4),
vC 
From Eq. (3),
vA 
From Eq. (2),
v1 
From Eq. (1),
v0  224.71 
(2)(0.25)(3.00)(0.5)
93.168  103
 2.8372 ft/s
(94.138  103 )(2.3166)
970.5  106
 224.71 ft/s
(93.138  103 )(2.8372)
970.5  106
v0  497 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.150
A 180-lb man and a 120-lb woman stand at opposite ends of a
300-lb boat, ready to dive, each with a 16-ft/s velocity relative
to the boat. Determine the velocity of the boat after they have
both dived, if (a) the woman dives first, (b) the man dives first.
SOLUTION
(a)
Woman dives first:
Conservation of momentum:

120
300  180
(16  v1 ) 
v1  0
g
g
v1 
(120)(16)
 3.20 ft/s
600
Man dives next. Conservation of momentum:
300  180
300
180
v1  
v2 
(16  v2 )
g
g
g
v2 
(b)
480v1  (180)(16)
 2.80 ft/s
480
v 2  2.80 ft/s

Man dives first:
Conservation of momentum:
180
300  120
(16  v1 ) 
v1  0
g
g
v1 
(180)(16)
 4.80 ft/s
600
Woman dives next. Conservation of momentum:

300  120
300
120
v1 
v2 
(16  v2 )
g
g
g
420v1  (120)(16)
v2 
 0.229 ft/s
420
v2  0.229 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.151
A 75-g ball is projected from a height of 1.6 m with a horizontal
velocity of 2 m/s and bounces from a 400-g smooth plate
supported by springs. Knowing that the height of the rebound is
0.6 m, determine (a) the velocity of the plate immediately after
the impact, (b) the energy lost due to the impact.
SOLUTION
Just after impact
Just before impact
vy 
vy 
2 g (1.6)  5.603 m/s
2 g (0.6)  3.431 m/s
(a) Conservation of momentum: ( y )
mballv y  0  mballvy  mplatevplate
(0.075)(5.603)  0  0.075(3.431)  0.4vplate
vplate  1.694 m/s 
(b)
Energy loss
Initial energy
Final energy
(T  V )1 
(T  V )2 
1
(0.075)(2)2  0.075g (1.6)
2
1
1
(0.075)(2)2  0.075g (0.6)  (0.4)(1.694)2
2
2
Energy lost  (1.3272  1.1653) J  0.1619 J 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.152
A ballistic pendulum is used to measure the speed of high-speed
projectiles. A 6-g bullet A is fired into a 1-kg wood block B
suspended by a cord of length l= 2.2 m. The block then swings
through a maximum angle of = 60o. Determine (a) the initial speed
of the bullet vo, (b) the impulse imparted by the bullet on the block,
(c) the force on the cord immediately after the impact
SOLUTION
mA  6 g  0.006 kg
Given:
Impulse-Momentum Diagram in x-direction:
mB  1 kg
l  2.2 m
 =60
m A v0   m A  mB  v1
Impulse-Momentum in the x-direction during impact:
(1)
Apply Work-Energy between the impact location and the maximum swing angle (Datum at the pivot, O):
T1  Vg1  Ve1  U1NC
 3  T2  Vg 2  Ve2
where:
1
 mA  mB  v12
2
Vg1    mA  mB  gl
T1 
Vg1    mA  mB  gl cos
Hence:
1
 mA  mB  v12   mA  mB  g    mA  mB  gl cos 
2
Solving for v1:
v1  2 gl 1  cos  
 4.646 m/s
(a) Substitute into (1) and solve for v0:
v0 
 mA  mB 
mA
v0  778.92 m/s 
v1
(b) Write the Impulse Moment Equation for the block during impact:

 Fdt  4.646 J 
0  Fdt  mB v1
(c) FBD of Block just after impact:
F
n
 man
T   mA  mB  g   mA  mB 
v12
l
T  19.74 N 
 v2

T   mA  mB   1  g 
 l

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.153
A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes
embedded in a 5-lb wooden block. The block can move vertically without friction.
Determine (a) the velocity of the bullet and block immediately after the impact, (b) the
horizontal and vertical components of the impulse exerted by the block on the bullet.
SOLUTION
Weight and mass.
Bullet: w  1 oz 
Block: W  5 lb
(a)
1
lb
16
m  0.001941 lb  s 2 /ft.
M  0.15528 lb  s 2 /ft.
Use the principle of impulse and momentum applied to the bullet and the block together.
m v1  Imp1 2  mv 2
mv0 cos30  0  (m  M )v
Components :
mv0 cos30 (0.001941)(1400)cos 30°

0.157221
mM
v  14.968 ft/s
v 
v  14.97 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.153 (Continued)
(b)
Use the principle of impulse and momentum applied to the bullet alone.
x-components:
mv0 sin 30  Rx t  0
Rx t  mv0 sin 30  (0.001941)(1400)sin 30
 1.3587 lb  s
y-components:
Rx t  1.359 lb  s 
mv0 cos30  Ry t  mv
Ry t  m(v0 cos30  v)
 (0.001941)(1400 cos 30°  14.968)
Ry t  2.32 lb  s 
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PROBLEM 13.154
In order to test the resistance of a chain to impact, the chain is suspended from
a 240-lb rigid beam supported by two columns. A rod attached to the last link
is then hit by a 60-lb block dropped from a 5-ft height. Determine the initial
impulse exerted on the chain and the energy absorbed by the chain, assuming
that the block does not rebound from the rod and that the columns supporting
the beam are (a) perfectly rigid, (b) equivalent to two perfectly elastic springs.
SOLUTION
Velocity of the block just before impact:
T1  0
V1  Wh  (60 lb)(5 ft)  300 lb  ft
1 2
mv V2  0
2
T1  V1  T2  V2
T2 
0  300 
1  60  2
 v
2 g 
(600)(32.2)
60
 17.94 ft/s
v
(a)
Rigid columns:
mv  F t  0
 60 
  (17.94)  F t
 g 
F t  33.43 lb  s on the block.
F t  33.4 lb  s 
All of the kinetic energy of the block is absorbed by the chain.
T
1  60 
2
  (17.94)
2 g 
 300 ft  lb
E  300 ft  lb 
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PROBLEM 13.154 (Continued)
(b)
Elastic columns:
Momentum of system of block and beam is conserved.
mv  ( M  m)v
60
m
(17.94 ft/s)
v 
v
(m  M )
300
v  3.59 ft/s
Referring to figure in part (a),
 mv  F t   mv
F t  m(v  v )
 60 
   (17.94  3.59)
 g 
1 2 1
1
mv  mv2  Mv2
2
2
2
60
240

[(17.94) 2  (3.59) 2 ] 
(3.59) 2
2g
2g
F t  26.7 lb  s 
E
E  240 ft  lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.155
The coefficient of restitution between the two collars is known to be
0.70. Determine (a) their velocities after impact, (b) the energy loss
during impact.
SOLUTION
Impulse-momentum principle (collars A and B):
mv1  Imp12  mv 2
Horizontal components
Using data,
: mAvA  mB vB  mAvA  mB vB
(5)(1)  (3)(1.5)  5vA  3vB
5vA  3vB  0.5
or
(1)
Apply coefficient of restitution.
vB  vA  e(v A  vB )
vB  vA  0.70[1  (0.5)]
vB  vA  1.75
(a)
(2)
Solving Eqs. (1) and (2) simultaneously for the velocities,
vA  0.59375 m/s
v A  0.594 m/s

vB  1.15625 m/s
v B  1.156 m/s

1
1
1
1
m A v A2  mB vB2  (5)(1) 2  (3)(1.5) 2  5.875 J
2
2
2
2
1
1
1
1
2
2
T2  m A (vA )  mB (vB )  (5)(0.59375) 2  (3)(1.15625) 2  2.8867 J
2
2
2
2
Kinetic energies: T1 
(b)
Energy loss:
T1  T2  2.99 J 
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PROBLEM 13.156
Collars A and B, of the same mass m, are moving toward each other
with identical speeds as shown. Knowing that the coefficient of
restitution between the collars is e, determine the energy lost in the
impact as a function of m, e and v.
SOLUTION
Impulse-momentum principle (collars A and B):
mv1  Imp12  mv 2
Horizontal components
: mAvA  mB vB  mAvA  mB vB
mv  m(v)  mvA  mvB
Using data,
vA  vB  0
or
(1)
Apply coefficient of restitution.
vB  vA  e(v A  vB )
vB  vA  e [v  (v)]
vB  vA  2ev
Subtracting Eq. (1) from Eq. (2),
2vA  2ev
v A  ev
Adding Eqs. (1) and (2),
(2)
v A  ev
2vB  2ev
vB  ev
v B  ev
1
1
1
1
m A v A2  mB vB2  mv 2  m( v)2  mv 2
2
2
2
2
1
1
1
1
T2  m A (vA )2  mB (vB ) 2  m(ev) 2  m(ev) 2  e 2 mv 2
2
2
2
2
Kinetic energies: T1 
Energy loss:
T1  T2  (1  e2 ) mv 2 
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PROBLEM 13.157
One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid
surface from a height of 100 in., the height of the first bounce of the ball must be in the range 53 in.  h  58 in.
Determine the range of the coefficient of restitution of the tennis balls satisfying this requirement.
SOLUTION
Uniform accelerated motion:
v  2 gh
v  2 gh
Coefficient of restitution:
e
e
Height of drop
Height of bounce
Thus,
v
v
h
h
h  100 in.
53 in.  h  58 in.
53
58
e
100
100
0.728  e  0.762 
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PROBLEM 13.158
Two disks sliding on a frictionless horizontal plane with opposite velocities of the
same magnitude v0 hit each other squarely. Disk A is known to have a weight of
6-lb and is observed to have zero velocity after impact. Determine (a) the weight of
disk B, knowing that the coefficient of restitution between the two disks is 0.5,
(b) the range of possible values of the weight of disk B if the coefficient of
restitution between the two disks is unknown.
SOLUTION
Total momentum conserved:
mAvA  mB vB  mAvA  mB v
(mA )v0  mB (v0 )  0  mB v
m

v   A  1 v0
 mB

(1)
Relative velocities:
vB  vA  e(v A  vB )
v  2ev0
(2)
Subtracting Eq. (2) from Eq. (1) and dividing by v0 ,
mA
 1  2e  0
mB
mA
 1  2e
mB
mB 
mA
1  2e
WB 
Since weight is proportional to mass,
(a)
(b)
WA
(3)
1  2e
With WA  6 lb and e  0.5,
WB 
6
1  (2)(0.5)
WB 
6
 2 lb
1  (2)(1)
WB 
6
 6 lb
1  (2)(0)
WB  3.00 lb 
With WA  6 lb and e  1,
With WA  6 lb and e  0,

Range:
2.00 lb  WB  6.00 lb 
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PROBLEM 13.159
To apply shock loading to an artillery shell, a 20-kg pendulum A is
released from a known height and strikes impactor B at a known
velocity v0. Impactor B then strikes the 1-kg artillery shell C.
Knowing the coefficient of restitution between all objects is e,
determine the mass of B to maximize the impulse applied to the
artillery shell C.
SOLUTION
mA  20 kg, mB  ?
First impact: A impacts B.
mv  Imp12  mv 2
Impulse-momentum:
mAv0  mAvA  mB vB
Components directed left:
20v0  20vA  mB vB
(1)
vB  vA  e(v A  vB )
Coefficient of restitution:
vB  vA  ev0
vA  vB  ev0
(2)
Substituting Eq. (2) into Eq. (1) yields
20v0  20(vB  ev0 )  mB vB
20v0 (1  e)  ( mB )vB
vB 
Second impact: B impacts C.
Impulse-momentum:
20v0 (1  e)
20  mB
mB  ?, mC  1 kg
mv 2  Imp23  mv3
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(3)
PROBLEM 13.159 (Continued)
Components directed left:
mB vB  mB vB  mC vC
mB vB  mB vB  vC
(4)
vC  vB  e(vB  vC )
Coefficient of restitution:
vC  vB  evB
vB  vC  evC
(5)
Substituting Eq. (4) into Eq. (5) yields
mB vB  mB (vC  evB )  mC vC
mB vB (1  e)  (1  mB )vC
vC 
mB vB (1  e)
1  mB
vC 
20mB v0 (1  e) 2
(20  mB )(1  mB )
(6)
Substituting Eq. (3) for vB in Eq. (6) yields
The impulse applied to the shell C is
mC vC 
(1)(20) mB v0 (1  e)2
(20  mB )(1  mB )
To maximize this impulse choose mB such that
Z
mB
(20  mB )(1  mB )
is maximum. Set dZ /dmB equal to zero.
(20  mB )(1  mB )  mB [(20  mB )  (1  mB )]
dZ

0
dmB
(20  mB ) 2 (1  mB )2
20  21mB  mB2  mB (21  2mB )  0
20  mB2  0
mB  4.47 kg 
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PROBLEM 13.160
Packages in an automobile parts supply house are transported to
the loading dock by pushing them along on a roller track with
very little friction. At the instant shown, packages B and C are
at rest and package A has a velocity of 2 m/s. Knowing that the
coefficient of restitution between the packages is 0.3, determine
(a) the velocity of package C after A hits B and B hits C, (b) the
velocity of A after it hits B for the second time.
SOLUTION
(a)
Packages A and B.
Total momentum conserved.
mAv A  mB vB  mAvA  mB vB
(8 kg)(2 m/s)  0  (8 kg)vA  (4 kg)vB
4  2vA  vB
(1)
Relative velocities.
(v A  vB )e  (vB  vA )
(2)(0.3)  vB  vA
(2)
Solving Equations (1) and (2) simultaneously,
vA  1.133 m/s 
vB  1.733 m/s 
Packages B and C.
mB vB  mC vC  mB vB  mC vC
(4 kg)(1.733 m/s)  0  4vB  6vc
6.932  4vB  6vC
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(3)
PROBLEM 13.160 (Continued)
Relative velocities.
(vB  vC )e  vC  vB
(1.733)(0.3)  0.5199  vC  vB
(4)
Solving equations (3) and (4) simultaneously,
vC  0.901 m/s  
(b)
Packages A and B (second time),
Total momentum conserved.
(8)(1.133)  (4)(0.381)  8vA  4vB
10.588  8vA  4vB
(5)
Relative velocities.
(vA  vB )e  vB  vA
(1.133  0.381)(0.3)  0.2256  vB  vA
(6)
Solving (5) and (6) simultaneously,
vA  0.807 m/s
vA  0.807 m/s  
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PROBLEM 13.161
Three steel spheres of equal mass are suspended from the ceiling by cords of
equal length which are spaced at a distance slightly greater than the diameter of
the spheres. After being pulled back and released, sphere A hits sphere B, which
then hits sphere C. Denoting by e the coefficient of restitution between the
spheres and by v0 the velocity of A just before it hits B, determine (a) the
velocities of A and B immediately after the first collision, (b) the velocities of B
and C immediately after the second collision. (c) Assuming now that n spheres are
suspended from the ceiling and that the first sphere is pulled back and released as
described above, determine the velocity of the last sphere after it is hit for the first
time. (d) Use the result of part c to obtain the velocity of the last sphere when
n  8 and e  0.9.
SOLUTION
(a) First collision (between A and B)
The total momentum is conserved
mvA  mvB  mvA  mvB
v0  vA  vB
(1)
Relative velocities
 vA  vB  e   vB  vA 
v0e  vB  vA
(2)
Solving equations (1) and (2) simultaneously



vA 
v0 1  e 

2
vB 
v0 1  e 

2
(b) Second collision (Between B and C)
The total momentum is conserved.
mvB  mvC  mvB  mvC
Using the result from (a) for vB
v0 1  e 
 0  vB  vC
2
(3)
Relative velocities
 vB  0  e  vC
 vB
Substituting again for vB from (a)
v0
1  e 
2
 e   vC
 vB
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(4)
PROBLEM 13.161 (Continued)
Solving equations (3) and (4) simultaneously
vC 
e 
1  v0 1  e 
 v0 1  e 


2
2
2 
vC 
vB 
 
v0 1  e 
2

4
v0 1  e 
2
4

(c) For n spheres
n Balls
n  1th collision
We note from the answer to part (b), with n  3
vn  v3  vC 
v3 
or
v0 1  e 
2
4
 3 1
v0 1  e 
2
3 1
Thus for n balls
vn 
 n 1
v0 1  e 
2
n 1

(d) For n  8, e  0.90
From the answer to part (c) with n  8
vB 
8 1
v0 1  0.9 
2
8 1

v0 1.9 
7
 2 7
v8  0.698v0 
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PROBLEM 13.162
At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg,
and 35 kg respectively. Car A is moving to the right with a velocity vA  2 m/s and car C has a velocity vB
 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8.
Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same
time, (b) car A hits car B before car C does.
SOLUTION
Assume that each car with its rider may be treated as a particle. The masses are:
m A  200  40  240 kg,
mB  200  60  260 kg,
mC  200  35  235 kg.
Assume velocities are positive to the right. The initial velocities are:
v A  2 m/s vB  0 vC  1.5 m/s
Let vA , vB , and vC be the final velocities.
(a)
Cars A and C hit B at the same time. Conservation of momentum for all three cars.
m A v A  mB vB  mC vC  m A vA  mB vB  mC vC
(240)(2)  0  (235)(1.5)  240vA  260vB  235vC
(1)
Coefficient of restition for cars A and B.
vB  vA  e(v A  vB )  (0.8)(2  0)  1.6
(2)
Coefficient of restitution for cars B and C.
vC  vB  e(vB  vC )  (0.8)[0  (1.5)]  1.2
(3)
Solving Eqs. (1), (2), and (3) simultaneously,
vA  1.288 m/s vB  0.312 m/s vC  1.512 m/s
vA  1.288 m/s

vB  0.312 m/s

vC  1.512 m/s

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PROBLEM 13.162 (Continued)
(b)
Car A hits car B before C does.
First impact. Car A hits car B. Let vA and vB be the velocities after this impact. Conservation of
momentum for cars A and B.
m A v A  mB vB  m A vA  mB vB
(240)(2)  0  240vA  260vB
(4)
Coefficient of restitution for cars A and B.
vB  vA  e(v A  vB )  (0.8)(2  0)  1.6
(5)
Solving Eqs. (4) and (5) simultaneously,
vA  0.128 m/s, vB  1.728 m/s
vA  0.128 m/s
vB  1.728 m/s
Second impact. Cars B and C hit. Let vB and vC be the velocities after this impact. Conservation of
momentum for cars B and C.
mB vB  mC vC  mB vB  mC vC
(260)(1.728)  (235)(1.5)  260vB  235vC
(6)
Coefficient of restitution for cars B and C.
vC  vB  e(vB  vC)  (0.8)[1.728  (1.5)]  2.5824
(7)
Solving Eqs. (6) and (7) simultaneously,
vB  1.03047 m/s vC  1.55193 m/s
vB  1.03047 m/s
vC  1.55193 m/s
Third impact. Cars A and B hit again. Let vA and vB be the velocities after this impact. Conservation of
momentum for cars A and B.
m A vA  mB vB  mA vA  mB vB
(240)(0.128)  (260)(1.03047)  240vA  260vB
(8)
Coefficient of restitution for cars A and B.
vB  vA  e(vA  vB )  (0.8)[0.128  (1.03047)]  0.926776
Solving Eqs. (8) and (9) simultaneously,
vA  0.95633 m/s
vB  0.02955 m/s
vA  0.95633 m/s
vB  0.02955 m/s
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(9)
PROBLEM 13.162 (Continued)
There are no more impacts. The final velocities are:
vA  0.956 m/s

vB  0.0296 m/s

vC  1.552 m/s

We may check our results by considering conservation of momentum of all three cars over all three
impacts.
mA v A  mB vB  mC vC  (240)(2)  0  (235)(1.5)
 127.5 kg  m/s
mA vA  mB vB  mC vC  (240)(0.95633)  (260)(0.02955)  (235)(1.55193)
 127.50 kg  m/s.
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PROBLEM 13.163
At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg,
and 35 kg respectively. Car A is moving to the right with a velocity vA  2 m/s when it hits stationary car B.
The coefficient of restitution between each car is 0.8. Determine the velocity of car C so that after car B
collides with car C the velocity of car B is zero.
SOLUTION
Assume that each car with its rider may be treated as a particle. The masses are:
mA  200  40  240 kg
mB  200  60  260 kg
mC  200  35  235 kg
Assume velocities are positive to the right. The initial velocities are:
vA  2 m/s, vB  0, vC  ?
First impact. Car A hits car B. Let vA and vB be the velocities after this impact. Conservation of momentum
for cars A and B.
m A v A  mB vB  m A vA  mB vB
(240)(2)  0  240vA  260vB
(1)
Coefficient of restitution for cars A and B.
vB  vA  e(v A  vB )  (0.8)(2  0)  1.6
(2)
Solving Eqs. (1) and (2) simultaneously,
vA  0.128 m/s
vB  1.728 m/s
vA  0.128 m/s
vB  1.728 m/s
Second impact. Cars B and C hit. Let vB and vC be the velocities after this impact. vB  0. Coefficient of
restitution for cars B and C.
vC  vB  e(vB  vC )  (0.8)(1.728  vC )
vC  1.3824  0.8vC
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PROBLEM 13.163 (Continued)
Conservation of momentum for cars B and C.
mB vB  mC vC  mB vB  mC vC
(260)(1.728)  235vC  (260)(0)  (235)(1.3824  0.8vC )
(235)(1.8)vC  (235)(1.3824)  (260)(1.728)
vC  0.294 m/s
vC  0.294 m/s
Note: There will be another impact between cars A and B.
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
PROBLEM 13.164
Two identical billiard balls can move freely on a horizontal
table. Ball A has a velocity v0 as shown and hits ball B, which is
at rest, at a Point C defined by   45°. Knowing that the
coefficient of restitution between the two balls is e  0.8 and
assuming no friction, determine the velocity of each ball after
impact.
SOLUTION
Ball A: t-dir
mv0 sin   mvAt  vAt  v0 sin 
Ball B: t-dir
0  mB vBt  vBt  0
Balls A + B: n-dir
mv0 cos  0  m vAn  m vBn
(1)
Coefficient of restitution
vBn
  vAn  e (v An  vBn )
vBn  vAn  e (v0 cos  0)
Solve (1) and (2)
1  e

1  e 
vAn  v0 
cos  ; vBn  v0 
 cos
 2

 2 
With numbers
e  0.8;   45
vAt  v0 sin 45 0.707 v0
 1  0.8

vAn  v0 
cos 45   0.0707 v0
 2

vBt  0
 1  0.8 
vBn  v0 
 cos 45  0.6364 v0
 2 
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(2)
PROBLEM 13.164 (Continued)
(A)
1
v" A  [(0.707 v0 ) 2  (0.0707v0 ) 2 ] 2  0.711v0

 0.711v0 
 0.0707 
  tan 1 
  5.7106
 0.707 
So
  45  5.7106  39.3
(B)
vA  0.711v0
39.3° 
vB  0.636 v0
45° 
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PROBLEM 13.165
Two identical pool balls of 2.37-in. diameter may move freely on a pool
table. Ball B is at rest and ball A has an initial velocity v  v0 i.
(a) Knowing that b  2 in. and e  0.7, determine the velocity of each ball
after impact. (b) Show that if e  1, the final velocities of the balls form a
right angle for all values of b.
SOLUTION
Geometry at instant of impact.
b
2

d 2.37
  57.552
sin  
Directions n and t are shown in the figure.
Principle of impulse and momentum.
Ball B
Ball A
Ball A, t-direction:
mv0 sin   0  m(v A )t
Ball B, t-direction:
0  0  m(vB )t
Balls A and B, n-direction:
Coefficient of restitution.
(a)
(v A )t  v0 sin 
(1)
( vB ) t  0
(2)
mv0 cos   0  m(v A )n  m(vB )n
(v A )n  (vB )n  v0 cos 
(3)
(vB )n  (v A )n  e [v0 cos  ]
(4)
e  0.7. From Eqs. (1) and (2),
(v A )t  0.84388v0
(1)
(vB )t  0
(2)
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PROBLEM 13.165 (Continued)
From Eqs. (3) and (4) ,
(vA )n  (vB )n  0.53653v0
(3)
(vB )n  (vA )n  (0.7)(0.53653v0 )
(4)
Solving Eqs. (5) and (6) simultaneously,
(v A ) n  0.08048v0
(vB )n  0.45605v0
v A  (v A ) 2n  (v A )t2
 (0.08048v0 ) 2  (0.84388v02
 0.84771v0
tan  
(v A ) n 0.08048v0

 0.09537
(v A )t 0.84388v0
  5.448
  90    
 90  57.552  5.448
 27.000
(b)
v A  0.848v0
27.0 
v B  0.456v0
57.6 
e  1. Eqs. (3) and (4) become
(vA )n  (vB )n  v0 cos 
(3)
(vB )n  (vA )n  v0 cos
(4)
Solving Eqs. (3) and (4) simultaneously,
(v A )n  0, (vB )t  v0 cos 
But
(vA )t  v0 sin  , and (vB )t  0
vA is in the t-direction and vB is in the n-direction; therefore, the velocity vectors form a right angle.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.166
A 600-g ball A is moving with a velocity of magnitude 6 m/s when it is
hit as shown by a 1-kg ball B which has a velocity of magnitude 4 m/s.
Knowing that the coefficient of restitution is 0.8 and assuming no
friction, determine the velocity of each ball after impact.
SOLUTION
Before
After
v A  6 m/s
(v A )n  (6)(cos 40)  4.596 m/s
(v A )t  6(sin 40°)  3.857 m/s
vB  (vB )n  4 m/s
(vB )t  0
t-direction:
Total momentum conserved:
mA (v A )t  mB (vB )t  mA(vB )t  mB (vB )t
(0.6 kg)( 3.857 m/s)  0  (0.6 kg)(vA )t  (1 kg)(vB )t
2.314 m/s  0.6 (vA )t  (vB )t
(1)
Ball A alone:
Momentum conserved:
mA (v A )t  mA (vA )t 3.857  (vA )t
(vA )t  3.857 m/s
Replacing (vA )t in (2) in Eq. (1)
2.314  (0.6)(3.857)  (vB )t
2.314  2.314  (vB )t
(vB )t  0
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(2)
PROBLEM 13.166 (Continued)
n-direction:
Relative velocities:
[(v A ) n  (vB )n ]e  (vB ) n  (vA ) n
[(4.596)  (4)](0.8)  (vB ) n  (vA )n
6.877  (vB ) n  (vA )n
(3)
Total momentum conserved:
mA (v A )n  mB (vB )n  mA (vA ) n  mB (vB ) n
(0.6 kg)(4.596 m/s)  (1 kg)(  4 m/s)  (1 kg)(vB )n  (0.6 kg)(vA )n
1.2424  (vB ) n  0.6(vA ) n
(4)
Solving Eqs. (4) and (3) simultaneously,
(vA )n  5.075 m/s
(vB )n  1.802 m/s
Velocity of A:
tan  
|(v A )t |
|(v A ) n |
3.857
5.075
  37.2

  40  77.2
vA  (3.857) 2  (5.075) 2
 6.37 m/s
vA  6.37 m/s
Velocity of B:
77.2 
vB  1.802 m/s
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40 
PROBLEM 13.167
Two identical hockey pucks are moving on a hockey rink at the same speed of
3 m/s and in perpendicular directions when they strike each other as shown.
Assuming a coefficient of restitution e  0.9, determine the magnitude and direction
of the velocity of each puck after impact.
SOLUTION
Use principle of impulse-momentum:
mv1  Imp12  mv 2
t-direction for puck A:
 mv A sin 20  0  m(vA )t
(vA )t  v A sin 20  3sin 20  1.0261 m/s
t-direction for puck B:
 mvB cos 20  0  m(vB )t
(vB )t  vB cos 20  3cos 20  2.8191 m/s
n-direction for both pucks:
mvA cos 20  mvB sin 20  m(vA )n  m(vB )n
(vA )n  (vB )n  vA cos 20  vB sin 20
 3cos 20  3sin 20
(1)
e  0.9
Coefficient of restitution:
(vB )n  (vA )n  e[(vA )n  (vB )n ]
 0.9[3cos 20  (3)sin 20]
Solving Eqs. (1) and (2) simultaneously,
(vA )n  0.8338 m/s
(vB )n  2.6268 m/s
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(2)
PROBLEM 13.167 (Continued)
Summary:
( vA )n  0.8338 m/s
20
( vA )t  1.0261 m/s
70
( vB )n  2.6268 m/s
20
( vB )t  2.8191 m/s
70
v A  (0.8338) 2  (1.0261) 2  1.322 m/s
tan  
1.0261
0.8338
  50.9
  20  70.9
vA  1.322 m/s
70.9 
vB  (2.6268) 2  (2.8191)2  3.85 m/s
tan  
2.8191
2.6268
  47.0
  20  27.0
vB  3.85 m/s
27.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.168
The coefficient of restitution is 0.9 between the two 60-mmdiameter billiard balls A and B. Ball A is moving in the
direction shown with a velocity of 1 m/s when it strikes ball B,
which is at rest. Knowing that after impact B is moving in the x
direction, determine (a) the angle  , (b) the velocity of B after
impact.
SOLUTION
(a) Since vB is in the x-direction and (assuming no friction), the
common tangent between A and B at impact must be parallel to the
y-axis
tan  
Thus
250
150  D
  tan 1
250
 70.20
150  60
  70.2 
(b) Conservation of momentum in x(n) direction
mv A cos   m  vB n  m  vA n  mvB
1 cos  70.20   0   vA n
 vB
0.3387   vA n   vB 
(1)
Relative velocities in the n direction
e  0.9
 vA cos   vB n  e  vB   vA n
 0.3387  0  0.9   vB   vA n
(2)
(1)  (2)
2vB  0.3387 1.9 
vB  0.322 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.169
A boy located at Point A halfway between the center O of a
semicircular wall and the wall itself throws a ball at the wall in a
direction forming an angle of 45 with OA. Knowing that after
hitting the wall the ball rebounds in a direction parallel to OA,
determine the coefficient of restitution between the ball and the
wall.
SOLUTION
Law of sines:
sin 
R
2

sin135
R
  20.705
  45  20.705
 24.295
Conservation of momentum for ball in t-direction:
v sin   v sin 
Coefficient of restitution in n:
Dividing,
v(cos  )e  v cos 
tan 
 tan 
e
e
tan 20.705
tan 24.295
e  0.837 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.170
The Mars Pathfinder spacecraft used large airbags to cushion
its impact with the planet’s surface when landing. Assuming
the spacecraft had an impact velocity of 18 m/s at an angle of
45° with respect to the horizontal, the coefficient of restitution
is 0.85 and neglecting friction, determine (a) the height of the
first bounce, (b) the length of the first bounce. (Acceleration of
gravity on the Mars  3.73 m/s2.)
SOLUTION
Use impulse-momentum principle.
mv1  Imp12  mv 2
The horizontal direction (x to the right) is the tangential direction and the vertical direction ( y upward) is the
normal direction.
Horizontal components:
mv0 sin 45  0  mvx
vx  v0 sin 45.
v x  12.728 m/s
Vertical components, using coefficient of restitution e  0.85
v y  0  e[0  (v0 cos 45)]
v y  (0.85)(18cos 45)
v y  10.819 m/s
The motion during the first bounce is projectile motion.
Vertical motion:
Horizontal motion:
1 2
gt
2
v y  (v y )0  gt
y  (v y ) 0 t 
x  vx t
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PROBLEM 13.170 (Continued)
(a)
Height of first bounce:
v y  0:
0  (v y )0  gt
(v y ) 0
10.819 m/s
 2.901 s
3.73 m/s 2
1
y  (10.819)(2.901)  (3.73)(2.901)2
2
t
g

y  15.69 m 
(b)
Length of first bounce:
y  0:
10.819t 
1
(3.73) t 2  0
2
t  5.801 s
x  (12.728)(5.801)
x  73.8 m 
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PROBLEM 13.171
A girl throws a ball at an inclined wall from a height of 3
ft, hitting the wall at A with a horizontal velocity v0 of
magnitude 25 ft/s. Knowing that the coefficient of
restitution between the ball and the wall is 0.9 and
neglecting friction, determine the distance d from the
foot of the wall to the Point B where the ball will hit the
ground after bouncing off the wall.
SOLUTION
Momentum in t direction is conserved
mv sin 30  mvt
(25)(sin 30)  vt
vt  12.5 ft/s
Coefficient of restitution in n-direction
(v cos30)e  vn
(25)(cos30)(0.9)  vn
vn  19.49 ft/s
Write v in terms of x and y components
(vx )0  vn (cos30)  vt (sin 30)  19.49(cos30)  12.5(sin 30)
 10.63 ft/s
(vy )0  vn (sin 30)  vt (cos30)  19.49(sin 30)  12.5(cos30)
 20.57 ft/s
Projectile motion
y  y0  (vy )0 t 
1 2
t2
gt  3 ft  (20.57 ft/s)t  (32.2 ft/s 2 )
2
2
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PROBLEM 13.171 (Continued)
At B,
y  0  3  20.57tB  16.1tB2 ;
tB  1.4098 s
xB  x0  (vx )0 t B  0  10.63(1.4098);
xB  14.986 ft
d  xB  3cot 60  (14.986 ft)  (3 ft)cot 60  13.254 ft
d  13.25 ft 
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PROBLEM 13.172
Rockfalls can cause major damage to roads and infrastructure. To design
mitigation bridges and barriers, engineers use the coefficient of restitution
to model the behavior of the rocks. The rock A falls a distance of 20 m
before striking an incline with slope  = 40o. Knowing that the coefficient
of restitution between A and the incline is 0.2, determine the velocity of
the rock after the impact.
SOLUTION
Given:
y  20 m
e  0.2
Sketch:
 =40
Apply Work-Energy to determine speed at impact:
T1  Vg1  Ve1  U1NC
 3  T2  Vg 2  Ve2
where: Vg1  mgy and T2 
1 2
mvB
2
vB  40 g m/s
Components in n and t directions:
 v B n  vB cos 
 v B t  vB sin 
Conservation of momentum in the t-direction:
n-components using coefficient of restitution:
Speed of rock after impact:
 v'B t   v B t  vB sin 
e   v B n  0   0   v'B n
 v'B n  e  v B n
v'B 
 v'B t2   v'B 2n
 e2  v B n   v B t
2
2
 vB e2 cos2   sin 2   13.09 m/s
Direction after impact:
tan  
 v'B n
 v'B t

e vB cos 
vB sin 
   13.4
v'B  13.09 m/s
26.6 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.173
From experimental tests, smaller boulders tend to have a greater
coefficient of restitution than larger boulders. The rock A falls a distance
of 20 meters before striking an incline with slope  = 45o. Knowing that
h= 30 m and d= 20 m, determine if a boulder will land on the road or
beyond the road for a coefficient of restitution of (a) e= 0.2, (b) e= 0.1.
SOLUTION
Given:
y  20 m
h  30 m
d  20 m
 =45
Sketch:
Apply Work-Energy to determine speed of the rock at impact:
T1  Vg1  Ve1  U1NC
 3  T2  Vg 2  Ve2
where: Vg1  mgy and T2 
1 2
mvB
2
vB  40 g  19.809 m/s
Components in n and t directions:
 v B n  vB cos 
 v B t  vB sin 
Conservation of momentum in the t-direction:
 v'B t   v B t  vB sin 
n-components using coefficient of restitution:
e  v B n  0  0   v'B n


 v'B n  e  v B n
Speed of rock after impact:
v'B 
 v'B t2   v'B 2n
 e 2  v B  n   v B t
2
2
 vB e2 cos2   sin 2 
 v'B n
 v'B t
e vB cos 
Direction after impact:
tan  
x and y components of velocity after impact:
 v'B  x  v 'B cos    
 v'B  y  v 'B sin    

vB sin 
   tan 1 e
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.173 (Continued)
Projectile Motion in y-dir:
Projectile Motion in x-dir:
(a) For e=0.2:
1 2
gt
2
1
0  30  (vB ) y t  gt 2
2
y  y0  (vB ) y t 
x  x0  (vB ) x t
(1)
(2)
v'B  14.285 m/s
 =11.31
 v'B  x  11.885 m/s
 v'B  y  7.924 m/s
Solve (1) for t:
t  1.7939 s or t  3.4094 s
Solve (2) for x:
x  21.32 m > d
(b) For e=0.1:
Rock will land beyond the road
v'B  14.077 m/s
 =5.711
 v'B  x  10.894 m/s
 v'B  y  8.914 m/s
Solve (1) for t:
t  1.7261 s or t  3.5434 s
Solve (2) for x:
x  18.81 m < d
Rock will land on the road
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PROBLEM 13.174
Two cars of the same mass run
head-on into each other at C. After
the collision, the cars skid with
their brakes locked and come to a
stop in the positions shown in the
lower part of the figure. Knowing
that the speed of car A just before
impact was 5 mi/h and that the
coefficient of kinetic friction
between the pavement and the tires
of both cars is 0.30, determine
(a) the speed of car B just before
impact, (b) the effective coefficient
of restitution between the two cars.
SOLUTION
(a)
At C:
Conservation of total momentum:
m A  mB  m
5 mi/h  7.333 ft/s
mAvA  mB vB  mAvA  mB vB
7.333  vB  vA  vB
Work and energy.
Care A (after impact):
1
mA (vA )2
2
T2  0
T1 
U1 2  F f (12)
U1 2  k m A g (12 ft)
T1  U1 2  T2
1
m A (vA ) 2  mk m A g (12)  0
2
(vA ) 2  (2)(12 ft)(0.3)(32.2 ft/s 2 )
 231.84 ft/s 2
vA  15.226 ft/s
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(1)
PROBLEM 13.174 (Continued)
Car B (after impact):
1
mB (vB ) 2
2
T2  0
T1 
U1 2   k mB g (3)
T1  U1 2  T2
1
mB (vB )2  k mB g (3)
2
vB2  (2)(3 ft)(0.3)(32.2 ft/s 2 )
(vB ) 2  57.96 ft/s 2
vB  7.613 ft/s
From (1)
vB  7.333  vA  vB
 7.333  15.226  7.613
vB  30.2 ft/s  20.6 mi/h 
(b)
Relative velocities:
(vA  vB ) e  vB  vA
(7.333  30.2) e  7.613  15.226
(7.613)
e
 0.2028
(37.53)
e  0.203 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.175
A 1-kg block B is moving with a velocity v0 of magnitude v0  2 m/s as
it hits the 0.5-kg sphere A, which is at rest and hanging from a cord
attached at O. Knowing that k  0.6 between the block and the
horizontal surface and e  0.8 between the block and the sphere,
determine after impact (a) the maximum height h reached by the sphere,
(b) the distance x traveled by the block.
SOLUTION
Velocities just after impact
Total momentum in the horizontal direction is conserved:
mA v A  mB vB  m A vA  mB vB
0  (1 kg)(2 m/s)  (0.5 kg)(vA )  (1 kg)(vB )
4  vA  2vB
(1)
Relative velocities:
(v A  vB ) e  (vB  vA )
(0  2)(0.8)  vB  vA
1.6  vB  vA
(2)
Solving Eqs. (1) and (2) simultaneously:
vB  0.8 m/s
vA  2.4 m/s
(a)
Conservation of energy:
1
m A v12 V1  0
2
1
T1  m A (2.4 m/s) 2  2.88 m A
2
T1 
T2  0
V2  mA gh
T1  V1  T2  V2
2.88 m A  0  0  m A (9.81)h
h  0.294 m 
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PROBLEM 13.175 (Continued)
(b)
Work and energy:
1
1
mB v12  mB (0.8 m/s)2  0.32mB
T2  0
2
2
  F f x   k Nx    x mB gx  (0.6)(mB )(9.81) x
T1 
U1 2
U1 2  5.886mB x
T1  U1 2  T2 :
0.32mB  5.886mB x  0
x  0.0544 m
x  54.4 mm 
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PROBLEM 13.176
A 0.25-lb ball thrown with a horizontal velocity v0 strikes a 1.5-lb plate attached to a vertical wall at a height
of 36 in. above the ground. It is observed that after rebounding, the ball hits the ground at a distance of 24 in.
from the wall when the plate is rigidly attached to the wall (Figure 1) and at a distance of 10 in. when a foamrubber mat is placed between the plate and the wall (Figure 2). Determine (a) the coefficient of restitution
e between the ball and the plate, (b) the initial velocity v0 of the ball.
SOLUTION
(a)
Figure (1), ball alone relative velocities
v0e  (vB )1
Projectile motion
t  time for the ball to hit ground
2 ft  v0et
(1)
Figure (2), ball and plate relative velocities
(vB  vA )e  vP  (vB )2
vB  v0 ,
vP  0
v0e  vP  (vB )2
Conservation of momentum
mBvB  mPvP  mBvB  mPvP
0.25
0.25
1.5
v0  0 
vP
(vB )2 
g
g
g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.176 (Continued)
0.25v0  0.25(vB )2  1.5vp  v0  (vB )2  6vp
(vB )2 
Solving (2) and (3) for (vB ) 2 ,
(3)
(6e  1)
v0
7
Projectile motion
0.8333 
(6e  1)
v0t
7
(4)
Dividing Equation (4) by Equation (1)
0.8333 6e  1

; 2.91655e  6e  1
2
7e
e  0.324 
(b)
From Figure (1)
h
Projectile motion,
1 2
1
gt ; 3  (32.2)t 2
2
2
6  32.2t 2
(5)
From Equation (1),
2  v0et  t 
2
6.1728

0.324v0
v0
Using Equation (5)
2
 6.1728 
2
6  32.2 
  6v0  1226.947
v
0


v02  204.49
v0  14.30 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.177
After having been pushed by an airline employee, an empty
40-kg luggage carrier A hits with a velocity of 5 m/s an
identical carrier B containing a 15-kg suitcase equipped with
rollers. The impact causes the suitcase to roll into the left wall
of carrier B. Knowing that the coefficient of restitution between
the two carriers is 0.80 and that the coefficient of restitution
between the suitcase and the wall of carrier B is 0.30, determine
(a) the velocity of carrier B after the suitcase hits its wall for
the first time, (b) the total energy lost in that impact.
SOLUTION
(a)
Impact between A and B:
Total momentum conserved:
mAvA  mB vB  mAvA  mB vB
mA  mB  40 kg
5 m/s  0  vA  vB
(1)
Relative velocities:
(vA  vB )eAB  vB  vA
(5  0)(0.80)  vB  vA
(2)
Adding Eqs. (1) and (2)
(5 m/s)(1  0.80)  2vB
vB  4.5 m/s
Impact between B and C (after A hits B)
Total momentum conserved:
mB vB  mC vC  mB vB  mC vC
(40 kg)(4.5 m/s)  0  (40 kg) vB  (15 kg) vC
4.5  vB  0.375 vC
(3)
Relative velocities:
(vB  vC ) eBC  vC  vB
(4.5  0)(0.30)  vC  vB
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(4)
PROBLEM 13.177 (Continued)
Adding Eqs. (4) and (3)
(4.5)(1  0.3)  (1.375)vC
vC  4.2545 m/s
vB  4.5  0.375(4.2545) vB  2.90 m/s
(b)
vB  2.90 m/s 
TL  (TB  TC )  (TB  TC )
1
 40 
kg  (4.5 m/s) 2  405 J
mB (vB )2  
2
 2

1
 40 
TC  0
TB  mB (vB )2  
kg  (2.90)2  168.72 J
2
 2

TB 
TC 
1
 15

mC (vC )2  
kg  (4.2545 m/s)2  135.76 J
2
 2

 TL  (405  0)  (168.72  135.76)  100.5 J
TL  100.5 J 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.178
Blocks A and B each weigh 0.8 lb
and block C weighs 2.4 lb. The
coefficient of friction between the
blocks and the plane is k  0.30.
Initially block A is moving at a
speed v0  15 ft/s and blocks B and
C are at rest (Fig. 1). After A
strikes B and B strikes C, all three
blocks come to a stop in the
positions
shown
(Fig.
2).
Determine (a) the coefficients of
restitution between A and B and
between B and C, (b) the
displacement x of block C.
SOLUTION
(a)
Work and energy
Velocity of A just before impact with B:
T1 
1 WA 2
v0
2 g
T2 
 
1 WA 2
vA
2 g
2
U1 2   kWA (1 ft)
T1  U1 2  T2
 
1 WA 2
1 WA 2
v0   kWA (1) 
vA
2 g
2 g
2
(v A2 ) 2  v02  2 k g  (15 ft/s) 2  2(0.3)(32.2 ft/s 2 )(1 ft)
(v A2 ) 2  205.68 ft/s 2 ,
(v A ) 2  14.342 ft/s
Velocity of A after impact with B:
(vA )2
1 WA
(vA ) 22
T3  0
2 g
  k WA (3/12)
T2 
U 2 3
T2  U 2 3  T3 ,
1 WA
(vA )22  ( k )(WA /4)  0
2 g
1 
(vA )22  2(0.3)(32.2 ft/s 2 )  ft   4.83 ft 2 /s2
4 

(v A )2  2.198 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.178 (Continued)
Conservation of momentum as A hits B:
(v A )2  14.342 ft/s
(vA )2  2.198 ft/s
mA (v A )2  mB vB  mB (vA ) 2  mB vB m A  mB
14.342  0  2.198  vB
vB  12.144 ft/s
Relative velocities A and B:
[(vA )2  vB ]eAB  vB  (vA )2
(14.342  0)eAB  12.144  2.198
eAB  0.694 
Work and energy.
Velocity of B just before impact with C:
W
1 WB
(vB ) 22  B (12.144) 2
2 g
2g
W
1 WB
T4 
(vB ) 24  B (vB )42
2 g
2g

U 2  4   kWB (1 ft)  (0.3) WB
(vB )  12.144 ft/s
T2 
F f  k WB
(v  ) 2
(12.144)2
 0.3  B 4
2g
2g
T2  U 2  4  T4 ,
(vB )4  11.321 ft/s
Conservation of momentum as B hits C:
0.8
g
2.4
mC 
g
mB 
(vB )4  11.321 ft/s
mB (vB ) 4  mC vC  mB (vB ) 4  mC vC
0.8
0.8
(2.4)
(11.321)  0 
(vB ) 4 
(vC )
g
g
g
11.321  (vB ) 4  3vC
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.178 (Continued)
Velocity of B after B hits C ,(vB )4  0.
(Compare Figures (1) and (2).)
vC  3.774 ft/s
Relative velocities B and C:
((vB ) 4  vC )eBC  vC  (vB ) 4
(11.321  0)eBC  3.774  0
eBC  0.333 
(b)
Work and energy, Block C:
1 WC
T5  0
U 4 5   kWC ( x)
(vC ) 2
2 g
1 WC
 T5
(3.774) 2  (0.3) WC ( x)  0
2 g
T4 
T4  U 4 5
x
(3.774) 2
 0.737 ft
2 (32.2)(0.3)
x  8.84 in. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.179
A 5 kg sphere is dropped from a height y=2 m to test newly designed spring
floors used in gymnastics. The mass of the floor section is 10 kg, and the
effective stiffness of the floor is k= 120 kN/m. Knowing that the coefficient of
restitution between the ball and the platform is 0.6, determine (a) the height h
reached by the sphere after rebound, (b) the maximum force in the springs.
SOLUTION
Given:
mA  5 kg
Sketch before/after impact:
mB  10 kg
y2m
e  0.6
k  120 kN/m
Apply Work-Energy to determine speed of the sphere just before impact:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: Vg1  mA gy and T2 
1
mAvA2
2
vA  2 gy  6.264 m/s 
Conservation of momentum in the y-direction:
mAvA  mB vB  mAvA  mB vB
Dividing by mA with ( m B /m A  2) :
- 6.264  vA  2vB
Coefficient of restitution:
vB  vA  e v A  vB

(1)

(2)
 6.264e
Solving Eqs. (1) and (2) simultaneously yields:
vA  0.4176 m/s
vB  3.341 m/s
(a) Apply Work-Energy to sphere A to determine height it rises:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
h
where: T1 
1
mAv '2A and Vg2  mA gh
2
v '2A
m/s
2g
h  8.89 mm 
(b) Let  0 be the initial compression of the spring and hB be the additional compression of the spring after
impact. In the initial equilibrium state:
Fy  0: k 0  WB  0 or k 0  mB g
Apply Work-Energy to determine maximum deflection of the springs (when the plate reaches its
lowest point and the speed is zero):
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.179 (Continued)
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
1
1
mB v '2B , Vg1  mB g 0 , Ve1  k 02
2
2
1
2
Vg2  mB g  0  hB  , Ve2  k  0  hB 
2
where: T1 
1
1
1
1
mB v '2B  mB g 0  k 02   mB g  0  hB   k 02  k 0 hB  khB2
2
2
2
2
1
1
Simplify:
mB v '2B   mB ghB  k 0 hB  khB2
2
2
1
1
Substitute k  0  mB g :
m B v '2B   mB ghB  mb ghB  khB2
2
2
Put in known values:
120000 hB2  111.616  0
Solving for hB:
Maximum force in spring
hB  0.0305 m
F  k  0  hB 
 k 0  khB
 mB g  khB
Fmax  3758 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.180
A 5 kg sphere is dropped from a height y=3 m to test a new spring floor used in
gymnastics. The mass of the floor section B is 12 kg, and the sphere bounces
back upwards a distance of 44 mm. Knowing that the maximum deflection of the
floor section is 33 mm from its equilibrium position, determine (a) the
coefficient of restitution between the sphere and the floor, (b) the effective
spring constant k of the floor section.
SOLUTION
Given:
mA  5 kg
Sketch before/after impact:
mB  12 kg
y 3m
h  44 mm
smax  33 mm
Apply Work-Energy to determine speed of the sphere just before impact:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: Vg1  mA gy and T2 
1
mAvA2
2
vA  2 gy  7.672 m/s  or v A  7.672 m/s
Apply Work Energy to determine the speed of the sphere just after impact:
T2  Vg1  Ve1  U 2NC
 3  T3  Vg 3  Ve3
where: T2 
1
mAv '2A and Vg3  mA gh
2
v ' A  2 gh  0.9291 m/s 
(a) Conservation of momentum in the y-direction: mAvA  mB vB  mAvA  mB vB
Put in known values and solve:
vB 
mA
 vA  vA 
mB
 3.584 m/s
Coefficient of restitution:

vB  vA  e v A  vB
e
Put in known values and solve:

vB  vA
vA
e  0.588 
(b) Let  0 be the initial compression of the spring and smax be the additional compression of the spring
m g
Fy  0: k 0  WB  0 or  0  B
after impact. In the initial equilibrium state:
k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.180 (Continued)
Apply Work-Energy to determine the stiffness of the springs (when the plate reaches its lowest point
and the speed is zero). Take the equilibrium position as the datum.
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: T1 
1
1
mB v '2B , Ve1  k 02
2
2
Vg2  mB g  smax  , Ve2 
m g
Substitute  0  B :
k
1
2
k  smax 
2
2
1
1 m g
1
2
mB v '2B  k  B    mB gsmax  k  smax 
2
2  k 
2
mB v '2B k   mB g   2 mB g  smax  k  k 2  smax 
2
2
k 2  smax    mB v '2B  2 mB gsmax  k   mB g   0
2
2
Put in known values and solve for k:
k  148.7 kN/m or k  85.5 N/m
k  148.7 kN/m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.181
The three blocks shown are identical. Blocks B and C are at rest
when block B is hit by block A, which is moving with a
velocity v A of 3 ft/s. After the impact, which is assumed to be
perfectly plastic (e  0), the velocity of blocks A and B
decreases due to friction, while block C picks up speed, until all
three blocks are moving with the same velocity v. Knowing
that the coefficient of kinetic friction between all surfaces is
k  0.20, determine (a) the time required for the three blocks
to reach the same velocity, (b) the total distance traveled by
each block during that time.
SOLUTION
(a)
Impact between A and B, conservation of momentum
mvA  mvB  mvC  mvA  mvB  mvC
3  0  vA  vB  0
Relative velocities (e  0)
(v A  vB )e  vB  vA
0  vB  vA
vA  vB
3  2vB
vB  1.5 ft/s
v  Final (common) velocity
Block C: Impulse and momentum
WC
v
g
F f  kWC
v
g
v  (0.2) gt
WC vC  F f t 
0  (0.2)t 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 13.181 (Continued)
Blocks A and B: Impulse and momentum
2
W
W
(1.5)  4(0.2)Wt  2 v
g
g
1.5  0.4 gt  v
(2)
Substitute v from Eq. (1) into Eq. (2)
1.5  0.4 gt  0.2 gt
t
(b)
(1.5 ft/s)
0.6(32.2 ft/s 2 )
t  0.0776 s 
Work and energy:
From Eq. (1)
v  (0.2)(32.2)(0.0776)  0.5 ft/s
Block C:
T1  0
T2 
1W
W
(v ) 2 
(0.5) 2
2 g
2g
U12  Ff xC  kWxC  0.2WxC
T1  U1 2  T2
xC 
0  (0.2)(W ) xC 
1W 2
v
2 g
(0.5 ft/s)2
 0.01941 ft
0.2(2)(32.2 ft/s2 )
xC  0.01941 ft 

Blocks A and B:
T1 
1 W 
2
 2  (1.5)  2.25W
2 g 
T2 
1 W 
2
 2  (0.5)  0.25W
2 g 
U1 2  4kWgxA  0.8WgxA
T1  U1 2  T2
2.25W  4(0.2)W (32.2) xA  0.25W

xA  0.07764 ft
xA  0.0776 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.182
Block A is released from rest and slides down the
frictionless surface of B until it hits a bumper on
the right end of B. Block A has a mass of 10 kg
and object B has a mass of 30 kg and B can roll
freely on the ground. Determine the velocities of A
and B immediately after impact when (a) e  0,
(b) e  0.7.
SOLUTION
Let the x-direction be positive to the right and the y-direction vertically upward.
Let (v A ) x , (v A ) y , (v A ) x and (vB ) y be velocity components just before impact and (vA ) x ,(vA )y ,(vB ) x , and
(vB ) y those just after impact. By inspection,
(vA ) y  (vB ) y  (vA ) y  (vB ) y  0
Conservation of momentum for x-direction:
While block is sliding down:
0  0  mA (v A ) x  mB (vB ) x
(vB ) x    (v A ) x
(1)
Impact:
0  0  mA (vA ) x  mB (vB ) x
(vB ) x    (vA ) x
(2)
  mA /mB
where
Conservation of energy during frictionless sliding:
Initial potential energies: mA gh for A,
0 for B.
Potential energy just before impact:
V1  0
Initial kinetic energy:
T0  0 (rest)
Kinetic energy just before impact:
1
1
T1  mAvA2  mB vB2
2
2
T0  V0  T1  V1
1
1
1
mA v A2  mB vB2  (mA  mB  2 )v A2
2
2
2
1
 mA (1   ) v A2
2
mA gh 
v A2  (v A )2x 
2 gh
1 
vA 
2 gh
1 
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(3)
PROBLEM 13.182 (Continued)
Velocities just before impact:
2 gh
1 
vA 
vB  
2 gh
1 
Analysis of impact. Use Eq. (2) together with coefficient of restitution.
(vB ) x  (vA ) x  e[(v A ) x  (vB ) x ]
  (vA ) x  (vA ) x  e[(v A ) x   (v A ) x ]
(vA ) x  e(v A ) x
(4)
mA  10 kg
Data:
mB  30 kg
h  0.2 m
g  9.81 m/s 2

From Eq. (3),
(a)
e  0:
10 kg
 0.33333
30 kg
(2)(9.81)(0.2)
1.33333
 1.71552 m/s
vA 
(vA ) x  0
(vB ) x  0
vA  0 
vB  0 
(b)
e  0.7:
(vA ) x  (0.7)(1.71552)
 1.20086 m/s
(vB ) x  (0.33333)(1.20086)
 0.40029 m/s
vA  1.201 m/s

vB  0.400 m/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.183
A 340-g ball B is hanging from an inextensible cord attached to a
support C. A 170-g ball A strikes B with a velocity v 0 of magnitude
1.5 m/s at an angle of 60 with the vertical. Assuming perfectly
elastic impact ( e  1 ) and no friction, determine the height h reached
by ball B.
SOLUTION
Ball A alone
Momentum in t-direction conserved
m A  v A t  m A  vA t
 v A t
 0   vA t
 vA n
Thus
 vA
60
Total momentum in the x-direction is conserved.
m Av A sin 60  mB  vB  x  m A  vA  sin 60  mB vB
v A  v0  1.5 m/s
 vB  x
0
0.17 1.5 sin 60   0    0.17  vA  sin 60    0.34  vB
0.2208   0.1472vA  0.34vB
(1)
Relative velocity in the n-direction
 v A   vB   e  vB cos 30  vA;
n

 1.5  01  0.866vB  vA
Solving Equations (1) and (2) simultaneously
vB  0.9446 m/s, vA  0.6820 m/s
Conservation of energy ball B
1
2
mB  vB 
2
1 WB
T1 
 3.0232 2
2 g
T1 
T2  0
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(2)
PROBLEM 13.183 (Continued)
V1  0
T1  V1  T2  V2 ;
h
V2  W B h
1 WB
 0.9446 2  0  WB h;
2 g
 0.9446 2
 2  9.81
 0.0455 m
h  45.5 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.184
A test machine that kicks soccer balls has a 5-lb simulated foot
attached to the end of a 6-ft long pendulum arm of negligible
mass. Knowing that the arm is released from the horizontal
position and that the coefficient of restitution between the foot
and the 1 lb ball is 0.8, determine the exit velocity of the ball
(a) if the ball is stationary, (b) if the ball is struck when it is
rolling towards the foot with a velocity of 10 ft/s.
SOLUTION
Given:
m A  5 / 32.2 slugs
Pendulum in two positions:
mB  1 / 32.2 slugs
l  6 ft
e  0.8
Apply Work-Energy to determine speed of foot before impact:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: Vg1  mA gl and T2 
1
mAvA2
2
vA  2 gl  19.66 ft/s  m/s
Impulse momentum diagrams for ball and foot:
Conservation of momentum of the ball in the t-direction:
 vB t   vB t
 vB cos60
(1)
Conservation of momentum of the systems in the x-direction:
mA  vA  x  mB  vB  x  mA  vA  x  mB  vB  x
where:
therefore:
 vA  x  vA
 vB  x  vB
 v 'A  x  v 'A
 v 'B  x   v 'B n sin 60   v 'B t cos 60
mAvA  mB vB  mAvA  mB  v 'B n sin 60  vB cos2 60
(2)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.184 (Continued)
Coefficient of Restitution:
 v'B n   vA n  e  vA n   vB n 
where:
 v 'A n  v 'A sin 60
 vA n  vA sin 60
 vB n  vB sin 60
therefore:
 v'B n  v ' A sin 60  e  vA sin 60  vB sin 60 
Solving equations (2) and (3)

(3)

 v 'B  n
 mAvA 1  e   emAvB  mB vB 1  cos2 60  sin 60


2
mA  mB sin 60
v 'B 
 v 'B t2   v 'B 2n
  v ' B t 

  v 'B  
n 

  30  tan 1 
(a) Evaluating equations (1), (4) for vB  0
(4)
(5)
(6)
 vB t  0
 vB n  26.65 ft/s
26.65 ft/s
Using equations (5) and (6) yields:
30°
(b) Evaluating equations (1), (4) for vB  10 ft/s
 vB t  5 ft/s
 vB n  31.54 ft/s
Using equations (5) and (6) yields:
31.93 ft/s
39.0°
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PROBLEM 13.185
Ball B is hanging from an inextensible cord. An identical ball A is released
from rest when it is just touching the cord and drops through the vertical
distance hA  8 in. before striking ball B. Assuming e  0.9 and no friction,
determine the resulting maximum vertical displacement hB of ball B.
SOLUTION
Ball A falls
T1  0 V2  0
T1  V1  T2  V2
(Put datum at 2)
h  8 in.  0.66667 ft
1
mgh  mv A2
2
v A  2 gh  (2)(32.2)(0.66667)  6.5524 ft/s
Impact
  sin 1
r
 30
2r
Impulse-Momentum
Unknowns:
vB , vAt , vAn
x-dir
0  0  mB vB  mAvAn sin 30  mA vAt cos 30
(1)
Noting that mA  mB and dividing by mA
vB  vAn sin 30  vAt cos 30  0
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(1)
PROBLEM 13.185 (Continued)
Ball A alone:
Momentum in t-direction:
mAv A sin 30  0  mAvAt
vAt  vA sin 30  6.5524 sin 30  3.2762 ft/s
(2)
Coefficient of restitution:
  vAn  e(v An  ven )
vBn
vB sin 30  vAn  0.9(vA cos 30  0)
(3)
With known value for vAt, Eqs. (1) and (3) become
vB  vAn sin 30  3.2762 cos 30
vB sin 30  vAn  (0.9)(6.5524) cos 30
Solving the two equations simultaneously,
vB  4.31265 ft/s
vAn  2.9508 ft/s
After the impact, ball B swings upward. Using B as a free body
T   V   TB  VB
where
1
mB (vB ) 2 ,
2
V   0,
T 
TB  0
and
VB  mB ghB
1
mB (vB )2  mB ghB
2
hB 
1 (vB )2
2 g
1 (4.31265)2
2
32.2
 0.2888 ft

hB  3.47 in. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.186
A 70 g ball B dropped from a height h0  1.5 m reaches a
height h2  0.25 m after bouncing twice from identical 210-g
plates. Plate A rests directly on hard ground, while plate C rests
on a foam-rubber mat. Determine (a) the coefficient of
restitution between the ball and the plates, (b) the height h1 of
the ball’s first bounce.
SOLUTION
(a)
Plate on hard ground (first rebound):
Conservation of energy:
1
1
1
mB v 2y  mB v02  mB gh0  mB vx2
2
2
2
v0  2 gh0
Relative velocities., n-direction:
v0 e  v1
v1  e 2 gh0
  vBx
vBx
t-direction
Plate on foam rubber support at C.
Conservation of energy:
V1  V3  0
Points  and :
1
1
1
1
 )2
 ) 2  mB v12  mB (v3 ) 2B  mB (vBx
mB (vBx
2
2
2
2
(v3 ) B  e 2 gh0
Conservation of momentum:
At :
mB (v3 ) B  mP vP  mB (v3 ) B  mP vP
mP 210

3
70
mB
 e 2 gh0  (v3 ) B  3vP
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 13.186 (Continued)
Relative velocities:
[(v3 ) B  (vP )]e  vP  (v3 ) B
e2 2 gh0  0  vP  (v3 ) B
(2)
Multiplying (2) by 3 and adding to (1)
4(v3 ) B  2 gh0 (3e 2  e)
Conservation of energy at ,
Thus,
(v3 ) B  2 gh2
4 2 gh2  2 gh0 (3e2  e)
3e2  e  4
h2
0.25
4
 1.63299
1.5
h0
3e2  e  1.633  0
(b)
e  0.923 
Points  and :
Conservation of energy.
1
1
1
 ) 2  mB v12  mB (vBx
 )2 ;
mB (vBx
2
2
2
1 2
e (2 gh0 )  gh1
2
h1  e2 h0  (0.923)2 (1.5)
h1  1.278 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.187
A 2-kg sphere moving to the right with a velocity of 5 m/s strikes at
A the surface of a 9-kg quarter cylinder which is initially at rest and
in contact with a spring of constant 20 kN/m. The spring is held by
cables so that it is initially compressed 50 mm. Neglecting friction
and knowing that the coefficient of restitution is 0.6, determine
(a) the velocity of the sphere immediately after impact, (b) the
maximum compressive force in the spring.
SOLUTION
Momentum:
mv1n  0  mv1n  Mv2 (0.7071)
(2 kg)(5 m/s)(0.7071)  2 kg v1n  9 kg v2 (0.7071)
Restitution:
v2 (0.7071)  v1n  0.6 v1n  0.6(5)(0.7071)
Solve for
 16 
v2    m/s,
 11 
 17 
v1n     (0.7071)
 11 
v1n  1.092801 m/s

(b) Conservation of energy – cylinder + spring:
1 2 1
1
kx0  M (v2 )2  kx22
2
2
2
2
20, 000
1  16 
20, 000 2
x2  34.52
(0.05) 2  (9)   
2
2  11 
2
x2  0.05875 m,
F  kx2  20,000
N
(0.0587 m) = 1175 N 
m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.188
When the rope is at an angle of   30 the 1-lb sphere A has
a speed v0  4 ft/s. The coefficient of restitution between A
and the 2-lb wedge B is 0.7 and the length of rope l  2.6 ft.
The spring constant has a value of 2 lb/in. and   20.
Determine (a) the velocities of A and B immediately after the
impact, (b) the maximum deflection of the spring assuming A
does not strike B again before this point.
SOLUTION
Masses:
mA  (1/32.2) lb  s 2 /ft
mB  (2/32.2) lb  s 2 /ft
Analysis of sphere A as it swings down:
Initial state:
  30, h0  l (1  cos  )  (2.6)(1  cos30)  0.34833 ft
V0  mA gh0  (1)(0.34833)  0.34833 lb  ft
T0 
Just before impact:
  0, h1  0, V1  0
T1 
Conservation of energy:
1
1  1.0  2
mAv02  
(4)  0.24845 lb  ft
2
2  32.2 
1
1  1.0  2  1.0  2
mAv A2  
vA  
 vA
2
2  32.2 
 64.4 
T0  V0  T1  V1
1 2
vA  0
64.4
v A2  38.433 ft 2 /s 2
0.24845  0.34833 
v A  6.1994 ft/s
Analysis of the impact. Use conservation of momentum together with the coefficient of restitution. e  0.7.
Note that the rope does not apply an impulse since it becomes slack.
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PROBLEM 13.188 (Continued)
Sphere A: Momentum in t-direction:
m A v A sin   0  m A (vA )t
(vA )t  v A sin   6.1994 sin 20  2.1203 m/s
( v A )t  2.1203 m/s 70°
Both A and B: Momentum in x-direction:
mA v A  0  mA (vA ) n cos   mA (vA )t sin   mB vB
(1/32.2)(6.1994)  (1/32.2)(v A )n cos 20  (1/32.2)(2.120323) sin 20  (2/32.2)vB
(1/32.2)(vA ) n cos 20  (2/32.2)vB  0.17001
(1)
Coefficient of restitution:
(vB ) n  (vA ) n  e[(v A ) n  (vB ) n ]
vB cos   (vA ) n  e[v A cos   0]
vB cos 20  (vA ) n  (0.7)(6.1994) cos 20
(2)
Solving Eqs. (1) and (2) simultaneously for (vA )n and vB ,
(vA ) n  1.0446 ft/s
vB  3.2279 ft/s
Resolve vA into horizontal and vertical components.
tan  
(vA )t
(vA ) n
2.1203
1.0446
  63.77

  20  83.8
vA  (2.1203) 2  (1.0446) 2
 2.3637 ft/s
(a)
vA  2.36 ft/s
Velocities immediately after impact.
83.8 
vB  3.23 ft/s
(b)
Maximum deflection of wedge B.
Use conservation of energy:
TB1  VB1  TB 2  VB 2
1
mB vB2
2
VB1  0
TB1 
TB 2  0
VB 2 
1
k (x) 2
2
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
PROBLEM 13.188 (Continued)
The maximum deflection will occur when the block comes to rest (ie, no kinetic energy)
1
1
mB vB2  k (x)2
2
2
m v2
(  x)  B B 
k
2

2 lb
32.2 ft/s2
( x)  0.1642118 ft
 (3.2279 ft/s)
2
2 lb/in (12 in/ft)
( x)  1.971 in. 
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PROBLEM 13.189
When the rope is at an angle of   30 the 1-kg sphere A has
a speed v0  0.6 m/s. The coefficient of restitution between A
and the 2-kg wedge B is 0.8 and the length of rope l  0.9 m
The spring constant has a value of 1500 N/m and   20.
Determine, (a) the velocities of A and B immediately after the
impact (b) the maximum deflection of the spring assuming A
does not strike B again before this point.
SOLUTION
m A  1 kg
Masses:
mB  2 kg
Analysis of sphere A as it swings down:
  30, h0  l (1  cos  )  (0.9)(1  cos30)  0.12058 m
Initial state:
V0  mA gh0  (1)(9.81)(0.12058)  1.1829 N  m
T0 
1 2 1
mv0  (1)(0.6) 2  0.180 N  m
2
2
  0, h1  0, V1  0
Just before impact:
T1 
Conservation of energy:
1
1
mAv A2  (1)v A2  0.5v A2
2
2
T0  V0  T1  V1
0.180  1.1829  0.5 v A2  0
v A2  2.7257 m 2 /s 2
v A  1.6510 m/s
Analysis of the impact: Use conservation of momentum together with the coefficient of restitution. e  0.8.
Note that the ball rebounds horizontally and that an impulse  Tdt is applied by the rope. Also, an impulse
 Ndt is applied to B through its supports.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.189 (Continued)
Both A and B:
Momentum in x-direction:
mA (v A ) x  0  m A (vA ) x  mB (vB ) x
(1)(1.6510)  (1)(vA ) x  (2)(vB ) x
(1)
(v A )n  (v A ) x cos 
Coefficient of restitution:
(vB ) n  0, (vA ) n  (vA ) x cos  , (vB ) x cos 30
(vB )n  (vA ) n  e[(v A ) n  (vB ) n ]
(vB ) x cos   (vA ) x cos   e[(v A ) x cos  ]
Dividing by cos  and applying e  0.8 gives
(vB ) x  (vA ) x  (0.8)(1.6510)
(2)
Solving Eqs. (1) and (2) simultaneously,
(vA ) x  0.33020 m/s
(vB ) x  0.99059 m/s
(a)
Velocities immediately after impact.
(b)
Maximum deflection of wedge B.
Use conservation of energy:
vA  0.330 m/s

vB  0.991 m/s

TB1  VB1  TB 2  VB 2
1
mB vB2
2
VB1  0
TB1 
TB 2  0
VB 2 
1
k (x) 2
2
The maximum deflection will occur when the block comes to rest (ie, no kinetic energy)
1
1
mB vB2  k ( x)2
2
2
( x)2 
mB vB2 (2)(0.99059 m/s) 2

1500 N/m)
k
( x)  0.0362 m
 x  36.2 mm 
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PROBLEM 13.190
A 32,000-lb airplane lands on an aircraft carrier
and is caught by an arresting cable. The cable is
inextensible and is paid out at A and B from
mechanisms located below dock and consisting
of pistons moving in long oil-filled cylinders.
Knowing that the piston-cylinder system
maintains a constant tension of 85 kips in the
cable during the entire landing, determine the
landing speed of the airplane if it travels a
distance d  95 ft after being caught by the
cable.
SOLUTION
Mass of airplaine:
m
W 32000 lb

 993.79 lb  s 2 /ft
2
g 32.2 ft/s
Work of arresting cable force.
Q  85 kips  85000 lb.
As the cable is pulled out, the cable tension acts paralled to the cable at the airplane hook. For a small
displacement
U  Q(l AC )  Q(lBC )
Since Q is constant,
U12  Q  AC  BC  AB 
d  95 ft,
For
AC  BC  (35)2  (95)2  101.24 ft
U12  (85000)(101.24  101.24  70)  11.261 ft  lb
Principle of work and energy:
T1  U12  T2
1 2
1
mv1  U12  mv22
2
2
Since v2  0, we get
v12  
Initial speed:
2U12
(2)(11.261)

 22.663  103 ft 2 /s 2
m
993.79
v1  150.54 ft/s
v1  102.6 mi/h 
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PROBLEM 13.191
A 2-oz pellet shot vertically from a spring-loaded pistol on the surface of the earth rises to a height of 300 ft.
The same pellet shot from the same pistol on the surface of the moon rises to a height of 1900 ft. Determine
the energy dissipated by aerodynamic drag when the pellet is shot on the surface of the earth. (The acceleration
of gravity on the surface of the moon is 0.165 times that on the surface of the earth.)
SOLUTION
Since the pellet is shot from the same pistol the initial velocity v0 is the same on the moon and on the earth.
Work and energy.
1 2
mv0
2
  mg E (300 ft)  EL
T1 
Earth:
U1 2
( EL  Loss of energy due to drag)
1
T1  mv02
2
Moon:
T2  0
U1 2  mg M (1900)
T1  300mg E  EL  0
(1)
T2  0
T1  1900mg M  0
Subtracting (1) from (2)
(2)
1900mg M  300mg E  EL  0
g M  0.165 g E
m
(2/16)
gE
EL  (1900)
(2/16)
(2/16)
(0.165 g E )  300
gE
gE
gE
EL  1.688 ft  lb 
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PROBLEM 13.192
A satellite describes an elliptic orbit about a planet of mass M. The
minimum and maximum values of the distance r from the satellite to
the center of the planet are, respectively, r0 and r1 . Use the
principles of conservation of energy and conservation of angular
momentum to derive the relation
1 1 2GM
  2
r0 r1
h
where h is the angular momentum per unit mass of the satellite and
G is the constant of gravitation.
SOLUTION
Angular momentum:
h  r0 ,
v0  r1 v1
b  r0 v0  r1v1
v0 
h
r0
v1 
h
r1
(1)
Conservation of energy:
1 2
mv0
2
GMm
VA  
r0
TA 
1 2
mv1
2
GMm
VB  
r1
TB 
TA  VA  TB  VB
1 2 GMm 1 2 GMm
 mv1 
mv0 
2
2
r0
r1
1 1
r  r 
v02  v12  2GM     2GM  1 0 
 r0 r1 
 r1r0 
Substituting for v0 and v1 from Eq. (1)
1
r  r 
1
h 2  2  2   2GM  1 0 
 r0 r1 
 r1r0 
 r2  r2 
r  r 
h2
h 2  1 2 20   2 2 (r1  r0 )(r1  r0 )  2GM  1 0 
 r1 r0  r1 r0
 r1r0 
1 1
h 2     2GM
 r0 r1 
 1 1  2GM
   2
h
 r0 r1 
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Q.E.D. 
PROBLEM 13.193
A 60-g steel sphere attached to a 200-mm cord can swing
about Point O in a vertical plane. It is subjected to its own
weight and to a force F exerted by a small magnet embedded
in the ground. The magnitude of that force expressed in
newtons is F  3000/r 2 where r is the distance from the
magnet to the sphere expressed in millimeters. Knowing that
the sphere is released from rest at A, determine its speed as it
passes through Point B.
SOLUTION
m  0.060 kg
Mass and weight:
W  mg  (0.060)(9.81)  0.5886 N
Vg  Wh
Gravitational potential energy:
where h is the elevation above level at B.
Potential energy of magnetic force:
3000
dV

( F , in newtons, r in mm)
2
dr
r
r 3000
3000

Vm  
N  mm
2
 r
r
F

Use conservation of energy:
T1  V1  T2  V2
Position 1: (Rest at A.)
v1  0
T1  0
h1  100 mm
(Vg )1  (0.5886 N)(100 mm)  58.86 N  mm
2
From the figure, AD  2002  1002 (mm 2 )
MD  100  12  112 mm
2
r12  AD  MD
2
 2002  1002  1122
 42544 mm 2
r1  206.26 mm
(Vr )1  
3000
 14.545 N  mm
r1
V1  58.86  14.545  44.3015 N  mm  44.315  103 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.193 (Continued)
Position 2. (Sphere at Point B.)
1 2 1
mv2  (0.060)v22  0.030 v22
2
2
(Vg )2  0
(since h2  0)
T2 
r2  MB  12 mm
(See figure.)
3000
 250 N  mm  250  103 N  mm
12
T1  V1  T2  V2
(Vm )2  
0  44.315  103  0.030v22  250  103
v22  9.8105 m 2 /s 2
v2  3.13 m/s 
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PROBLEM 13.194
A 50-lb sphere A of radius 4.5 in. moving with a velocity of magnitude
v0  6 ft/s strikes a 4.6-lb sphere B of radius 2 in. which is hanging from
an inextensible cord and is initially at rest. Knowing that sphere B
swings to a maximum height h = 0.75 ft, determine the coefficient of
restitution between the two spheres.
SOLUTION
Angle of impulse force from geometry of A and B
 6 
  cos 1 
  22.62
 6.5 
Total momentum conserved
Ball A:
Ball B:
(1)
Restitution
e
e
vB cos    vA  x cos    vA  y sin 
v A cos 
vB   vA  x   vA  y tan 
vA

v A  v0  6 ft/s
vB   vA  x   vA  y
 256 
2
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PROBLEM 13.194 (Continued)
A:
mAv A sin   mA  vA  x  sin    mA  vA  y  cos 
 2.5 
 2.5 
v A tan   (vA ) x tan   (vA ) y ; 6 
  (vA ) x 
  (vA ) y
6


 6 
15  2.5  vA  x  6  vA  y
(2)
 50 
 50 
 4.6 
A  B : mAv A  mA  vA  x  mB vB ;   (6 ft/s)    (vA ) x  
 vB
 g 
 g 
 g 
(3)
g’s cancel
From equation (1)
From equation (3)
vB  2(32.2 ft/s 2 )(0.75 ft)  6.9498 ft/s
(50)(6)  50(vA ) x  4.6(6.9498)
(vA ) x  5.3606 ft/s
From equation (2)
15  2.5(5.3606)  6(vA ) y
(vA ) y  0.2664 ft/s
 2.5 
6.9498  5.3606  0.2664 

 6   0.2834
e
6
e  0.283 
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PROBLEM 13.195
A 300-g block is released from rest after a spring of constant
k  600 N/m has been compressed 160 mm. Determine the force
exerted by the loop ABCD on the block as the block passes through
(a) Point A, (b) Point B, (c) Point C. Assume no friction.
SOLUTION
Conservation of energy to determine speeds at locations A, B, and C.
Mass: m  0.300 kg
Initial compression in spring: x1  0.160 m
Place datum for gravitational potential energy at position 1.
Position 1: v1  0
V1 
T1 
1 2
mv1  0
2
1 2 1
kx1  (600 N/m)(0.160 m)2  7.68 J
2
2
Position 2: T2 
1 2 1
mvA  (0.3)vA2  0.15vA2
2
2
V2  mgh2  (0.3 kg)(9.81 m/s 2 )(0.800 m)  2.3544 J
T1  V1  T2  V2 : 0  7.68  0.15v A2  2.3544
v A2  35.504 m 2 /s 2
Position 3:
T3 
1 2 1
mvB  (0.3)vB2  0.15vB2
2
2
V3  mgh3  (0.3 kg)(9.81 m/s 2 )(1.600 m)  4.7088 J
T1  V1  T3  V3 : 0  7.68  0.15vB2  4.7088
vB2  19.808 m 2 /s 2
Position 4: T2 
1 2 1
mvC  (0.3)vC2  0.15vC2
2
2
V4  mgh4  (0.3 kg)(9.81 m/s)(0.800 m)  2.3544 J
T1  V1  T4  V4 : 0  7.68  0.15vC2  2.3544
vC2  35.504 m 2 /s 2
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PROBLEM 13.195 (Continued)
(a)
Newton’s second law at A:
an 
v A2


35.504 m 2 /s 2
 44.38 m/s 2
0.800 m
an  44.38 m/s 2
F  man : N A  man
N A  (0.3 kg)(44.38 m/s 2 )
(b)
N A  13.31 N

Newton’s second law at B:
an 
vB2


19.808 m 2 /s 2
 24.76 m/s 2
0.800 m
an  24.76 m/s2
F  man : N B  mg  man
N B  m(an  g )  (0.3 kg)(24.76 m/s2  9.81 m/s2 )
(c)
N B  4.49 N 
Newton’s second law at C:
an 
vC2


35.504 m 2 /s 2
 44.38 m/s 2
0.800 m
an  44.38 m/s2
F  man : NC  man
NC  (0.3 kg)(44.38 m/s2 )
NC  13.31 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 13.196
A soccer kicking machine has a 5 lb “simulated foot” attached to A
“kicking” attachment goes on the front of a wheelchair, allowing athletes
with mobility impairments to play soccer. The athletes load up the spring
shown through a ratchet mechanism that pulls the 2 kg “foot” back to the
position 1. They then release the “foot” to impact the 0.45 kg soccer ball,
which is rolling towards the “foot” with a speed of 2 m/s at an angle
= 30º as shown. The impact occurs with a coefficient of restitution e =
0.75 when the foot is at position 2, where the spring is unstretched.
Knowing that the effective friction coefficient during rolling is k = 0.1,
determine (a) the necessary spring coefficient to make the ball roll 30
meters, (b) the direction the ball will travel after it is kicked.
SOLUTION
mA  2 kg, mB  0.45 kg, vB  2 m/s, e  0.75, k =0.1, s 2  0
Given:
Conservation of momentum of the ball in the t-direction:
 vB t   vB t
 vB sin 30
 1 m/s
Apply Work-Energy to the ball after impact to when it comes to rest:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where:
1
T2  mB v '2B
2
NC
U12  k mB g  30 
1
mB v '2B  k mB g  30  =0
2
v 'B  7.672 m/s
Therefore:
Find  v 'B n using:
v 'B 
 v 'B  n 
 v 'B 2n   v 'B t2
v '2B   v 'B t
2
 7.607 m/s 
Conservation of momentum of the systems in the n-direction Impulse momentum diagrams for ball and
foot:
mA  vA n  mB  vB n  mA  vA n  mB  vB n
 vA n 
1
mA  vA   mB  vB   mB  vB  
n
n
n
mA 
(1)
Coefficient of Restitution:
 v'B n   vA n  e  vA n   vB n 
(2)
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PROBLEM 13.196 (Continued)
Sub (1) into (2) and solve for  vA n :
1
mA  vA   mB  vB   mB  vB    e  vA    vB  
n
n
n
n
n

mA 
 v'B n 
 vA  n
=
Sub in known quantities:  vA n =
 mA  mB  v'B n  mB  vB n  emA  vB n
mA  e  1
 2  0.45 7.607   0.45  1.732  0.75  2 1.732
2  0.75  1
 4.805 m/s
(a) Apply Work-Energy to determine speed of foot before impact:
“Foot in two positions:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where:
1
2
1

Ve1  2  ks12  , and T2  mA  v A n
2
2

s1  0.42  0.32  0.3
 0.2 m
mA  v A  n
2
Therefore:
k

2s12
2  4.805
2  0.2 
2
2
k  577.3 N/m 
(b) Direction of the ball:
tan  
 v 'B  n
 v 'B t

7.607
1
  82.5 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.197
A 300-g collar A is released from rest, slides down a frictionless rod,
and strikes a 900-g collar B which is at rest and supported by a spring
of constant 500 N/m. Knowing that the coefficient of restitution
between the two collars is 0.9, determine (a) the maximum distance
collar A moves up the rod after impact, (b) the maximum distance
collar B moves down the rod after impact.
SOLUTION
After impact
Velocity of A just before impact, v0
v0 
2 gh 

2(9.81 m/s 2 )(1.2 m)sin 30
2(9.81)(1.2)(0.5)  3.431 m/s
Conservation of momentum
mAv0  mBvB  mAv A : 0.3v0  0.9vB  0.3v A (1)
Restitution
(vA  vB )  e(v0  0)  0.9v0 (2)
Substituting for vB from (2) in (1)
0.3v0  0.9(0.9v0  v A )  0.3vA
1.2vA  0.51v0
vA  1.4582 m/s, vB  1.6297 m/s
(a)
A moves up the distance d where:
1
mAv A2  mA gd sin 30;
2
1
(1.4582 m/s)2  (9.81 m/s2 )d (0.5)
2
d A  0.21675 m  217 mm 
(b)
Static deflection  x0 , B moves down
Conservation of energy (1) to (2)
Position (1) – spring deflected, x0
k x0  mB g sin 30
T1  V1  T2  V2: T1 
1
mBvB2 , T2  0
2
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PROBLEM 13.197 (Continued)
V1  Ve  Vg 
x  dB
V2  Ve  Vg  0 0
1 2
kx0  mB gd B sin 30
2
kxdx 

1
k d B2  2d B x0  x02
2



1 2
1
1
kx0  mgd B sin 30  mBvB2  k d B2  2d B x0  x02  0  0
2
2
2
 kd B2  mBvB2 ;
500d B2  0.9(1.6297)2
d B  0.0691 m
d B  69.1 mm 
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PROBLEM 13.198
Blocks A and B are connected by a cord which passes over pulleys
and through a collar C. The system is released from rest when x
 1.7 m. As block A rises, it strikes collar C with perfectly plastic
impact (e  0). After impact, the two blocks and the collar keep
moving until they come to a stop and reverse their motion. As A and
C move down, C hits the ledge and blocks A and B keep moving
until they come to another stop. Determine (a) the velocity of the
blocks and collar immediately after A hits C, (b) the distance the
blocks and collar move after the impact before coming to a stop,
(c) the value of x at the end of one compete cycle.
SOLUTION
(a)
Velocity of A just before it hits C:
Conservations of energy:
Datum at :
Position :
(v A )1  (vB )1  0
T1  0
v1  0
Position :
1
1
mA (v A )2  mB vB2
2
2
v A  vB (kinematics)
T2 
1
11
(5  6)v A2  v A2
2
2
V2  mA g (1.7)  mB g (1.7)
T2 
 (5  6)( g )(1.7)
V2  1.7 g
T1  V1  T2  V2
00
11 2
v A  1.7 g
2
 3.4 
v A2  
 (9.81)
 11 
 3.032 m 2 /s 2
v A  1.741 m/s
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PROBLEM 13.198 (Continued)
Velocity of A and C after A hits C:
vA  vC (plastic impact)
Impulse-momentum A and C:
mA vA  T t  (mA  mC ) vA
(5)(1.741)  T t  8vA
(1)
vB  v A ; vB  vA (cord remains taut)
B alone:
mB v A  T  t  mB vA
(6)(1.741)  T t  6vA
Adding Equations (1) and (2),
(2)
11(1.741)  14vA
vA  1.3679 m/s
vA  vB  vC  1.368 m/s 
(b)
Distance A and C move before stopping:
Conservations of energy:
Datum at :
Position :
1
(m A  mB  mC )(vA )
2
 14 
T2    (1.3681) 2
 2
T2  13.103 J
T2 
V2  0
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PROBLEM 13.198 (Continued)
Position :
T3  0
V3  (mA  mC ) gd  mB gd
V3  (8  6) gd  2 gd
T2  V2  T3  V3
13.103  0  0  2gd
d  (13.103)/(2)(9.81)  0.6679 m
(c)
d  0.668 m 
As the system returns to position  after stopping in position , energy is conserved, and the
velocities of A, B, and C before the collar at C is removed are the same as they were in Part (a) above
with the directions reversed. Thus, vA  vC  vB  1.3679 m/s. After the collar C is removed, the
velocities of A and B remain the same since there is no impulsive force acting on either.
Conversation of energy:
Datum at :
1
(m A  mB )(vA ) 2
2
1
T2  (5  6)(1.3679)2
2
T2  10.291 J
T2 
V2  0
T4  0 V4  mB gx  mA gx
V4  (6  5) gx
T2  V2  T4  V4
10.291  0  (1)(9.81) x
x  1.049 m 
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PROBLEM 13.199
A 2-kg ball B is traveling horizontally at
10 m/s when it strikes 2-kg ball A. Ball A is
initially at rest and is attached to a spring with
constant 100 N/m and an unstretched length
of 1.2 m. Knowing the coefficient of
restitution between A and B is 0.8 and friction
between all surfaces is negligible, determine
the normal force between A and the ground
when it is at the bottom of the hill.
SOLUTION
Ball B impacts on ball A. Use the principle of impulse and momentum.
mv1  Imp12  mv 2
v0  10 m/s
Velocity components:
(v0 ) x  v0
(v0 ) n  v0 cos 40 (v0 )t  v0 sin 40
(v A ) x  v A
(v A ) n  v A cos 40
(vB ) x  (vB ) n cos 40  (vB )t sin 40
Impulse-momentum for ball B alone.
t-direction:
mB (v0 )t  mB (vB )t
(vB )t  (v0 )t  10sin 40  6.4279 m/s
(1)
Impulse-momentum for balls A and B.
x-direction
mB v0  0  mAv A  mB (vB ) x  mB (vB )t
(2)(10)  0  2vA  2[(vB )n cos 40  6.4279sin 40]
2vA  2(vB )n cos 40  11.7365
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(1)
PROBLEM 13.199 (Continued)
(e  0.8)
Coefficient of restitution.
(vB )n  (vA )n  e[0  (v0 )n ]
(vB )n  v A cos 40  (0.8)(10) cos 40
(2)
Solving Eqs. (1) and (2) simultaneously,
v A  6.6566 m/s
(vB )n  1.0291 m/s
As ball A moves from the impact location to the lowest point on the path, the spring compresses and the
elevation decreases. Since friction is negligible, energy is conserved.
T1  V1  T2  V2
1
1
mAv A2  (Ve )1  (Vg )1  mAv22  (Ve )2  (Vg )2
2
2
Position 1: (Just after impact.)
1
1
mA v A2  (2)(6.6566)2  44.3101 J
2
2
(Ve )1  0 (The spring is unstretched.)
T1 
(Vg )1  0 (Datum)
Position 2: (Lowest point on path.)
T2 
For the spring,
1
1
mAv22  (2)v22  v22
2
2
x2  l2  l0  0.4 m  1.2 m  0.8 m
Fe  kx2  (100)(0.8)  80 N
(V2 )e 
1 2 1
kx2  (100)(0.8)2  32 J
2
2
h2  0.4 m
Elevation above datum:
(V2 ) g  mA g h2  (2)(9.81)(0.4)  7.848
Conservation of energy:
44.310  0  0  v22  32  7.848
v22  20.158 m2 /s2
v2  4.489 m/s
Normal acceleration at lowest point on path:
an 
v22


20.158
 28.798 m/s 2
0.7
an  28.8 m/s2
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PROBLEM 13.199 (Continued)
Apply Newton’s second law to the ball.
F  man : N  mg  Fe  man
N  mg  Fe  man
 (2)(9.81)  80  (2)(28.798)
N  157.2 N 
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PROBLEM 13.200
A 2-kg block A is pushed up against a spring compressing it a
distance x. The block is then released from rest and slides down
the 20 incline until it strikes a 1-kg sphere B which is
suspended from a 1 m inextensible rope. The spring constant
k  800 N/m, the coefficient of friction between A and the
ground is 0.2, the distance A slides from the unstretched length
of the spring d  1.5 m and the coefficient of restitution
between A and B is 0.8. Knowing the tension in the rope is 20 N
when   30, determine initial compression x of the spring.
SOLUTION
Data:
mA  2 kg, mB  1 kg, k  800 N/m, d  1.5 m, T  20 N
k  0.2, e  0.8,   20,   30, l  1.0 m
Block slides down the incline:
Fy  0
N  mA g cos   0
N  mA g cos 
 (2)(9.81) cos 20
 18.4368 N
F f  k N  (0.2)(18.4368)
 3.6874 N
Use work and energy. Datum for Vg is the impact point near B.
T1  0, (V1 )e 
1 2 1
k x  (800) x 2  400 x 2 J
2
2
(V1 ) g  mA gh1  mA g ( x  d )sin   (2)(9.81)( x  1.5)sin 20  6.7104( x  1.5) J
U12   Ff ( x  d )  (3.6874)( x  1.5) J
T2 
1
1
mAvA2  (1)(vA2 )  1.000 v A2
2
2
V2  0
T1  V1  U12  T2  V2 : 0  400 x 2  6.7104( x  1.5)  (3.6874)( x  1.5)  1.000 v A2  0
400 x 2  3.0231x  4.5346  vA2
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(1)
PROBLEM 13.200 (Continued)
Impact: Conservation of momentum.
Both A and B, horizontal components
:
mA v A cos   0  mAvA cos   mB vB
2v A cos 20  2vA cos 20  vB
(2)
(vB )n  (vA )n  e[(v A )n  (vB )n ]
Relative velocities:
vB cos   vA  e[v A  0]
vB cos 20  vA  (0.8)v A
(3)
Solving Eqs. (2) and (3) simultaneously,
3.6cos 20
vA
2 cos 2 20  1
 1.2230v A
vB 
(4)
Sphere B rises: Use conservation of energy.
1
mB (vB ) 2 V1  0
2
1
T2  mB v22
V2  mB gh2  mB gl (1  cos  )
2
T1 
1
1
mB (vB ) 2  0  mB v22  mB g (1  cos)
2
2
2
2
v2  (1.2230v A )  2 gl (1  cos  )
T1  V1  T2  V2 :
Tension in the rope:
  1.00 m
an 
v22

 (1.2230v A )2  2 gl (1  cos  )
Fn  mB an :
T  mB g cos   mB an
T  mB (an  g cos  )
T  mB ((1.2230vA )2  2 gl (1  cos  )  g cos  )
20  (1.0)((1.2230v A )2  2(9.81)(1)(1  cos30)  9.81cos30)
Solving (5)
Sub (6) into (1)
Solve for x:
(5)
vA  3.0739 m/s
(6)
400 x 2  3.0231x  4.9141  0
x  0.107 m 
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PROBLEM 13.201*
The 2-lb ball at A is suspended by an inextensible cord and given an initial
horizontal velocity of v0. If l  2 ft, xB  0.3 ft and yB  0.4 ft determine the
initial velocity v so that the ball will enter in the basket. Hint: use a computer
to solve the resulting set of equations.
SOLUTION
v1  v0
Let position 1 be at A.
Let position 2 be the point described by the angle  where the path of the ball changes from circular to
parabolic. At position 2 the tension Q in the cord is zero.
Relationship between v2 and  based on Q  0. Draw the free body diagram.
F  0: Q  mg sin   man 
With Q  0,
v22  g  sin 
or
v2 
mv22

g  sin 
(1)
Relationship among v0 , v2 , and  based on conservation of energy.
T1  V1  T2  V2
1 2
1
mv0  mg   mv22  mg  sin 
2
2
v02  v22  2 g (1  sin  )
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 13.201* (Continued)
x and y coordinates at position 2:
x2   cos 
(3)
y2   sin 
(4)
Let t2 be the time when the ball is in position 2.
Motion on the parabolic path. The horizontal motion is
x  v2 sin 
x  x2  (v2 sin  )(t  t2 )
At Point B,
x  xB
(t B  t2 ) 
Vertical motion:
and t  tB .
(5)
From Eq. (5),
 cos   xB
v sin 
(6)
y  v2 cos   g (t  t2 )
y  y2  (v2 cos  )(t  t2 ) 
1
g (t  t2 )2
2
At Point B,
yB   sin   (v2 cos  )(tB  t2 ) 
  2 ft, xB  0.3 ft,
Data:
1
g (tB  t2 )2
2
(7)
yB  0.4 ft, g  32.2 ft/s2
With the numerical data,
Eq. (1) becomes
Eq. (6) becomes
Eq. (7) becomes
v2  64.4 sin 
t B  t2 
2 cos   0.3
v2 sin 
yB  2sin   (v2 cos  )(tB  t2 )  16.1(tB  t2 )2
Method of solution. From a trial value of , calculate v2 from Eq. (1), tB  t2 from Eq. (6), and yB
from Eq. (7). Repeat until yB  0.4 ft as required.
Try   30.
v2  64.4sin 30  5.6745 ft/s
2cos30  0.3
 0.50473s
5.6745sin 30
yB  2sin 30  (5.6745cos30)(0.50473)  (16.1)(0.50473)2
t B  t2 
 0.62116 ft
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(1)
(6)
(7)
PROBLEM 13.201* (Continued)
v2  64.4sin 45  6.7482
Try   45.
2cos 45  0.3
 0.23351 s
6.7482sin 45
yB  2sin 45  (6.7482cos 45)(0.23351)  (16.1)(0.23351)2
t B  t2 
 1.65060 ft
Try   37.5.
v2  64.4sin 37.5  6.2613 ft/s
t B  t2 
2 cos 37.5  0.3
 0.33757 s
6.2613 sin 37.5
yB  2sin 37.5  (6.2613cos 37.5)(0.33757)  (16.1)(0.33757) 2
 1.05972 ft
Let u    30.
The following sets of data points have be determined:
(u, yB )  (0, 0.62114 ft), (7.5, 1.05972 ft), (15, 1.65060 ft)
The quadratic curve fit of this data gives
yB  0.62114  0.29678 u  0.009688711 u 2
Setting yB  0.4 ft gives the quadratic equation
0.009688711 u 2  0.29678 u  1.02114  0
Solving for u,
u  3.95 and 26.68
Rejecting the second value gives
  30  u  33.95.
Try   33.95.
v2  64.4sin 33.95  5.997 ft/s
t B  t2 
2 cos 33.95  0.3
 0.40578 s
5.9971 sin 33.95
yB  2sin 33.95  (5.997 cos 33.95)(0.40578)  (16.1)(0.40578) 2
 0.48462 ft
The new quadratic curve-fit is based on the data points
(u, yB )  (0, 0.62114 ft), (3.95, 0.48462 ft), (7.5, 1.05972 ft).
The quadratic curve fit of this data is
yB  0.62114  0.342053907 u  0.015725232 u 2
Setting yB  0.4 ft gives
0.015725232 u 2  0.342053907 u  1.02114  0
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PROBLEM 13.201* (Continued)
Solving for u,
u  3.572
Try   33.572.
  30  3.572  33.572
v2  64.4 sin 33.572  5.9676 ft/s
t B  t2 
2 cos 33.572  0.3
 0.41406 s
5.9676 sin 33.572
yB  2sin 33.572  (5.9676 cos 33.572)(0.41406)  (16.1)(0.41406) 2
 0.40445 ft
which is close enough to 0.4 ft.
Substituting   33.572 and v2  5.9676 ft/s into Eq. (2) along with other data gives
v02  (5.9676)2  (2)(32.2)(2)(1  sin 33.572)  235.64 ft 2 /s2
v0  15.35 ft/s
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
PROBLEM 13.CQ1
Block A is traveling with a speed v0 on a smooth surface when the
surface suddenly becomes rough with a coefficient of friction of 
causing the block to stop after a distance d. If block A were traveling
twice as fast, that is, at a speed 2v0, how far will it travel on the rough
surface before stopping?
(a) d/2
(b) d
(c)
2d
(d) 2d
(e) 4d
SOLUTION
1 2
mv ). The work
2
done by friction to stop the block is U  Fd . This work must equal the amount of kinetic energy before the
block hits the rough patch. Since the kinetic energy is 4x as much it will take 4x the distance to stop.
If the block were traveling twice as fast, its kinetic energy would be 4x as much ( T 
Answer: (e)
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PROBLEM 13.CQ2
Two small balls A and B with masses 2m and m respectively are released
from rest at a height h above the ground. Neglecting air resistance, which of
the following statements are true when the two balls hit the ground?
(a)
The kinetic energy of A is the same as the kinetic energy of B.
(b)
The kinetic energy of A is half the kinetic energy of B.
(c)
The kinetic energy of A is twice the kinetic energy of B.
(d)
The kinetic energy of A is four times the kinetic energy of B.
SOLUTION
Since each ball accelerates due to gravity at the same rate, they have the same speed when they hit the ground.
1
Since kinetic energy, T  mv2 , and ball A has twice the mass as ball B, then A will have twice the kinetic
2
energy. Answer: (c)
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PROBLEM 13.CQ3
Block A is released from rest and slides down the frictionless
ramp to the loop. The maximum height h of the loop is the same
as the initial height of the block. Will A make it completely
around the loop without losing contact with the track?
(a) Yes
(b) No
(c) need more information
SOLUTION
Answer: (b) In order for A to not maintain contact with the track, the normal force must remain greater than
zero, which requires a non-zero speed at the top of the loop.
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PROBLEM 13.CQ4
A large insect impacts the front windshield of a sports car traveling down a road. Which of the following
statements is true during the collision?
(a) The car exerts a greater force on the insect than the insect exerts on the car.
(b) The insect exerts a greater force on the car than the car exerts on the insect.
(c) The car exerts a force on the insect, but the insect does not exert a force on the car.
(d) The car exerts the same force on the insect as the insect exerts on the car.
(e) Neither exerts a force on the other; the insect gets smashed simply because it gets in the way of the car.
SOLUTION
Newton’s third law states that when a body exerts a force on a second body, the second body exerts a force
equal in magnitude and opposite in direction on the first body. Therefore the answer is: (d)
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PROBLEM 13.CQ5
The expected damages associated with two types of perfectly plastic collisions are to be
compared. In the first case, two identical cars traveling at the same speed impact each
other head on. In the second case, the car impacts a massive concrete wall. In which case
would you expect the car to be more damaged?
(a) Case 1
(b) Case 2
(c) The same damage in each case
SOLUTION
In both cases the car will come to a complete stop, so the applied impulse will be the same. With the same
impulse, the same damage should occur. Answer: (c)
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PROBLEM 13.CQ6
A 5 kg ball A strikes a 1 kg ball B that is initially at rest. Is it
possible that after the impact A is not moving and B has a speed
of 5v?
(a) Yes
(b) No
Explain your answer.
SOLUTION
Answer: (b) No.
Conservation of momentum is satisfied, but the coefficient of restitution equation is not. The coefficient of
restitution must be less than 1.
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PROBLEM 13.F1
The initial velocity of the block in position A is 30 ft/s. The coefficient
of kinetic friction between the block and the plane is k  0.30. Draw
impulse-momentum diagrams that could be used to determine the time
it takes for the block to reach B with zero velocity, if   20°.
SOLUTION
Answer:
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PROBLEM 13.F2
A 4-lb collar which can slide on a frictionless vertical rod is
acted upon by a force P which varies in magnitude as shown.
Knowing that the collar is initially at rest, draw impulsemomentum diagrams that could be used to determine its
velocity at t  3 s.
SOLUTION
Answer:
Where
ò
t2  3
t1  0
Pdt is the area under the curve.
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PROBLEM 13.F3
The 15-kg suitcase A has been
propped up against one end of a
40-kg luggage carrier B and is
prevented from sliding down by
other luggage. When the luggage
is unloaded and the last heavy
trunk is removed from the carrier,
the suitcase is free to slide down,
causing the 40-kg carrier to move
to the left with a velocity vB of
magnitude 0.8 m/s. Neglecting
friction, draw impulse-momentum
diagrams that could be used to
determine (a) the velocity of A as
it rolls on the carrier and (b) the
velocity of the carrier after the
suitcase hits the right side of the
carrier without bouncing back.
SOLUTION
Answer:
(a)
(b)
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PROBLEM 13.F4
Car A was traveling west at a speed of 15 m/s and car B was traveling north at an
unknown speed when they slammed into each other at an intersection. Upon
investigation it was found that after the crash the two cars got stuck and skidded
off at an angle of 50° north of east. Knowing the masses of A and B are mA and mB
respectively, draw impulse-momentum diagrams that could be used to determine
the velocity of B before impact.
SOLUTION
Answer:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.F5
Two identical spheres A and B, each of mass m, are attached to an inextensible
inelastic cord of length L and are resting at a distance a from each other on a
frictionless horizontal surface. Sphere B is given a velocity v0 in a direction
perpendicular to line AB and moves it without friction until it reaches B' where the
cord becomes taut. Draw impulse-momentum diagrams that could be used to
determine the magnitude of the velocity of each sphere immediately after the cord
has become taut.
SOLUTION
Answer:
Where v A¢ y = vB¢ y since the cord is inextensible.
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PROBLEM 13.F6
A sphere with a speed v0 rebounds after striking a frictionless inclined plane as
shown. Draw impulse-momentum diagrams that could be used to find the velocity
of the sphere after the impact.
SOLUTION
Answer:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.F7
An 80-Mg railroad engine A
coasting at 6.5 km/h strikes a 20Mg flatcar C carrying a 30-Mg
load B which can slide along the
floor of the car (k  0.25). The
flatcar was at rest with its brakes
released. Instead of A and C
coupling as expected, it is observed
that A rebounds with a speed of 2
km/h after the impact. Draw
impulse-momentum diagrams that
could be used to determine (a) the
coefficient of restitution and the
speed of the flatcar immediately
after impact, and (b) the time it
takes the load to slide to a stop
relative to the car.
SOLUTION
Answer:
(a)
Look at A and C (the friction force between B and C is not impulsive) to find the velocity after impact.
(b)
Consider just B and C to find their final velocity.
Consider just B to find the time.
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PROBLEM 13.F8
Two frictionless balls strike each other as shown. The coefficient of restitution
between the balls is e. Draw the impulse-momentum diagrams that could be used
to find the velocities of A and B after the impact.
SOLUTION
Answer:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.F9
A 10-kg ball A moving horizontally at 12 m/s strikes a 10-kg block B. The
coefficient of restitution of the impact is 0.4 and the coefficient of kinetic
friction between the block and the inclined surface is 0.5. Draw impulsemomentum diagrams that could be used to determine the speeds of A and B
after the impact.
SOLUTION
Answer:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 13.F10
Block A of mass mA strikes ball B of mass mB with a speed of vA as shown.
Draw impulse-momentum diagrams that could be used to determine the speeds
of A and B after the impact and the impulse during the impact.
SOLUTION
Answer:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
CHAPTER 14
PROBLEM 14.1
A 30-g bullet is fired with a horizontal velocity of 450 m/s and
becomes embedded in block B which has a mass of 3 kg. After
the impact, block B slides on 30-kg carrier C until it impacts
the end of the carrier. Knowing the impact between B and C is
perfectly plastic and the coefficient of kinetic friction between
B and C is 0.2, determine (a) the velocity of the bullet and B
after the first impact, (b) the final velocity of the carrier.
SOLUTION
For convenience, label the bullet as particle A of the system of three particles A, B, and C.
(a)
Impact between A and B: Use conservation of linear momentum of A and B. Assume that the time
period is so short that any impulse due to the friction force between B and C may be neglected.
 mv1  Imp12   mv 2
Components
:
mAv0  0  (mA  mB )v
v 
mAv0
(30  103 kg)(450 m/s)

 4.4554 m/s
mA  mB
(30  103 kg  3 kg
v  4.46 m/s
(b)

Final velocity of the carrier: Particles A, B, and C have the same velocity v to the left. Use
conservation of linear momentum of all three particles. The friction forces between B and C are internal
forces. Neglect friction at the wheels of the carrier.
 mv 2  Imp 23   mv3
Components
:
(mA  mB )v  0  (mA  mB  mC )v
v 

mAv0
(mA  mB )v

mA  mB  mC mA  mB  mC
(30  103 kg)(450 m/s)
 0.4087 m/s
30  103 kg  3 kg  30 kg
v  0.409 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.2
Two identical 1350-kg automobiles A and B are at
rest with their brakes released when B is struck by a
5400-kg truck C which is moving to the left at 8
km/h. A second collision then occurs when B strikes
A. Assuming the first collision is perfectly plastic
and the second collision is perfectly elastic,
determine the velocities of the three vehicles just
after the second collision.
SOLUTION
Given: mA  mB  1350 kg and mC  5400 kg.
Let vA , vB , and vC be the sought after final velocities, positive to the left.
 v A  0   v B 0
Initial velocities:
 0,
 vC 0
 8 km/h  2.2222 m/s
First collision. Truck C strikes car B. Plastic impact: e  0
Let  vBC 0 be the common velocity of B and C after impact.
Conservation of momentum for B and C :
 mB
 mC  vBC  m B  vB 0  mC  vC 0
6750 vBC  0   5400  2.2222 
vBC  1.77778 m/s
Second collision. Car-truck BC strikes car A.
Elastic impact. e  1
v A  vBC   e  v A 0   vBC 0   1.77778 m/s
(1)
Conservation of momentum for A, B, and C.
m Av A   m B  mC  v BC   m A  v A 0   m B  mC  v BC 0
1350 v A  6750 vBC  0   6750 1.77778 
(2)
Solving (1) and (2) simultaneously for vA and vBC ,
v A  2.9630 m/s, vBC  1.18519 m/s  vB  vC
v A  10.67 km/h

v B  4.27 km/h

vC  4.27 km/h

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PROBLEM 14.3
An airline employee tosses two suitcases, of weight 30 lb and
40 lb, respectively, onto a 50-lb baggage carrier in rapid
succession. Knowing that the carrier is initially at rest and that
the employee imparts a 9-ft/s horizontal velocity to the 30-lb
suitcase and a 6-ft/s horizontal velocity to the 40-lb suitcase,
determine the final velocity of the baggage carrier if the first
suitcase tossed onto the carrier is (a) the 30-lb suitcase,
(b) the 40-lb suitcase.
SOLUTION
The weights are WA  30 lb, WB  40 lb, and WC  50 lb.
 v A 0
Initial velocities:
 9 ft/s
 v B 0
,
 6 ft/s
,
and
 vC 0
 0.
There are no horizontal external forces acting during the impacts, and the baggage carrier is free to coast
between the impacts.
(a) Suitcase A is thrown first.
Let v1 be the common velocity of suitcase A and the carrier after the first impact and v2 be the common
velocity of the two suitcases and the carrier after the second impact.
WA
 v A 0 ,
g
Initial momenta:
WB
 vB  0 ,
g
and
0.
Suitcase A impacts carrier. Conservation of momentum:
WA
W W
 v A 0  0  A C v1
g
g
v1 
W A  v A 0
W A  WC

 30  9 
80
 3.375 ft/s
Suitcase B impacts on suitcase A and carrier. Conservation of momentum:
WB
W W
W W W
 vB 0  A C v1  A B C v2
g
g
g
v2 
WB  vB 0  W A  WC  v1
W A  WB  WC

 40  6   80  3.375 
120
 4.25 ft/s
v 2  4.25 ft/s
(b) Suitcase B is thrown first.
Let v3 be the common velocity of suitcase B and the carrier after the first impact and v4 be the common
velocity of all after the second impact.
Suitcase B impacts the carrier. Conservation of momentum:
WB
W W
 vB 0  0  B C v3
g
g
v3 
WB  vB 0
WB  WC

 40  6 
90
 2.6667 ft/s
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
PROBLEM 14.3 (Continued)
Suitcase A impacts on suitcase B and carrier. Conservation of momentum:
WA
W W
W W W
 vA 0  B C v3  A B C v4
g
g
g
v4 
W A  v A 0  WB  WC  v3
W A  WB  WC

 30  9    90  2.6667 
120
 4.25 ft/s
v 4  4.25 ft/s
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
PROBLEM 14.4
A bullet is fired with a horizontal velocity of 1500 ft/s through
a 6-lb block A and becomes embedded in a 4.95-lb block B.
Knowing that blocks A and B start moving with velocities of
5 ft/s and 9 ft/s, respectively, determine (a) the weight of the
bullet, (b) its velocity as it travels from block A to block B.
SOLUTION
The masses are m for the bullet and mA and mB for the blocks.
(a)
The bullet passes through block A and embeds in block B. Momentum is conserved.
Initial momentum:
mv0  mA (0)  mB (0)  mv0
Final momentum:
mvB  mAvA  mB vB
Equating,
mv0  mvB  mAv A  mB vB
m
mAv A  mB vB (6)(5)  (4.95)(9)

 0.0500 lb
v0  vB
1500  9
m  0.800 oz 
(b)
The bullet passes through block A. Momentum is conserved.
Initial momentum:
mv0  mA (0)  mv0
Final momentum:
mv1  mAvA
Equating,
mv0  mv1  mAvA
v1 
mv0  mA v A (0.0500)(1500)  (6)(5)

 900 ft/s
0.0500
m
v1  900 ft/s
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
PROBLEM 14.5
Two swimmers A and B, of weight 190 lb and 125 lb,
respectively, are at diagonally opposite corners of a floating raft
when they realize that the raft has broken away from its anchor.
Swimmer A immediately starts walking toward B at a speed of 2
ft/s relative to the raft. Knowing that the raft weighs 300 lb,
determine (a) the speed of the raft if B does not move, (b) the
speed with which B must walk toward A if the raft is not to
move.
SOLUTION
(a)
The system consists of A and B and the raft R.
Momentum is conserved.
(mv )1  (mv ) 2
0  m A vA  mB vB  mR vR
v A  v A/R  v R
v B  v B/R  v R
B
v A  2 ft/s   v R
A
(1)
B
vB/R  0
vB  vR
0  mA [2   v R ]  mB v R  mR v e
A
vR 
(b)
2 m A
(2 ft/s)(190 lb)

(mA  mB  mR ) (190 lb  125 lb  300 lb)
vR  0.618 ft/s 
From Eq. (1),
0  mA v A  mB vB  0
(vR  0)
vB  
mAv A
mB
vB  
(2 ft/s)(190 lb)
 3.04 ft/s
(125 lb)
0
v A  v A/R 
v R  2 ft/s
vB  3.04 ft/s 
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PROBLEM 14.6
A 180-lb man and a 120-lb woman stand side by side at the
same end of a 300-lb boat, ready to dive, each with a 16-ft/s
velocity relative to the boat. Determine the velocity of the
boat after they have both dived, if (a) the woman dives first,
(b) the man dives first.
SOLUTION
(a)
Woman dives first.
Conservation of momentum:
120
300  180
(16  v1 ) 
v1  0
g
g
v1 
(120)(16)
 3.20 ft/s
600
Man dives next. Conservation of momentum:

300  180
300
180
v1  
v2 
(16  v2 )
g
g
g
v2 
(b)
480v1  (180)(16)
 9.20 ft/s
480
v 2  9.20 ft/s

v2  9.37 ft/s

Man dives first.
Conservation of momentum:
180
300  120
(16  v1 ) 
v1  0
g
g
v1 
(180)(16)
 4.80 ft/s
600
Woman dives next. Conservation of momentum:

300  120
300
120
v1  
v2 
(16  v2 )
g
g
g
v2 
420v1  (120)(16)
 9.37 ft/s
420
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PROBLEM 14.7
A 40-Mg boxcar A is moving in a railroad switchyard with a velocity of 9 km/h toward cars B and C, which
are both at rest with their brakes off at a short distance from each other. Car B is a 25-Mg flatcar supporting a
30-Mg container, and car C is a 35-Mg boxcar. As the cars hit each other they get automatically and tightly
coupled. Determine the velocity of car A immediately after each of the two couplings, assuming that the
container (a) does not slide on the flatcar, (b) slides after the first coupling but hits a stop before the second
coupling occurs, (c) slides and hits the stop only after the second coupling has occurred.
SOLUTION
Each term of the conservation of momentum equation is mass times velocity. As long as the same units are
used in all terms, any unit may be used for mass and for velocity. We use Mg for mass and km/h for velocity
and apply conservation of momentum.
Note: Only moving masses are shown in the diagrams.
Initial momentum:
(a)
mAv0  (40)(9)  360
Container does not slide
360  95v1  130v2
(b)
v1  3.79 km/h

v 2  2.77 km/h

v1  5.54 km/h

v 2  2.77 km/h

Container slides after 1st coupling, stops before 2nd
360  65v1  130v2

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PROBLEM 14.7 (Continued)
(c)
Container slides and stops only after 2nd coupling
360  65v1  100v2
v1  5.54 km/h

v 2  3.60 km/h

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PROBLEM 14.8
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different
style but of the same weight, has been pushed by dockworkers and hits car B with a velocity of 1.5 m/s.
Knowing that the coefficient of restitution is 0.8 between B and C and 0.5 between A and B, determine the
velocity of each car after all collisions have taken place.
SOLUTION
mA  mB  mC  m
Collision between B and C:
The total momentum is conserved:
mvB  mvC  mvB  mvC
vB  vC  0  1.5
(1)
Relative velocities:
(vB  vC )(eBC )  (vC  vB )
(1.5)(0.8)  (vC  vB )
1.2  vC  vB
(2)
Solving (1) and (2) simultaneously,
vB  1.35 m/s
vC  0.15 m/s
vC  0.150 m/s

Since vB  vC , car B collides with car A.
Collision between A and B:
mvA  mvB  mv A  mvB
vA  vB  0  1.35
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(3)
PROBLEM 14.8 (Continued)
Relative velocities:
(v A  vB )eAB  (vB  vA )
(0  1.35)(0.5)  vB  vA
vA  vB  0.675
(4)
Solving (3) and (4) simultaneously,
2vA  1.35  0.675

vA  1.013 m/s

vB  0.338 m/s

Since vC  vB  vA , there are no further collisions.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.9
A 20-kg base satellite deploys three sub-satellites, each which
has its own thrust capabilities, to perform research on tether
propulsion. The weights of sub-satellite A, B, and C are 4 kg,
6 kg, and 8 kg, respectively, and their velocities expressed in
m/s are given by vA = 4i - 2j + 2k, vB = i + 4j, vC = 2i + 2j +
4k. At the instant shown, what is the angular momentum HO
of the system about the base satellite?
SOLUTION
Given:
m A  4 kg, mB  6 kg, mC  8 kg
v A  4i - 2 j  2k m/s, v B  i  4 j m/s, v C  2i  2 j  4k m/s
Linear momentum of each particle expressed in kg  m/s :
m A v A  16.0i  8.0 j  8.0k
mB v B  6.0i  24.0 j  0k
mC v C  16.0i  16.0 j  32.0k
Position Vectors from point O to each satellite in meters:
rA  30 j
rB  15i  25 j  35k
rC  40i
Angular Momentum about O, kg  m 2 /s :
HO  rA  (mA v A )  rB  (mB v B )  rC  (mC vC )
i
j
k
i
j
k
i
j
k
 0
30 0  15 25 35  40 0
0
16.0 8.0 8.0
6.0 24.0 0
16.0 16.0 32.0
 (240.0i  480.0k )  (840.0i  210.0 j  210.0k )  (1280.0 j  640.0k )
 600.0i  1070.0 j  370.0k
HO  (600 kg  m2 /s)i  (1070.0 kg  m2 /s) j  (370.0 kg  m2 /s)k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.10
For the satellite system of Prob. 14.9, assuming that the
velocity of the base satellite is zero, determine (a) the position
vector r of the mass center G of the system, (b) the linear
momentum mv of the system, (c) the angular momentum HG
of the system about G. Also, verify that the answers to this
problem and to Prob. 14.9 satisfy the equation given in Prob.
14.27.
PROBLEM 14.9 A 20-kg base satellite deploys three subsatellites, each which has its own thrust capabilities, to
perform research on tether propulsion. The weights of subsatellite A, B, and C are 4 kg, 6 kg, and 8 kg, respectively, and
their velocities expressed in m/s are given by vA = 4i - 2j +
2k, vB = i + 4j, vC = 2i + 2j + 4k. At the instant shown, what
is the angular momentum HO of the system about the base
satellite?
SOLUTION
Given:
m A  4 kg,
mB  6 kg,
v A  4i - 2 j  2k m/s,
rA  30 j m,
mC  8 kg
v B  1i + 4 j m/s,
v C  2i  2 j  4k m/s
rB  15i  25 j  35k m, rC  40i m
(a) The mass center must satisfy:
n
r 
 miri
i 1
n
 mi
i 1
r 
mArA  mBrB  mC rC
(mA  mB  mC )
r 
(4)(30 j)  (6)(15i  25j  35k )  (8)(40i)
18
r  22.778i  15.000 j  11.667k m 
(b) Linear momentum of the system:
mv  m A v A  mB v B  mC v C
 16.0i  8.0 j  8.0k  6.0i  24.0 j  0k  16.0i  16.0 j  32.0k
mv  (38.0 kg  m/s)i  (32.0 kg  m/s)j  (40.0 kg  m/s)k 
(c) Angular Momentum about the mass center, G:
n
H G   ri'  mi v i
i 1
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PROBLEM 14.10 (Continued)
rA  rA  r  22.778i  15.000 j  11.667k ft
where:
rB  rB  r  7.778i  10.000 j  23.333k ft
rC  rC  r  17.222i  15.000 j  11.667k ft
HG  rA  mA v A  rB  mB v B  rC  mC vC
i
j
k
i
j
k
i
j
k
 22.778 15.000 11.667  7.778 10.000 23.333  17.222 15.000 11.667
8.0
16.0
8.0
6.0
24.0
0
16.0
16.0
32.0
 (26.667i  4.444 j  57.778k )  (560.0i  140.0 j  246.667k )  (293.333i  737.778j  515.556k )
 826.667i  602.222 j  211.111k
H G  826.667i  602.222 j  211.111k kg  m 2 /s 
From Prob. 14.27 show that:
HO  HG  r  mv
i
j
k
r  mv  22.778 15.000 11.667
38.0
32.0
40.0
 226.667i  467.778 j  158.889k kg  m 2 /s
HG  r  mv  600.0i  1070.0 j  370.0k kg  m 2 /s
This is the identical answer to Problem 14.9.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.11
A system consists of three identical 19.32-lb particles A, B, and C. The
velocities of the particles are, respectively, v A  v A j, v B  vB i, and
vC  vC k. Knowing that the angular momentum of the system about
O, expressed in ft  lb  s is HO  1.2k , determine (a) the velocities of
the particles, (b) the angular momentum of the system about its mass
center G.
SOLUTION
19.32
 0.6 lb  s 2 /ft.
32.2
The masses are mA  mB  mC 
rA  3k,
Position vectors (ft):
In units of ft  lb  s,
rB  2i  2 j  3k,
rC  i  4 j
HO  rA   mA v A   rB   mB v B   rC   mC vC 
i
 0
0
j
k
0
3 
 0.6vA  0
i
j k
i j
2
2 3  1 4
0 0
 0.6vB  0 0
k
0
 0.6vC 
  1.8v Ai   1.8vB j  1.2vBk    2.4vC i  0.6vC j
  1.8v A  2.4vC  i  1.8vB  0.6vC  j   1.2vB  k
But, HO is given as 1.2k ft  lb  s.
Equating the two expressions for HO and resolving into components,
i:
 1.8v A  2.4vC  0
(1)
j: 1.8vB  0.6vC  0
(2)
k:  1.2vB  1.2
(3)
(a) Solving,
v A  4 ft/s
v A   4.00 ft/s  j 
vB  1 ft/s
v B  1.000 ft/s  i 
vC  3 ft/s
vC   3.00 ft/s  k 
Coordinates of mass center G in ft.
r 

mArA  mBrB  mC rC
mA  mB  mC
 0.6  3k    0.6  2i  2 j  3k    0.6  i  4 j
1.8
 i  2 j  2k
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PROBLEM 14.11 (Continued)
Position vectors relative to the mass center in ft.
rA  rA  r   3k    i  2 j  2k   i  2 j  k
rB  rB  r   2i  2 j  3k    i  2 j  2k   i  k
rC  rC  r   i  4 j   i  2 j  2k   2 j  2k
m A v A   0.6  4 j   2.4 lb  s  j
mB v B   0.6  i    0.6 lb  s  i
mC v C   0.6  3k   1.8 lb  s  k
(b)
H G  rA   m A v A   rB   mB v B   rC   mC v C 
  i  2 j  k    2.4 j   i  k    0.6i    2 j  2k   1.8k 
  2.4i  2.4k    0.6 j   3.6i   1.2i  0.6 j  2.4k
HG  1.20 ft  lb  s  i   0.60 ft  lb  s  j   2.40 ft  lb  s  k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.12
A system consists of three identical 19.32-lb particles A, B, and C. The
velocities of the particles are, respectively, v A  v A j, v B  vB i, and
vC  vC k, and the magnitude of the linear momentum L of the system
is 9 lb  s. Knowing that HG  HO , where HG is the angular
momentum of the system about its mass center G and HO is the
angular momentum of the system about O, determine (a) the velocities
of the particles, (b) the angular momentum of the system about O.
SOLUTION
The masses are mA  mB  mC 
19.32
 0.6 lb  s2 /ft.
32.2
rA  3k,
Position vectors (ft):
rB  2i  2 j  3k,
rC  i  4 j
Coordinates of mass center G expressed in ft.
r 

m ArA  mBrB  mC rC
mA  mB  mC
 0.6  3k    0.6  2i  2 j  3k    0.6  i  4 j
1.8
 i  2 j  2k
Position vectors relative to the mass center expressed in ft.
rA  rA  r   3k    i  2 j  2k   i  2 j  k
rB  rB  r   2i  2 j  3k    i  2 j  2k   i  k
rC  rC  r   i  4 j   i  2 j  2k   2 j  2k
Angular momenta.
H O  rA   m A v A   rB   mB v B   rC   mC v C 
H G  rA   m A v A   rB   mB v B   rC   mC v C 
Subtracting,
HO  HG   rA  rA    mA v A    rB  rB   mB v B   rC  rC   mC vC
0  r   m A v A   r   mB v B   r   mC v C 
 r   m A v A  mB v B  mC v C   r  L
L  r
L is parallel to r .
L  L 9
 

 32 ,
2
rr
 3
L  L   2r  r
2
2
  3 lb  s/ft
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PROBLEM 14.12 (Continued)
mA v A  mB v B  mC vC   r
 0.6 vA j   0.6  vBi    0.6  vCk   3  i  2j  2k 
(a) Resolve into components and solve for vA , vB , and vC .
vA  10 ft/s
v A  10.00 ft/s  j 
vB  5 ft/s
v B   5.00 ft/s  i 
vC  10.00 ft/s  k 
vC  10 ft/s
(b) Angular momentum about O expressed in ft  lb  s.
HO  rA   mA v A   rB   mB v B   rC   mC vC 
i
 0
0
j
k
0
3 
 0.6vA  0
i
j k
i j
2
2 3  1 4
0 0
 0.6vB  0 0
k
0
 0.6vC 
  1.8v Ai   1.8vB j  1.2vBk    2.4vC i  0.6vC j
  1.8v A  2.4vC  i  1.8vB  0.6vC  j   1.2vB  k
 6i  3j  6k
HO   6.00 ft  lb  s  i   3.00 ft  lb  s  j   6.00 ft  lb  s  k 
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PROBLEM 14.13
A system consists of three particles A, B, and C. We know
that mA  3 kg, mB  2 kg, and mC  4 kg and that the
velocities of the particles expressed in m/s are, respectively,
v A  4i  2 j  2k, v B  4i  3j, and vC  2i  4 j  2k.
Determine the angular momentum HO of the system about O.
SOLUTION
Linear momentum of each particle expressed in kg  m/s.
m A v A  12i  6 j  6k
mB v B  8i  6 j
mC v C  8i  16 j  8k
rA  3j,
Position vectors, (meters):
rB  1.2i  2.4 j  3k ,
rC  3.6i
Angular momentum about O,  kg  m2 /s  .
H O  rA   mA v A   rB   mB v B   rC   mC vC 
i j k
i
j k
i
j k
 0 3 0  1.2 2.4 3  3.6 0 0
8 16 8
12 6 6
8 6 0
 18i  36k    18i  24 j  12k    28.8j  57.6k 
 0i  4.8j  9.6k

 

H O   4.80 kg  m 2 /s j  9.60 kg  m 2 /s k 
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PROBLEM 14.14
For the system of particles of Prob. 14.13, determine (a) the
position vector r of the mass center G of the system, (b) the
linear momentum mv of the system, (c) the angular
momentum HG of the system about G. Also verify that he
answers to this problem and to problem 14.13 satisfy the
equation given in Prob. 14.27.
SOLUTION
Position vectors, (meters):
(a) Mass center:
rA  3j,
rB  1.2i  2.4 j  3k ,
 mA  mB  mC  r
rC  3.6i
 mArA  mBrB  mC rC
9 r   3 3j   2 1.2i  2.4 j  3k    4  3.6i 
r  1.86667i  1.53333j  0.66667k
r  1.867 m  i  1.533 m  j   0.667 m  k 
Linear momentum of each particle,  kg  m2/s  .
m A v A  12i  6 j  6k
mB v B  8i  6 j
mC v C  8i  16 j  8k
(b) Linear momentum of the system,  kg  m/s.
mv  mA v A  mB v B  mC vC  12i  28j  14k
mv  12.00 kg  m/s  i   28.0 kg  m/s  j  14.00 kg  m/s  k 
Position vectors relative to the mass center, (meters).
rA  rA  r  1.86667i  1.46667 j  0.66667k
rB  rB  r  0.66667i  0.86667 j  2.33333k
rC  rC  r  1.73333i  1.53333j  0.66667k
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PROBLEM 14.14 (Continued)
(c) Angular momentum about G,  kg  m2/s  .
HG  rA  mA v A  rB  mB v B  rC  mC vC
i
j
k
i
j
k
 1.86667 1.46667 0.66667  0.66667 0.86667 2.33333
12
6
6
8
6
0
i
j
k
 1.73333 1.53333 0.66667
8
16
8
 12.8i  3.2 j  28.8k    14i  18.6667 j  10.9333k 
  1.6i  8.5333j  15.4667k 
 2.8i  13.3333j  24.2667k

 
 

H G   2.80 kg  m 2 /s i  13.33 kg  m 2 /s j  24.3 kg  m 2 /s k 
i
j
k
r  mv  1.86667 1.53333 0.66667
12
28
14
  2.8 kg  m 2 /s  i  18.1333 kg  m 2 /s  j   33.8667 kg  m 2 /s  k
HG  r  mv    4.8 kg  m2/s  j   9.6 kg  m2/s  k
Angular momentum about O.
HO  rA   mA v A   rB   mB v B   rC   mC vC 
i j k
i
j k
i
j k
 0 3 0  1.2 2.4 3  3.6 0 0
8 16 8
12 6 6
8 6 0
 18i  36k    18i  24 j  12k    28.8j  57.6k 
   4.8 kg  m 2 /s  j   9.6 kg  m 2 /s  k
Note that
HO  HG  r  mv
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PROBLEM 14.15
A 13-kg projectile is passing through the origin O with a velocity v0  (35 m/s)i when it explodes into two
fragments A and B, of mass 5 kg and 8 kg, respectively. Knowing that 3 s later the position of fragment A is
(90 m, 7 m, –14 m), determine the position of fragment B at the same instant. Assume a y   g  9.81 m/s2
and neglect air resistance.
SOLUTION
Motion of mass center:
It moves as if projectile had not exploded.
r  v0ti 
1 2
gt j
2
1
 (35 m/s)(3 s)i  (9.81 m/s 2 )(3 s) 2 j
2
 (105 m)i  (44.145 m) j
Equation (14.12):
m r  mi ri :
m r  m ArA  mB rB
13(105i  44.145 j)  5(90i  7 j  14k )  8rB
8rB  (13  105  5  90)i
 (13  44.145  5  7) j  (5  14)k
 915i  608.89 j  70k
rB  (114.4 m)i  (76.1 m) j  (8.75 m)k 
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PROBLEM 14.16
A 300-kg space vehicle traveling with a velocity v0  (360 m/s)i passes through the origin O at t  0.
Explosive charges then separate the vehicle into three parts A, B, and C, with mass, respectively, 150 kg, 100 kg,
and 50 kg. Knowing that at t  4 s, the positions of parts A and B are observed to be A (1170 m, –290 m,
–585 m) and B (1975 m, 365 m, 800 m), determine the corresponding position of part C. Neglect the effect
of gravity.
SOLUTION
Motion of mass center:
Since there is no external force,
r  v0t  (360 m/s)i (4 s)  (1440 m)i
Equation (14.12):
m r  mi ri :
(300)(1440i)  (150)(1170i  290 j  585k )
 (100)(1975i  365 j  800k )
 (50)rC
50rC  (300  1440  150  1170  100  1975)i
 (150  290  100  365) j  (150  585  100  800)k
 59, 000i  7, 000 j  7, 750k
rC  (1180 m)i  (140 m) j  (155 m)k 
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PROBLEM 14.17
A 2-kg model rocket is launched vertically and reaches an
altitude of 70 m with a speed of 30 m/s at the end of powered
flight, time t  0. As the rocket approaches its maximum
altitude it explodes into two parts of masses mA  0.7 kg and
mB  1.3 kg. Part A is observed to strike the ground 80 m west
of the launch point at t  6 s. Determine the position of part B at
that time.
SOLUTION
Choose a planar coordinate system having coordinates x and y with the origin at the launch point on the
ground and the x-axis pointing east and the y-axis vertically upward.
Let subscript E refer to the point where the explosion occurs, and A and B refer to the fragments A and B.
Let t be the time elapsed after the explosion.
Motion of the mass center:
x  xE  (vx )t  0
y  yE  (v y )t0 
1 2
gt
2
yE  70 m and (v y )0  30 m/s
where
t  6 s,
At
x 0
1
y  70  (30)(6)  (9.81)(6)2  73.42 m
2
Definition of mass center:
mA x  mA x A  mB xB
0  (0.7 kg)(80 m)  (1.3 kg) xB
xB  43.1 m
my  mA y A  mB yB
(2 kg)(73.42 m)  (0.7 kg)(0)  (1.3 kg) yB
yB  113.0 m
Position of part B:
43.1 m (east), 113.0 m (up) 
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PROBLEM 14.18
An 18-kg cannonball and a 12-kg cannonball are chained
together and fired horizontally with a velocity of 165 m/s from
the top of a 15-m wall. The chain breaks during the flight of the
cannonballs and the 12-kg cannonball strikes the ground
at t  1.5 s, at a distance of 240 m from the foot of the wall, and
7 m to the right of the line of fire. Determine the position of the
other cannonball at that instant. Neglect the resistance of the
air.
SOLUTION
Let subscript A refer to the 12-kg cannonball and B to the 18-kg cannonball.
The motion of the mass center of A and B is uniform in the x-direction, uniformly accelerated with
acceleration  g  9.81 m/s 2 in the y-direction, and zero in the z-direction.
x  (v0 ) x t  (165 m/s)(1.5 s)  247.5 m
y  y0  (v0 ) y t 
1 2
gt
2
1
 15 m  0  (9.81 m/s 2 )(1.55)2  3.964 m
2
z 0
r  (247.5 m)i  (3.964 m)j
Definition of mass center:
mr  mArA  mB rB
Data:
mA  12 kg,
mB  18 kg, m  mA  mB  30 kg
t  1.5 s, xA  240 m, y A  0,
zA  7 m
(30)(247.5i  3.964 j)  (12)(240i  7k )  (18)( xB i  yB j  zB k )

i: (30)(247.5)  (12)(240)  18 xB
xB  253 m 
j: (30)(3.964)  (12)(0)  18 yB
yB  6.61 m 
k: (30)(0)  (12)(7)  18 zB
zB  4.67 m 
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PROBLEM 14.19
Car A was traveling east at high speed
when it collided at Point O with car B,
which was traveling north at 45 mi/h. Car
C, which was traveling west at 60 mi/h,
was 32 ft east and 10 ft north of Point O
at the time of the collision. Because the
pavement was wet, the driver of car C
could not prevent his car from sliding into
the other two cars, and the three cars,
stuck together, kept sliding until they hit
the utility pole P. Knowing that the
weights of cars A, B, and C are,
respectively, 3000 lb, 2600 lb, and 2400
lb, and neglecting the forces exerted on
the cars by the wet pavement, solve the
problems indicated.
Knowing that the speed of car A was 75
mi/h and that the time elapsed from the
first collision to the stop at P was 2.4 s,
determine the coordinates of the utility
pole P.
SOLUTION
Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system
consisting of cars A, B, and C during the impacts with one another. The mass center of the system moves at
the velocity it had before the collision.
Setting the origin at O, we can find the initial mass center r0 : at the moment of the first collision:
(mA  mB  mC )( x0 i  y0 j)  mA (0)  mB (0)  mC ( xC i  yC j)
x0  0.3xC  (0.3)(32)  9.6 ft,
y0  0.3 yC  (0.3)(10)  3 ft
Given velocities:
vA  (75 mi/h)i  (110 ft/s)i, v B  (45 mi/h)i  (66 ft/s)j,
vC  (60 mi/h)i  (88 ft/s)i
Velocity of mass center:
(mA  mB  mC ) v  mA v A  mB v B  mC vC
v  0.375vA  0.325vB  0.3vC
Since the collided cars hit the pole at
rP  xP i  yP j
x P i  yP j  x0 i  y0 j  vt
Resolve into components.
x : xP  x0  0.375vAtP  0.3vC tP
(1)
y : yP  y0  0.325vB t P
(2)
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PROBLEM 14.19 (Continued)
Data:
t P  2.4 s
From (1),
xP  9.6  (0.375)(110)(2.4)  (0.3)(88)(2.4)  45.240
From (2),
yP  3.0  (0.325)(66)(2.4)  54.480 ft
xP  45.2 ft 
yP  54.5 ft 
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PROBLEM 14.20
Car A was traveling east at high speed
when it collided at Point O with car B,
which was traveling north at 45 mi/h. Car
C, which was traveling west at 60 mi/h,
was 32 ft east and 10 ft north of Point O
at the time of the collision. Because the
pavement was wet, the driver of car C
could not prevent his car from sliding into
the other two cars, and the three cars,
stuck together, kept sliding until they hit
the utility pole P. Knowing that the
weights of cars A, B, and C are,
respectively, 3000 lb, 2600 lb, and 2400
lb, and neglecting the forces exerted on
the cars by the wet pavement, solve the
problems indicated. Knowing that the
coordinates of the utility pole are
xP  46 ft and yP  59 ft, determine
(a) the time elapsed from the first
collision to the stop at P, (b) the speed of
car A.
SOLUTION
Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system
consisting of cars A, B, and C during the impacts with one another. The mass center of the system moves at
the velocity it had before the collision.
Setting the origin at O, we can find the initial mass center r0 : at the moment of the first collision:
(mA  mB  mC )( x0 i  y0 j)  mA (0)  mB (0)  mC ( xC i  yC j)
x0  0.3xC  (0.3)(32)  9.6 ft,
y0  0.3 yC  (0.3)(10)  3 ft
Given velocities:
vA  vAi, v B  (45 mi/h)i  (66 ft/s)j,
vC  (60 mi/h)i  (88 ft/s)i
Velocity of mass center:
(mA  mB  mC ) v  mA v A  mB v B  mC vC
v  0.375vA  0.325vB  0.3vC
Since the collided cars hit the pole at
rP  xP i  yP j
xP i  yP j  x0 i  y0 j  vt
Resolve into components.
x : xP  x0  0.375vAtP  0.3vC tP
(1)
y : yP  y0  0.325vB t P
(2)
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PROBLEM 14.20 (Continued)
Data:
x P  46 ft, y P  59 ft
( a) From (2),
59  3  0.32566 t P 
t P  2.611 s
(b) From (1),
t P  2.61 s 
46  9.6  0.375 v A 2.611  0.388 2.611
v A  107.58 ft/s
v A  73.3 mi/h 
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PROBLEM 14.21
An expert archer demonstrates his ability by hitting tennis balls thrown by an assistant. A 2-oz tennis ball has
a velocity of (32 ft/s)i – (7 ft/s)j and is 33 ft above the ground when it is hit by a 1.2-oz arrow traveling with a
velocity of (165 ft/s)j  (230 ft/s)k where j is directed upwards. Determine the position P where the ball and
arrow will hit the ground, relative to Point O located directly under the point of impact.
SOLUTION
Assume that the ball and arrow move together after the hit.
Conservation of momentum of ball and arrow during the hit.
mA 
1.2/16
 2.3292  103 slug
32.2
mB 
2/16
 3.8820  103 slug
32.2
mA vA  mB v B  (mA  mB ) v
3
(2.3292  10 )(165 j  230k )  (3.8820  103 )(32i  7 j)  (2.3292  103  3.8820  103 )v
v  (20.0 ft/s)i  (57.5 ft/s) j  (86.25 ft/s)k
After the hit, the ball and arrow move as a projectile.
1 2
gt
2
1
y  33  57.5t  (32.2)t 2
2
y  y0  (v y )0 t 
Vertical motion:
y  0 at ground.
16.1t  57.5t  33  0
2
After rejecting the negative root,
t  4.0745 s
Horizontal motion:
x  x0  (vx )0 t
Solve for t.
z  z0  (v2 )0 t
x  0  (20)(4.07448)
 81.490 ft
z  0  (86.25)(4.07448)
 351.42 ft

rP  (81.5 ft)i  (351 ft)k 
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PROBLEM 14.22
Two spheres, each of mass m, can slide freely on a frictionless,
horizontal surface. Sphere A is moving at a speed v0  16 ft/s when it
strikes sphere B which is at rest and the impact causes sphere B to
break into two pieces, each of mass m/2. Knowing that 0.7 s after the
collision one piece reaches Point C and 0.9 s after the collision the
other piece reaches Point D, determine (a) the velocity of sphere A
after the collision, (b) the angle  and the speeds of the two pieces
after the collision.
SOLUTION
Velocities of pieces C and D after impact and fracture.
(vC ) x 
xC
6.3

 9 ft/s,
tC
0.7
(vC ) y  9 tan 30 ft/s
(vD ) x 
xD
6.3

 7 ft/s,
tD
0.9
(vD ) y  7 tan  ft/s
Assume that during the impact the impulse between spheres A and B is directed along the x-axis. Then, the
y component of momentum of sphere A is conserved.
0  m(vA ) y
Conservation of momentum of system:
: mAv0  mB (0)  mAvA  mC (vC ) x  mD (vD ) x
m (16)  0  mvA 
m
m
(9)  (7)
2
2
vA  8.00 ft/s
(a )

: mA (0)  mB (0)  mA (vA ) y  mC (vC ) y  mD (vD ) y
000
(b)
tan  
m
m
(9 tan 30)  (7 tan  )
2
2
9
tan 30  0.7423
7
vC  (vC )2x  (vC )2y  (9)2  (9 tan 30)2
vD  (vD )2x  (vD )2y  (7)2  (7 tan 36.6)2
  36.6 
vC  10.39 ft/s 
vD  8.72 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.23
In a game of pool, ball A is moving with a velocity v0 when it strikes
balls B and C which are at rest and aligned as shown. Knowing that after
the collision the three balls move in the directions indicated, and that
v0  12 ft/s and vC  6.29 ft/s, determine the magnitude of the velocity
of (a) ball A, (b) ball B.
SOLUTION
Conservation of linear momentum. In x direction:
m(12 ft/s) cos 45  mv A sin 4.3  mvB sin 37.4
 m(6.29) cos 30
0.07498vA  0.60738vB  3.0380

(1)
In y direction:
m(12 ft/s)sin 45  mv A cos 4.3  mvB cos 37.4
 m(6.29)sin 30
0.99719vA  0.79441vB  5.3403
(a)
Multiply (1) by 0.79441, (2) by 0.60738, and add:
0.66524vA  5.6570
(b)

(2)
vA  8.50 ft/s 
Multiply (1) by 0.99719, (2) by –0.07498, and add:
0.66524vB  2.6290 
vB  3.95 ft/s 
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PROBLEM 14.24
A 6-kg shell moving with a velocity v0
 (12 m/s)i  (9 m/s) j  (360 m/s)k explodes
at Point D into three fragments A, B, and C
of mass, respectively, 3 kg, 2 kg, and 1 kg.
Knowing that the fragments hit the vertical
wall at the points indicated, determine the
speed of each fragment immediately after the
explosion. Assume that elevation changes
due to gravity may be neglected.
SOLUTION
rD  4k
Position vectors (m):
rA  1.5i
rA/D  1.5i  4k
rA/D  4.272
rB  4i  2 j
rB/D  4i  2 j  4k
rB/D  6
rC  3j
rC/D  3 j  4k
rC/D  5
1
(1.5i  4k )
4.272
1
 B  (4i  2 j  4k )
Along rB/D ,
6
1
λ C  (3 j  4k )
Along rC/D ,
5
Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explosion
have the directions of the unit vectors.
Unit vectors:
Along rA/D ,
λA 
vA  vA λ A
vB  vB λ B
vC  vC λ C
Conservation of momentum:
mv0  mA vA  mB vB  mC vC
 v 
v 
6(12i  9 j  360k )  3  A  (1.5i  4k )  2  B  (4i  2 j  4k )
 4.272 
 6 
v 
 1 C  (3j  4k )
 5 
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PROBLEM 14.24 (Continued)
Resolve into components.
72  1.0534v A  1.3333vB
54  0.66667vB  0.60000vC
2160  2.8090v A  1.3333vB  0.80000vC
Solving,
vA  431 m/s 
vB  395 m/s 
vC  528 m/s 
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PROBLEM 14.25
A 6-kg shell moving with a velocity
v0  (12 m/s)i  (9 m/s) j  (360 m/s)k explodes at
Point D into three fragments A, B, and C of mass,
respectively, 2 kg, 1 kg, and 3 kg. Knowing that the
fragments hit the vertical wall at the points
indicated, determine the speed of each fragment
immediately after the explosion. Assume that
elevation changes due to gravity may be neglected.
SOLUTION
rD  4k
Position vectors (m):
rA  1.5i
rA/D  1.5i  4k
rA/D  4.272
rB  4i  2 j
rB/D  4i  2 j  4k
rB/D  6
rC  3 j
rC/D  3 j  4k
rC/D  5
1
(1.5i  4k )
4.272
1
λ B  (4i  2 j  4k )
Along rB/D ,
6
1
λ C  (3j  4k )
Along rC/D ,
5
Assume that elevation changes due to gravity may be neglected. Then the velocity vectors after the explosion
have the directions of the unit vectors.
vA  vA A
Along rA/D ,
Unit vectors:
λA 
vB  vB  B
vC  vC  C
Conservation of momentum:
mv0  mA vA  mB vB  mC vC
Resolve into components.
 v 
v
6(12i  9 j  360k )  2  A  (1.5i  4k )  1 B
 4.272 
 6
v 
 3  C  (3j  4k )
 5 

 (4i  2 j  4k )

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PROBLEM 14.25 (Continued)
72  0.70225v A  0.66667vB
54  0.33333vB  1.8000vC
2160  1.8727v A  0.66667vB  2.40000vC
Solving,
vA  646 m/s 
vB  789 m/s 
vC  176 m/s 
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PROBLEM 14.26
In a scattering experiment, an alpha particle A is projected with
the velocity u0  (600 m/s)i  (750 m/s) j  (800 m/s)k into a
stream of oxygen nuclei moving with a common velocity
v0  (600 m/s) j . After colliding successively with nuclei B and
C, particle A is observed to move along the path defined by the
Points A1(280, 240, 120)and A2(360, 320, 160), while nuclei B
and C are observed to move along paths defined, respectively,
by B1(147, 220, 130), B2(114, 290, 120),and by C1(240, 232, 90)
and C2(240, 280, 75). All paths are along straight lines and all
coordinates are expressed in millimeters. Knowing that the mass
of an oxygen nucleus is four times that of an alpha particle,
determine the speed of each of the three particles after the
collisions.
SOLUTION
Position vectors (mm):
Unit vectors:

A1 A2  80i  80 j  40k

B1 B2  33i  70 j  10k

C1C2  48 j  15k
( A1 A2 )  120
( B1 B2 )  78.032
(C1C2 )  50.289
Along A1 A2 ,
A  0.66667i  0.66667 j  0.33333k
Along B1 B2 ,
B  0.42290i  0.89707 j  0.12815k
Along C1C2 ,
C  0.95448 j  0.29828k
Velocity vectors after the collisions:
vA  vA  A
vB  vB B
vC  vC C
Conservation of momentum:
mu0  4mv 0  4mv0  mvA  4mvB  4mvC
Divide by m and substitute data.
(600i  750 j  800k )  2400 j  2400 j  vAA  4vB B  4vC C
Resolving into components,
i :  600  0.66667vA  1.69160vB
j: 5550  0.66667vA  3.58828vB  3.81792vC
k :  800  0.33333vA  0.51260vB  1.19312vC
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PROBLEM 14.26 (Continued)
Solving the three equations simultaneously,
vA  919.26 m/s
vB  716.98 m/s
vC  619.30 m/s
vA  919 m/s 
vB  717 m/s 
vC  619 m/s 
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PROBLEM 14.27
Derive the relation
HO  r  mv  HG
between the angular momenta HO and HG defined in Eqs. (14.7) and (14.24), respectively. The vectors r
and v define, respectively, the position and velocity of the mass center G of the system of particles relative to
the newtonian frame of reference Oxyz, and m represents the total mass of the system.
SOLUTION
From Eq. (14.7),
HO 
n
 (r  m v )
i
i i
i 1

n
  r  r  m v 
i
i i
i 1
r
n

(mi vi ) 
i 1
n
  r  m v 
i
i i
i 1
 r  mv  HG
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PROBLEM 14.28
Show that Eq. (14.23) may be derived directly from Eq. (14.11) by substituting for HO the expression given
in Problem 14.27.
SOLUTION
HO 
From Eq. (14.7),
n
 (r  m v )
i
i i
i 1

n
  r  r  m v 
i
i i
i 1
r
n

(mi vi ) 
i 1
n
  r  m v 
i
i i
i 1
 r  mv  HG
  r  mv  r  mv  H

H
O
G
Differentiating,

M O  r  mv  r  mv  H
G
Using Eq. (14.11),

 v  mv  r  ma  H
G
 n 

 0  r   Fi   H
G


 i 1 

n
n
n

(1)

n
 M  M  r    F 
But
O
i 1

G
i 1
i 1
 i 1
i


Subtracting r  in1Fi from each side of Eq. (1) gives

M G  H
G
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PROBLEM 14.29
Consider the frame of reference Axy z  in translation with respect to the
newtonian frame of reference Oxyz. We define the angular momentum HA of a
system of n particles about A as the sum
H A 
n
 r  m v
i
i
i
(1)
i 1
of the moments about A of the momenta mi vi of the particles in their
motion relative to the frame Axyz . Denoting by H A the sum
HA 
n
 r  m v
i
i
i
(2)
i 1
of the moments about A of the momenta mi vi of the particles in their
motion relative to the newtonian frame Oxyz, show that HA  HA at a
given instant if, and only if, one of the following conditions is satisfied at
that instant: (a) A has zero velocity with respect to the frame Oxyz,
(b) A coincides with the mass center G of the system, (c) the velocity vA
relative to Oxyz is directed along the line AG.
SOLUTION
vi  vA  vi
HA 




n
 r  m v
i
i i
i 1
n
 r  m  v
i
i
A
i 1
n
n
  r  m v
i
i A
i 1
n
   ri  mi vi
i 1
  m r   v
i i
 vi 
A
 H A
i 1
n
 m (r  r )  v
i
i
A
A
 HA
i 1
 m( r  rA )  vA  HA
H A  HA if, and only if,
m(r  rA )  vA  0
This condition is satisfied if
(a) vA  0
Point A has zero velocity.
or
(b)
r  rA
Point A coincides with the mass center.
or
(c )
vA is parallel to r  rA.
Velocity vA is directed along line AG.
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PROBLEM 14.30
  , where H is defined by Eq. (1) of
Show that the relation MA  H
A
A
Problem 14.29 and where M A represents the sum of the moments about A of
the external forces acting on the system of particles, is valid if, and only if,
one of the following conditions is satisfied: (a) the frame Axy z  is itself a
newtonian frame of reference, (b) A coincides with the mass center G,
(c) the acceleration aA of A relative to Oxyz is directed along the line AG.
SOLUTION
From equation (1),
H A 
n
  r   m v 
i
i i
i 1
H A 
n
[(r  r )  m ( v
i
A
i
i
 vA )]
i 1
Differentiate with respect to time.
 
H
A
n
 [(r  r
i
A )  mi ( v i
 vA )] 
i 1
But
ri  v i
v i  ai
rA  vA
and
vA  aA
  0
H
A
Hence,
n
 [(r  r )  m ( v
i
A
i
i
 v A )]
i 1
n
[(r  r )  m (a
i
A
i
i
 aA )]
i 1

n
[(r  r )  (F  m a )]
i
A
i
i A
i 1

n
n
[(r  r )  F ]  [m (r  r )]  a
i
i 1
A
i
i
i
A
A
i 1
 MA  m( r  rA )  aA

H A  MA
if, and only if,
m( r  rA )  aA  0
This condition is satisfied if
(a ) aA  0
The frame is newtonian.
or
(b)
r  rA
Point A coincides with the mass center.
or
(c )
aA is parallel to r  rA.
Acceleration aA is directed along line AG.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.31
Determine the energy lost due to friction and the impacts for
Problem 14.1.
PROBLEM 14.1 A 30-g bullet is fired with a horizontal
velocity of 450 m/s and becomes embedded in block B which
has a mass of 3 kg. After the impact, block B slides on 30-kg
carrier C until it impacts the end of the carrier. Knowing the
impact between B and C is perfectly plastic and the coefficient
of kinetic friction between B and C is 0.2, determine (a) the
velocity of the bullet and B after the first impact, (b) the final
velocity of the carrier.
SOLUTION
From the solution to Problem 4.1 the velocity of A and B after the first impact is v  4.4554 m/s and the
velocity common to A, B, and C after the sliding of block B and bullet A relative to the carrier C has ceased in
v  0.4087 m/s.
Friction loss due to sliding:
N  WA  WB  (m A  mB ) g
Normal force:
 (0.030 kg  3 kg)(9.81 m/s)  29.724 N
F f  k N  (0.2)(29.724)  5.945 N
Friction force:
Assume d  0.5 m.
Relative sliding distance:
Ff d  (5.945)(0.5)
Energy loss due to friction:
F f d  2.97 J 
Kinetic energy of block with embedded bullet immediately after first impact:
 
TAB
1
1
(m A  mB )(v) 2  (3.03 kg)(4.4554 m/s) 2  30.07 J
2
2
Final kinetic energy of A, B, and C together
 
TABC
1
1
(mA  mB  mC )(v) 2  (33.03 kg)(0.4087 m/s)2  2.76 J
2
2
  TABC
  30.07  2.76  27.31 J
TAB
Loss due to friction and stopping impact:
Since 27.31 J  2.97 J, the block slides 0.5 m relative to the carrier as assumed above.
Impact loss due to AB impacting the carrier:
27.31  2.97  24.34
Loss  24.3 J 
Initial kinetic energy of system ABC.
T0 
Impact loss at first impact:
1
1
mAv02  (0.030 kg)(450 m/s) 2  3037.5 J
2
2
  3037.5  30.07
T0  TAB
Loss  3007 J 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.32
Assuming that the airline employee of Prob. 14.3 first tosses
the 30-lb suitcase on the baggage carrier, determine the energy
lost (a) as the first suitcase hits the carrier, (b) as the second
suitcase hits the carrier.
PROBLEM 14.3 An airline employee tosses two suitcases, of
weight 30 lb and 40 lb, respectively, onto a 50-lb baggage
carrier in rapid succession. Knowing that the carrier is initially
at rest and that the employee imparts a 9-ft/s horizontal
velocity to the 30-lb suitcase and a 6-ft/s horizontal velocity to
the 40-lb suitcase, determine the final velocity of the baggage
carrier if the first suitcase tossed onto the carrier is (a) the 30-lb
suitcase,(b) the 40-lb suitcase.
SOLUTION
WA  30 lb,
The weights are
 v A 0
Initial velocities:
 9 ft/s
WB  40 lb,
,
 v B 0
WC  50 lb.
and
 6 ft/s
,
 vC 0
and
 0.
There are no horizontal external forces acting during the impacts, and the baggage carrier is free to coast
between the impacts.
Let v1 be the common velocity of suitcase A and the carrier after the first impact and v2 be the common
velocity of the two suitcases and the carrier after the second impact.
WA
WB
Initial momenta:
and
0.
 v A 0 ,
 vB  0 ,
g
g
Suitcase A impacts carrier. Conservation of momentum.
WA
W W
 v A 0  0  A C v1
g
g
v1 
W A  v A 0
W A  WC

 30  9 
 3.375 ft/s
80
Kinetic energies:
T0 
Before:
T1 
After:
1 WA
1 30
 v A 02 
 9 2  37.733 ft  lb
2 g
2 32.2
1 WA  WC 2 1 30  50
v1 
 3.3752  14.150 ft  lb
2
g
2 32.2
T0  T1  23.583 ft  lb
(a) Lost:
energy lost  23.6 ft  lb 
Suitcase B impacts on suitcase A and carrier. Conservation of momentum.
WB
W W
W W W
 vB 0  A C v1  A B C v2
g
g
g
v2 
WB  vB 0  W A  WC  v1
W A  WB  WC

 40  6   80  3.375 
120
 4.25 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.32 (Continued)
Kinetic energies:
Before:
After:
(b) Lost:
T1 
1 WA  WC 2 1 WB
1 40
v1 
 vB 02  14.150 
 6 2  36.510 ft  lb
2
2 g
2 32.2
g
T2 
1 WA  WB  WC 2 1 120
v2 
 4.252  33.657 ft  lb
2
2 32.2
g
T1  T2  2.853 ft  lb
energy lost  2.85 ft  lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.33
In Problem 14.6, determine the work done by the woman and
by the man as each dives from the boat, assuming that the
woman dives first.
PROBLEM 14.6 A 180-lb man and a 120-lb woman stand
side by side at the same end of a 300-lb boat, ready to dive,
each with a 16-ft/s velocity relative to the boat. Determine
the velocity of the boat after they have both dived, if (a) the
woman dives first, (b) the man dives first.
SOLUTION
Woman dives first.
120
300  180
(16  v1 ) 
v1  0
g
g
Conservation of momentum:
(120)(16)
 3.20 ft/s  and 16  v1  12.80 ft/s
600
T0  0
Kinetic energy before dive:
1 300  180
1 120
Kinetic energy after dive:
T1 
(3.20)2 
(12.80)2
2 32.2
2 32.2
 381.61 ft  lb
v1 
T1  T0  381.61 ft  lb
Work of woman:
T1  T0  382 ft  lb 
Man dives next. Conservation of momentum:

300  180
300
180
v1  
v2 
(16  v2 )
g
g
g
480v1  (180)(16)
v2 
 9.20 ft/s
480
16  9.20  6.80 ft/s
1 300  180
(3.20)2
2 32.2
 76.323 ft  lb
Kinetic energy before dive:
T1 
Kinetic energy after dive:
T2 
Work of man:
1 300
1 180
(9.20) 2 
(6.80) 2
2 32.2
2 32.2
 523.53 ft  lb
T2  T1  447.2 ft  lb
T2  T1  447 ft  lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.34
Determine the energy lost as a result of the series of collisions described in Problem 14.8.
PROBLEM 14.8 Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a
slightly different style but of the same weight, has been pushed by dockworkers and hits car B with a velocity
of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between B and C and 0.5 between A and B,
determine the velocity of each car after all collisions have taken place.
SOLUTION
From the solution to Problem 14.8
v A  2 m/s, vB  vC  0,
vB  1.35 m/s, vC  0.15 m/s
vA  1.013 m/s
vB  0.338 m/s,
mA  mB  mC  m
C hits B:
1
1
mC vC2  m(1.5 m/s) 2  1.125m J
2
2
1
1
T2  mB (vB )2  mC (vC )2
2
2
1
1
T  m(1.35 m/s) 2  m (0.15 m/s)2  0.9m J
2
2
T1 
Loss  T1  T2 :
B hits A:
Loss  0.225m J 
1
1
mB (vB ) 2  m(1.35)2  0.9113m J
2
2
1
1
2
T4  mB (vB )  mA (vA ) 2
2
2
1
1
 m(0.338 m/s)2  m(1.013 m/s) 2  0.5702 m J
2
2
T3 
Loss  T3  T4 :
Loss  0.341m J 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.35
Two automobiles A and B, of mass mA and mB , respectively, are traveling in opposite directions when they
collide head on. The impact is assumed perfectly plastic, and it is further assumed that the energy absorbed by
each automobile is equal to its loss of kinetic energy with respect to a moving frame of reference attached to
the mass center of the two-vehicle system. Denoting by EA and EB, respectively, the energy absorbed by
automobile A and by automobile B, (a) show that EA /EB  mB /mA , that is, the amount of energy absorbed by
each vehicle is inversely proportional to its mass, (b) compute EA and EB , knowing that mA  1600 kg and
mB  900 kg and that the speeds of A and B are, respectively, 90 km/h and 60 km/h.
SOLUTION
Velocity of mass center:
(mA  mB ) v  mA vA  mB vB
v
mA vA  mB vB
mA  mB
Velocities relative to the mass center:
vA  vA  v  vA 
mA vA  mB vB mB ( vA  vB )

mA  mB
mA  mB
vB  vB  v  vB 
mA vA  mB vB mA ( vA  vB )

mA  mB
mA  mB
Energies:
(a)
(b)
EA 
m m 2 ( v  v B )  ( vA  vB )
1
mA v A  v A  A B A
2
2( m A  mB ) 2
EB 
m 2 m ( v  vB )  ( vA  vB )
1
mB vB  vB  A B A
2
2( mA  mB ) 2
EA mB


EB mA
Ratio:
vA  90 km/h  25 m/s
vB  60 km/h  16.667 m/s
vA  vB  41.667 m/s
EA 
(1600)(900) 2 (41.667) 2
 180.0  103 J
2
(2)(2500)
EB 
(1600) 2 (900)(41.667)2
 320  103 J
2
(2)(2500)
EA  180.0 kJ 
EB  320 kJ 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.36
It is assumed that each of the two automobiles involved in the collision described in Problem 14.35 had been
designed to safely withstand a test in which it crashed into a solid, immovable wall at the speed v0. The
severity of the collision of Problem 14.35 may then be measured for each vehicle by the ratio of the energy it
absorbed in the collision to the energy it absorbed in the test. On that basis, show that the collision described
in Problem 14.35 is (mA /mB )2 times more severe for automobile B than for automobile A.
SOLUTION
Velocity of mass center:
(mA  mB ) v  mA vA  mB vB
v
mA vA  mB vB
mA  mB
Velocities relative to the mass center:
vA  vA  v  v A 
mA vA  mB vB mB ( vA  vB )

mA  mB
mA  mB
vB  vB  v  vB 
mA vA  mB vB
m (v  v )
 A A B
mA  mB
mA  mB
Energies:
Energies from tests:
Severities:
Ratio:
EA 
m m 2 ( v  vB )  ( vA  vB )
1
mA v A  v A  A B A
2
2(mA  mB ) 2
EB 
m 2 m ( v  vB )  ( vA  vB )
1
mB vB  vB  A B A
2
2(mA  mB ) 2
( E A )0 
1
1
mA v02 , ( EB )0  mB v02
2
2
SA 
EA
m 2 ( v  vB )  ( vA  vB )
 B A
( E A )0
(m A  mB ) 2 v02
SB 
EB
m 2 ( v  vB )  ( vA  vB )
 A A
( EB )0
(mA  mB ) 2 v02
SA mB2


SB m A2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.37
Solve Sample Problem 14.5, assuming that cart A is given an initial horizontal velocity v0 while ball B is at rest.
SOLUTION
(a)
Velocity of B at maximum elevation: At maximum elevation, ball B is at rest relative to cart A. vB  vA
Use impulse-momentum principle.
mA v0  0  m A vA  mB vB
x components:
 (mA  mB )vB
vB 
(b)
mA v0
mA  mB

Conservation of energy:
1
m A v02 , V1  0
2
1
1
T2  m A v A2  mB vB2
2
2
1
 (m A  mB )vB2
2
m A2 v02

2(mA  mB )
T1 
V2  mB gh
T2  V2  T1  V1
m A2 v02
2(mA  mB )
 mB gh 
h
1
m A v02
2
mA2 v02 
1 
2
 mAv0 

mA  mB 
2mB g 
h
v02
mA

mA  mB 2 g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.38
Two hemispheres are held together by a cord which maintains a spring under
compression (the spring is not attached to the hemispheres). The potential
energy of the compressed spring is 120 J and the assembly has an initial
velocity v 0 of magnitude v0  8 m/s. Knowing that the cord is severed when
  30, causing the hemispheres to fly apart, determine the resulting velocity
of each hemisphere.
SOLUTION
Use a frame of reference moving with the mass center.
Conservation of momentum:
0  mAvA  mB vB
vA 
mB
vB
mA
V
1
1
mA (vA )2  mB (vB )2
2
2
Conservation of energy:
2
m

1
1
mA  B vB   mB (vB )2
m
2
2
 A 
m (m  mB )
 B A
(vB )2
2m A

vB 
2mAV
mB (mA  mB )
mA  2.5 kg mB  1.5 kg
Data:
V  120 J
vB 
vA 
(2)(2.5)(120)
 10
(1.5)(4.0)
1.5
(10)  6
2.5
vB  10 m/s
vA  6 m/s
30
30
Velocities of A and B.
vA  [8 m/s
]  [6 m/s
vB  [8 m/s
]  [10 m/s
30]
30]
vA  4.11 m/s
vB  17.39 m/s
46.9 
16.7 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.39
A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is
supported by a horizontal surface. Neglecting friction, determine (a) the velocity
of B relative to A after it has slid 3 ft down the inclined surface of the wedge,
(b) the corresponding velocity of A.
SOLUTION
Kinematics:
v B  v A  v B/A
Law of cosines:
vB2  vA2  vB2/A  2vAvB/A cos30
Principle of impulse and momentum:
mv0  Ft  mv
Components
:
0  0  mA v A  mB (v A  vB/A cos 30)
vA 
mB vB /A cos 30 15cos 30

vB/A
m A  mB
25  15
 0.32476 vB/A
From Eq. (1)
vB2  (0.32476) 2 vB2/A  vB2/A  (2)(0.32476) cos 30vB2/A
 0.54297 vB2/A
Principle of conservation of energy:
T0  V0  T1  V1
1
1
mA v A2  mB vB2  WB d sin 30
2
2
W
1 A
1 WB
(0.32476vB/A ) 2 
(0.54297) vB2/A  WB d sin 30
2 g
2 g
00
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 14.39 (Continued)
1 15
 1 25
 2
2
 2 32.2 (0.32476)  2 32.2 (0.54297)  vB/A  (15)(3)sin 30


0.16741vB2/A  22.5
v B/A  11.59 ft/s
(a)
(b)
vA  (0.32476)(11.59)
v A  3.76 ft/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
30 

PROBLEM 14.40
A 40-lb block B is suspended from a 6-ft cord attached to a 60-lb cart A,
which may roll freely on a frictionless, horizontal track. If the system is
released from rest in the position shown, determine the velocities of A and B
as B passes directly under A.
SOLUTION
Conservation of linear momentum:
Since block and cart are initially at rest,
L0  0
Thus, as B passes under A,
L  mA vA  mB vB  0
mAv A  mB vB  0
mB
vB
mA
vA  
(1)
Conservation of energy:
Initially,
T0  0
V0  mB gl (1  cos  )
As B passes under A,
Thus,
1
1
m A v A2  mB vB2
2
2
V 0
T
T0  V0  T  V : mB gl (1  cos  ) 
1
1
m A v A2  mB vB2
2
2
Substituting for vA from (1) and multiplying by 2:
 m2 
2mB gl (1  cos  )  mA  B2 vB2   mB vB2
m

 A 
 m2

m  mA 2
  B  mB  vB2  mB B
vB
m

mA
 A

vB 
2m A
gl (1  cos  )
mA  mB
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 14.40 (Continued)
Given data:
wA  60 lb wB  40 lb, l  6 ft
g  32.2 ft/s 2   25
2mA
2 wA
(2)(60)


 1.2
mA  mB wA  wB 60  40
From Eq. (2),
v B  (1.2)(32.2)(6)(1  cos 25)
v B  4.66 ft/s

From Eq. (1),
vA  
wB
40
vB   (4.66)
wA
60
vA  3.11 ft/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.41
In a game of pool, ball A is moving with a velocity v0 of magnitude
v0  15 ft/s when it strikes balls B and C, which are at rest and aligned
as shown. Knowing that after the collision the three balls move in the
directions indicated and assuming frictionless surfaces and perfectly
elastic impact (that is, conservation of energy), determine the
magnitudes of the velocities v A , v B , and vC .
SOLUTION
v 0  v0  cos30 i  sin 30j
Velocity vectors:
v0  15 ft/s
v A  v A j
v B  vB  sin 30 i  cos 30 j
v C  vC  cos 30 i  sin 30 j
Conservation of momentum:
mv0  mv A  mv B  mvC
Divide by m and resolve into components.
i: v0 cos30  vB sin 30  vC cos30
j: v0 sin 30  v A  vB cos30  vC sin 30
Solving for vB and vC ,
vB 
1
2
vC 
1
 v0  vA 
2
1 2 1 2 1 2 1 2
mv0  mv A  mvB  mvC
2
2
2
2
Conservation of energy:
Divide by
3
 v0  vA 
2
m and substitute for vB and vC .
v02  vA2 
3
1
 v0  vA 2   v0  vA 2
4
4
 2vA2  v02  v0vA
vA 
1
v0  7.5 ft/s
2
3
15  7.5  6.4952 ft/s
2
1
vC  15  7.5   11.25 ft/s
2
vB 
v A  7.50 ft/s 
vB  6.50 ft/s 
vC  11.25 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.42
In a game of pool, ball A is moving with a velocity v0 of magnitude
v0  15 ft/s when it strikes balls B and C, which are at rest and aligned
as shown. Knowing that after the collision the three balls move in the
directions indicated and assuming frictionless surfaces and perfectly
elastic impact (that is, conservation of energy), determine the
magnitudes of the velocities vA, vB and vC.
SOLUTION
v 0  v0  cos 45 i  sin 45j
Velocity vectors:
v0  15 ft/s
v A  vA j
v B  vB  sin 60 i  cos 60 j
v C  vC  cos 60 i  sin 60 j
Conservation of momentum:
mv0  mv A  mv B  mvC
Divide by m and resolve into components.
i: v0 cos 45  vB sin 60  vC cos 60
j: v0 sin 45  vA  vB cos 60  vC sin 60
Solving for vB and vC ,
vB  0.25882v0  0.5vA
Conservation of energy:
vC  0.96593v0  0.86603vA
1 2 1 2 1 2 1 2
mv0  mv A  mvB  mvC
2
2
2
2
Divide by m and substitute for vB and vC .
v02  v A2   0.25882v0  0.5v A    0.96593v0  0.86603v A 
2
2
 v02  1.4142v0v A  2v A2
vA  0.70711v0  10.607 ft/s
vA  10.61 ft/s 
vB  0.61237v0  9.1856 ft/s
vB  9.19 ft/s 
vC  0.35356v0  5.303 ft/s
vC  5.30 ft/s 
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PROBLEM 14.43
Three spheres, each of mass m, can slide freely on a frictionless, horizontal
surface. Spheres A and B are attached to an inextensible, inelastic cord of
length l and are at rest in the position shown when sphere B is struck
squarely by sphere C which is moving with a velocity v 0. Knowing that the
cord is taut when sphere B is struck by sphere C and assuming perfectly
elastic impact between B and C, and thus conservation of energy for the
entire system, determine the velocity of each sphere immediately after
impact.
SOLUTION
sin  
1
,
3
8
,
3
cos 
  19.471
v 0  v0 j
Velocity vectors
v A  v A  cos i  sin  j
v B/ A  u B   sin  i  cos  j 
v B  v A  v B/ A
vC  vC j
mv 0  mv A  mv B  mvC  2mv A  mv B/ A  mvC
Conservation of momentum:
Divide by m and resolve into components.
i: 0  2v A cos  uB sin 
 j : v0  2vA sin   uB cos  vC
vA 
Solving for v A and uB ,
Conservation of energy:
1
 v0  vC  uB  0.94281 v0  vC 
6
1 2 1 2 1 2 1 2
mv0  mv A  mvB  mvC
2
2
2
2

Divide by
1
2


1 2 1
1
mv A  m v A2  u B2  mvC2
2
2
2
m and substitute for v A and uB .
2
2
2
2
1
v02  2    v0  vC    0.94281  v0  vC   vC2
6
v02  vC2  0.94445  v0  vC 
v A  0.17143v0
2
vC  0.02857v0
v A   0.17143v0
u B  0.96975v0
19.471 ,
v B/ A  0.96975v0
v B  v A  v B/ A
vC  0.0286v0 
v A  0.1714v0
19.5 
19.471
v B  0.985
80.1 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.44
In a game of pool, ball A is moving with the velocity v0  v0 i
when it strikes balls B and C, which are at rest side by side.
Assuming frictionless surfaces and perfectly elastic impact (i.e.,
conservation of energy), determine the final velocity of each
ball, assuming that the path of A is (a) perfectly centered and
that A strikes B and C simultaneously, (b) not perfectly centered
and that A strikes B slightly before it strikes C.
SOLUTION
(a)
A strikes B and C simultaneously:
During the impact, the contact impulses make 30 angles with the velocity v0.
v B  vB (cos 30i  sin 30 j)
Thus,
v C  vC (cos 30i  sin 30 j)
vA  vAi
By symmetry,
Conservation of momentum:
mv0  mv A  mv B  mvC
y component:
0  0  mvB sin 30  mvC sin 30
x component:
mv0  mvA  mvB cos30  mvC cos30
vC  vB
v0  vA
2
(v0  vA )

cos30
3
v v
vB  vC  0 A
3
vB  vC 
Conservation of energy:
1 2 1 2 1 2 1 2
mv0  mvA  mvB  mvC
2
2
2
2
2
v02  v A2  (v0  vA ) 2
3
2
v02  v A2  (v0  vA )(v0  vA )  (v0  vA )2
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.44 (Continued)
2
1
5
(v0  vA )
v0   v A
3
3
3
6
2 3
vB  vC 
v0 
v0
5
5 3
v0  vA 
1
vA   v0
5
vA  0.200v0
(b)
vB  0.693v0
30 
vC  0.693v0
30 
A strikes B before it strikes C:
First impact: A strikes B.
During the impact, the contact impulse makes a 30 angle with the velocity v 0 .
Thus,
Conservation of momentum:
v B  vB (cos30i  sin 30 j)
mv 0  mvA  mvB
y component:
0  m(vA ) y  mvB sin 30
x component:
v0  m(vA ) x  mvB cos30

(vA ) y  vB sin 30
(vA ) x  v0  vB cos30
Conservation of energy:
1 2 1
1
1
mv0  m(vA )2x  m(vA ) 2y  mvB2
2
2
2
2
1
1
1
 m(v0  vB cos 30)2  (vB sin 30) 2  vB2
2
2
2
1
 m v02  2v0 vB  vB2 cos 2 30  vB2 sin 2 30  vB2
2


3
1
v0 , (vA ) x  v0 sin 2 30  v0 ,
2
4
3
(vA ) y  v0 cos30 sin 30  
v0
4
vB  v0 cos30 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.44 (Continued)
Second impact: A strikes C.
During the impact, the contact impulse makes a 30 angle with the velocity v 0 .
vC  vC (cos30i  sin 30 j)
Thus,
Conservation of momentum:
x component:
y component:
mvA  mvA  mvC
m(vA ) x  m(vA ) x  mvC cos 30,
1
(vA ) x  (vA ) x  vC cos 30  v0  vC cos 30
4
m(vA ) y  m(vA ) y  mvC sin 30
(vA ) y  (vA ) y  vC sin 30  
3
v0  vC sin 30
4
Conservation of energy:
1
1
1
1
1
m(vA )2x  m(vA )2y  m(v A ) 2x  m(v A ) 2y  mvC2
2
2
2
2
2
2
2


1  1 2 3 2  1  1
3
 
m  v0  v0   m  v0  vC cos 30    
v0  vC sin 30   vC2 


2 16
16  2  4
  4



1 1 2 1
m  v0  v0 vC cos 30  vC2 cos 2 30
2 16
2


3 2
3
v0 
v0 vC sin 30  vC2 sin 2 30  vC2 
16
2

1

3
0  v0 vC  cos 30 
sin 30   2vC2
2

2


1

3
3
vC  v0  cos 30 
v0
sin 30  
4

4

 4
1
3
1
v0 cos 30   v0
(v A ) x  v0 
4
4
8
3
3
3
v0 
v0 sin 30  
v0
(v A ) y  
4
4
8
vA  0.250v0
60° 
vB  0.866v0
30° 
vC  0.433v0
30° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.45
The 2-kg sub-satellite B has an initial velocity vB = 3 m/s j. It is connected to the
20-kg base-satellite A by a 500-m space tether. Determine the velocity of the base
satellite and sub-satellite immediately after the tether becomes taut (assuming no
rebound).
SOLUTION
mA  20 kg, mB  2 kg
Given:
 v A 0  0,  v B 0  3j m/s
Because the cord is inextensible, after it becomes taut, the component of vB along AB must be equal to vA.
Impulse-Momentum Diagram:
Linear momentum is conserved:
x-components:
0  mAv A cos   mBv A cos   mBvB / A sin 
0  13.2v A  1.6vB / A
y-components:
2 kg  3 m/s   mAv A sin   mBv A sin   mBvB / A cos 
6  17.6v A  1.2vB / A
Solving (1) and (2) simultaneously:
(1)
(2)
v A  0.2182 m/s, vB / A  1.8 m/s
v A  0.218 m/s
53.1 
vB  vA  vB /A
  v A cos   vB / A sin   i   v A sin   vB / A cos   j
 1.3091i  1.2545 j m/s
v B  1.813 m/s
43.8 
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PROBLEM 14.46
A 900-lb space vehicle traveling with a velocity v 0  (1500 ft/s) k passes through the origin O. Explosive
charges then separate the vehicle into three parts A, B, and C, with masses of 150 lb, 300 lb, and 450 lb,
respectively. Knowing that shortly thereafter the positions of the three parts are, respectively, A(250, 250, 2250),
B (600, 1300, 3200), and C (–475, –950, 1900), where the coordinates are expressed in ft, that the velocity of B
is vB  (500 ft/s)i  (1100 ft/s) j  (2100 ft/s) k , and that the x component of the velocity of C is 400 ft/s,
determine the velocity of part A.
SOLUTION
rA  250i  250 j  2250 k
Position vectors (ft):
rB  600i  1300 j  3200 k
rC  475i  950 j  1900 k
Since there are no external forces, linear momentum is conserved.
(mA  mB  mC ) v 0  mA v A  mB v B  mC vC
vA 
mA  mB  mC
m
m
v0  B vB  C vC  6 v0  2 vB  3vC
mA
mA
mA
 (6)(1500k )  (2)(500i  1100 j  2100k )  (3)[400i  (vC ) y j  (vC ) z k ]
 3(vC ) y j  3(vC ) z k  200i  2200 j  4800 k
(v A ) x  200, (vC ) y  3(vC ) y  2200, (v A ) z  3(vC ) z  4800
Conservation of angular momentum about O:
(HO )2  (HO )1
Since the vehicle passes through the origin, (HO )1  0.
(HO )2  rA  (mA vA )  rB  (mB vB )  rC  (mC vC )  0
Divide by mA.
rA  vA 
m
mB
rB  vB  C rC  vC  rA  vA  2rB  vB  3rC  vC
mA
mA
 rA  (6 v 0  2 v B  3vC )  2rB  v B  3rC  vC
 3(rC  rA )  vC  6rA  v0  2(rB  rA )  vB
 (2175 i  3600 j  1050 k )  vC  (1500 i  1500 j  13500 k )  v0
 (700i  2100 j  1900k )  v B  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 14.46 (Continued)
i
j
k
i
j
k
i
j
k
2175 3600 1050  1500 1500 13500  700 2100 1900  0
(vC ) x (vC ) y (vC ) z
0
0
1500
500 1100 2100
Resolve into components.
i:
1050(vC ) y  3600(vC ) z  2, 250, 000  2,320,000  0
(2)
j:
2175(vC ) z  1050(vC ) x  2, 250,000  520,000  0
(3)
k:
3600(vC ) x  2175(vC ) y  0  280, 000  0
(4)
Set
(vC ) x  400 ft/s
From Eq. (4),
(vC ) y  790.80 ft/s
From Eq. (3),
(vC ) z  1080.5 ft/s
From Eq. (1),
vA  (6)(1500k )  (2)(500 i  1100 j  2100 k )  (3)[400 i  (790.80) j  (1080.5) k ] 
 200 i  172.4 j  1558.6 k

vA  (200 ft/s) i  (172 ft/s)j  (1560 ft/s) k 
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PROBLEM 14.47
Four small disks A, B, C, and D can slide freely on a frictionless horizontal
surface. Disks B, C, and D are connected by light rods and are at rest in the
position shown when disk B is struck squarely by disk A, which is moving to
the right with a velocity v0  (38.5 ft/s)i. The weights of the disks are
WA  WB  WC  15 lb, and WD  30 lb. Knowing that the velocities of the
disks immediately after the impact are v A  v B  (8.25 ft/s)i, vC  vC i, and
v D  vD i, determine (a) the speeds vC and vD , (b) the fraction of the
initial kinetic energy of the system which is dissipated during the collision.
SOLUTION
There are no external forces. Momentum is conserved.
(a) Moments about D
vC 
3mA
3(mA  mB )
v0 
vB  (0.5)(38.5)  (8.25)  11
6mC
6mC
Moments about C
vD 
3mAv0  6mC vC  3(mA  mB )vB
:
vC  11.00 ft/s 
3mAv0  3(mA  mB )vB  6mDvD
:
3mAv0 3(mA  mB )

vB  (0.25)(38.5)  (0.5)(8.25)  5.5 ft/s
6mD
6mD
vD  5.50 ft/s 
(b) Initial kinetic energy:
T1 
1 WA 2 1 15
v0 
(38.5) 2  345.24 ft  lb
2 g
2 32.2
Final kinetic energy:
T2 

1 WA  WB 2 1 WC 2 1 WD 2
vB 
vC 
vD
2
g
2 g
2 g
1 30
1 15
1 30
(8.25) 2 
(11.00) 2 
(5.50) 2  73.98 ft  lb
2 32.2
2 32.2
2 32.2
345.24  73.98  271.26 ft  lb
Energy lost:
Fraction of energy lost 
271.26
 0.786
345.24
(T1  T2 )
 0.786 
T1
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.48
In the scattering experiment of Problem 14.26, it is known that
the alpha particle is projected from A0 (300, 0, 300) and that it
collides with the oxygen nucleus C at Q(240, 200, 100), where
all coordinates are expressed in millimeters. Determine the
coordinates of Point B0 where the original path of nucleus B
intersects the zx plane. (Hint: Express that the angular
momentum of the three particles about Q is conserved.)
PROBLEM 14.26 In a scattering experiment, an alpha particle A
is projected with the velocity u0  (600 m/s)i  (750 m/s) j 
(800 m/s)k into a stream of oxygen nuclei moving with a
common velocity v0  (600 m/s) j . After colliding successively
with nuclei B and C, particle A is observed to move along the
path defined by the Points A1 (280, 240, 120) and A2 (360, 320,
160), while nuclei B and C are observed to move along paths
defined, respectively, by B1 (147, 220, 130), B2 (114, 290, 120),
and by C1(240, 232, 90) and C2 (240, 280, 75). All paths are
along straight lines and all coordinates are expressed in
millimeters. Knowing that the mass of an oxygen nucleus is
four times that of an alpha particle, determine the speed of each
of the three particles after the collisions.
SOLUTION
Conservation of angular momentum about Q:






QA0  ( mu 0 )  QB 0  (4 m v0 )  QC 0  (4 mv0 )  QA1  ( mvA )  QB1  (4 mvB )  QC 1  (4 mvC )



QA0  ( m u 0 )  QB 0  (4 mv 0)  0  0  QB1  (4 mvB )  0

QA0  rA0  rQ  (300i  300 k )  (240 i  200 j  100 k )
where
 (60 mm)i  (200 mm) j  (200 mm)k

QB 0  (x)i  (y ) j  (z )k

QB1  rB1  rQ  (147i  220 j  130 k )  (240i  200 j  100 k )
 (93 mm)i  (20 mm)j  (30 mm)k
u 0  (600 m/s)i  (750 m/s)j  (800 m/s)k
v 0  (600 m/s)j
and from the solution to Problem 14.26,
vB  vB  B  (716.98)(0.42290i  0.89707 j  0.12815 k )
 (303.21 m/s) i  (643.18 m/s) j  (91.88 m/s)k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 14.48 (Continued)
Calculating each term and dividing by m,
i
j
k

QA0  u 0 60 200 200  10, 000 i  72, 000 j  75, 000 k
600 750 800

QB 0  (4 v 0 )  [( x )i  ( y ) j  ( z ) k ]  (2400 j)
 2400( z )i  2400( x ) k

QB1  (4 vB ) 
i
j
k
93
 84,532 i  70,565 j  215, 006 k
20
30
1212.84 2572.72 367.52
Collect terms and resolve into components.
Coordinates:
i:
10,000  2400(z )  84,532
k:
 75,000  2400(x)  215, 006
z  39.388 mm
x  58.336 mm
xB0  xQ  x  240  58.336
xB0  181.7 mm 
yB0  0 
zB0  zQ  z  100  39.388
zB0  139.4 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.49
Three identical small spheres, each of weight 2 lb, can slide
freely on a horizontal frictionless surface. Spheres B and C
are connected by a light rod and are at rest in the position
shown when sphere B is struck squarely by sphere A which is
moving to the right with a velocity v0  (8 ft/s)i. Knowing
that θ = 45° and that the velocities of spheres A and B
immediately after the impact are v A  0 and v B  (6 ft/s)i 
( v B ) y j, determine (v B ) y and the velocity of C immediately
after impact.
SOLUTION
Let m be the mass of one ball.
Conservation of linear momentum: (mv)  (mv)0
mv A  mv B  mvC  m( v A )0  (mv B )0  (mvC )0
Dividing by m and applying numerical data,
0  [(6 ft/s)i  (vB ) y j]  [(vC ) x i  (vC ) y j]  (8 ft/s)i  0  0
Components:
x: 6  (vC ) x  8
(vC ) x  2 ft/s 
y: (vB ) y  (vC ) y  0 

(1)
Conservation of angular momentum about O:
[r  (mv)]  [r  (mv0 )]
where rA = 0,
rB = 0,
rC  (1.5 ft)(cos 45i  sin 45j)
(1.5)(cos 45i  sin 45 j)  [m(vC ) x i  m(vC ) y j]  0
Since their cross product is zero, the two vectors are parallel.
(vC ) y  (vC ) x tan 45  2 tan 45  2 ft/s
From (1), (vB ) y   2 ft/s
(vB ) y   2.00 ft/s 


vC  (2.00 ft/s)i + (2.00 ft/s) j 
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PROBLEM 14.50
Three small spheres A, B, and C, each of mass m, are connected to a
small ring D of negligible mass by means of three inextensible,
inelastic cords of length l. The spheres can slide freely on a frictionless
horizontal surface and are rotating initially at a speed v0 about ring D
which is at rest. Suddenly the cord CD breaks. After the other two
cords have again become taut, determine (a) the speed of ring D,
(b) the relative speed at which spheres A and B rotate about D, (c) the
fraction of the original energy of spheres A and B that is dissipated
when cords AD and BD again became taut.
SOLUTION
Let the system consist of spheres A and B.
State 1: Instant cord DC breaks.

3
1 
m ( v A )1  mv0  
i  j
 2
2 

 3
1 
i  j
m( v B )1  mv0 
 2
2 

L1  m ( v A )1  m( v B )1  mv0 j
v
L1
1
  v0 j
2m
2
Mass center lies at Point G midway between balls A and B.
3
3
lj  (mv A )1  
lj  (mv B )1
2
2
3
 lmv0k
2
1
1
T1  mv02  mv02  mv02
2
2
(H G )1 
State 2: The cord is taut. Conservation of linear momentum:
1
v D  v   v0 j
2
(a)
Let
( v A )2  v  u A
vD  0.500v0 
and
v B  v  uB
L2  2mv  mu A  mu B  L1
u B  u A
uB  u A
(HG )2  lmu Ak  lmuBk  2lmu Ak
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.50 (Continued)
(b)
Conservation of angular momentum:
(HG )2  (HG )1
2lmu Ak 
3
lmv0k
2
u A  uB 
3
v0
4
u  0.750v0 
1
1
1
(2m)v 2  mu A2  muB2
2
2
2
1
9
9  13 2
1
 mv02  
 
mv0
2
2
16
16

 16
T2 
(c)
Fraction of energy lost:
13
T1  T2 1  16
3


T1
1
16
T1  T2
 0.1875 
T1
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.51
In a game of billiards, ball A is given an initial velocity v0 along
the longitudinal axis of the table. It hits ball B and then ball C,
which are both at rest. Balls A and C are observed to hit the sides
of the table squarely at A and C , respectively, and ball B is
observed to hit the side obliquely at B. Knowing that
v0  4 m/s, v A  1.92 m/s, and a  1.65 m, determine (a) the
velocities v B and vC of balls B and C, (b) the Point C where
ball C hits the side of the table. Assume frictionless surfaces and
perfectly elastic impacts (that is, conservation of energy).
SOLUTION
Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.
( v A )0  v0i  4i,
Before impacts:
v A  1.92 j,
After impacts:
(v B )0  ( vC )0  0
v B  (vB ) x i  (vB ) y j,
vC  vC i
v 0  v A  v B  vC
Conservation of linear momentum:
i: 4  0  (vB ) x  vC
j: 0  1.92  (vB ) y  0
(vB ) x  4  vC
(vB ) y  1.92
1 2 1 2 1 2 1 2
v0  v A  vB  vC
2
2
2
2
Conservation of energy:
1 2 1
1
1
1
(4)  (1.92) 2  (1.92) 2  (4  vC ) 2  vC2
2
2
2
2
2
vC2  4vC  3.6864  0
vC 
4  (4) 2  (4)(3.6864)
 2  0.56  2.56
2
or
1.44
Conservation of angular momentum about B :
0.75v0  (1.8  a)v A  cvC
cvC  (0.75)(4)  (1.8  1.65)(1.92)  2.712
c
2.712
vC
If vC  1.44,
c  1.8833
If vC  2.56,
c  1.059
off the table. Reject.
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PROBLEM 14.51 (Continued)
(vB ) x  4  2.56  1.44,
Then,
v B  1.44i  1.92j
Summary.
v B  2.40 m/s
(a)

(b)

53.1 
vC  2.56 m/s

c  1.059 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.52
For the game of billiards of Problem 14.51, it is now assumed
that v0  5 m/s, vC  3.2 m/s, and c  1.22 m. Determine
(a) the velocities v A and v B of balls A and B, (b) the Point A
where ball A hits the side of the table.
PROBLEM 14.51 In a game of billiards, ball A is given an initial
velocity v0 along the longitudinal axis of the table. It hits ball B
and then ball C, which are both at rest. Balls A and C are
observed to hit the sides of the table squarely at A and C ,
respectively, and ball B is observed to hit the side obliquely at B.
Knowing that v0  4 m/s, v A  1.92 m/s, and a  1.65 m,
determine (a) the velocities v B and vC of balls B and C, (b) the
Point C where ball C hits the side of the table. Assume
frictionless surfaces and perfectly elastic impacts (that is,
conservation of energy).
SOLUTION
Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.
( v A )0  v0i  5i,
Before impacts:
v A  v A j,
After impacts:
(v B )0  ( vC )0  0
v B  (vB ) x i  (vB ) y j,
vC  3.2i
v 0  v A  v B  vC
Conservation of linear momentum:
i:
5  0  (vB ) x  3.2
(vB ) x  1.8
j:
0  v A  (vB ) y  0
(vB ) y  v A
1 2 1 2 1 2 1 2
v0  v A  vB  vC
2
2
2
2
Conservation of energy:
1 2 1
1
1
1
(5)  (v A )2  (1.8) 2  (v A )2  (3.2) 2
2
2
2
2
2
(a)
v A2  11.52
(vB ) y  2.4
v A  2.40 m/s 
v A  2.4
v B  1.8i  2.4 j
v B  3.00 m/s
53.1 
Conservation of angular momentum about B :
0.75v0  (1.8  a)vA  cvC
avA  1.8v A  cvC  0.75v0
 (1.8)(2.4)  (1.22)(3.2)  (0.75)(5)  4.474
(b)
a
4.474 4.474

2.4
vA
a  1.864 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.53
Two small disks A and B, of mass 3 kg and
1.5 kg, respectively, may slide on a
horizontal, frictionless surface. They are
connected by a cord, 600 mm, long, and
spin counterclockwise about their mass
center G at the rate of 10 rad/s. At t  0, the
coordinates of G are x0  0, y0  2 m, and
its velocity is v0  (1.2 m/s)i  (0.96 m/s) j .
Shortly thereafter, the cord breaks; disk A is
then observed to move along a path parallel
to the y axis and disk B along a path which
intersects the x axis at a distance b  7.5 m
from O. Determine (a) the velocities of A
and B after the cord breaks, (b) the distance
a from the y-axis to the path of A.
SOLUTION
Initial conditions.
Location of G:
AG BG AG  GB AB 0.6 m




mB mA mB  mA
m 4.5 kg
 0.6 
AG  1.5 
  0.2 m
 4.5 
BG  0.4 m
Linear momentum:
L0  m v0  (4.5 kg)(1.2 i  0.96 j)
 5.4i  4.32 j
Angular momentum:
About G:


(H G )0  GA  m A vA  GB  mB vB
 (0.2 m)(3 kg)(0.2 m  10 rad/s)k
 (0.4 m)(1.5 kg)(0.4 m  10 rad/s)k
(H G )0  (3.6 kg  m 2 /s)k
About O: Using formula derived in Problem 14.27,
(HO )0  r  mv 0  (HG )0
 2 j  (5.4i  4.32 j)  3.6k
 10.8k  3.6k  (7.2 kg  m2 /s)k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.53 (Continued)
Kinetic energy: Using Eq. (14.29),
1
1
1
1
1
mv02   m : vi2  mv02  m A vA2  mB vB2
2
2i
2
2
2
1
1
1
 (4.5)[(1.2) 2  (0.96) 2 ]  (3)(0.2  10)2  (1.5)(0.4  10) 2
2
2
2
 (5.3136  6  12)  23.314 J
T0 
(a)
Conservation of linear momentum:
L0  L
5.4i  4.32 j  mA v A  mB v B
 3(v A j)  1.5[(vB ) z i  (vB ) y j]
Equating coefficients of i:
5.4  1.5(vB ) x
(vB ) x  3.6 m/s
(1)
Equating coefficients of j: 4.32  3v A  1.5(vB ) y
(vB ) y  2.88  2vA
(2)
Conservation of energy:
T0  T : T0 
23.314 J 
1
1
m Av A2  mB vB2
2
2
1
1
(3)  v A2   (1.5)[(vB ) 2x  (vB )2y ]
2
2
Substituting from Eqs. (1) and (2):
23.314  1.5v A2  0.75(3.6) 2  0.75(2.88  2v A ) 2
4.5v A2  8.64vA  7.373  0
v A2  1.92vA  1.6389  0
vA  0.96  1.60  2.56 m/s
vA  2.56 m/s 
and vA  0.96  1.60  0.64 m/s (rejected, since vA is shown directed up)
From Eqs. (1) and (2):
(vB ) x  3.6 m/s
(vB ) y  2.88  2(2.56)  2.24 m/s
vB  3.6 i  2.24 j
(b)
vB  4.24 m/s
31.9° 
Conservation of angular momentum about O:
(HO )0  HO :  7.2k  ai  mA vA  bi  mB vB
 7.2k  ai  3(2.56 j)  7.5i  1.5(3.6i  2.24 j)
7.2k  7.68ak  25.2k
a  2.34 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.54
Two small disks A and B, of mass 2 kg and 1 kg,
respectively, may slide on a horizontal and
frictionless surface. They are connected by a cord
of negligible mass and spin about their mass center
G. At t  0, G is moving with the velocity v 0 and
its coordinates are x0  0, y0  1.89 m. Shortly
thereafter, the cord breaks and disk A is observed to
move with a velocity v A  (5 m/s) j in a straight
line and at a distance a  2.56 m from the y-axis,
while
B
moves
with
a
velocity
v B  (7.2 m/s)i  (4.6 m/s) j
along
a
path
intersecting the x-axis at a distance b  7.48 m
from the origin O. Determine (a) the initial velocity
v 0 of the mass center G of the two disks, (b) the
length of the cord initially connecting the two disks,
(c) the rate in rad/s at which the disks were spinning
about G.
SOLUTION
Initial conditions.
Location of G:
AG BG AG  GB l



mB m A mB  m A m
m
1
AG  B l  l
m
3
mA
2
BG 
l l
m
3
Linear momentum:
L0  mv0  3 v0
Angular momentum about G:


(HG )0  GA  mA vA  GB  mB vB
1 
1 
2 
2 
  l  (2 kg)  l  k   l  (1 kg)  l  k
3
3
3
 


 
3 
2
 l 2 k 2
3
Kinetic energy: Using Eq. (14.29),
T0 
1
1
1
1
1
mv02   mi vi2  mv02  mA vA2  mB vB2
2
2i
2
2
2
2
1
1
1 2 
1 
(3)v02  (2)  l   (1)  l 
2
2 3 
2 3 
3
1
T0  v02  l 2 2
2
3
2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.54 (Continued)
Conservation of linear momentum:
m v 0  m A v A  mB v B
3v 0  (2)(5 j)  (1)(7.2i  4.6 j)  7.2i  5.4 j
v0  (2.4 m/s)i  (1.8 m/s) j 
(a)
Conservation of angular momentum about O:
: (1.89 j)  mv0  (HG )0  (2.56i )  mA vA  (7.48 i)  mB vA
Substituting for v0 , (HG )0 , vA , vB and masses:
2
(1.89 j)  3(2.4i  1.8 j)  l 2 k  (2.56i )  2(5 j)  (7.48i )  (7.2i  4.6 j)
3
2
13.608k  l 2 k  25.6k  34.408k
3
2 2
l   4.80
3
l 2  7.20
(1)
Conservation of energy:
T0  T :
3 2 1 2 2 1
1
v0  l   mA v A2  mB vB2
2
3
2
2
3
1
1
1
[(2.4)2  (1.8)2 ]  l 2 2  (2)(5)2  (1)[(7.2)2  (4.6)2 ]
2
3
2
2
1
13.5  l 2 2  25  36.5
3
l 2 2  144.0
(2)
Dividing Eq. (2) by Eq. (1), member by member:

144.0
 20.0 rad/s
7.20
Original rate of spin  20.0 rad/s 
(c)
Substituting for  into Eq. (1):
l 2 (20.0)  7.20
(b)
l 2  0.360
l  0.600 m
Length of cord  600 mm 
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PROBLEM 14.55
Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to
three 9-in-long strings, which are tied to a ring G. Initially, the spheres rotate clockwise about the ring with
a relative velocity of 2.6 ft/s and the ring moves along the x-axis with a velocity v0  (1.3 ft/s)i. Suddenly,
the ring breaks and the three spheres move freely in the xy plane with A and B, following paths parallel to the
y-axis at a distance a  1.0 ft from each other and C following a path parallel to the x-axis. Determine
(a) the velocity of each sphere, (b) the distance d.
SOLUTION
Conservation of linear momentum:
L0  (3 m) v  3m (1.3i )  m (3.9 ft/s)i
Before break:
L  mv A j  mvB j  mvC i
After break:
L  L0 :
mvC i  m(vA  vB ) j  m(3.9 ft/s)i
Therefore,
vA  vB
(1)
vC  3.9000 ft/s vC  3.90 ft/s
(2)
Conservation of angular momentum:
Before break:
( H O )0  3mlv  3m (0.75ft)(2.6 ft/s)
 5.85m
After break:
H O  mv A x A
 mv A ( x A  0.346)
 mvC d
H O  mA xA  mv A ( xA  1.0)  mvC d
H O  ( H O )0 :
1.0mvA  mvC d  5.85m
Recalling Eq. (2):
vA  3.9d  5.85
d  1.5  0.25641vA
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(3)
PROBLEM 14.55 (Continued)
Conservation of energy.
Before break:
1
1

(3 m)v 2  3  mv2 
2
2


3
3
 m v02  v2  [(1.3)2  (2.6) 2 ]m  12.675m
2
2
T0 


After break:
T
1 2 1 2 1 2
mv A  mvB  mvC
2
2
2
T  T0 : Substituting for vB from Eq. (1) and vC from Eq. (2),
1 2
v A  v A2  (3.900)2   12.675

2
v A2  5.0700
vA  vB  2.2517 ft/s
(a)
Velocities:
vA  2.25 ft/s ; v B  2.25 ft/s ; vC  3.9 ft/s
(b)

Distance d:
From Eq. (3):
d  1.5  0.25641(2.2517)  0.92265 ft
d  11.1 in. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.56
Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached
to three strings of length l which are tied to a ring G. Initially, the spheres rotate clockwise about the ring
which moves along the x axis with a velocity v0. Suddenly the ring breaks and the three spheres move freely in
the xy plane. Knowing that vA  (3.5 ft/s) j, vC  (6.0 ft/s)i, a  16 in. and d  9 in., determine (a) the initial
velocity of the ring, (b) the length l of the strings, (c) the rate in rad/s at which the spheres were rotating
about G.
SOLUTION
Conservation of linear momentum:
(3m) v  mvA  mvB  mv C
3mv0 i  m(3.5 ft/s) j  mvB j  m(6.0 ft/s)i
Equating coefficients of unit vectors:
3v0  6.00 ft/s
0  3.5 ft/s  vB
vB  3.5 ft/s
(1)
v0  2.00 ft/s
(a)

Conservation of angular momentum:
Before break:
After break:
( H O )0  3ml 2
H O   mv A xA  mv A ( x A  16/12)  mvC (9/12)
 m(3.5)(16/12)  m(6.0)(9/12)
 m(9.1667)
( H O )0  H 0 :
3ml 2  m(9.1667)
l 2  3.0556
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 14.56 (Continued)
Conservation of energy:
Before break:
1
3
1
 3
(3m)v 2  3  mv2   mv02  m(l) 2
2
2
2
 2
3
3
 m(2.0)2  ml 2 2
2
2
T0 
After break:
1 2 1 2 1 2
mv A  mvB  mvC
2
2
2
1
1
 m [(3.5)2  (6.0)2 ]  m (60.5)
2
2
T
T  T0 :
1
3
3
m(60.5)  m(2.0) 2  ml 2 2
2
2
2
l 2 2  16.167
(3)
16.167
 5.2909
Dividing Eq. (3) by Eq. (2):  
3.0556
(b)
From Eq. (2):
l2 
3.0556
5.2909
l  0.75994 ft
l  0.76 ft 
(c)
Rate of rotation:
  5.29 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.57
A stream of water with density  = 1000 kg/m3 is
discharged from a nozzle at the rate of 0.06 m3/s. Using
Bernoulli’s equation the gage pressure P in the pipe just
upstream from the nozzle is P 

v
2
2
2

 v12 . Knowing
the nozzle is held to the pipe by six flange bolts, determine
the tension in each bolt, neglecting the initial tension
caused by the tightening of the nuts.
SOLUTION
Given:
  1000 kg/m3
FBD of pipe:
Q  0.06 m /s
d1  0.150 m
3
d 2  0.05 m
Fp  Force in the pipe wall
Mass flow rates.
dm1 dm2

  Q  60 kg/s
dt
dt
Velocity of fluid upstream and downstream from the nozzle:
Q  A1v1
v1 
Q  A2 v2
4Q
 d12
v2 
v1  3.395 m/s
4Q
 d 22
v2  30.56 m/s
The gauge pressure in the pipe upstream of the nozzle is:
P

2
v
2
2

 v12  461.12 kN/m2
Apply principle of Impulse-momentum in x-direction
F 
dm
 v2  v1  where:
dt
Fp  461.12

4
 6519.0 N
Force per bolt:
Fb 
Fp
6
 F PA  F
1
p
 0.15 2  60  30.56  3.395 
Fb  1086.5 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.58
A jet ski is placed in a channel and
is tethered so that it is stationary.
Water enters the jet ski with
velocity v1 and exits with velocity
v2. Knowing the inlet area is A1
and the exit area is A2, determine
the tension in the tether.
SOLUTION
Mass flow rates. Consider a cylindrical portion of the fluid lying in a section of pipe of cross sectional area
A and length l.
The volume and mass are
m   A(l )
m
l
  A   Av
t
t
Then
At the pipe inlet and outlet, we get
m1
  A1v1 ,
t
m2
  A2 v2
t
Impulse and momentum principle:
mv1  Imp12  mv 2
Using horizontal components ( ),
(m1 )v1 cos   P(t )  (m2 )v2
m1
m
v2  1 v1 cos
t
t
2
  A2 v2   A1v12 cos 
P
P   A2 v22   A1v12 cos  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.59
The nozzle shown discharges a stream of water at a flow rate
Q  475 gal/min with a velocity v of magnitude 60 ft/s. The stream is split
into two streams with equal flow rates by a wedge that is kept in a fixed
position. Determine the components (drag and lift) of the force exerted by the
stream on the wedge. (1 ft3  7.48 gal.)
SOLUTION
Let F be the force that the wedge exerts on the stream. Assume that the fluid speed is constant.
v  60 ft/s.
Q  475 gal/min   ft 3/s
Volumetric flow rate:


dm
 62.4

slug/ft 3  1.0584 ft 3/s  2.051 slug/s
 Q  
32.2
dt


Mass flow rate:
v  v i,
Velocity vectors:
v1  v  cos30 i  sin 30 j
v 2  v  cos 45 i  sin 45 j
Impulse – momentum principle:
 m  v  F  t  
F

m
m
v1 
v2
2
2
m  1
1

 v1  v 2  v 
t  2
2

dm  1
1

v   cos 30 i  sin 30 j   cos 45 i  sin 45 j  i 
dt  2
2

  2.051 60 ft/s  0.21343i  0.10355 j
   26.26 lb  i  12.74 lb  j
Force that the stream exerts on the wedge:
F   26.26 lb  i  12.74 lb  j
drag  26.3 lb

lift  12.74 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.60
The nozzle shown discharges a stream of water at a flow rate
Q  500 gal/min with a velocity v of magnitude 48 ft/s. The stream is split
into two streams of equal flow rates by a wedge that is moving to the left at a
constant speed of 12 ft/s. Determine the components (drag and lift) of the
force exerted by the stream on the wedge. (1ft3 = 7.48 gal.)
SOLUTION
For a fixed observer, the upstream velocity is v   48 ft/s  i.
Volumetric flow rate: Q  500 gal/min   ft 3/s


dm
 62.4

 Q  
slug/ft 3  1.1141 ft 3/s  2.1590 slug/s
dt
 32.2

Use a frame of reference that is moving with the wedge to the left at 12 ft/s. In this frame of reference the
upstream velocity vector is
u  48i   12i    60 ft/s  i.
Mass flow rate:
For the moving frame of reference the mass flow rate is
u  u i,
Velocity vectors:
dm u dm 60


 2.1590   2.6987 slug/s.
dt
v dt
48
u1  u  cos30 i  sin 30 j
u 2  u  cos 45 i  sin 45 j
Let F be the force that the wedge exerts on the stream.
Impulse-momentum principle:
 m  u  F  t  
F

m
m
u1 
u2
2
2
m  1
1

 u1  u 2  u 
t  2
2

dm  1
1

u   cos30 i  sin 30 j   cos 45 i  sin 45 j  i 
dt  2
2

  2.6987  60  0.21343 i  0.10355 j
   34.6 lb  i  16.76 lb  j
Force that the stream exerts on the wedge
F   34.6 lb  i  16.76 lb  j
drag  34.6 lb

lift  16.76 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.61
A rotary power plow is used to remove snow
from a level section of railroad track. The plow
car is placed ahead of an engine that propels it
at a constant speed of 20 km/h. The plow car
clears 160 Mg of snow per minute, projecting it
in the direction shown with a velocity of 12 m/s
relative to the plow car. Neglecting friction,
determine (a) the force exerted by the engine
on the plow car, (b) the lateral force exerted by
the track on the plow.
SOLUTION
Velocity of the plow:
vP  20 km/h  5.5556 m/s
Velocity of thrown snow:
v s  (12 m/s)(cos30i  sin 30 j)  (5.5556 m/s)k
Mass flow rate:
dm (160000 kg/min)

 2666.7 kg/s
dt
(60 s/min)
Let F be the force exerted on the plow and the snow.
Apply impulse-momentum, noting that the snow is initially at rest and that the velocity of the plow is
constant. Neglect gravity.
F(t )  (m) v s
 dm 
F
 vs  (2.666.7)(12cos30i  12sin 30 j  5.5556k )
 dt 
 (27713 N)i  (16000 N) j  (14815 N)k
(a)
Force exerted by engine.
Fz  14.8 kN 
(b)
Lateral force exerted by track.
Fx  27.7 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.62
Tree limbs and branches are being fed at A at the rate of
5 kg/s into a shredder which spews the resulting wood
chips at C with a velocity of 20 m/s. Determine the
horizontal component of the force exerted by the shredder
on the truck hitch at D.
SOLUTION
Eq. (14.38):
( m) vA  F t  ( m) vC
F 
Force exerted on chips   F  100 N
m
vC  (5 kg/s)(20 m/s
t
25)
25
Free body: shredder:
Fx  0: Fx  (100 N)cos 25  0
Fx  90.6 N
On hitch:
Fx  90.6 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.63
Sand falls from three hoppers onto
a conveyor belt at a rate of 90 lb/s
for each hopper. The sand hits the
belt with a vertical velocity v1  10
ft/s and is discharged at A with a
horizontal velocity v2  13 ft/s.
Knowing that the combined mass
of the beam, belt system, and the
sand it supports is 1300 lb with a
mass center at G, determine the
reaction at E.
SOLUTION
Principle of impulse and momentum:
Moments about F:
(m)v1(3a)  mv1 (2a)  (m) v1 a  (W t )c  ( Rt ) L  3(m)v2h
R
Data: L  20 ft,
R
1
L
c  13 ft,
m
m 

cW  6av1 t  3hv2 t  


a  5 ft,
h  2.5 ft,
m W /g
1 dW


 2.7950 slug/s
t
dt
g dt
1
13(1300)  (6)(5)(10)(2.7950)  (3)(2.5)(13)(2.7950)
20
R  873 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.64
The stream of water shown flows at a rate of 550 liters/min and moves
with a velocity of magnitude 18 m/s at both A and B. The vane is
supported by a pin and bracket at C and by a load cell at D that can exert
only a horizontal force. Neglecting the weight of the vane, determine the
components of the reactions at C and D.
SOLUTION
dm
(1000 kg/m3 )(550 liters/min)(1 min)
 Q 
dt
(1000 liters/m3 )(60 sec)
dm
 9.1667 kg/s
dt
Mass flow rate:
vA  18 m/s
Velocity vectors:
v B  18 m/s
40
Apply the impulse-momentum principle.
Moments about C : 0.040(m)v A  0.150 D(t )  0.200(m)vB cos 40  0.165(m)vB sin 40
1  m 
[0.200vB cos 40  0.165vB sin 40  0.040v A ]
0.150  t 
1

(9.1667)[(0.200)(18) cos 40  0.165(18)sin 40  0.040(18)]
0.150
Dx  329 N 
 329.20 N
D
Dy  0 
x components:
Cx (t )  D(t )  (m)vB cos 40
 m 
Cx  
 vB cos 40  D  (9.1667)(18cos 40)  329.20  202.79 N
 t 
y components:
Cx  203 N 
(m)v A  C y (t )  (m)vB sin 40
m
 m 
Cy  
vA 
vB sin 40  (9.1667)(18  18sin 40)  271.06 N

t
 t 
C y  271 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.65
The nozzle shown discharges water at the rate of 40 ft 3/min. Knowing
that at both A and B the stream of water moves with a velocity of
magnitude 75 ft/s and neglecting the weight of the vane, determine the
components of the reactions at C and D.
SOLUTION
dm 
62.4 lb/ft 3 40 ft 3/min
 Q
 1.29193 lb  s/ft
dt
g
32.2 ft/s 2 60 s/min
Mass flow rate:
v A  vB  75 ft/s
Use impulse - momentum principle.
moments about D :
 15 
 23 
 15 
  m  v A sin 60      m  v A cos 60     C  t   
12
12
 
 
 12 
 3
  m  vB  
 12 
15
23
3
 m   15
cos 60    1.29193 75  0.37420 
C 
 v A   sin 60 
12
12
12 
 t   12
C  29.006 lb
Cx  0, C y  29.0 lb 
 m  vA cos 60  Dx  t    m  vB
x component:
 m 
Dx  
  vB  v A cos 60   1.29193 75  75cos 60 
 t 
 m  v A sin 60  C  t   D y  t  
y component:
Dy  C 
Dx  48.4 lb 
0
m
v A sin 60  29.006  1.29193 75  sin 60
t
D y  54.9 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.66
A stream of water flowing at a rate of 1.2 m3/min and moving
with a velocity of magnitude 30 m/s at both A and B is deflected
by a vane welded to a hinged plate. Knowing that the combined
mass of the vane and plate is 20 kg with the mass center at point
G, determine (a) the angle  , (b) the reaction at C.
SOLUTION
dm
kg 

  Q  1000 3 
dt
m 

Mass flow rate:

m3   1 min 
1.2
 
  20 kg/s
 min   60 s 
vA  vB  30 m/s
Use the impulse-momentum principle.
 m  v A a  m  g  t 
Moments about C:
cos  
(a)
l cos    m  vB b
 30  0.110    30  0.030   0.81549
m v A a  v B b
  20 
m p gl
t
 20  9.81 0.300 
  35.364
  35.4 
 m  vA  Cx  t    m  vB cos
x components:
Cx 
0  C y  t   m p g  t     m  vB sin 
y components:
Cy  mp g 
(b)
m
 vB cos  vA    20  30cos  30   110.70 N
t
m
vB sin    20  9.81   20  30  sin   151.06 N
t
C  110.70 N
  151.06 N 
C  187.3 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
53.8 
PROBLEM 14.67
A stream of water flowing at a rate of 1.2 m3/min and
moving with a velocity of magnitude v at both A and B is
deflected by a vane welded to a hinged plate. The combined
mass of the vane and plate is 20 kg with the mass center at
point G. Knowing that   45, determine (a) the speed v of
the flow, (b) the reaction at C.
SOLUTION
dm
kg 

  Q  1000 3 
dt
m 

Mass flow rate:
vA  vA  v,

m3   1 min 
1.2
 
  20 kg/s
 min   60 s 
  45
Use the impulse-momentum principle.
 m  v Aa  m p g  t  l cos
moments about C:
v A  vB 
(a)
m p g l cos
 a  b  m  /  t 

  m  vBb
 20  9.81 0.300  cos 45
 0.110  0.030  20 
 26.013m/s
v  26.0 m/s 
 m  vA  Cx  t    m  vB cos
x components:
Cx 
m
 vB cos  vA    20   26.013cos 45  26.013  152.38 N
t
0  C y  t   m p g  t     m  vB sin 
y components:
Cy  mp g 
(b)
m
vB sin    20  9.81   20  26.013 sin 45  171.68 N
t
C   152.38 N
   171.68 N 
C  230 N
48.4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.68
Coal is being discharged from a first conveyor
belt at the rate of 120 kg/s. It is received at A by
a second belt that discharges it again at B.
Knowing that v1  3 m/s and v 2  4.25 m/s
and that the second belt assembly and the coal
it supports have a total mass of 472 kg,
determine the components of the reactions at C
and D.
SOLUTION
Velocity before impact at A:
(vA ) x  v1  3 m/s
(vA )2y  2 g (y)  (2)(9.81)(0.545)  10.693 m2 /s2
tan  
Slope of belt:
2.4  1.2
,
2.25
(vA ) y  3.270 m/s
  28.07
v 2  4.25 (cos i  sin  j)
Velocity of coal leaving at B:
Apply the impulse-momentum principle.
(m)(vA ) x  Cx (t )  (m)v2 cos 
x components:
Cx 
m
[v2 cos   (v A ) x ]  (120)(4.25cos 28.07  3)
t
Cx  90.0 N

moments about C:

(m)[1.2(v A ) x  0.75(v A ) y ]  3.00 D(t )  1.8W (t )  (m)[2.4 v2 cos   3v2 sin  ]
m
[(1.2)(3)  (0.75)(3.270)]  3D  (1.8)(472)(9.81)
t

m
[(2.4)(4.25cos  )  (3)(4.25sin  )]
t
D  2775  1.0168
dm
 2775  (1.0168)(120)  2897 N
dt
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.68 (Continued)
y components:
(m)(v A ) y  (C y  D  W )(t )  (m)v2 sin 
Cy  D  W 
m
(3.270  4.25sin  )
t
 (120)(5.268)  632.2 N
C y  4625.6  2897  632.2
C y  2361 N 
Cx  90.0 N, C y  2360 N 

Dx  0, 
Dy  2900 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.69
The total drag due to air friction on a jet airplane traveling at 900 km/h is 35 kN. Knowing that the exhaust
velocity is 600 m/s relative to the airplane, determine the mass of air that must pass through the engine per
second to maintain the speed of 900 km/h in level flight.
SOLUTION
dm
 mass flow rate
dt
u  exhaust relative to the airplane
v  speed of airplane
D  drag force
Symbols:
Principle of impulse and momentum:
(m)v  D(t )  (m)u
m dm
D


t
dt
uv
Data: v  900 km/h = 250 m/s
u  600 m/s
D  35 kN  35000 N
dm
35000

dt
600  250
dm
 100 kg/s 
dt
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.70
While cruising in level flight at a speed of 600 mi/h, a jet plane scoops in air at the rate of 200 lb/s and
discharges it with a velocity of 2100 ft/s relative to the airplane. Determine the total drag due to air friction on
the airplane.
SOLUTION
Flight speed: v  600 mi/h  880 ft/s
Mass flow rate:
dm 200 lb/s

 6.2112 slug/s
dt 32.2 ft/s 2
F  D 
dm
(u  v )
dt
or
D
dm
(v  u )
dt
where, for a frame of reference moving with the plane, v is the free stream velocity (equal to the air speed)
and u is the relative exhaust velocity.
D  (6.2112)(2100  880)  7577.6 lb
D  7580 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.71
In order to shorten the distance required for landing, a jet airplane is equipped
with movable vanes, that partially reverse the direction of the air discharged by
each of its engines. Each engine scoops in the air at a rate of 120 kg/s and
discharges it with a velocity of 600 m/s relative to the engine. At an instant when
the speed of the airplane is 270 km/h, determine the reverse thrust provided by
each of the engines.
SOLUTION
Apply the impulse-momentum principle to the moving air. Use a frame of reference that is moving with the
airplane. Let F be the force on the air.
v  270 km/h  75 m/s
u  600 m/s
(m)v  F(t )  2
(m)
u sin 20
2
m
dm
(v  u sin 20) 
(v  u sin 20)
t
dt
F  (120)(75  600sin 20)  33.6  103 N
F
Force on airplane is  F.
F  33.6 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.72
The helicopter shown can produce a maximum downward air
speed of 80 ft/s in a 30-ft-diameter slipstream. Knowing that
the weight of the helicopter and its crew is 3500 lb and
assuming   0.076 lb/ft 3 for air, determine the maximum
load that the helicopter can lift while hovering in midair.
SOLUTION
F
The thrust is
Calculation of
dm
.
dt
dm
( vB  v A )
dt
mass  density  volume  density  area  length
m   AB (l )   AB vB (t )
m
dm

  AB vB  AB vB 
t
g
dt
where AB is the area of the slipstream well below the helicopter and vB is the corresponding velocity in the
slipstream. Well above the blade, vA  0.
F
Hence,

g
AB vB2
 0.076 lb/ft 3    
2
2
 
2 
  4  (30 ft) (80 ft/s)
32.2
ft/s


 10,678 lb
F  10, 678 lb
The force on the helicopter is 10,678 lb .
Weight of helicopter:
WH  3500 lb
Weight of payload:
WP  WP
Statics:
Fy  F  WH  WP  0
WP  F  WH  10,678  3500  7178 lb
W = 7180 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.73
Prior to take-off the pilot of a 3000-kg twin-engine airplane
tests the reversible-pitch propellers by increasing the reverse
thrust with the brakes at point B locked. Knowing that point G
is the center of gravity of the airplane, determine the velocity
of the air in the two 2.2-meter-diameter slipstreams when
the nose wheel A begins to lift off the ground. Assume
  1.21 kg/m3 and neglect the approach velocity of the air.
SOLUTION
Let F be the force exerted on the slipstream of one engine. Then, the force exerted on the airplane is 2F as
shown.
Statics.
M B  0
 0.3 mg  1.6  2F   0
F 
 0.3 3000  9.81
 2 1.6 
 2.5791  103 N
Calculation of
dm
.
dt
mass  density  volume  density  area  length
m   AB   l    ABvB   t 
m dm

  ABvB
t
dt
F 
Force exerted on the slipstream:
dm
 vB  vA 
dt
Assume that v A , the speed far upstream, is negligible.


F   ABvB  vB  0     D 2  vB2
4

vB2


 4  2.5791  103
4F


 599.8 m 2 /s 2
2
 D2 
  2.2  1.21
v B  24.5 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.74
The jet engine shown scoops in air at A at a rate of 200 lb/s
and discharges it at B with a velocity of 2000 ft/s relative to
the airplane. Determine the magnitude and line of action of
the propulsive thrust developed by the engine when the speed
of the airplane is (a) 300 mi/h, (b) 600 mi/h.
SOLUTION
Use a frame of reference moving with the plane. Apply the impulse-momentum principle. Let F be the force
that the plane exerts on the air.
x components:
(m)u A  F (t )  (m) uB
F
moments about B:
m
dm
(u B  u A ) 
(u B  u A )
t
dt
(1)
e(m)u A  M B (t )  0
MB  e
dm
uA
dt
(2)
Let d be the distance that the line of action is below B.
Fd  M B
d
MB
eu A

F
uB  u A
(3)
dm
200
 200 lb/s 
 6.2112 slugs/s, uB  2000 ft/s, e  12 ft
dt
32.2
Data:
u A  300 mi/h  440 ft/s
(a)
From Eq. (1),
F  (6.2112)(2000  440)
From Eq. (3),
d
(12)(440)
2000  440
F  9690 lb 
d  3.38 ft 
u A  600 mi/h  880 ft/s
(b)
From Eq. (1),
F  (6.2112)(2000  880)
From Eq. (3),
d
(12)(880)
2000  880
F  6960 lb 
d  9.43 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.75
A jet airliner is cruising at a speed of 900 km/h with each of its
three engines discharging air with a velocity of 800 m/s relative
to the plane. Determine the speed of the airliner after it has lost
the use of (a) one of its engines, (b) two of its engines. Assume
that the drag due to air friction is proportional to the square of
the speed and that the remaining engines keep operating at the
same rate.
SOLUTION
Let v be the airliner speed and u be the discharge relative velocity.
u  800 m/s.
dm
(u  v)
dt
Thrust formula for one engine:
F
Drag formula:
D  kv 2
Three engines working.
Cruising speed  v0  900 km/h  250 m/s
3F  D  3
dm
(u  v0 )  kv02  0
dt
kv02
dm
k (250) 2


 37.879k
dt 3(u  v0 ) 3(800  250)
(a)
One engine fails. Two engines working. Cruising speed  v1
2F  D  2
dm
(u  v1 )  kv12  0
dt
(2)(37.879k )(800  v1 )  kv12  0
v12  75.758v1  60.606  103  0
v1  211.20 m/s
(b)
v1  760 km/h 
Two engines fail. One engine working. Cruising speed  v2
FD
dm
(u  v2 )  kv22  0
dt
(37.879k )(800  v2 )  kv22  0
v22  37.879v2  30.303  103  0
v2  156.17 m/s
v2  562 km/h 
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PROBLEM 14.76
A 16-Mg jet airplane maintains a constant speed of 774 km/h while
climbing at an angle   18. The airplane scoops in air at a rate of
300 kg/s and discharges it with a velocity of 665 m/s relative to the
airplane. If the pilot changes to a horizontal flight while maintaining
the same engine setting, determine (a) the initial acceleration of the
plane, (b) the maximum horizontal speed that will be attained.
Assume that the drag due to air friction is proportional to the square
of the speed.
SOLUTION
Calculate the propulsive force using velocities relative to the airplane.
F
dm
( vB  v A )
dt
dm
 300 kg/s
dt
v A  774 km/h
Data:
 215 m/s
vB  665 m/s
F  (300)(665  215)
 135, 000 N
Since there is no acceleration while the airplane is climbing, the forces are in equilibrium.

18F  0:
F  D  mg sin   0
F  D  mg sin   (16,000)(9.8)sin18  48, 454 N
(a)
Initial acceleration of airplane in horizontal flight:
ma  F  D : 16, 000a  48.454  103

Corresponding drag force:
D  135, 000  48, 454
 86,546 N

Drag force factor:
D  k v A2
or
k

a  3.03 m/s2
D
v A2
86,546
(215)2
 1.87228 N  s 2 /m 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
18°
PROBLEM 14.76 (Continued)
(b)
Maximum speed in horizontal flight:
Since the acceleration is zero, the forces are in equilibrium.
F D0
dm
dm
dm
(vB  v A )  k v A2  0 k v A2 
vA 
vB  0
dt
dt
dt
1.87228v A2  300v A  (300)(665)  0
vA  256.0 m/s
v A  922 km/h 
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PROBLEM 14.77
The propeller of a small airplane has a 2-m-diameter slipstream and produces a thrust of 3600 N when the
airplane is at rest on the ground. Assuming   1.225 kg/m3 for air, determine (a) the speed of the air in the
slipstream, (b) the volume of air passing through the propeller per second, (c) the kinetic energy imparted per
second to the air in the slipstream.
SOLUTION
Calculation of
dm
dt
at a section in the airstream:
mass  density  volume
 density  area  length
m   A(l )   Av t
m dm

  Av
t
dt
(a)
Thrust 
dm
dt
( vB  vA ) where vB is the velocity just downstream of propeller and vA is the velocity far
upstream. Assume vA is negligible.


Thrust  (  Av)v    D 2  v 2
4

 
3600  1.225   (2)2 v 2
4
v 2  935.44
v  30.585 m/s
(b)
(c)
Q
1 dm
 dt



 Av   D 2  v  (2)2 (30.585)  96.086
4
4


v  30.6 m/s 
Q  96.1 m3 /s 
Kinetic energy of mass m :
1
1
1
(m)v 2   A(l )v 2   Av(t )v3
2
2
2
T dT

dt
t
1
  Av3
2
1 

   D 2  v3
2 4 
1
 
 (1.225)   (2) 2 (30.585)3
2
4
 55,053 N  m/s
T 
dT
 55,100 N  m/s 
dt
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.78
The wind turbine-generator shown has an output-power rating of 1.5 MW for a
wind speed of 36 km/h. For the given wind speed, determine (a) the kinetic energy
of the air particles entering the 82.5-m-diameter circle per second, (b) the
efficiency of this energy conversion system. Assume   1.21 kg/m3 for air.
SOLUTION
(a)
Rate of kinetic energy in the slipstream.
Let m be the mass moving through the slipstream of area A in the time t. Then,
m   A(l )   Av(t )
The kinetic energy carried by this mass is
t 
1
1
(m)v 2   Av3 (t )
2
2
dT T 1

  Av3
dt
t 2
Data:
A

d2 

(82.5 m) 2  5345.6 m 2
4
4
v  36 km/h  10 m/s
dT 1
 (1.21 kg/m3 )(5345.6 m 2 )(10 m/s)3
dt 2
 3.234  106 kg  m 2 /s3
dT
 3.234 MW 
dt
(b)
Efficiency n:

output power
1.5 MW

available input power 3.234 MW
  0.464 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.79
A wind turbine-generator system having a diameter of 82.5 m produces 1.5 MW at
a wind speed of 12 m/s. Determine the diameter of blade necessary to produce
10 MW of power assuming the efficiency is the same for both designs and
  1.21 kg/m3 for air.
SOLUTION
Rate of kinetic energy in the slipstream.
Let m be the mass moving through the slipstream of area A in time t. Then
m   A(l )   Av(t )
The kinetic energy carried by this mass is
1
1
(m) v 2   Av3 (t )
2
2
dT T 1

  Av3
dt
t 2
T 
This is the available input power for the wind turbine. For a wind turbine of efficiency  , the output
power P is
P 
dT 
  Av3
dt 2
We want to compare two turbines having P1  1.5 MW and P2  10 MW, respectively. Then
P2 2 2 A2 v23

P1 1 1 A1v13
Since 2  1 , 2  1 , and v2  v1 , we get
P2 A2 d 22 10



 6.6667
P1 A1 d12 1.5
d 22  6.6667 d12  (6.6667)(82.5 in) 2
d 2  213 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.80
While cruising in level flight at a speed of 570 mi/h, a jet airplane scoops in air at a rate of 240 lb/s and
discharges it with a velocity of 2200 ft/s relative to the airplane. Determine (a) the power actually used to
propel the airplane, (b) the total power developed by the engine, (c) the mechanical efficiency of the airplane.
SOLUTION
dm 240

 7.4534 slugs/s
dt 32.2
u  2200 ft/s
Data:
v  570 mi/h  836 ft/s
dm
F
(u  v)
dt
 (7.4534)(2200  836)
 10,166 lb
(a)
Power used to propel airplane:
P1  Fv
 (10,166)(836)
 8.499  106 ft  lb/s
Propulsion power  15, 450 hp 
Power of kinetic energy of exhaust:
1
(m)(u  v)2
2
1 dm
(u  v)2
P2 
2 dt
1
 (7.4534)(2200  836)2
2
 6.934  106 ft  lb/s
P2 (t ) 
(b)
Total power:
P  P1  P2
 15.433  106 ft  lb/s
Total power  28, 060 hp 
(c)
Mechanical efficiency:
P1 8.499  106

P 15.433  106
 0.551
Mechanical efficiency  0.551 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.81
In a Pelton-wheel turbine, a stream of water is deflected by a series of
blades so that the rate at which water is deflected by the blades is equal
to the rate at which water issues from the nozzle (m/t  A v A ).
Using the same notation as in Sample Problem 14.8, (a) determine the
velocity V of the blades for which maximum power is developed,
(b) derive an expression for the maximum power, (c) derive an
expression for the mechanical efficiency.
SOLUTION
Let u be the velocity of the stream relative to the velocity of the blade.
u  (v  V )
m
  AvA
t
Mass flow rate:
Principle of impulse and momentum:
(m)u  Ft (t )  (m)u cos 
m
Ft 
u (1  cos  )
t
  Av A (v A  V )(1  cos  )
where Ft is the tangential force on the fluid.
The force Ft on the fluid is directed to the left as shown. By Newton’s law of action and reaction, the
tangential force on the blade is Ft to the right.
Output power:
Pout  FV
t
  Av A (v A  V )V (1  cos  )
(a)
V for maximum power output:
dPout
  A (v A  2V )(1  cos  )  0
dV
(b)
vA 
1
V 
2
Maximum power:
1  1 

( Pout ) max   Av A  v A  v A  v A  (1  cos  )
2  2 

( Pout ) max 
1
 Av3A (1  cos  ) 
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.81 (Continued)
Input power  rate of supply of kinetic energy of the stream
1 1

(m)v A2 
t  2

1 m 2

vA
2 t
1
  Av3A
2
Pin 
(c)
Efficiency:

Pout
Pin

 Av A (v A  V )V (1  cos  )
1
 Av3A
2

  2 1 

V
vA
V
 (1  cos  ) 
 vA
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.82
A circular reentrant orifice (also called Borda’s mouthpiece) of
diameter D is placed at a depth h below the surface of a tank.
Knowing that the speed of the issuing stream is v  2 gh and
assuming that the speed of approach v1 is zero, show that the
diameter of the stream is d  D/ 2. (Hint: Consider the section of
water indicated, and note that P is equal to the pressure at a depth h
multiplied by the area of the orifice).
SOLUTION
From hydrostatics, the pressure at section 1 is p1   h   gh.
The pressure at section 2 is p2  0.
Calculate the mass flow rate using section 2.
mass  density  volume
 density  area  length
m   A2 (l )   A2 v(t )
dm m

  A2 v
dt
t
Apply the impulse-momentum principle to fluid between sections 1 and 2.
(m)v1  p1 A1 (t )  (m)v
dm
dm
v1  p1 A1 
v
dt
dt
dm
p1 A1 
(v  v1 )
dt
  A2 v(v  v1 )
p1   gh and v  2 gh
But v1 is negligible,
 ghA1   A2 (2 gh) or A1  2 A2



D2  2  d 2 
4
4 
d
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
D
2

PROBLEM 14.83
A railroad car of length L and mass m0 when empty is moving
freely on a horizontal track while being loaded with sand from
a stationary chute at a rate dm/dt  q. Knowing that the car
was approaching the chute at a speed v0 , determine (a) the
mass of the car and its load after the car has cleared the chute,
(b) the speed of the car at that time.
SOLUTION
Consider the conservation of the horizontal component of momentum of the railroad car of mass m0 and the
sand mass qt.
m0v0   m0  qt  v
v
m0v0
m0  qt
(1)
dx
m0v0
v
dt
m0  qt
x0  0
Integrating, using
x  L when t  t L ,
and
t
t
L   0L vdt   0L

ln
m0v0
mv
dt  0 0 ln  m0  qt L   ln m0 
m0  qt
q
m0v0 m0  qt L
ln
q
m0
m0  qtL
qL

m0
m0v0
m0  qt L
 eqL/m0v0
m0
m0  qt L  m0eqL/m0v0 
(a) Final mass of railroad car and sand
(b) Using (1),
vL 
m0v0
mv
 0 0 e qL/m0v0
m0  qt L
m0
vL  v0eqL/m0v0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.84*
The depth of water flowing in a rectangular channel of width b
at a speed v1 and a depth d1 increases to a depth d2 at a
hydraulic jump. Express the rate of flow Q in terms of b, d1,
and d2.
SOLUTION
mass  density  volume
 density  area  length
m   bd (l )   bdv(t )
Mass flow rate:
dm m

  bdv
dt
t
1 dm
 bdv
Q
 dt
Q1  Q2  Q
Continuity of flow:
v1 
Q
bd1
v2
Q
bd 2
Resultant pressure forces:
p1   d1
p2   d 2
1
1
p1bd1   bd12
2
2
1
1
F2  p2 bd 2   bd 22
2
2
F1 
Apply impulse-momentum principle to water between sections 1 and 2.
(m)v1  F1 (t )  F2 (t )  (m)v2
m
(v1  v2 )  F2  F1
t
 Q 2 (d 2  d1 )
bd1d 2
Noting that    g ,
 Q
Q  1
2
2

   b d 2  d1
 bd1 bd 2  2
Q  


1
  b(d1  d 2 )(d 2  d1 )
2
Qb
1
gd1d 2 (d1  d 2 ) 
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.85*
Determine the rate of flow in the channel of Problem 14.84,
knowing that b  12 ft, d1  4 ft, and d2  5 ft.
PROBLEM 14.84 The depth of water flowing in a rectangular
channel of width b at a speed v1 and a depth d1 increases to a
depth d2 at a hydraulic jump. Express the rate of flow Q in
terms of b, d1, and d2.
SOLUTION
mass  density  volume
 density  area  length
m   bd (l )   bdv(t )
Mass flow rate:
dm m

  bdv
dt
t
1 dm
 bdv
Q
 dt
Q1  Q2  Q
Continuity of flow:
v1 
Resultant pressure forces:
Q
bd1
p1   d1
v2
Q
bd 2
p2   d 2
1
1
p1bd1   bd12
2
2
1
1
F2  p2 bd 2   bd 22
2
2
F1 
Apply impulse-momentum principle to water between sections 1 and 2.
(m)v1  F1 (t )  F2 (t )  (m)v2
m
(v1  v2 )  F2  F1
t
 Q 2 (d 2  d1 )
bd1d 2
 Q
Q  1
2
2

   b d 2  d1
bd
bd
2
2 
 1
Q  


1
  b(d1  d 2 )(d 2  d1 )
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.85* (Continued)
1
gd1d 2 (d1  d 2 )
2
Noting that    g ,
Qb
Data:
g  32.2 ft/s,2
Q  12
b  12 ft, d1  4 ft, d 2  5 ft
1
(32.2)(4)(5)(9)
2
Q  646 ft 3 /s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.86
A chain of length l and mass m lies in a pile on the floor. If its end A is raised
vertically at a constant speed v, express in terms of the length y of chain that is off
the floor at any given instant (a) the magnitude of the force P applied at A, (b) the
reaction of the floor.
SOLUTION
Let  be the mass per unit length of chain. Apply the impulse-momentum to the entire chain. Assume that the
reaction from the floor is equal to the weight of chain still in contact with the floor.
Calculate the floor reaction.
R   g (l  y )
y

R  mg 1  
l

Apply the impulse-momentum principle.
 yv  P(t )  R(t )   gL(t )   ( y  y)v
Pt   (y )v   gL(t )  R(t )
P
(a)
y
v   gL   ( L  y ) g
t
  v 2   gy
Let
(b)
From above,
P
m 2
(v  gy ) 
l
y dy

v
t dt
y

R  mg 1  
l

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.87
Solve Problem 14.86, assuming that the chain is being lowered to the floor at a
constant speed v.
PROBLEM 14.86 A chain of length l and mass m lies in a pile on the floor. If its
end A is raised vertically at a constant speed v, express in terms of the length y of
chain that is off the floor at any given instant (a) the magnitude of the force P
applied at A, (b) the reaction of the floor.
SOLUTION
(a)
Let  be the mass per unit length of chain. The force P supports the weight of chain still off the floor.
P   gy
(b)
P
mgy

l
Apply the impulse-momentum principle to the entire chain.
  yv  P (t )  R (t )   gL(t )    g ( y  y )v
R (t )   gL(t )  P (t )   g (y )v
R   gL   gy  
Let t  0. Then
y
v
t
y dy

 v
t dt
R   g ( L  y)   v2
R
m
[ g ( L  y )  v 2 ] 
l
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.88
The ends of a chain lie in piles at A and C. When released from rest at time
t  0, the chain moves over the pulley at B, which has a negligible mass.
Denoting by L the length of chain connecting the two piles and neglecting
friction, determine the speed v of the chain at time t.
SOLUTION
Let m be the mass of the portion of the chain between the two piles. This is the portion of the chain that is
moving with speed v. The remainder of the chain lies in either of the two piles. Consider the time period
between t and t  t and apply the principle impulse and momentum. Let m be the amount of chain that is
picket up at A and deposited at C during the time period t.
At time t, m is still in pile A while m has a downward at speed v just above pile C. The remaining mass
(m  m) is moving with speed v.
At time t  t , m is moving with speed v  v just above pile A and m is at rest in pile C.
Over the time period an unbalanced weight of chain acts on the system. The weight is
mgh
L
Apply the impulse-momentum principle to the system.
W
Consider moments about the pulley axle.
r[(m)v  (m  m)v]  rW (t )
 r[(m)(v  v)  (m  m)(v  v)]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.88 (Continued)
Dividing by r and canceling the terms (m)v and (m  m)v
0
mgh
(t )  (m)(v  v)  (m  m)(v)
L
 v(m)  m(v)
m
v ( t )
L
But
m 
Hence,
mgh
mv 2
(t ) 
(t )  m(v)
L
L
Solving for t ,
t 
L(v)
gh  v 2
Letting c 2  gh, and considering the limit as t and v become infinitesimal, gives
dt 
L dv
c  v2
2
Integrate, noting that v  0 when t  0
tL
tanh

v
0
v
dv
L
v
 tanh 1
2
2
c
c0
c v
ct v

L c
v
 gh 
gh tanh 
t 
 L 


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.89
A toy car is propelled by water that squirts from an internal
tank at a constant 6 ft/s relative to the car. The weight of the
empty car is 0.4 lb and it holds 2 lb of water. Neglecting other
tangential forces, determine the top speed of the car.
SOLUTION
Consider a time interval t. Let m be the mass of the car plus the water in the tank at the beginning of the
interval and (m  m) the corresponding mass at the end of the interval. m0 is the initial value of m. Let v be
the velocity of the car. Apply the impulse and momentum principle over the time interval.
Horizontal components
:
mv  0  (m)( v  u cos 20)  ( m  m)( v  v)
m
v  u cos 20
m  m
Let v be replaced by differential dv and m be replaced by the small differential dm, the minus sign
meaning that dm is the infinitesimal increase in m.
dv  u cos 20
dm
m
Integrating,
v  v0  u cos 20 ln
Since v0  0,
v  u cos 20 ln
m
m0
m0
m
The velocity is maximum when m  m f , the value of m when all of the water is expelled.
vmax  u cos 20 ln
m0
mf
vmax  (6 ft/s) cos 20 ln
0.4  2
0.4
vmax  10.10 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.90
A toy car is propelled by water that squirts from an internal
tank. The weight of the empty car is 0.4 lb and it holds 2 lb of
water. Knowing the top speed of the car is 8 ft/s, determine the
relative velocity of the water that is being ejected.
SOLUTION
Consider a time interval t. Let m be the mass of the car plus the water in the tank at the beginning of the
interval and (m  m) the corresponding mass at the end of the interval. m0 is the initial value of m. Let v be
the velocity of the car. Apply the impulse and momentum principle over the time interval.
Horizontal components
:
mv  0  (m)( v  u cos 20)  ( m  m)( v  v)
m
v  u cos 20
m  m
Let v be replaced by differential dv and m be replaced by the small differential  dm, the minus sign
meaning that dm is the infinitesimal increase in m.
dv  u cos 20
dm
m
Integrating,
v  v0  u cos 20 ln
Since v0  0,
v  u cos 20 ln
m
m0
m0
m
The velocity is maximum when m  m f , the value of m when all of the water is expelled.
vmax  u cos 20 ln
8 ft/s  u cos 20 ln
m0
mf
0.4  2
0.4
u  4.75 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.91
The main propulsion system of a space shuttle consists of three identical
rocket engines which provide a total thrust of 6 MN. Determine the rate at
that the hydrogen-oxygen propellant is burned by each of the three
engines, knowing that it is ejected with a relative velocity of 3750 m/s.
SOLUTION
Thrust of each engine:
Eq. (14.44):
1
P  (6 MN)  2  106 N
3
dm
u
dt
dm
2  106 N 
(3750 m/s)
dt
P
dm 2  106 N

dt 3750 m/s
dm
 533 kg/s 
dt
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.92
The main propulsion system of a space shuttle consists of three identical
rocket engines, each of which burns the hydrogen-oxygen propellant at
the rate of 750 lb/s and ejects it with a relative velocity of 12000 ft/s.
Determine the total thrust provided by the three engines.
SOLUTION
From Eq. (14.44) for each engine:
dm
u
dt
(750 lb/s)

(12000 ft/s)
32.2 ft/s 2
 279.50  103 lb
P
For the 3 engines:
Total thrust  3(279.50  103 lb)
Total thrust  839, 000 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.93
A rocket sled burns fuel at the constant rate of 120
lb/s. The initial weight of the sled is 1800 lb,
including 360 lb of fuel. Assume that the track is
lubricated and the sled is aerodynamically designed
so that air resistance and friction are negligible. (a)
Derive a formula for the acceleration a of the sled as
a function of time t and the exhaust velocity vex of the
burned fuel relative to the sled. Plot the ratio a/vex
versus time t for the range 0 < t < 4 s, and check the
slope of the graph at t = 0 and t = 4 s using the
formula for a. (b) Determine the ratio of the velocity
vb of the sled at burnout to the exhaust velocity vex.
SOLUTION
Given:
dm
 120 / 32.2  3.727 slug/s
dt
mi  1800 / 32.2  55.90 slugs, m f  1800  360  / 32.2  44.72 slugs
Mass of rocket sled over time: m  mi 
dm
t  55.90  3.727t
dt
From Equation 14.42 applied to the direction of motion:
dm
 F  ma  dt u
or a 
(a) Putting in known values:
a
where:
 F  0, u  v
ex
1 dm
vex
m dt
3.727vex
 55.90  3.727t 
a / vex 
1

15  t
Plot of a/vex vs. time:
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.93 (Continued
Slope of graph from figure:
Find slope analytically:
at t=0 s :
slope  0.004
at t=4 s:
slope  0.008
d  a 
1


dt  vex  15  t 2
Evaluating the function:
at t=0 s :
slope  0.00444
at t=4 s:
slope  0.00826
(b) Burnout of fuel occurs when m=mf:
dm
tf
dt
44.72  55.90  3.727t f
m f  mi 
t f  3.0 s
Velocity of sled:
a

tf
0
dv
 adt  dv
dt
adt 

vb
dv
0
vb
vex
dt 
dv
0 15  t
0
 15 
vb
 ln 

 15  t f 
vex



Evaluate with t f  3.0 s
tf

vb
 0.223 
vex
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.94
A space vehicle describing a circular orbit at a speed of 24  103 km/h
releases its front end, a capsule that has a gross mass of 600 kg,
including 400 kg of fuel. If the fuel is consumed at the rate of 18 kg/s
and ejected with a relative velocity of 3000 m/s, determine (a) the
tangential acceleration of the capsule as the engine is fired, (b) the
maximum speed attained by the capsule.
SOLUTION
dm
u
dt
 (18 kg/s)(3000 m/s)
P
Thrust:
 54  103 N
(at )0 
(a )
(b)
P 54  103

 90 m/s 2
m0
600
(at )0  90.0 m/s 2 
Maximum speed is attained when all the fuel is used up:
v1  v0 
 v0 


t1
0
at dt  v0 
t1
u  dm
dt 
0
m

t1
0
P
dt
m
dt  v0  u

m1 
m0
dm 
 m 



m
m 
v1  v0  u   ln 1   v0  u ln 0
m0 
m1

Data:
v0  24  103 km/h
 6.6667  103 m/s
u  3000 m/s
m0  600 kg
m1  600  400  200 kg
v1  6.6667  103  3000 ln
 9.9625  103 m/s
600
200
v1  35.9  103 km/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.95
A 540-kg spacecraft is mounted on top of a rocket with a mass of 19 Mg, including 17.8 Mg of fuel.
Knowing that the fuel is consumed at a rate of 225 kg/s and ejected with a relative velocity of 3600
m/s, determine the maximum speed imparted to the spacecraft if the rocket is fired vertically from the
ground.
SOLUTION
See sample Problem 14.8 for derivation of
v  u ln
m0
 gt
m0  qt
(1)
u  3600 m/s q  225 kg/s, mfuel  17,800 kg
Data:
m0  19,000 kg  540 kg  19,540 kg
We have
mfuel  qt , 17,800 kg  (225 kg/s)t
t
17,800 kg
 79.111 s
225 kg/s
Maximum velocity is reached when all fuel has been consumed, that is, when qt  mfuel. Eq. (1) yields
vm  u ln
m0
 gt
m0  mfuel
19,540
 (9.81 m/s 2 )(79.111 s)
19,540  17,800
 (3600 m/s) ln 11.230  776.1 m/s
vm  7930 m/s 
 7930.8 m/s
 (3600 m/s) ln
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.96
The rocket used to launch the 540-kg spacecraft of Problem 14.95 is redesigned to include two
stages A and B, each of mass 9.5 Mg, including 8.9 Mg of fuel. The fuel is again consumed at a rate
of 225 kg/s and ejected with a relative velocity of 3600 m/s. Knowing that when stage A expels its
last particle of fuel its casing is released and jettisoned, determine (a) the speed of the rocket at that
instant, (b) the maximum speed imparted to the spacecraft.
SOLUTION
dm
 uq
dt
Thrust force:
Pu
Mass of rocket  unspent fuel:
m  m0  qt
Corresponding weight force:
W  mg
a
Acceleration:
F P W P
uq

 g
g
m
m
m
m0  qt
Integrating with respect to time to obtain the velocity,
v  v0 

t
0
adt  v0  u
 v0  u ln
For each stage,

qdt
 gt
0 m  qt
0
t
m0  qt
 gt
m0
mfuel  8900 kg
q  225 kg/s
For the first stage,
v0  0
(a)
v1  0  3600 ln
(1)
u  3600 m/s
mfuel 8900

 39.556 s
225
q
t
m0  540  (2)(9500)  19,540 kg
19,540  8900
 (9.81)(39.556)  1800.1 m/s
19,540
v1  1800 m/s 
For the second stage,
(b)
v0  1800.1 m/s,
m0  540  9500  10,040 kg
v2  1800.1  3600 ln
10,040  8900
 (9.81)(39.556)  9244 m/s
10, 040
v2  9240 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.97
The weight of a spacecraft, including fuel, is 11,600 lb when the rocket engines are
fired to increase its velocity by 360 ft/s. Knowing that 1000 lb of fuel is consumed,
determine the relative velocity of the fuel ejected.
SOLUTION
Apply conservation of momentum to the rocket plus the fuel.
mv   m  m  v  v    m  v  v  v 
 mv  m  v    m  v   m  v    m  v   m  v    m  v
m  v   u  m   0
m  
dm
 t 
dt
v
dv
u dm


t
dt
m dt
u dm
dm
v1
t1
m1
 v0 dv   0 m dt dt   m0 u m
v1  v0   u ln
Data:
m1
m
W
 u ln 0  u ln 0
m0
m1
W1
v1  v0  360 ft/s
W0  11, 600 lb,
360  u ln
11, 600
10, 600
W1  11, 600  1000   10,600 lb
u  3990 ft/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.98
The rocket engines of a spacecraft are fired to increase its velocity by 450 ft/s.
Knowing that 1200 lb of fuel is ejected at a relative velocity of 5400 ft/s, determine
the weight of the spacecraft after the firing.
SOLUTION
Apply conservation of momentum to the rocket plus the fuel.
mv   m  m  v  v    m  v  v  v 
 mv  m  v    m  v   m  v    m  v   m  v    m  v
m  v   u  m   0
m  
dm
 t 
dt
v
dv
u dm


t
dt
m dt
u dm
dm
v1
t1
m1
 v0 dv   0 m dt dt   m0 u m
v1  v0   u ln
m1
m
 u ln 0
m0
m1
m0
 v  v0  W0
 exp  1

m1
 u  W1
Data:
v1  v0  450 ft/s,
u  5400 ft/s
W0  W1  Wfuel  W1  1200 lb
W1  1200
450
 exp
 1.08690
W1
5400
1200
 0.08690
W1
W1  13810 lb 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.99
Determine the distance traveled by the spacecraft of Prob. 14.97 during the rocket
engine firing, knowing that its initial speed was 7500 ft/s and the duration of the
firing was 60 s.
SOLUTION
dm
 uq
dt
m  m0  qt
P
uq
a

m m0  qt
Pu
Thrust force
Mass of rocket plus unspent fuel
Acceleration:
Integrating with respect to time to obtain the velocity,
v  v0 
q
t
t
 0 adt  v0  u  0 m  qt dt
0
 v0  u ln  m0  qt   ln m0   v0  u ln
m0  qt
m0
(1)
Integrating again to obtain the displacement,
m  qt
t
s  s0  v0t  u  0 ln 0
dt
m0
Let
z 
m0  qt
m0
s  s0  v0t 
dz  
q
dt
m0
or
dt  
m0
dz
q
m0u z
m0u
z
 ln zdz  q  z ln z  z  z0
q z0
 s0  v0t 
 m0  m0

m0u  m0  qt  m0  qt
 1 
 1 

 ln
 ln
q  m0 
m0
 m0  m0
 
 s0  v0t 


m0u 
qt  m0  qt
 1  1
 1 
 ln
q 
mo 
m0


 s0  v0t 

 m  qt

m0u  m0  qt
 1  1  ut ln 0
 1
ln
q 
m0
m
0



m u
 m  qt
 s0  v0t  ut   0  ut  ln 0
m0
 q

 m

m0 
s  s0  v0t  u t   0  t  ln

 m0  qt 
  q
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 14.99 (Continued)
v0  7500 ft/s
Data:
t  60 s
m0 
m0  qt 
v  7500  360  7860 ft/s
q
mfuel Wfuel
1000


 0.5176 slug/s
t
gt
 32.2  60 
W0 11600

 360.25 slugs
g
32.2
11600 1000

 329.19 slugs
32.2
32.2
s0  0
From (1),
7860  7500  u ln
360  u ln
From (2),
329.19
360.25
360.25
329.19
u  3993 ft/s

 360.25
 360.25 
 60  ln
s  0   7500  60   3993 60  

 0.5176
 329.19 

 460.6  106 ft
s  87.2 mi 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.100
A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with
a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine (a) its
acceleration as it is fired, (b) its acceleration as the last particle of fuel is being consumed, (c) the altitude at
which all the fuel has been consumed, (d) the velocity of the rocket at that time.
SOLUTION
Given initial conditions:
W  W0  2600 lb
g  32.2 ft/s 2
m
2600
 80.745 slug
32.2
dm
u
dt
(25 lb/s)

(13000 ft/s)
32.2 ft/s 2
 10.093  103 lb
 F  ma
P  mg  ma
P
a g
m
10.093  103 lb
(a) Initial Acceleration:
a
 32.2  92.80 ft/s 2
80.745 slug
(b) As the last particle of fuel is consumed:
W  2600  2200  400 lb
400
m
 12.422 slug
32.2
10.093  103 lb
a
 32.2  780.30 ft/s 2
From (1):
12.422
See Sample Problem 14.8 for derivation of
m0
m  qt
v  u ln
 gt  u ln 0
 gt
m0  qt
m0
P
Thrust Force:
Set
dy
dt
(1)
a  92.8 ft/s 2 
a  780 ft/s 2 
(2)
 v in Eq. (2) and integrate with respect to time:

m0
 gt  dt
 u ln
m0  qt


t
m  qt
1
 u ln 0
dt  gt 2
0
m0
2
h

h
dy 
0

t
0
vdt 

t
0

Making the following substitutions:
z
m0  qt
m0
dz  
q
dt
m0
or
dt  
m0
dz
q
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.100 (Continued)
h
m0u
q

z
z0
ln z dz 
1 2 m0 u
1
gt 
[( z ln z  z )]zz0  gt 2
q
2
2

 1
m0u  m0  qt  m0  qt  m0  m0
 1 
 1   gt 2

 ln
 ln
q  m0 
m0
 m0  m0
  2

m0u 
qt  m0  qt   1 2
 1  1  gt
 1 
 ln
q  m0 
m0
  2


 m  qt  1 2
m0u  m0  qt
 1  1  ut ln 0
 1  gt
ln
q 
m0
m0


 2
m u
 m  qt 1 2
 ut   0  ut  ln 0
 gt
m0
2
 q

 m

m0  1 2
h  u t   0  t  ln
  gt
 m0  qt  2
  q
(c) From Eq. (3),
(3)

 1
2600
 2600

2
 88  ln
h  (13000) 88  
  (32.2)(88)
 25
 2600  2200  2

 (13000)(88  16 ln 6.5)  124680
 630, 000 ft
h  119.3 mi 
(d) From Eq. (2),
2600  2200
 (32.2)(88)
2600
 13000 ln 6.5  2834
v  13000 ln
 21500 ft/s
v  14, 660 mi/h 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.101
Determine the altitude reached by the spacecraft of Problem 14.95 when all the fuel of its launching
rocket has been consumed.
SOLUTION
See Sample Problem 14.8 for derivation of
v  u ln
m0
m  qt
 gt  u ln 0
 gt
m0  qt
m0
(1)
Note that g is assumed to be constant.
Set
dy
dt
 v in Eq. (1) and integrate with respect to time.

m0
 gt  dt
 u ln
0
0
0
m0  qt


t
m0  qt
1 2
dt  gt
 u ln
0
2
m0
h

h
dy 

t
vdt 

t

Let
z
m0  qt
m0
q
dt
m0
or
dt  
m0
dz
q
1 2 m0u
1
gt 
( z ln z  z )zz0  2 gt 2
2
q
 m  m
 1
m u  m  qt  m0  qt
 0  0
 1  0  ln 0  1   gt 2
 ln
q  m0 
m0
 m0  m0
  2
h
m0u
q
dz  

z
z0
ln z dz 

  1
m0u 
qt  m0  qt
 1  1  gt 2
 1 
 ln
q  m0 
m0
  2


 m  qt  1 2
m0u  m0  qt
 1  1  ut ln 0
 1  gt
ln
q 
m0
m0


 2
m u
 m  qt 1 2
 ut   0  ut  ln 0
 gt
2
m0
 q

 m

m0  1 2
h  u t   0  t  ln
  gt
 m0  qt  2
  q
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 14.101 (Continued)
u  3600 m/s
Data:
q  225 kg/s
m0  19, 000  540  19,540 kg
mfuel  17,800 kg
mfuel 17,800

 79.111 s
225
q
m0  qt  1740 kg
t
From Eq. (2),
g  9.81 m/s 2

 19,540
 19,540  1
2
 79.111 ln
h  (3600) 79.111  
  (9.81)(79.111)
1740  2
 225


 186,766 m
h  186.8 km 
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PROBLEM 14.102
For the spacecraft and the two-stage launching rocket of Problem 14.96, determine the altitude at
which (a) stage A of the rocket is released, (b) the fuel of both stages has been consumed.
SOLUTION
dm
 uq
dt
Thrust force:
Pu
Mass of rocket  unspent fuel:
m  m0  qt
Corresponding weight force:
W  mg
a
Acceleration:
F P W P
uq

 g
g
m
m
m
m0  qt
Integrating with respect to time to obtain velocity,
v  v0 

t
0
adt  v0  u
 v0  u ln
Integrating again to obtain the displacement,
s  s0  v0t  u
m0  qt
m0
Let
z
Then
s  s0  v0t 

t
ln
0

qdt
 gt
0 m  qt
0
t
m0  qt
 gt
m0
(1)
m0  qt
1
dt  gt 2
m0
2
dz  
q
dt
m0
dt  
m0
dz
q
m0u z
1
ln z dz  gt 2
q z0
2
mu
1
z
 s0  v0t  0 ( z ln z  z ) z  gt 2
0
q
2
 s0  v0t 

m0u  m0  qt m0  qt m0  qt m0
m
m  1


ln
ln 0  0   gt 2

q  m0
m0
m0
m0
m0 m0  2
 m
 m  qt  1 2
 s0  v0t  u t   0  t  ln 0
  gt
m0  2

  q
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 14.102 (Continued)
For each stage,
mfuel  8900 kg
u  3600 m/s
q  225 kg/s
For the first stage,
v0  0
t
mfuel 8900

 39.556 s
q
225
s0  0
m0  540  (2)(9500)  19,540 kg
From Eq. (1),
v1  0  3600 ln
19,540  8900
 (9.81)(39.556)
19,540
 1800.1 m/s
From Eq. (2),
(a)

 19,540
 19,540  8900  1
2
s1  0  0  3600 39.556  
 39.556  ln
  (9.81)(39.556)
225
19,540
2




h1  31.2 km 
 31, 249 m
For the second stage,
v0  1800.1 m/s s0  31, 249 m
m0  540  9500  10, 040 kg
From Eq. (2),
(b)

 10, 040
 10,040  8900 
s2  31, 249  (1800.1)(39.556)  3600 39.556  
 39.556  ln

10,040
 225



1
 (9.81)(39.556)2
2
 197,502 m
h2  197.5 km 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.103
In a jet airplane, the kinetic energy imparted to the exhaust gases is wasted as far as propelling the airplane is
concerned. The useful power is equal to the product of the force available to propel the airplane and the speed
of the airplane. If v is the speed of the airplane and u is the relative speed of the expelled gases, show that the
mechanical efficiency of the airplane is   2v/(u  v). Explain why   1 when u  v.
SOLUTION
Let F be the thrust force, and
dm
dt
be the mass flow rate.
Absolute velocity of exhaust:
ve  u  v
Thrust force:
F
Power of thrust force:
P1  Fv 
Power associated with exhaust:
Total power supplied by engine:
dm
(u  v)
dt
dm
(u  v)v
dt
1
1
(m)ve2  (m)(u  v) 2
2
2
1 dm
(u  v)2
P2 
2 dt
P2 (t ) 
P  P1  P2
dm 
1

(u  v)v  (u  v)2 
dt 
2

1 dm 2
(u  v 2 )

2 dt
P
Mechanical efficiency:
useful power P1

P
total power
2(u  v)v
 2 2
u v


2v

(u  v)
  1 when u  v. The exhaust, having zero velocity, carries no power away.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.104
In a rocket, the kinetic energy imparted to the consumed and ejected fuel is wasted as far as propelling the
rocket is concerned. The useful power is equal to the product of the force available to propel the rocket and
the speed of the rocket. If v is the speed of the rocket and u is the relative speed of the expelled fuel, show that
the mechanical efficiency of the rocket is   2uv/(u 2  v 2 ). Explain why   1 when u  v.
SOLUTION
Let F be the thrust force and
dm
dt
be the mass flow rate.
Absolute velocity of exhaust:
ve  u  v
Thrust force:
F
Power of thrust force:
P1  Fv 
Power associated with exhaust:
Total power supplied by engine:
Mechanical efficiency:
dm
u
dt
dm
uv
dt
1
1
(m)ve2  (m)(u  v) 2
2
2
1 dm
(u  v)2
P2 
2 dt
P2 (t ) 
P  P1  P2
P
dm 
1
 1 dm 2
uv  (u  v) 2  
(u  v 2 )
dt 
dt
2
2


useful power P1

total power
P

2uv

(u  v 2 )
2
  1 when u  v. The exhaust, having zero velocity, carries no power away.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.105
Three identical cars are being unloaded from an
automobile carrier. Cars B and C have just been
unloaded and are at rest with their brakes off when car A
leaves the unloading ramp with a velocity of 5.76 ft/s
and hits car B, which hits car C. Car A then again hits
car B. Knowing that the velocity of car B is 5.04 ft/s
after the first collision, 0.630 ft/s after the second
collision, and 0.709 ft/s after the third collision,
determine (a) the final velocities of cars A and C, (b) the
coefficient of restitution for each of the collisions.
SOLUTION
There are no horizontal forces acting. Horizontal momentum is conserved.
(a)
Velocities:
Event 1
2: Car A hits car B.
m(5.76)  0  m(v A )2  m(5.04)
Event 2
3: Car B hits car C.
m (5.04)  0  m (0.630)  m(vC )3
Event 3
( vC )3  4.41 ft/s

(v A )4  0.641 ft/s

4: Car A hits car B again.
m(0.720)  m(0.630)  m(vA )4  m(0.709)
(b)
( v A )2  0.720 ft/s
Coefficients of restitution:
Event 1
2: e1 2 
Event 2
3:
Event 3
4: e3 4 
e2 3 
(v A ) 2  (v B ) 2
5.04  0.720

(v A )1  (vB )1
5.76  0
e1 2  0.750 
(vB )3  (vC )3
4.41  0.630

(vB )2  (vC )2
5.04  0
e2 3  0.750 
(vA )4  (vB )4
0.709  0.641

(vA )3  (vB )3
0.720  0.630
e3 4  0.756 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.106
A 30-g bullet is fired with a velocity of 480 m/s into block A,
which has a mass of 5 kg. The coefficient of kinetic friction
between block A and cart BC is 0.50. Knowing that the cart
has a mass of 4 kg and can roll freely, determine (a) the final
velocity of the cart and block, (b) the final position of the
block on the cart.
SOLUTION
(a)
Conservation of linear momentum:
m0 v0  (m0  mA )v  (m0  mA  mC )vf
(0.030 kg)(480 m/s)  (5.030 kg)v  (9.030 kg)vf
0.030
(480 m/s)  2.863 m/s
5.030
0.030
(480 m/s)  1.5947 m/s
vf 
9.030
v 
(b)
v f  1.595 m/s 
Work-energy principle:
Just after impact:
Final kinetic energy:
Work of friction force:
1
(m0  m A )v2
2
1
 (5.030 kg)(2.863 m/s)2
2
 20.615 J
1
1
Tf  (m0  m A  mC ) v 2f
2
2
1
 (9.030 kg)(1.5947 m/s)2
2
 11.482 J
T 
F  k N
 k (m0  mA ) g
 0.50(5.030)(9.81)
 24.672 N
Work  U   Fx  24.672 x
T   U  T f : 20.615  24.672 x  11.482
x  0.370 m 
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PROBLEM 14.107
An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B that can
slide along the floor of the car ( k  0.25). Knowing that the car was at rest with its brakes released and that
it automatically coupled with the engine upon impact, determine the velocity of the car (a) immediately after
impact, (b) after the load has slid to a stop relative to the car.
SOLUTION
The masses are the engine (mA  80  103 kg), the load (mB  30  103 kg), and the flat car (mC  20  103 kg).
(v A )0  6.5 km/h
Initial velocities:
 1.80556 m/s
(vB )0  (vC )0  0.
No horizontal external forces act on the system during the impact and while the load is sliding relative to the
flat car. Momentum is conserved.
mA (vA )0  mB (0)  mC (0)  mA (v A )0
Initial momentum:
(a)
(1)
Let v be the common velocity of the engine and flat car immediately after impact. Assume that the
impact takes place before the load has time to acquire velocity.
Momentum immediately after impact:
mAv  mB (0)  mC v  (mA  mC )v
(2)
Equating (1) and (2) and solving for v,
v 
m A (v A ) 0
mA  mC
(80  103 )(1.80556)
(100  103 )
 1.44444 m/s

v  5.20 km/h
(b)

Let vf be the common velocity of all three masses after the load has slid to a stop relative to the car.
Corresponding momentum:
mAv f  mB v f  mC v f  (mA  mB  mC )vf
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(3)
PROBLEM 14.107 (Continued)
Equating (1) and (3) and solving for vf ,
vf 
m A (v A ) 0
mA  mB  mC
(80  103 )(1.80556)
(130  103 )
 1.11111 m/s

vf  4.00 km/h
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.108
In a game of pool, ball A is moving with a velocity v0 when it strikes
balls B and C which are at rest and aligned as shown. Knowing that
after the collision the three balls move in the directions indicated and
that v0  12 ft/s and vC  6.29 ft/s, determine the magnitude of the
velocity of (a) ball A, (b) ball B.
SOLUTION
Conservation of linear momentum. In x direction:
m(12 ft/s) cos 30  mv A sin 7.4  mvB sin 49.3
 m(6.29) cos 45
0.12880vA  0.75813vB  5.9446
(1)
In y direction:

m(12 ft/s)sin 30  mvA cos 7.4  mvB cos 49.3
 m(6.29)sin 45
0.99167vA  0.65210vB  1.5523
(a)
Multiply (1) by 0.65210, (2) by 0.75813, and add:
0.83581 vA  5.0533
(b)

(2)
vA  6.05 ft/s 
Multiply (1) by 0.99167, (2) by –0.12880, and add:
0.83581 vB  5.6951 
vB  6.81 ft/s 
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PROBLEM 14.109
Mass C, which has a mass of 4 kg, is suspended from a cord attached to cart
A, which has a mass of 5 kg and can roll freely on a frictionless horizontal
track. A 60-g bullet is fired with a speed v0  500 m/s and gets lodged in
block C. Determine (a) the velocity of C as it reaches its maximum elevation,
(b) the maximum vertical distance h through which C will rise.
SOLUTION
Consider the impact as bullet B hits mass C. Apply the principle of impulse-momentum to the two particle
system.
mv1  Imp12  mv 2
Using both B and C and taking horizontal components gives
mB v0 cos   O  (mB  mC )v  mBC v
v 

mB v0 cos 
mBC
(0.060 kg)(500 m/s) cos 20
 6.9435 m/s
(4.06 kg)
Now consider the system of mA and mBC after the impact, and apply to impulse momentum principle.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.109 (Continued)
mv 2  Imp 23  mv3
Horizontal components: 
mBC v  0  mA vA  mBC vcx
vA 

mBC
(v  vcx )
mA
4.06
(6.9435  vcx )
5
vA  5.6381  0.812vcx
(a)
in m/s
(1)
At maximum elevation.
Both particles have the same velocity, thus
vcx  v A
v A  5.6381  0.812 v A
vA  3.1115 m/s
(b)
Conservation of energy:
v A  3.11 m/s 
T2  V2  T3  V3
1
1
mA (0)  mBC (v)2
2
2
1
 (4.06)(6.9435)2  97.871 J
2
(datum)
V2  0
T2 
1
1
2
2 2
)
m A v A2  mBC (vBx
 vBy
2
2
1
1
 (5)(3.1115)2  (4.06)[(3.1115)2  0]  43.857 J
2
2
V3  mBC gh  (4.06)(9.81) h  39.829 h
T3 
97.871  0  43.857  39.829 h
h  1.356 m 
Another method: We observe that no external horizontal forces are exerted on the system consisting of A, B,
and C. Thus the horizontal component of the velocity of the mass center remains constant.
m  mA  mB  mC  5  0.06  4  9.06 kg
vx 
(a)
mB v0 cos 
(0.060 kg)(500 m/s) cos 20°

 3.1115 m/s
mA  mB  mC
9.06 kg
At maximum elevation, vA and vBC are equal.
vA  3.1115 m/s
v A  3.11 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.109 (Continued)
Immediately after the impact of B on C, the velocity vA is zero.
(mB  mC )v  (mA  mB  mC )vx
v 
(b)
Principle of work and energy:
mA  mB  mC
9.06
vx 
(3.1115 m/s)  6.9435 m/s
mB  mC
4.06
T2  V2  T3  V3
T2 , V2 , and V3 are calculated as before.
For T3 we note that the velocities vA and v BC relative to the mass center are zero. Thus, T3 is given by
T3 
As before, h is found to be
1
1
mv 2  (9.06)(3.1115)2  43.857 J
2
2
h  1.356 m 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.110
A 15-lb block B is at rest and a spring of constant
k  72 lb/in. is held compressed 3 in. by a cord. After
5-lb block A is placed against the end of the spring, the
cord is cut causing A and B to move. Neglecting
friction, determine the velocities of blocks A and B
immediately after A leaves B.
SOLUTION
5
 0.15528 lb  s 2 /ft
32.2
15
mB 
 0.46584 lb  s 2 /ft
32.2
k  72 lb/in  864 lb/ft
e  3 in.  0.25 ft
h  6 in.  0.5 ft
mA 
Conservation of linear momentum:
Horizontal components
0  0  mAvA  mB vB
:
vB 
mA
1
vA  vA
3
mB
Conservation of energy:
State 1:
1 2 1
ke  (864)(0.25) 2  27 ft  lb
2
2
V1g  0
V1e 
T1  0
State 2:
V2e  0
V2 g  WA h  (5)(0.5)  2.5 ft  lb
T2 
1
1
m A v A2  mB vB2
2
2
2
1
1
v 
 (0.15528)v A2  (0.46584)  A   0.10352v A2
2
2
 3 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.110 (Continued)
T1  V1  T2  V2 :
0  27  0.10352v A2  2.5
v A2  236.67 ft 2
vA  15.38 ft/s

vB  5.13 ft/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.111
Car A of mass 1800 kg and car B of mass 1700 kg
are at rest on a 20-Mg flatcar which is also at rest.
Cars A and B then accelerate and quickly reach
constant speeds relative to the flatcar of 2.35 m/s
and 1.175 m/s, respectively, before decelerating to
a stop at the opposite end of the flatcar. Neglecting
friction and rolling resistance, determine the
velocity of the flatcar when the cars are moving at
constant speeds.
SOLUTION
The masses are mA  1800 kg, mB  1700 kg, and mF  20  103 kg.
Let vA , vB , and vF be the sought after velocities in m/s, positive to the right.
 v A 0
Initial values:
  vB 0   vF 0  0.
m A  v A 0  m B  vB 0  mF  vF 0  0.
Initial momentum of system:
There are no horizontal external forces acting during the time period under consideration. Momentum is
conserved.
0  mAv A  mBvB  mF vF
1800v A  1700vB  20  103 vF  0
(1)
The relative velocities are given as
v A/F  v A  vF  2.35 m/s
(2)
vB/F  vB  vF  1.117 m/s
(3)
Solving (1), (2), and (3) simultaneously,
v A  2.085 m/s, vB  0.910 m/s, vF  0.265 m/s
v F  0.265 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 14.112
The nozzle shown discharges water at the rate of 200 gal/min. Knowing
that at both B and C the stream of water moves with a velocity of
magnitude 100 ft/s, and neglecting the weight of the vane, determine the
force-couple system that must be applied at A to hold the vane in place
(1 ft 3  7.48 gal).
SOLUTION
Q
200 gal/min
(7.48 gal/ft 3 )(60 s/min)
 0.44563 ft 3 /s
dm  Q

dt
g
(62.4 lb/ft 3 )(0.44563 ft 3 /s)
32.2 ft/s 2
 0.8636 lb  s/ft
vB  (100 ft/s) j
vC  (100 ft/s)(sin 40 i  cos 40 j)

Apply the impulse-momentum principle.
x components:
0  Ax (t )  (m)(100 sin 40)
m
(100sin 40)
t
 (0.8636)(100 sin 40)
Ax 
y components:
Ax  55.5 lb
(m)(100)  Ay (t )  (m)(100 cos 40)
m
(100)(cos 40  1)
t
 (0.8636)(100)(cos 40  1)
 20.2 lb
Ay 
Ay  20.2 lb
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.112 (Continued)
Moments about A:
 3
 9 
 12  (m)(100)  M A (t )   12  (m)(100 cos 40)
 
 
 15 
   (m)(100 sin 40)
 12 
 m 
MA  
 (75 cos 40  125 sin 40  25)
 t 
 (0.8636)(47.895)
 41.36 lb  ft
MA  41.4 lb  ft

A  59.1 lb

20.0° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.113
An airplane with weight W and total wing
span b flies horizontally at a constant speed v.
Use the airplane as a reference frame; that is,
consider the airplane to be motionless and the
air to flow past it with speed v. Suppose that a
cylinder of air with diameter b is deflected
downward by the wing (the cross section of
the cylinder is the dashed circle in in the
figure). Show that the angle through which the
cylinder stream is deflected (called the
downwash angle) is determined by the
formula
4 /
where  is the
mass density of the air.
SOLUTION
Given:
W  Weight, b  wing span, v  speed,   mass density of air
At constant speed, mass flow is steady:
dm
  vA
dt
where: A 
 b2
4
dm  v b 2

dt
4
Velocity of the airstream after the plane in the vertical direction:
v2, y  v sin 
For equilibrium of the plane in vertical direction the lift provided by the wings equals the weight of the
plane. This is also the downward force on the airstream
F
y
 W
Apply Impulse-Momentum in the vertical direction:

dm
v2, y  v1, y
dt
  v b 2 v sin 
W 
4
F
y


or
sin  
4W

 b2  v2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.114
The final component of a conveyor system
receives sand at a rate of 100 kg/s at A and
discharges it at B. The sand is moving horizontally
at A and B with a velocity of magnitude
vA  vB  4.5 m/s. Knowing that the combined
weight of the component and of the sand it
supports is W  4 kN, determine the reactions at
C and D.
SOLUTION
Apply the impulse-momentum principle.
  0.9   m v A  3D t   1.8W  t    1.65  m  vB
Moments about C :
D

1.8
1 m
W 
 0.9vA  1.65vB 
3
3 t
1.8 4000   1
3
3
100   0.9  4.5  1.65 4.5  2287.5 N
D  2.29 kN 
x components:
  m vA  Cx  t     m  vB
 m 
Cx  
  vB  v A   100  4.5  4.5   0
 t 
y components:
0  C y  t   D  t   W  t   0
C y  W  D  4000  2287.5  1712.5 N
C  1.712 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 14.115
A garden sprinkler has four rotating arms, each of which
consists of two horizontal straight sections of pipe forming
an angle of 120° with each other. Each arm discharges
water at a rate of 20 L/min with a velocity of 18 m/s
relative to the arm. Knowing that the friction between the
moving and stationary parts of the sprinkler is equivalent to
a couple of magnitude M  0.375 N  m, determine the
constant rate at which the sprinkler rotates.
SOLUTION
The flow through each arm is 20 L/min.
Q
20 L/min 1min

 333.33  106 m3 /s
3
60 s
1000 L/m
dm
  Q  (1000 kg/m3 )(333.33  106 )
dt
 0.33333 kg/s
Consider the moment about O exerted on the fluid stream of one arm. Apply the impulse-momentum
principle. Compute moments about O. First, consider the geometry of triangle OAB. Using first the law of
cosines,
(OA) 2  1502  1002  (2)(150)(100) cos120
OA  217.95 mm  0.21795 m
Law of sines:
sin  sin120

100
217.95
  23.413,   60    36.587
Moments about O:
(m)(vO )(0)  M O (t )  (OA)(m)vs sin   (OA)(m)(OA)
m
[(OA)vs sin   (OA) 2  ]
t
 (0.33333)[(0.21795)(18)sin 36.587  (0.21795) 2  ]
 0.77945  0.015834
MO 
Moment that the stream exerts on the arm is M O .
Balance of the friction couple and the four streams
MF  4 M O  0
0.375  4(0.77945  0.015834 )  0
  43.305 rad/s
  414 rpm 
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PROBLEM 14.116
A chain of length l and mass m falls through a small hole in a
plate. Initially, when y is very small, the chain is at rest. In
each case shown, determine (a) the acceleration of the first
link A as a function of y, (b) the velocity of the chain as the
last link passes through the hole. In case 1, assume that the
individual links are at rest until they fall through the hole; in
case 2, assume that at any instant all links have the same
speed. Ignore the effect of friction.
SOLUTION
Let  be the mass per unit length of chain. Assume that the weight of any chain above the hole is supported by
the floor. It and the corresponding upward reaction of the floor are not shown in the diagrams.
Case 1: Apply the impulse-momentum principle to the entire chain.
 yv   gy t   ( y  y )(v  v)
  yv   (y )v   y (v)   (y )(v)
 gy  
dy
dv
v y
dt
dt
d
  ( yv )
dt
 gy  
Let t  0.
Multiply both sides by yv.
Let v 
dy
on left hand side.
dt
Integrate with respect to time.
y
v
(y )(v)
v y

t
t
t
 gy 2 v   yv
d
( yv)
dt
 gy 2
dy
d
  yv ( yv)
dt
dt


 g y 2 dy   ( yv) d ( yv)
1
1
 gy 3   ( yv)2
3
2
Differentiate with respect to time.
2v
or
v2 
2
gy
3
dv 2 dy 2
 g
 gv
dt 3 dt 3
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(1)
PROBLEM 14.116 (Continued)
a
(a)
(b)
v2 
Set y  l in Eq. (1).
dv 1
 g
dt 3
a  0.333g 
2
gl
3
v  0.817 gl

Case 2: Apply conservation of energy using the floor as the level from which the potential energy is
measured.
T1  0
V1  0
1 2
y
mv V2    gy
2
2
T1  V1  T2  V2
T2 
0
Differentiating with respect to y,
(a)
Acceleration:
(b)
Setting y  l in Eq. (2),
2v
1 2 1
mv   gy 2
2
2
v2 
 gy 2
m

gy 2
l
(2)
dv 2 gy

dy
l
av
dv gy

dy
l
v 2  gl
a
gy
l

gl

v
Note: The impulse-momentum principle may be used to obtain the force that the edge of the hole
exerts on the chain.
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CHAPTER 15
PROBLEM 15.1
The brake drum is attached to a larger flywheel that is not shown. The
motion of the brake drum is defined by the relation   36t  1.6t 2 ,
where  is expressed in radians and t in seconds. Determine (a) the
angular velocity at t  2 s, (b) the number of revolutions executed by the
brake drum before coming to rest.
SOLUTION
  36t  1.6t 2
Given:
radians
Differentiate to obtain the angular velocity.

d
 36  3.2t
dt
(a)
At t  2 s,
  36  (3.2)(2)
(b)
When the rotor stops,
  0.
0  36  3.2t
rad/s
  29.6 rad/s 
t  11.25 s
  (36)(11.25)  (1.6)(11.25)2  202.5 radians
In revolutions,

202.5
2
  32.2 rev 
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PROBLEM 15.2
The motion of an oscillating flywheel is defined by the relation
   0 e 3 t cos 4 t , where  is expressed in radians and t in seconds. Knowing
that  0  0.5 rad, determine the angular coordinate, the angular velocity, and the
angular acceleration of the flywheel when (a) t  0, (b) t  0.125 s.
SOLUTION
  0.5 e3 t cos 4 t
 
d
 0.5 3 e 3 t cos 4 t  4 e 3 t sin 4 t
dt
 
d
 0.5 9 2e3 t cos 4 t  12 2e3 t sin 4 t  12 2e3 t sin 4 t  16 2e3 t cos 4 t
dt


 0.5  24 e
2 3 t
(a)

sin 4 t  7 2e3 t cos 4 t

   0.5 
t  0,
  0.500 rad 
   0.5  3   4.71

  4.71 rad/s 

   0.5  7 2  34.5
(b) t  0.125 s,
cos 4 t  cos

2
sin 4 t  sin
 0,


2
  34.5 rad/s 2 
1
e 3 t  0.30786
   0.5  0.30786  0   0
   0.5  0.30786  4   1.93437


   0.5  0.30786  24 2  36.461
  0
  1.934 rad/s 
  36.5 rad/s 2 
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PROBLEM 15.3
The motion of an oscillating flywheel is defined by the relation
   0 e 7 t / 6 sin 4 t , where  is expressed in radians and t in seconds. Knowing
that  0  0.4 rad, determine the angular coordinate, the angular velocity, and the
angular acceleration of the flywheel when (a) t  0.125 s, (b) t  .
SOLUTION
   0e 7 t/6 sin 4 t
 
 
d
 7 7 t/6

e
sin 4 t  4 e 7 t/6 cos 4 t 
 0  
dt
 6

 49 2 7 t/6

d
28 2 7 t/6
28 2 7 t/6
e
sin 4 t 
e
cos 4 t 
e
cos 4 t  16 2e7 t/6 sin 4 t 
  0 
dt
6
6
 36



49  2
28 2
  0e7 t/6 16 
 cos 4 t 
  sin 4 t 
36 
3


(a)  0  0.4 rad,
t  0.125 s
e 7 (0.125)/6  0.63245, 4 t 

2
, sin

2
 1, cos
   0.4  0.63245 1  0.25298 radians
 7 
 1  0.92722 rad/s
 6 
   0.4  0.63245   


    0.4  0.63245   16 
(b)
t  ,
e 7 t/6  0
49  2
 1  36.551 rad/s 2

36 

2
0
  0.253 rad 
  0.927 rad/s 
  36.6 rad/s 2 
  0
  0
  0
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PROBLEM 15.4
The rotor of a gas turbine is rotating at a speed of 6900 rpm when the turbine is shut down. It is observed that
4 min is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine (a) the
angular acceleration, (b) the number of revolutions that the rotor executes before coming to rest.
SOLUTION
 0  6900 rpm
 722.57 rad/s
t  4 min  240 s
(a)
   0   t ; 0  722.57   (240)
  3.0107 rad/s
(b)
  3.01 rad/s 2 
1
1
2
2
  173, 416  86, 708  86, 708 rad
   0 t   t 2  (722.57)(240)  (3.0107)(240) 2
 1 rev 

 2 rad 
  86, 708 rad 
  13,800 rev 
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PROBLEM 15.5
A small grinding wheel is attached to the shaft of an electric motor
which has a rated speed of 3600 rpm. When the power is turned on, the
unit reaches its rated speed in 5 s, and when the power is turned off, the
unit coasts to rest in 70 s. Assuming uniformly accelerated motion,
determine the number of revolutions that the motor executes (a) in
reaching its rated speed, (b) in coasting to rest.
SOLUTION
For uniformly accelerated motion,
  0   t
(1)
1
2
   0  0 t   t 2
(a)
Data for start up:
 0  0, 0  0,
At t  5 s,
  3600 rpm 
From Eq. (1),
(b)
120   (5)
(2)
2 (3600)
 120 rad/s
60
  24 rad/s 2
1
2
From Eq. (2),
  0  0  (24 )(5)2  300 radians
In revolutions,

300
2
  150 rev 
Data for coasting to rest:
 0  0, 0  120 rad/s
At t  70 s,
 0
From Eq. (1),
0  120   (70)
From Eq. (2),
  0  (120 )(70) 
In revolutions,

4200
2

120
rad/s
70
(120 )(70) 2
 4200 radians
2(70)
  2100 rev 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.6
A connecting rod is supported by a knife-edge at Point A. For small oscillations the
angular acceleration of the connecting rod is governed by the relation   6 where  is
expressed in rad/s2 and  in radians. Knowing that the connecting rod is released from rest
when   20, determine (a) the maximum angular velocity, (b) the angular position
when t  2 s.
SOLUTION
Angular motion relations:
d  d

 6
dt
d

(1)
Separation of variables  and  gives
 d  6 d
Integrating, using   0 when    0 ,


0
 d  6

  d
0
1 2
  3( 2   02 )  3( 02   2 )
2
 2  6(02   2 )
(a)
  6(02   2 )
 is maximum when   0.
 0  20  0.34907 radians
Data:
2
max
 6(0.34907 2  0)  0.73108 rad 2 /s
(b)
From  
d
we get
dt
dt 
d


1
6
max  0.855 rad/s 
d
02
 2
Integrating, using t  0 when    0 ,

t
0
dt 
t
d

1
6 
1
6
0
02
cos
2 1
 2

0


0

1 
1
1  
cos 1
0  cos

0 
0
6
6
   0 cos( 6t )  0.34907 cos[( 6)2]  (0.34907)(0.18551)  0.064756 radians
  3.71 
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PROBLEM 15.7
When studying whiplash resulting from rear end collisions, the rotation of the head is of
primary interest. An impact test was performed, and it was found that the angular
acceleration of the head is defined by the relation   700cos   70sin  where  is
expressed in rad/s2 and  in radians. Knowing that the head is initially at rest, determine the
angular velocity of the head when   30°.
SOLUTION
Angular motion relations:

d  d

 700 cos   70sin 
dt
d
Separating variables  and  gives
 d   (700cos   70sin  )d
Integrating, using   0 when   0,


0
 d 


0
(700 cos   70sin  ) d

1 2
  (700sin   70 cos  )
2
0
 700sin   70(1  cos  )
  1400sin   140(1  cos  )
Data:
  30 

6
rad
With calculator set to “degrees” for trigonometric functions,
  1400sin 30  140(1  cos 30)  26.8 rad/s
  26.8 rad/s 
With calculator set to “radians” for trigonometric functions,
  1400sin( /6)  140(1  cos( /6))  26.8 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.8
The angular acceleration of an oscillating disk is defined by the relation   k . Determine (a) the value of
k for which   12 rad/s when   0 and   6 rad when   0, (b) the angular velocity of the disk when
  3 rad.
SOLUTION
  k and  =
Given:
 d    d

d
d
 d   k d
0
6
 12  d  k  0  d
Integrating,
0
k 
(a)
 62

122
 k 
 0 
2
 2

122
 9 s 2
62
k  9.00 s 2 

3
 12  d  k  0  d
2
2
(b)

 32

122
 9   0 
2
 2

 2  122   9  3  63 rad 2/s2
2
  7.94 rad/s 
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PROBLEM 15.9
The angular acceleration of a shaft is defined by the relation   0.5 , where  is expressed in rad/s 2
and  in rad/s. Knowing that at t  0 the angular velocity of the shaft is 30 rad/s, determine (a) the number
of revolutions the shaft will execute before coming to rest, (b) the time required for the shaft to come to rest,
(c) the time required for the angular velocity of the shaft to reduce to 2 percent of its initial value.
SOLUTION
(a) Given:
  0.5 ,

Integrating,

d
 0.5
d
d
 0.5
dt
Integrating,
d   0.5d
 30  0.5
60
 9.55 rev

dt  2
  9.55 rev 
d

d
t
0
 0 dt  2 30 
t  2 ln
(c)

0

 30 d  0.5 0 d
  60 radians 
(b)
d

d
0

30
t  
   0.02  30   0.6 rad/s
d
t
0.6
 0 dt  2 30 
t  2 ln
0.6
 2 ln 50
30
t  7.82 s 
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PROBLEM 15.10
The bent rod ABCDE rotates about a line joining Points A and
E with a constant angular velocity of 9 rad/s. Knowing that
the rotation is clockwise as viewed from E, determine the
velocity and acceleration of corner C.
SOLUTION
EA2  0.42  0.42  0.22
EA  0.6 m
rC/E  (0.4 m)i  (0.15 m) j

EA  (0.4 m)i  (0.4 m) j  (0.2 m)k

EA
1
1
 EA 
(0.4i  0.4 j  0.2k )  (2i  2 j  k )

EA 0.6
3
1
   AE  EA  (9 rad/s) (2i  2 j  k )
3
  (6 rad/s)i  (6 rad/s) j  (3 rad/s)k
vC    rC/E
i
j
k
 6
6
3  0.45i  1.2 j  ( 0.9  2.4)k
0.4 0.15 0
v C  (0.45 m/s)i  (1.2 m/s) j  (1.5 m/s)k 
aC   AC  rC/E    (  rC/E )   AC  rC/E    v C
i
j
k
6
3
a C  0  6
0.45 1.2 1.5
 (9  3.6)i  (1.35  9) j  (7.2  2.7)k
aC  (12.60 m/s 2 )i  (7.65 m/s 2 ) j  (9.90 m/s 2 )k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.11
In Problem 15.10, determine the velocity and acceleration of
corner B, assuming that the angular velocity is 9 rad/s and
increases at the rate of 45 rad/s 2 .
PROBLEM 15.10 The bent rod ABCDE rotates about a line
joining Points A and E with a constant angular velocity of 9
rad/s. Knowing that the rotation is clockwise as viewed from
E, determine the velocity and acceleration of corner C.
SOLUTION
EA2  0.42  0.42  0.22
EA  0.6 m
rB/A  (0.25 m) j

EA  (0.4 m)i  (0.4 m) j  (0.2 m)k

EA  0.4i  0.4 j  0.2k  1
 EA 

  (2i  2 j  k )
EA 
0.6
 3
1
   AE  EA  (9 rad/s) (2 i  2 j  k )
3
  (6 rad/s)i  (6 rad/s) j  (3 rad/s)k
v B    rB/A  (6i  6 j  3k )  (0.25) j  1.5k  0.75i
v B  (0.75 m/s)i  (1.5 m/s)k 
1
   AE  EA  (45 rad/s 2 ) (2i  2 j  k )
3
2
  (30 rad/s )i  (30 rad/s2 ) j  (15 rad/s 2 )k
a B    rB/A    (  rB/A )    rB/A    v B
i
j
k
i
j k
a B  30
30 15  6 6 3
0 0.25 0
0.75 0 1.5
 3.75i  7.5k  9i  (2.25  9) j  4.5k
a B  (12.75 m/s 2 )i  (11.25 m/s 2 ) j  (3.00 m/s 2 )k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.12
The rectangular block shown rotates about the diagonal OA with a constant
angular velocity of 6.76 rad/s. Knowing that the rotation is counterclockwise
as viewed from A, determine the velocity and acceleration of point B at the
instant shown.
SOLUTION
  6.76 rad/s
Given:
rA/O   5 in. i   31.2 in. j  12 in. k
rB/O   5 in. i  15.6 in. j
lOA 
 52   31.2 2  12 2
 33.8 in.
Angular velocity:


lOA
rA/O 
6.76
 5i  31.2 j  12k 
33.8
  1.0 rad/s  i   6.24 rad/s  j   2.4 rad/s  k
Velocity of point B:
v B    rB/O
i
j
k
v B  1.0 6.24 2.4   37.44i  12 j  15.6k
5 15.6 0
v B    37.4 in./s  i  12.00 in./s  j  15.60 in./s  k 
Acceleration of point B:
aB    vB
i
j
k
2.4  126.1i  74.26 j  245.6k
a B  1.0 6.24
 37.4 12 15.60

 
 

a B   126.1 in./s 2 i  74.3 in./s 2 j  246 in./s 2 k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.13
The rectangular block shown rotates about the diagonal OA with an
angular velocity of 3.38 rad/s that is decreasing at the rate of 5.07 rad/s2.
Knowing that the rotation is counterclockwise as viewed from A,
determine the velocity and acceleration of point B at the instant shown.
SOLUTION
  3.38 rad/s,   5.07 rad/s2
Given:
rA/O   5 in. i   31.2 in. j  12 in. k
rB/O   5 in. i  15.6 in. j
 52   31.2 2  12 2
lOA 
 33.8 in.
Angular velocity:


lOA
rA/O 
3.38
 5i  31.2 j  12k 
33.8
   0.5 rad/s  i   3.12 rad/s  j  1.2 rad/s  k
Velocity of point B:
v B    rB/O
i
j
k
v B  0.5 3.12 1.2  18.72i  6 j  7.80k
5 15.6 0
v B   18.72 in./s  i   6.00 in./s  j   7.80 in./s  k 
Angular Acceleration:
 

lOA

rA/O 
5.07
 5i  31.2 j  12k 
33.8
 
 

   0.75 rad/s 2 i  4.68 rad/s 2 j  1.8 rad/s 2 k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.13 (Continued)
Acceleration of point B:
a B    rB/O    v B
i
j
k
i
a B   0.75 4.68 1.8
5
15.6
0

j
k
0.5
3.12 1.2
18.72 6 7.8
 28.08i  9 j  11.7k  31.536i  18.564 j  61.406k

 
 

a B   3.46 in./s 2 i  27.6 in./s 2 j  73.1 in./s 2 k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.14
A circular plate of 120 mm radius is supported by two
bearings A and B as shown. The plate rotates about the
rod joining A and B with a constant angular velocity of
26 rad/s. Knowing that, at the instant considered, the
velocity of Point C is directed to the right, determine the
velocity and acceleration of Point E.
SOLUTION

BA  (100 mm) j  (240 mm)k
BA  260 mm

BA (100) j  (240)k
BA 
 0

,
BA
260
 1 
  BA  26 
 ((100) j  (240)k )
 260 
Point E:
rE /A  (120 mm)i  (80 mm) j  (120 mm)k
v E    rE /A
i
j
k
 0 10 24
120 80 120
 (3120 mm/s)i  (2880 mm/s) j  (1200 mm/s)k
v E  (3.12 m/s)i  (2.88 m/s) j  (1.200 m/s)k 
a E    rE/A    (  rE/A )  0  ω  v B
aE    v B

i
0
j
10
k
24
3120 2880 1200
 (81120 mm/s 2 )i  (74880 mm/s 2 ) j  (31200 mm/s 2 )
a E  (81.1 m/s 2 )i  (74.9 m/s 2 ) j  (31.2 m/s 2 )k 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.15
In Problem 15.14, determine the velocity and acceleration of Point E, assuming that the angular velocity is
26 rad/s and increases at the rate of 65 rad/s2.
SOLUTION
See Problem 15.14 for  BA and 
(100) j  (240)k
260
  (10 rad/s)j  (24 rad/s)k
 BA 
α  65 rad/s 2 ;
 1 
α  αBA  (65 rad/s 2 ) 
  (100) j  (240)k
 260 
α  (25 rad/s 2 ) j  (60 rad/s 2 )k
rE /A  (120 mm)i  (80) j  (120)k
Point E:
v E    rE /A
i
j
k
0 10 24
120 80 120
 (3120 mm/s)i  (2880 mm/s) j  (1200 mm/s)k

v E  (3.12 m/s)i  (2.88 m/s) j  (1.200 m/s)k 
a D    rE/A    (  rE/A )    rE/A    v E
aD 
i
j
k
i
j
k
0 25 60  0
10
24
120 80 120
3120 2880 1200
 (7800 mm/s 2 )i  (7200 mm/s 2 ) j  (3000 mm/s 2 )k
 (81120 mm/s 2 )i  (74880 mm/s 2 ) j  (31200 mm/s 2 )k
a B  (73320 mm/s 2 )i  (82080 mm/s 2 ) j  (34200 mm/s 2 )k
 (73.3 m/s 2 )i  (82.1 m/s 2 ) j  (34.2 m/s2 )k
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.16
The earth makes one complete revolution around the sun in 365.24 days. Assuming that the orbit of the earth
is circular and has a radius of 93,000,000 mi, determine the velocity and acceleration of the earth.
SOLUTION

2 rad
(365.24 days)
 
24 h
day
3600 s
h

 199.11  109 rad/s
v  r
 5280 ft 
9
 (93  106 mi) 
 (199.11  10 rad/s)
 mi 
v  97,770 ft/s
v  66,700 mi/h 
a  r 2
 (93  106 )(5280)(199.11  109 rad/s)2
a  19.47  10 3 ft/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.17
The earth makes one complete revolution on its axis in 23 h 56 min. Knowing that the mean radius of the
earth is 3960 mi, determine the linear velocity and acceleration of a point on the surface of the earth
(a) at the equator, (b) at Philadelphia, latitude 40° north, (c) at the North Pole.
SOLUTION
23 h 56 m  23.933 h
2 rad

s
(23.933 h)  3600
h 
 72.925  106 rad/s
 5280 ft 
R  (3960 mi) 

 mi 
 20.91  106 ft
r  radius of path
 R cos 
(a)
Equator: Latitude    0
v  r
 R (cos 0)
 (20.91  106 ft)(1)(72.925  106 rad/s)
v  1525 ft/s 
a  r 2
 R(cos 0) 2
 (20.91  106 ft)(1)(72.925  106 rad/s) 2
(b)
a  0.1112 ft/s 2 
Philadelphia: Latitude    40
v  r
 R (cos 40)
 (20.91  106 ft)( cos 40)(72.925  10 6 rad/s)
v  1168 ft/s 
a  r 2
 R(cos 40) 2
 (20.91  106 ft)(cos 40°)(72.925  106 rad/s) 2
(c)
a  0.0852 ft/s 2 
North Pole: Latitude    0
r  R cos 0  0
v  a  0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.18
A series of small machine components being moved by a
conveyor belt pass over a 120 mm radius idler pulley. At the
instant shown, the velocity of Point A is 300 mm/s to the left
and its acceleration is 180 mm/s2 to the right. Determine (a) the
angular velocity and angular acceleration of the idler pulley,
(b) the total acceleration of the machine component at B.
SOLUTION
vB  v A  300 mm/s
rB  120 mm
( a B )t  a A  180 mm/s
(a)
(b)
vB   rB ,

vB 300

 2.5 rad/s
rB 120
( a B )t   rB ,

(aB )t 180

 1.5 rad/s
rB
120
  2.50 rad/s

  1.500 rad/s 2

( aB ) n  rB 2  (120)(2.5)2  750 mm/s 2
aB  (aB )t2  (aB ) 2n  (180) 2  (750)2  771 mm/s 2
tan  
750
,
180
  76.5
a B  771 mm/s 2
76.5 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.19
A series of small machine components being moved by a
conveyor belt pass over a 120-mm-radius idler pulley. At the
instant shown, the angular velocity of the idler pulley is 4 rad/s
clockwise. Determine the angular acceleration of the pulley for
which the magnitude of the total acceleration of the machine
component at B is 2400 mm/s 2.
SOLUTION
 B  4 rad/s ,
rB  120 mm
(aB ) n  rB B2  (120)(4) 2  1920 mm/s 2
aB  2400 mm/s 2
(a B )t 
aB2  (aB ) 2n 
(aB )t  rB ,
 
24002  19202  1440 mm/s 2
( a B )t
1440

 12 rad/s 2
rB
120
12.00 rad/s 2
or
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.20
The belt sander shown is initially at rest. If the driving drum B
has a constant angular acceleration of 120 rad/s2 counterclockwise, determine the magnitude of the acceleration of the
belt at Point C when (a) t  0.5 s, (b) t  2 s.
SOLUTION
at  r  (0.025 m)(120 rad/s 2 )
at  3 m/s 2
(a)
t  0.5 s:
   t  (120 rad/s2 )(0.5 s)  60 rad/s
an  r 2  (0.025 m)(60 rad/s)2
a n  90 m/s 2
aB2  at2  an2  32  902
(b)
t  2 s:
aB  90.05 m/s 2 
   t  (120 rad/s2 )(2 s)  240 rad/s
an  r 2  (0.025 m)(240 rad/s)2
an  1440 m/s 2
aB2  at2  an2  32  14402
aB  1440 m/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.21
The rated speed of drum B of the belt sander shown is 2400 rpm.
When the power is turned off, it is observed that the sander
coasts from its rated speed to rest in 10 s. Assuming uniformly
decelerated motion, determine the velocity and acceleration of
Point C of the belt, (a) immediately before the power is turned
off, (b) 9 s later.
SOLUTION
0  2400 rpm
 251.3 rad/s
r  0.025 m
vC  r  (0.025 m)(251.3 rad/s)
(a)
aC  r 2  (0.025 m)(251.3 rad/s) 2
(b)
When t  10 s:
vC  6.28 m/s 
aC  1579 m/s 2 
  0.
  0   t
0  251.3 rad/s   (10 s)
  25.13 rad/s 2
When t  9 s:
  0   t
9  251.3 rad/s  (25.13 rad/s 2 )(9 s)
 25.13 rad/s
vC  r9
 (0.025 m)(25.13 rad/s)
v9  0.628 m/s 
(aC )t  r n
 (0.025 m)(25.13 rad/s 2 )
(aC )t  0.628 m/s 2
(aC )n  r92
 (0.025 m)(25.13 rad/s)2
(aC )n  15.79 m/s 2
aC2  (aC )t2  (aC )2n
 (0.628 m/s 2 ) 2  (15.79 m/s 2 ) 2
aC  15.80 m/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.22
The two pulleys shown may be operated with the V belt in any of three
positions. If the angular acceleration of shaft A is 6 rad/s2 and if the
system is initially at rest, determine the time required for shaft B to
reach a speed of 400 rpm with the belt in each of the three positions.
SOLUTION
Angular velocity of shaft A:
Belt speed:
Angular speed of shaft B:
Solving for t,
Data:  A  6 rad/s,
 A   At
v  rA A  rB B
B 
t
v rA At

rB
rB
rB B
rA A
 B  400 rpm  41.889 rad/s
t
rB 41.889
r
d

 6.9813 B  6.9813 B
6
rA
rA
dA
Belt at left:
d B 2 in.

d A 4 in.
t  3.49 s 
Belt in middle:
d B 3 in.

d A 3 in.
t  6.98 s 
Belt at right:
d B 4 in.

d A 2 in.
t  13.96 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.23
Three belts move over two pulleys without slipping in the speed
reduction system shown. At the instant shown the velocity of
Point A on the input belt is 2 ft/s to the right, decreasing at the
rate of 6 ft/s2. Determine, at this instant, (a) the velocity and
acceleration of Point C on the output belt, (b) the acceleration
of Point B on the output pulley.
SOLUTION
Left pulley.
Inner radius
r1  2 in.
Outer radius
r2  4 in.
v A  2 ft/s
(a A )t  6 ft/s 2  6 ft/s 2
 1
vA 2
  6 rad/s
r2 124
1 
( a A )t
6
 4  18 rad/s 2
r2
12
Intermediate belt.
 2 
v1  r11    (6)  1 ft/s
 12 
 2
(a1)t  r11    (18)  3 ft/s 2
 12 
Right pulley.
Inner radius
r3  2 in.
Outer radius
r4  4 in.
2 
v1
1

 3 rad/s
r4  124 
2 
( a1 )t
3
 4  9 rad/s 2
r4
 12 
PROBLEM 15.23 (Continued)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(a)
Velocity and acceleration of Point C.
 2
vC  r32    (3)  0.5 ft/s
 12 
 2
(aC )t  r3 2    (9)  1.5 ft/s 2
 12 
(b)
v C  0.5 ft/s

aC  1.5 ft/s 2

Acceleration of Point B.
 4
(aB ) n  r422    (3) 2  3 ft/s 2
 12 
 4 
( aB )t  r4 2    (9)  3 ft/s 2
 12 
(a B ) n  3 ft/s 2
(a B )t  3 ft/s 2
a B  4.24 ft/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
45 

PROBLEM 15.24
A gear reduction system consists of three gears A, B, and C. Knowing
that gear A rotates clockwise with a constant angular velocity
 A  600 rpm, determine (a) the angular velocities of gears B and C,
(b) the accelerations of the points on gears B and C which are in contact.
SOLUTION
 A  600 rpm 
(a)
(600)(2 )
 20 rad/s.
60
Let Points A, B, and C lie at the axles of gears A, B, and C, respectively.
Let D be the contact point between gears A and B.
vD  rD/ A A  (2)(20 )  40 in./s
B 
vD
40
60

 10 rad/s  10 
 300 rpm
rD/B
4
2
 B  300 rpm

Let E be the contact point between gears B and C.
vE  rE/BB  (2)(10 )  20  in./s
C 
vE
20
60

 3.333 rad/s  (3.333 )
 100 rpm
rE/C
6
2
C  100 rpm

(b) Accelerations at Point E.
On gear B :
On gear C :
aB 
aC 
vE2
(20 )2

 1973.9 in./s 2
rE/B
2
a B  1974 in./s 2

aC  658 in./s 2

vE2
(20 ) 2

 658 in./s2
rE/C
6
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.25
A belt is pulled to the right between cylinders A and B. Knowing that the
speed of the belt is a constant 5 ft/s and no slippage occurs, determine (a) the
angular velocities of A and B, (b) the accelerations of the points which are in
contact with the belt.
SOLUTION
(a)
(b)

Angular velocities.
Disk A:
A 
vP
5 ft/s

rA (4/12) ft
ω A  15.00 rad/s
Disk B:
B 
vP
5 ft/s

rB (8/12) ft
ω B  7.50 rad/s


Accelerations of contact points.
Disk A:
a A   A2 rA  (15.00 rad/s) 2 ((4/12) ft)
Disk B:
aB   B2 rB  (7.50 rad/s) 2 ((8/12) ft)
a A  75.00 rad/s 2

a B  37.5 rad/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.26
Ring C has an inside radius of 55 mm and an outside radius of 60 mm and is
positioned between two wheels A and B, each of 24-mm outside radius.
Knowing that wheel A rotates with a constant angular velocity of 300 rpm
and that no slipping occurs, determine (a) the angular velocity of the ring C
and of wheel B, (b) the acceleration of the Points on A and B that are in
contact with C.
SOLUTION
 2 

 60 
 31.416 rad/s
 A  300 rpm 
rA  24 mm
rB  24 mm
r1  60 mm
r2  55 mm
[We assume senses of rotation shown for our computations.]
(a)
Velocities:
Point 1 (Point of contact of A and C)
v1  rA A  r1C
C 

rA
A
r1
24 mm
(300 rpm)
60 mm
 120 rpm
C  120 rpm 
Point 2 (Point of contact of B and C)
v2  rB B  r2C
B 
r2
C
rB

r2  rA 
 A
rB  r1 

55 mm
24 mm
 24 mm 

 300 rpm
 60 mm 
 B  275 mm
 B  275 rpm 
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PROBLEM 15.26 (Continued)
(b)
Accelerations:
Point on rim of A:
rA  24 mm  0.024 m
a A  rA A2
 (0.024 m)(31.416 rad/s) 2
 23.687 m/s 2
Point on rim of B:
a A  23.7 m/s 2 
 2 

 60 
 28.798 rad/s
B  275 rpm 
aB  rB B2
 (0.024 m)(28.798 rad/s) 2
 19.904 m/s 2
a B  19.90 m/s 2
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
PROBLEM 15.27
Ring B has an inside radius r2 and hangs from the horizontal
shaft A as shown. Shaft A rotates with a constant angular velocity
of 25 rad/s and no slipping occurs. Knowing that r1  12 mm,
r2  30 mm, and r3  40 mm, determine (a) the angular velocity
of ring B, (b) the accelerations of the points of shaft A and ring B
which are in contact, (c) the magnitude of the acceleration of
point D.
SOLUTION
Let Point C be the point of contact between the shaft and the ring.
vC  r1 A
B 

vC
r2
r1 A
r2
B 
On shaft A:
a A  r1 A2
On ring B:
 r 
aB  r2B2  r2  1 A 
 r2 
r1 A
r2
a A  r1 A2
2
aB 
r12 A2
r2

Acceleration of Point D on outside of ring.
r

aD  r3B2  r3  1  A 
 r2

2
2
r 
aD  r3  1   A2
 r2 
Data:
 A  25 rad/s
r1  12 mm
r2  30 mm
r3  40 mm
(a)
r1
A
r2
12 mm

(25 rad/s)
30 mm
B 
 B  10 rad/s
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
PROBLEM 15.27 (Continued)
(b)
a A  r1 A2
 (12 mm)(25 rad/s)2
 7.5  103 mm/s 2
aB 

a A  7.50 m/s 2

a B  3.00 m/s 2

a D  4.00 m/s 2

r12 2
A
r2
(12 mm)2
(25 rad/s)2
(30 mm)
 3  103 mm/s 2
2
(c)
r 
aD  r3  1   A2
 r2 
2
 12 mm 
2
 (40 mm) 
 (25 rad/s)
30
mm


aD  4  103 mm/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.28
A plastic film moves over two drums. During a 4-s interval the
speed of the tape is increased uniformly from v0  2 ft/s to
v1  4 ft/s. Knowing that the tape does not slip on the drums,
determine (a) the angular acceleration of drum B, (b) the
number of revolutions executed by drum B during the 4-s
interval.
SOLUTION
Belt motion:
v  v0  a t
4 ft/s  2 ft/s  a (4 s)
a
4 ft/s  2 ft/s
 0.5 ft/s 2  6 in./s 2
4s
Since the belt does not slip relative to the periphery of the drum, the tangential acceleration at the periphery of
the drum is
at  6 in./s 2
(a)
Angular acceleration of drum B.
B 
(b)
at 6 in./s 2

15 in.
rB
α B  0.400 rad/s 2

Angular displacement of drum B.
At t  0,
0 
v0 24 in./s

 1.6 rad/s
15 in.
rB
At t  4 s,
1 
v1 48 in./s

 3.2 rad/s
rB
15 in.
12  02  2 B B
In revolutions,
B 
12  02 (3.2)2  (1.6)2

 9.6 radians
2 B
(2)(0.400)
B 
9.6
2
 B  1.528 rev 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.29
Cylinder A is moving downward with a velocity of 3 m/s when the brake is
suddenly applied to the drum. Knowing that the cylinder moves 6 m
downward before coming to rest and assuming uniformly accelerated motion,
determine (a) the angular acceleration of the drum, (b) the time required for
the cylinder to come to rest.
SOLUTION
Given:
 v A 0
 3 m/s,
Constant Acceleration:
 vA 12

2
aA 
(a) At the surface of the drum,
 vA 02
2
 vA 1  0,  s A 1   s A 0
 6m
 a A  s A 1   s A 0 
 vA 12   vA 02
2  s A 1   s A 0 

0  32
 0.75 m/s
 2  6 
r  250 mm  0.25 m
at  r
 
at
0.75

 3.00 rad/s 2
r
0.25
  3.00 rad/s 2
(b)
0 
 v A 0
r


3
 12 rad/s
0.25
1  0   0   t
t 
1   0 0  12

 4.00 s

3.00
t  4.00 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.30
The system shown is held at rest by the brake-and-drum system shown. After
the brake is partially released at t  0, it is observed that the cylinder moves
5 m in 4.5 s. Assuming uniformly accelerated motion, determine (a) the
angular acceleration of the drum, (b) the angular velocity of the drum at
t  3.5 s.
SOLUTION
sA 
(a) Assume uniformly accelerated motion.
aA 
For the drum,
1
aA t 2
2
 2  5  0.49383 m/s2
2s A

2
4.52
t
at  a A  0.49383 m/s 2 ,
at  r ,
 
r  250 mm  0.25 m
at
0.49383

r
0.25
 1.97531 rad/s 2 ,
(b)
  1.975 rad/s 2

  6.91 rad/s

   0   t  0  1.97531 3.5 
 6.91 rad/s,
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PROBLEM 15.31
A load is to be raised 20 ft by the hoisting system shown. Assuming gear A
is initially at rest, accelerates uniformly to a speed of 120 rpm in 5 s, and
then maintains a constant speed of 120 rpm, determine (a) the number of
revolutions executed by gear A in raising the load, (b) the time required to
raise the load.
SOLUTION
The load is raised a distance h  20 ft  240 in.
For gear-pulley B, radius to rope groove is r1  15 in.
Required angle change for B:
B 
h 240

 16 radians
r1 15
Circumferential travel of gears A and B:
s  r2 B  rA A
where r2  18 in. and rA  3 in.
s  (18 in.)(16 radians)  288 in.
(a)
(b)
Angle change of gear A:
A 
s 288

 96 radians
rA
3
In revolutions,
A 
96
2
Motion of gear A.
0  0,  f  120 rpm  4 rad/s
 A  15.28 rev 
Gear A is uniformly accelerated over the first 5 seconds.

 f  0
t
1
2

4 rad/s
 2.5133 rad/s 2
5s
1
2
 A   t 2  (2.5133)(5) 2  31.416 radians
The angle change over the constant speed phase is
   A    96  31.416  64.584 radians
For uniform motion,
   f (t )
t 
Total time elapsed:

f

64.584
 5.139 s
4
t f  5 s  t
t f  10.14 s 
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PROBLEM 15.32
A simple friction drive consists of two disks A and B. Initially, disk B has a
clockwise angular velocity of 500 rpm, and disk A is at rest. It is known that
disk B will coast to rest in 60 s. However, rather than waiting until both
disks are at rest to bring them together, disk A is given a constant angular
acceleration of 3 rad/s2 counterclockwise. Determine (a) at what time the
disks can be brought together if they are not to slip, (b) the angular velocity
of each disk as contact is made.
SOLUTION
Motion of disk B:
 B 0  500 rpm  52.360 rad/s
Assume that the angular acceleration of disk B is constant.
 B   B 0   B t
At
t  60 s,
B  0
B 
 B   B  0
t

0  52.360
  0.87266 rad/s 2
60
vB  rB B   3 52.360  0.87266 t   157.08  2.618 t in./s
Motion of disk A:
 A 0  0,
 A  3 rad/s 2
 A   A 0   At  3 t
v A  rA A   2.5  3 t   7.5 t in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.32 (Continued)
If disks are not to slip,
v A  vB
7.5 t  157.08  2.618 t
t  15.52 s 
(a)
(b)
 A   315.52   46.6 rad/s
 B  52.360   0.87266 15.52   38.8 rad/s
 A  445 rpm
,
 B  371 rpm
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
PROBLEM 15.33
Two friction wheels A and B are both rotating freely at 300 rpm
counterclockwise when they are brought into contact. After 12 s of
slippage, during which time each wheel has a constant angular acceleration,
wheel B reaches a final angular velocity of 75 rpm counterclockwise.
Determine (a) the angular acceleration of each wheel during the period of
slippage, (b) the time at which the angular velocity of wheel A is equal to
zero.
SOLUTION
 B 0  300 rpm  31.416 rad/s
Motion of disk B:
t  12s,  B  75 rpm = 7.854 rad/s
At
Angular acceleration.
 B   B 0
7.854  31.416
 1.9635 rad/s
12
t
Velocity at contact point with disk A at t  12 s:
B 

vB  rB B   3 7.854   23.562 in./s
Motion of disk A:
 A 0  300 rpm = 31.416 rad/s
Assume that slipping ends when t  12 s.
v A  23.562 in./s
Then,
A 
v A 23.562

 9.4248 rad/s
rA
2.5
 A  9.4248 rad/s
A 
(a)
 A   A 0
t
  9.4248 rad/s

 9.4248  31.416
  3.4034 rad/s 2
12
 A  3.40 rad/s 2

 B  1.963 rad/s 2

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PROBLEM 15.33 (Continued)
(b)
Time when A is zero.
 A   A  0   A t  0
t
 A   A 0
A

0  31.416
 3.4034
t  9.23 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.34
Two friction disks A and B are to be brought into contact without slipping when the
angular velocity of disk A is 240 rpm counterclockwise. Disk A starts from rest at
time t  0 and is given a constant angular acceleration of magnitude  . Disk B starts
from rest at time t  2 s and is given a constant clockwise angular acceleration, also
of magnitude  . Determine (a) the required angular acceleration magnitude ,
(b) the time at which the contact occurs.
SOLUTION
When contact is made:
 A  240 rpm  8 rad/s
Let C be the contact point between the two gears.
vC  rA A   0.15  8   1.2 m/s 2
B 
vC
1.2

 6 rad/s
rB
0.2
 A  8   t rad/s
 B  6    t  2  rad/s
Subtracting,
2    2 
  3.14 rad/s 2 
(a)
(b)
   rad/s 2
t 
8


8

 8s
t  8.00 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.35
Two friction disks A and B are brought into contact when the angular velocity of disk
A is 240 rpm counterclockwise and disk B is at rest. A period of slipping follows and
disk B makes 2 revolutions before reaching its final angular velocity. Assuming that
the angular acceleration of each disk is constant and inversely proportional to the
cube of its radius, determine (a) the angular acceleration of each disk, (b) the time
during which the disks slip.
SOLUTION
 A  0
Given:
 A 1  8
 240 rpm  8 rad/s
  A t1
3
 B  4 
3
1
1r 
1  0.15 
 B t12   A   A t12  
 A t12
2
2  rB 
2  0.2 
3
 A t12
 B 1
 0.2 
  8  
  59.574 radians
 0.15 
3
 0.15 
  B t1  
  A t1  0.421875 A t1
 0.2 
Let vC be the velocity at the contact point.
vC  rA A   0.15  8   A t1   1.2  0.15 A t1
vC  rB B   0.2  0.421875 A t1   0.084375 A t1
and
Equating the two expressions for vC ,
1.2  0.15 A t1  0.084375 A t1
t1 
Then,
A 
(a)
or
 A t1  16.0850 rad/s
 A t12
59.574

 3.7037 s
 A t1 16.0850
16.0850
 4.3429 rad/s 2
3.7037
 0.15 

 B  1.832 rad/s 2

3
2
B  
  4.3429   1.83218 rad/s
 0.2 
(b) From above,
 A  4.34 rad/s 2
t1  3.70 s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.36*
Steel tape is being wound onto a spool that rotates with a constant angular
velocity  0 . Denoting by r the radius of the spool and tape at any given
time and by b the thickness of the tape, derive an expression for the
acceleration of the tape as it approaches the spool.
SOLUTION
Let one layer of tape be wound and let v be the tape speed.
vt  2 r and r  b
r
bv
b


t 2 r 2
For the spool:
d  d  v  1 dv
d 1
  
v  
dt
dt  r  r dt
dt  r 
a v dr a v b

 
r r 2 dt r r 2 2
1
b 2 
 a 
0
2 
r

a
b02
2
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
PROBLEM 15.37*
In a continuous printing process, paper is drawn into the presses
at a constant speed v. Denoting by r the radius of the paper roll
at any given time and by b the thickness of the paper, derive an
expression for the angular acceleration of the paper roll.
SOLUTION
Let one layer of paper be unrolled.
vt  2 r and r  b
r bv dr


t 2 r dt
d

dt
d v
  
dt  r 
1 dv
d 1

v  
r dt
dt  r 
v dr
0 2
r dt
 v  bv 
   2 

 r  2 r 

bv 2
2 r 3

bv 2
2 r 3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.38
An automobile travels to the right at a constant speed of
48 mi/h. If the diameter of a wheel is 22 in., determine the
velocities of Points B, C, D, and E on the rim of the wheel.
SOLUTION
vA  48 mi/h  70.4 ft/s
d  22 in. r 

vC  0 
d
 11 in.  0.91667 ft
2
vA
70.4

 76.8 rad/s
0.91667
r
vB/A  vD/A  vE/A  r
 (0.91667)(76.8)  70.4 ft/s
vB  vA  vB/A  [70.4 ft/s
]  [70.4 ft/s
]
vB  140.8 ft/s
vD  vA  vD/A  [70.4 ft/s
vE  vA  vE/A  [70.4 ft/s

]  [70.4 ft/s

30]
vD  136.0 ft/s
15.0 
vE  99.6 ft/s
45.0 
]  [70.4 ft/s ]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.39
The motion of rod AB is guided by pins attached at A and B
which slide in the slots shown. At the instant shown,   40
and the pin at B moves upward to the left with a constant
velocity of 6 in./s. Determine (a) the angular velocity of the
rod, (b) the velocity of the pin at end A.
SOLUTION
vA  vB  vA/B
[ vA]  [6 in./s
15]  [v A/B
40]
Law of sines.
v
vA
6 in./s
 A/ B 
sin 55 sin 75 sin 50
vA  6.42 in./s
(b)
vA/B  7.566 in./s
vA/B  ( AB ) AB

40
AB  20 in.
7.566 in./s  (20 in.) AB
(a)
 AB  0.3783 rad/s
 AB  0.378 rad/s
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
PROBLEM 15.40
A painter is halfway up a 10-m ladder when the bottom starts
sliding out from under him. Knowing that point A has a velocity
vA = 2 m/s directed to the left when = 60°, determine (a) the
angular velocity of the ladder, (b) the velocity of the painter.
SOLUTION
v A  2i m/s,  =60
Given:
rB / A  10 cos 60i  10sin 60 j
Geometry:
rP / A  5cos 60i  5sin 60 j
v B  v A  v B / A  v A   AB k  rB /A
Relative Velocity:
vB j  2i   AB k   5i  8.660 j
   2  8.660 AB  i  5 AB j
Equate Components:
(a)
i:
0    2  8.660 AB 
 AB  0.231 rad/s
 AB  0.231 rad/s 
Velocity of Painter (point P):
v P  v A  v P / A  v A   AB k  rP/A
 2i  0.231k   2.5i  4.330 j
(b)
v P   1.00 m/s  i   0.577 m/s  j 
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PROBLEM 15.41
Rod AB can slide freely along the floor and the inclined plane. At
the instant shown the velocity of end A is 1.4 m/s to the left.
Determine (a) the angular velocity of the rod, (b) the velocity of end
B of the rod.
SOLUTION
Geometry:
sin  
tan  
0.3
,
0.5
  36.87
0.3
,
0.125
  67.38
Velocity analysis:
v A  1.4 m/s
v B/ A  r AB  0.5 AB
v B  vB


Plane motion  Translation with A  Rotation about A.
v B  v A  v B/ A
Draw velocity vector diagram.
  180     90     59.49
Law of sines:
vB/ A
sin 
(a)
(b)

vB
v
 A
sin  90    sin 
vB/ A 
v A sin 
1.4sin 67.38

 1.5 m/s
sin 
sin 59.49
 AB 
1.5
 3.00 rad/s
0.5
vB 
 AB  3.00 rad/s

v A cos 
1.4 cos36.87

 1.3 m/s
sin 
sin 59.49
v B  1.300 m/s
67.4 
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PROBLEM 15.42
Rod AB can slide freely along the floor and the inclined plane. At
the instant shown the angular velocity of the rod is 4.2 rad/s
counterclockwise. Determine (a) the velocity of end A of the rod,
(b) the velocity of end B of the rod.
SOLUTION
Geometry:
sin  
0.3
,
0.5
0.3
,
0.125
tan  
  36.87
  67.38
Velocity analysis:
 AB  4.2 rad/s
v B/ A  rB/ A AB   0.5  4.2   2.1 m/s
v B  vB


v A  vA
Plane motion  Translation with A  Rotation about A.
v B  v A  v B/ A
Draw velocity vector diagram.
  180     90     59.49
Law of sines:
(a)
vB/ A
vA
vB


sin 
sin  90    sin 
vA 
vB/ A sin 
sin 

2.1sin 59.49
 1.96 m/s
sin 67.38
v A  1.960 m/s
(b)
vB 
vB/ A cos 
sin 


2.1cos 36.87
 1.82 m/s
sin 67.38
v B  1.820 m/s
67.4 
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PROBLEM 15.43
Rod AB moves over a small wheel at C while end A moves to the
right with a constant velocity of 25 in./s. At the instant shown,
determine (a) the angular velocity of the rod, (b) the velocity of
end B of the rod.
SOLUTION
Slope angle of rod.
tan  
AC 
7
 0.7,
10
  35
10
 12.2066 in.
cos 
CB  20  AC  7.7934 in.
Velocity analysis.
v A  25 in./s
,
v C /A  AC AB

v C  vC

v C  v A  vC/A
Draw corresponding vector diagram.
vC/A  v A sin   25sin 35  14.34 in./s
(a)
 AB 
vC/A
AC

14.34
 1.175 rad/s
12.2066
 AB  1.175 rad/s

vC  v A cos   25cos   20.479 in./s
v B/ C  CB AB  (7.7934)(1.175)  9.1551 in./s
vB/C has same direction as vC/A.
v B  v C  v B/C
Draw corresponding vector diagram.
vB/C 9.1551
tan  
,   24.09

vC
20.479
(b)
vB 
vC
20.479

 22.4 in./s  1.869 ft/s
cos  cos 24.09
    59.1
v B  1.869 ft/s
59.1 
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PROBLEM 15.44
The disk shown moves in the xy plane. Knowing that
(v A ) y  7 m/s, (vB ) x = 7.4 m/s, and (vC ) x  1.4 m/s,
determine (a) the angular velocity of the disk, (b) the velocity
of point B.
SOLUTION
In units of m/s,
v B/ A   k  rB/ A   k   0.6i  0.6 j  0.6 i  0.6 j
v C/ A   k  rC/ A   k  1.2i  1.2 j
v B  v A  v B/ A
7.4i   vB  y j   v A  x i  7 j  0.6 i  0.6 j
i :  7.4   v A  x  0.6
Components.
j:
 vB  y
(1)
 7  0.6
(2)
v C  v A  v C/ A
1.4i   vC  y j   v A  x i  7 j  1.2 j
i : 1.4   v A  x
Components.
j:
From (3),
(a) From (1),
From (2),
(b)
 
 vB  y
 vC  y
 7  1.2
 vA x
 1.4 m/s
7.4   1.4 
 10 rad/s,
0.6
(3)
(4)
  10.00 rad/s

 7   0.6 10   1 m/s
v B    7.40 m/s  i  1.000 m/s  j  
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PROBLEM 15.45
The disk shown moves in the xy plane. Knowing that
(v A ) y  7 m/s, (vB ) x = 7.4 m/s, and (vC ) x  1.4 m/s, ,
determine (a) the velocity of point O, (b) the point of the disk
with zero velocity.
SOLUTION
In units of m/s,
v B/ A   k  rB/ A   k   0.6i  0.6 j  0.6 i  0.6 j
v C/ A   k  rC/ A   k  1.2i  1.2 j
v B  v A  v B/ A
7.4i   vB  y j   v A  x i  7 j  0.6 i  0.6 j
i :  7.4   v A  x  0.6
Components.
j:
 vB  y
(1)
 7  0.6
(2)
v C  v A  v C/ A
1.4i   vC  y   v A  x i  7 j  1.2 j
i : 1.4   v A  x
Components.
j:
 vA  x
From (3),
From (1),
 
 vC  y
(3)
 7  1.2
 1.4 m/s,
(4)
v A  1.4i  7 j
7.4   1.4 
 10 rad/s,
0.6
  10.00 rad/s  k
vO  v A  vO/ A  v A    rO/ A  v A  10k   0.6i 
(a)
 1.4i  7 j  6 j  1.4i  1j
v O   1.400 m/s  i  1.000 m/s  j 
0  v O     xi  yj
(b)
0  1.4i  1j  10k   xi  yj  1.4i  1j  10 xj  10 yj
Components.
i : 0  1.4  10 y,
y   0.14 m
j : 0  1  10 x,
x  0.1 m
y  140.0 mm 
x  100.0 mm  
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PROBLEM 15.46
The plate shown moves in the xy plane. Knowing that
(v A ) x  250 mm/s, (vB ) x  450 mm/s, and (vC ) x  500 mm/s,
determine (a) the angular velocity of the plate, (b) the velocity of
Point A.
SOLUTION
 k
Angular velocity:
rB/A  (150 mm)i
Relative position vectors:
rC/A  (200 mm)i  (150 mm) j
v A  (250 mm/s)i  (v A ) y j
Velocity vectors:
v B  (vB ) x i  (450 mm/s) j
vC  (500 mm/s)i  (vC ) y j
Unknowns are ω, (v A ) y , (vB ) x , and (vC ) y .
v B  v A  v B/A  v A   k  rB/A
(vB ) x i  450 j  250i  (v A ) y j   k  150i
 250i  (v A ) y j  150 j
i:
(vB ) x  250
(1)
j:
450  (v A ) y  150
(2)
v C  v A  v C/A  v A   k  rC/A
500i  (vC ) y j  250i  (v A ) y j   k  (200i  150 j)
 250i  (v A ) y j  200 j  150 i
(a)
i:
500  250  150
(3)
j:
(vC ) y  (v A ) y  150
(4)
Angular velocity of the plate.
From Eq. (3),

750
 5
150
ω  (5.00 rad/s)k  5.00 rad/s
(b)

Velocity of Point A.
From Eq. (2), (v A ) y  450  150  450  (150)(5)  300 mm/s
v A  (250 mm/s)i  (300 mm/s) j 
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PROBLEM 15.47
Velocity sensors are placed on a satellite that is
moving only in the xy plane. Knowing that at the
instant shown the uni-directional sensors measure
(vA)x = 2 ft/s, (vB)x = 0.333 ft/s, and (vC)y = 2 ft/s,
determine (a) the angular velocity of the satellite,
(b) the velocity of point B.
SOLUTION
Given:
 vA  x  2 ft/s,  vB  x  0.333 ft/s,  vC  y  2 ft/s
Geometry:
rB / A   2 ft  i   4 ft  j
rB / C    6 ft  i   2 ft  j
v B  v A   ABC k  rB /A
Relative Velocity:
0.333i   vB  y j  2i   v A  y j   ABC k   2i  4 j
Equate Components:
(c)
i:
0.333  2  4 ABC
 ABC  0.583 rad/s
 ABC  0.583 rad/s

v B  v C   ABC k  rB/C
Relative Velocity:
0.333i   vB  y j   vC  x i  2 j  0.583k   6i  2 j
Equate Components:
(d)
j:
 vB  y  1.500 ft/s
v B    0.333 ft/s  i  1.500 ft/s  j
v B  1.537 ft/s
77.48º 
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PROBLEM 15.48
In the planetary gear system shown, the radius of gears A, B, C, and D is a
and the radius of the outer gear E is 3a. Knowing that the angular velocity
of gear A is A clockwise and that the outer gear E is stationary, determine
(a) the angular velocity of each planetary gear, (b) the angular velocity of
the spider connecting the planetary gears.
SOLUTION
Gear E is stationary.
vE  0
Let A be the center of gear A and the spider. Since the motions of gears B, C, and D are similar, only gear B is
considered. Let H be the effective contact point between gears A and B.
Gear A:
v H  a A
(a )
v H  v E  v H /E
Planetary gears B, C, and D:
1
2
B   A
: a A  0  (2 a ) B
1
 B  ωC  ω D   A
2

v B  v E  v B/E
1

: vB  0  a   A 
2

(b)
Spider.
vB 
1
a A
2
(1)
v B  (2 a ) s
(2)
Equating expressions (1) and (2) for vB,
1
a A  (2a )s
2
1
4
s   A
1
s   A
4
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
PROBLEM 15.49
In the planetary gear system shown, the radius of gears A, B, C, and D is
30 mm and the radius of the outer gear E is 90 mm. Knowing that gear E
has an angular velocity of 180 rpm clockwise and that the central gear A
has an angular velocity of 240 rpm clockwise, determine (a) the angular
velocity of each planetary gear, (b) the angular velocity of the spider
connecting the planetary gears.
SOLUTION
Since the motions of the planetary gears B, C, and D are similar, only gear B is considered. Let Point H
be the effect contact point between gears A and B and let Point E be the effective contact point between
gears B and E.
Given angular velocities:
Outer gear E:
ω E  180 rpm
 6 rad/s
ω A  240 rpm
 8 rad/s
radius  rE  90 mm
vE  rE  E  (90 mm)(6 rad/s)  540 mm/s
v E  540 mm/s
Gear A:
radius  rA  30 mm
vH  rA A  (30 mm)(8 rad/s)  240 mm/s
v H  240 mm/s
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PROBLEM 15.49 (Continued)
Planetary gear B:
radius  rB  30 mm,
ω B  B
v H  v E  v H /E
]  [540 mm/s
[(30 mm) A
]  [(60 mm) B
]
30 A  540  60B
B 
(a)
540  30 A
1
1
 9   A  9  (8 )  5 rad/s
60
2
2
Angular velocity of planetary gears:
ω B  ω C  ω D  5 rad/s
v B  v H  vB/H  [(30 mm) A
]  [30 mm  B
 150 rpm

 195 rpm

]
vB  (30 mm)(8 rad/s)  (30 mm)(5 rad/s)  390 mm/s
(b)
Spider:
arm  rs  60 mm, ω s   s
vB  rss
s 
vB 390 mm/s

 6.5 rad/s
60 mm
rs
 s  6.5 rad/s
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PROBLEM 15.50
Arm AB rotates with an angular velocity of 20 rad/s counterclockwise. Knowing that the outer gear C is stationary, determine
(a) the angular velocity of gear B, (b) the velocity of the gear tooth
located at Point D.
SOLUTION
Arm AB:
Gear B:
(a)
BE  0.05 m:
v B  vD  vB/E  0  ( BE ) B
2.4 m/s  0  (0.05 m) B
 B  48 rad/s
(b)
DE  (0.05 2 ):
 B  48 rad/s

vD  vE  vD/E  0  ( DE )B
vD  0  (0.05 2)(48)
vD  3.39 m/s
vD  3.39 m/s
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45° 
PROBLEM 15.51
In the simplified sketch of a ball bearing shown, the diameter of the inner
race A is 60 mm and the diameter of each ball is 12 mm. The outer race B
is stationary while the inner race has an angular velocity of 3600 rpm.
Determine (a) the speed of the center of each ball, (b) the angular velocity
of each ball, (c) the number of times per minute each ball describes a
complete circle.
SOLUTION
Data:  A  3600 rpm  376.99 rad/s,
rA 
B  0
1
d A  30 mm
2
d  diameter of ball  12 mm
Velocity of point on inner race in contact with a ball.
v A  rA A  (30)(376.99)  11310 mm/s
Consider a ball with its center at Point C.
v A  v B  v A /B
v A  0  C d
C 
v A 11310

d
12
 942.48 rad/s
vC  vB  vC/B
0
1
d   (6)(942.48)  5654.9 mm/s
2
vC  5.65 m/s 
(a)
(b )
Angular velocity of ball.
C  942.48 rad/s
(c )
C  9000 rpm 
Distance traveled by center of ball in 1 minute.
lC  vC t  5654.9(60)  339290 mm
Circumference of circle:
2 r  2 (30  6)
 226.19 mm
Number of circles completed in 1 minute:
n
l
2 r

339290
226.19
n  1500 
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PROBLEM 15.52
A simplified gear system for a mechanical watch is
shown. Knowing that gear A has a constant angular
velocity of 1 rev/h and gear C has a constant angular
velocity of 1 rpm, determine (a) the radius r, (b) the
magnitudes of the accelerations of the points on gear B
that are in contact with gears A and C.
SOLUTION
Point where A contacts B:
v1  rA A  r B
B 
rA A
r
(1)
v2  rB  B  rC
C 
rB
B
r
(2)
Point where B contacts C:
From Eqs. (1) and (2),
C 
rA rB
r2
r 2  rA rB
r2 
 A 1 rev/h
1


C 1 rev/m 60
(0.6 in.)(0.36 in.)
 0.0036 in 2
60
r  0.0600 in. 
Radius r:
Angular velocity of B.
(b)
A
C
rA  0.6 in., rB  0.36 in.
Data:
(a)
A
Point where B contacts A.
2
rad/s
60
r
0.060 2
B  C 
 0.017453 rad/s
rB
0.36 60
C  1 rpm 
an  r B2  (0.0600 in.)(0.017453 rad/s) 2
an  18.28  10 6 in./s 2 

Point where B contacts C.
an  rB B2  (0.36 in.)(0.017453 rad/s) 2
an  109.7  10 6 in./s 2 
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PROBLEM 15.53
Arm ACB rotates about Point C with an angular velocity of
40 rad/s counterclockwise. Two friction disks A and B are
pinned at their centers to arm ACB as shown. Knowing that the
disks roll without slipping at surfaces of contact, determine the
angular velocity of (a) disk A, (b) disk B.
SOLUTION

Arm ACB : Fixed axis rotation.

rA/C  24 mm,
vA  rA/C  AB  (24)(40)  960 mm/s
rB/C  18 mm,
vB  rB/C  AB  (18)(40)  720 mm/s
Disk B: Plane motion  Translation with B  Rotation about B.
rB  30 mm,

v D  v B  v D/B
0  720  30  B
B 
mm/s
720
 24 rad/s
30
vE  vB  vE/B
 720  (30)(24)  1440 mm/s

Disk A: Plane motion  Translation with A  Rotation about A.
rA  12 mm,
vE  vA  vE/ A
1440  960  12 A
A 
1440  960
 200 rad/s
12
(a )
ω A  200 rad/s

(b)
ω B  24.0 rad/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.54
Arm ACB rotates about Point C with an angular velocity of
40 rad/s counterclockwise. Two friction disks A and B are
pinned at their centers to arm ACB as shown. Knowing that the
disks roll without slipping at surfaces of contact, determine the
angular velocity of (a) disk A, (b) disk B.
SOLUTION
Arm ACB : Fixed axis rotation.


rA/C  0.3 in.,
vA  rA/C  AB  (0.3)(40)  12 in./s
rB/C  1.8 in.,
vB  rB/C  AB  (1.8)(40)  72 in./s
Disk B: Plane motion  Translation with B  Rotation about B.
rB  0.6 in.,
vD  vB  vB/A
0  72  0.6 B
B 
72
 120 rad/s
0.6
vE  vB  vE/B
 72  (0.6)(120)  144 in./s

Disk A: Plane motion  Translation with A  Rotation about A.
rA  1.5 in.,
vE  vA  vE/A
144  12 1.5 A
A 
144  12
 104 rad/s
1.5
(a )
 A  104.0 rad/s

(b)
 B  120.0 rad/s

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PROBLEM 15.55
Knowing that at the instant shown the velocity of collar A is
900 mm/s to the left, determine (a) the angular velocity of rod ADB,
(b) the velocity of Point B.
SOLUTION
Consider rod ADB.
v D  vD j, v A  (900 mm/s)i
rD /A  (80 mm)i  (150 mm) j
v D /A  ω AD  rD /A   AD k  (80i  150 j)
 150 AD i  80 AD j
v 0  v A  v D /A
vD j  900i  150 AD i  80 AD j
Equate components.
i: 0  900  150 AD
(a)
 AD  6 rad/s
ω AD  (6.00 rad/s)k  6.00 rad/s
Angular velocity of ADB.
By proportions,

150  60
rD/A  1.4 rD/A
150
 (112 mm)i  (210 mm) j
rB /A 
v B  v A   AD k  rB/A
 900i  6k  (112i  210 j)
 900i  672 j  1260i
(b)
Velocity of B.
v B  (360 mm/s)i  (672 mm/s) j  762 mm/s
61.8° 
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PROBLEM 15.56
Knowing that at the instant shown the angular velocity of rod DE is
2.4 rad/s clockwise, determine (a) the velocity A, (b) the velocity of
Point B.
SOLUTION
ω DE  2.4 rad
Rod DE: Point E is fixed.
vD   DB rDE  (2.4 rad/s)(120 mm)  288 mm/s
v D  288 mm/s  (288 mm/s) j
rA /D  (80 mm)i  (150 mm) j, ω AD  ω AD k , v A  v A i
Rod ADB:
v A  v D  v D/A  v D   AD k  rA/D
v A i  (288 mm/s) j   AD k  [(80 mm)i  (150 mm) j]
v A i  288 j  80 AD j  150 AD i
Equate components.
i:
j:
From Eq. (2),
From Eq. (1),
(a)
Velocity of collar A.
(b)
Velocity of Point B.
By proportions
v A  150 AD
(1)
0  288  80 AD
 AD  
288
80
(2)
ω AD  ( 3.6 rad/s)k
v A  (150)( 3.6)  540 mm/s
v A  540 mm/s
rB/D  

60
rA/D  (32 mm)i  60 mm j
150
v B  v D  v B/D  v D  ω AD  rB/D
 (288 mm/s) j  [(3.6 rad/s)k ]  [(32 mm)i  (60 mm) j]
 (288 mm/s) j  (115.2 mm/s) j  (216 mm/s)i
v B  (216 mm/s)i  (403.2 mm/s) j
v B  457 mm/s
61.8° 
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PROBLEM 15.57
Knowing that the disk has a constant angular velocity of 15 rad/s clockwise,
determine the angular velocity of bar BD and the velocity of collar D when
(a)   0, (b)   90, (c)   180.
SOLUTION
 A  15 rad/s ,
Disk A
AB  2.8 in.
Rotation about a fixed axis.
vB   AB   A   2.8 15   42 in./s
(a)   0.
v B  42 in./s
2.8
,
10
sin  
  16.260
v D  v B  v D/B
  42
vD
Bar BD :
vD/B 
 DB 
vD/B
DB

  vD/B
 
42
 43.75 in./s
cos 
43.75
,
10
 DB  4.38 rad/s
vD  vB tan  ,
(b)   90.
v D  12.25 in./s 
v B  42 in./s
sin  
5.6
,   34.06
10
v D  v B  v D/B
Bar BD :
vD
Components:

  42
  vD/B
 
: vD/B  0
: vD  vB
 DB  0 
v D  42.0 in./s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.57 (Continued)
(c)   180.
vB  42 in./s
sin  
2.8
,
10
  16.26
v D  v B  v D/B
Bar BD :
v D   42
  vD/B
vD/B 
 DB 
vD/B
DB

 
42
 43.75
cos 
43.75
10
 DB  4.38 rad/s
vD  vB tan 

v D  12.25 in./s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.58
The disk has a constant angular velocity of 20 rad/s clockwise.
(a) Determine the two values of the angle  for which the velocity of collar D is
zero. (b) For each of these values of  , determine the corresponding value of the
angular velocity of bar BD.
SOLUTION
rA  2.8 in.,
lBD  10 in.
lBD sin   rA  rA sin 
From geometry,
(1)
v B is tangent to the circular path of B,
v B  rA A
thus
v B/D  lBD BD
For rod BD


v B  v D  v B/D  0  v B/D
v B/D  v B
(a) For matching direction   
For    , sin   sin 
sin  
or
so that
rA
2.8

,
lBD  rA 10  2.8
For   180   ,
  180  
lBD sin   rA  rA sin 
  22.9,
  22.9 
sin    sin  ,
lBD sin   rA  rA sin 
sin  
rA
2.8

,
lBD  rA 10  2.8
  12.6
  192.6 
(b) For matching magnitudes vD/B  vB
lBD BD  rA A ,
 BD 
 2.8 20   5.6 rad/s
rA A

lBD
10
For   22.9,
 BD  5.60 rad/s

For   192.6,
 BD  5.60 rad/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.59
The test rig shown was developed to perform fatigue testing
on fitness trampolines. A motor drives the 9-in radius
flywheel AB, which is pinned at its center point A, in a
counterclockwise direction. The flywheel is attached to slider
CD by the 18-in connecting rod BC. Knowing that the “feet”
at D should hit the trampoline twice every second, at the
instant when = 30°, determine (a) the angular velocity of the
connecting rod BC, (b) the velocity of D, (c) the velocity of
midpoint CB.
SOLUTION
rB / A  9 in, rC / B  18 in,  =30
Given:
Re v 2 rad

 4 rad/s
s
Re v
 0, v D  vC
 AB  2
 vC  x
9cos   18cos    =64.341
Geometry:
rB / A   9cos in  i   9sin   j
 7.794i  4.5 j
rC / B   18cos in  i  18sin in  j
 7.794i  16.225 j
rE / C   9cos in  i   9sin in  j
 3.897i  8.112 j
Relative Velocity:
v B  v A   AB k  rB /A
(1)
v C  v B   BC k  rC/B
(2)
v C   AB k  rB /A  BC k  rC/B
Sub (1) into (2):
 vC  y j  4 k   7.794i  4.5 j  BC k   7.794i  16.225 j
(a) Equate Components:
i:
0  18 i  16.225 BC j
(b)
j:
 vC  y  97.945  3.485 7.794 
 vC  y  125.110 in/s
(c)
Relative Velocity:
BC  3.485 rad/s 
v D  125.11 in/s  
v E  v C   BC k  rE/C
 125.11j  3.485k   3.897i  8.112 j
 28.274i  111.527 j in/s
v E  115.056 in/s
75.77º 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.60
In the eccentric shown, a disk of 2-in.-radius revolves about shaft O
that is located 0.5 in. from the center A of the disk. The distance
between the center A of the disk and the pin at B is 8 in. Knowing that
the angular velocity of the disk is 900 rpm clockwise, determine the
velocity of the block when   30.
SOLUTION
(OA)sin   ( AB )sin 
Geometry.
(OA)sin 
AB
0.5 sin 30
,

8
sin  
  1.79
Shaft and eccentric disk. (Rotation about O)
OA  900 rpm  30 rad/s
vA  (OA) OA  (0.5)(30 )  15 in/s
Rod AB.
(Plane motion  Translation with A  Rotation about A.)
vB  vA  vB/ A
Draw velocity vector diagram.
Law of sines.

[v B
]  [v A
60]  [v A/B
]
90    88.21
  180  60  88.21
 31.79
vB
vA

sin  sin(90   )
v A sin 
vB 
sin (90   )
(15 )sin 31.79

sin 88.21
 24.837 in./s 
vB  24.8 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.61
In the engine system shown, l  160 mm and b  60 mm. Knowing that the crank
AB rotates with a constant angular velocity of 1000 rpm clockwise, determine the
velocity of the piston P and the angular velocity of the connecting rod when
(a)   0, (b)   90.
SOLUTION
 AB  1000 rpm
(a)
  0. Crank AB. (Rotation about A)

(1000)(2 )
 104.72 rad/s
60
rB/A  0.06 m
vB  vB/A AB  (0.06)(104.72)  6.2832 m/s
Rod BD.
(Plane motion = Translation with B + Rotation about B)
vD  vB  vD/B
vD
 [6.2832
]  [ v D/ B
]
vD  0
vD/B  6.2832 m/s
vP  vD
 BD 
(b)
vP  0 
vB 6.2832

0.16
l
  90. Crank AB. (Rotation about A)
 BD  39.3 rad/s

rB/ A  0.06 m
vB  rB/ A AB  (0.06)(104.72)  6.2832 m/s
Rod BD. (Plane motion  Translation with B + Rotation about B.)
vD  vB  vD/B
[vD ] ]  [6.2832]   vD /B
vD/B  0,
BD 


vD  6.2832 m/s
vD/B
l
v P  v D  6.2832 m/s
 BD  0 
vP  6.28 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.62
In the engine system shown l  160 mm and b  60 mm. Knowing that crank AB
rotates with a constant angular velocity of 1000 rpm clockwise, determine the
velocity of the piston P and the angular velocity of the connecting rod when
  60.
SOLUTION
 AB  1000 rpm 
  60. Crank AB. (Rotation about A)
(1000)(2 )
 104.72 rad/s
60
rB/A  3 in.
30°
vB  rB /A AB  (0.06)(104.72)  6.2832 m/s
60°
Rod BD. (Plane motion  Translation with B  Rotation about B.)
Geometry.
l sin   r sin 
r
0.06
sin   sin  
sin 60
l
0.16
  18.95
vD  vB  vD/B
[vD ]  [314.16
60]  [vD /B
]
Draw velocity vector diagram.
  180  30  (90   )  78.95
Law of sines.
vD/B
vD
vB


sin  sin 30 sin (90   )
vB sin 
cos 
6.2832 sin 78.95

cos18.95
 6.52 m/s
vD 
vP  vD
vP  6.52 m/s 



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PROBLEM 15.62 (Continued)
vB sin 30
cos 
6.2832sin 30

cos18.95
 3.3216 m/s
vD/B 
BD 
vD/B
l

3.3216
0.16
 BD  20.8 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.63
Knowing that at the instant shown the angular velocity of rod AB
is 15 rad/s clockwise, determine (a) the angular velocity of rod
BD, (b) the velocity of the midpoint of rod BD.
SOLUTION
ω AB  15 rad/s
Rod AB:
vB  ( AB ) AB  (0.200)(15)  3 m/s
vB   (3 m/s)i,
Rod BD:
vB  3 m/s
v D  vD j, ω BD   BD k
rB /D  (0.6 m)i  (0.25 m) j
v B  v D  v B /D  v D  ω BD  rB /D
3i  vD j  BD k  (0.6i  0.25 j)
 vD j  0.6BD j  0.25BD i
Equate components.
i: 3  0.25 BD
0  vD  0.6 BD
j:
(a)
(2)
Angular velocity of rod BD.
From Eq. (1),
From Eq. (2),
(b)
(1)
3
0.25
 BD  12.00 rad/s
vD  0.6 BD
vD  7.2 m/s
 BD 

Velocity of midpoint M of rod BD.
1
rB /D  (0.3 m)i  (0.125 m) j
2
 v D  v M /D  vD j   BD k  rM /D
rM /D 
vM
 7.2 j  12.00k  (0.3i  0.125 j)
 (1.500 m/s)i  (3.60 m/s) j
vM  3.90 m/s
67.4 
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PROBLEM 15.64
In the position shown, bar AB has an angular velocity of 4 rad/s clockwise.
Determine the angular velocity of bars BD and DE.
SOLUTION
 AB  4 rad/s
Bar AB: (Rotation about A)
 (4 rad/s)k
rB /A  (175 mm)i
vB   AB  rB /A  ( 4k )  (175i )
vB  (700 mm/s)j
Bar BD: (Plane motion  Translation with B  Rotation about B.)
 BD  BD k rD /B  (200 mm)j
vD  vB   BD  rD /B  700 j  (BD k )  (200 j)
vD  700 j  200BD i
 DE  DE k
Bar DE: (Rotation about E )
rD /E  (275 mm)i  (75 mm)j
vD   DE  rD /E  (DE k )  (275i  75 j)
vD  275DE j  75DE i
Equating components of the two expressions for vD ,
700  275 DE
 DE  2.5455 rad/s
i : 200 BD  75 DE
 BD    BD
j:
3
 
 DE  2.55 rad/s

3
8
 BD     (2.5455)  0.95455 rad/s
8
 BD  0.955 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.65
Linkage DBEF is part of a windshield
wiper mechanism, where points O, F and D
are fixed pinned connections. At the
position shown, = 60° and link EB is
horizontal. Knowing that link EF has a
counterclockwise angular velocity of
4 rad/s at the instant shown, determine the
angular velocity of links EB and DB.
SOLUTION
Given:
 =60
EF  4 rad/s
Geometry:
30cos   20cos    =41.41
rE / F   20sin  i   20cos   j
rB / E
 13.229i  15.0 j mm
 85.0i mm
rB / D   30sin  i   30cos  j
 25.981i  15.0 j mm
Relative Velocity:
v E  v F   EF k  rE/F
(1)
v B  v E  EB k  rB/E
(2)
v B  v D  BD k  rB/D
(3)
Sub (1) into (2) = (3):
 EF k  rE/F   EB k  rB/E = BD k  rB/D
Put in known values:
4k  13.229i  15.0 j  EB k  85.0i =BD k   25.981i  15.0 j
Equate Components:
i:
60  15 BD   BD  4 rad/s
j:
52.915  85.0EB  25.981 4 
 EB  0.600 rad/s
BD  4.000 rad/s 
EB  0.600 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.66
Robert’s linkage is named after Richard Robert (1789–1864) and can be used
to draw a close approximation to a straight line by locating a pen at Point F.
The distance AB is the same as BF, DF and DE. Knowing that the angular
velocity of bar AB is 5 rad/s clockwise in the position shown, determine
(a) the angular velocity of bar DE, (b) the velocity of Point F.
SOLUTION
Bar AB:
ω AB  5 rad/s
In inches,
rB /A  3i  122  32 j  3i  135 j
  (5 rad/s)k
v B  ω AB  rB /A  5k  (3i  135 j)
 5 135i  15 j
rD /B  (6 in.)i, rF /B  3i  135 j (in.), ω BD   BD k
Object BDF:
v D  v B  v B /D  v B  BD k  rD /B
 5 135i  15 j  BD k  6i
 5 135i  15 j  6BD j
(1)
ω DE   DE k , rD /E  (3 in.)i  ( 135 in.) j,
Bar DE:
Point E is fixed so v E  0
v D  ω DE  rD /E  DE k  (3i  135 j)
  135DE i  3DE j
(2)
Equating like components of vD from Eqs. (1) and (2),
i:
j:
(a)
5 135   135 DE
(3)
15  6BD  3DE
(4)
Angular velocity of bar DE.
From Eq. (3),
DE  5 rad/s
From Eq. (4),
BD  (15  3DE )  (15  15)
1
6
DE  5.00 rad/s
1
6

ωBD  5.00 rad/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.66 (Continued)
(b)
Velocity of Point F.
vF /B  3i  135 j
vF  v B  v F /B  v B  BD k  rF /B
 5 135i  15 j  5k  (3i  135 j)
 5 135i  15 j  15 j  5 135i  10 135i
vF  116.2 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.67
Robert’s linkage is named after Richard Robert (1789–1864) and can be used
to draw a close approximation to a straight line by locating a pen at Point F.
The distance AB is the same as BF, DF and DE. Knowing that the angular
velocity of plate BDF is 2 rad/s counterclockwise when   90°, determine
(a) the angular velocities of bars AB and DE, (b) the velocity of Point F.
When   90, point F may be assumed to coincide with point E, with
negligible error in the velocity analysis.
SOLUTION
When   90, the configuration of the linkage is close to
that shown at the right.
Bar AB:
ω AB  ABk
In inches,
rB/A  6i  6 3 j
v B  ω AB  rB/A   AB k  (6i  6 3 j)  6 3 AB i  6 AB j
Object BDF:
rD/B  6i  6(2  3) j
rF /B  6i  6 3 j
ω BD  (2 rad/s)k
v D  v B  v D/B  v B  BD k  rD/B
 6 3 AB i  6 AB j  2k  [6i  6(2  3) j]
 6 3 AB i  6 AB j  24i  12 3i  12 j
Bar DE:
ωDE  DE k,
(1)
rD/E  12 j
v D  ω DE  rD/E   DE k  12 j  12 DE i
(2)
Equating like components of vD from Eqs. (1) and (2),
i:
j:
6 3 AB  12  6 3  12 DE
6AB  12  0
AB  2 rad/s
From Eq. (3),
(3)
 AB  2.00 rad/s

DE  1.464 rad/s

12 DE  (6 3)( 2)  24  12 3
DE  2  2 3  1.4641



Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.67 (Continued)
v D  (12)(1.4641)i  17.569i
v F  v D  BD k  rF /D
 17.569i  2k  (12 j)  (41.569 in./s)i
v F  41.6 in./s

Note: The exact configuration of the linkage when   90 may be calculated from trigonometry using the
figure given below.
Applying the law of cosines to triangle ADB gives
  13.5
so that angle EAB is
45  13.5  58.5.
We used 60° in the approximate analysis.
Point F then lies about 0.53 in. to the right of Point E.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.68
In the position shown, bar DE has a constant angular
velocity of 10 rad/s clockwise. Knowing that h  500 mm,
determine (a) the angular velocity of bar FBD, (b) the
velocity of Point F.
SOLUTION
Bar DE:
(Rotation about E)
DE  10 rad/s
 (10 rad/s)k
rD/E  (0.1 m)i  (0.2 m)j
vD  DE  rD/E  (10k )  (0.1i  0.2 j)
 (1 m/s)j  (2 m/s)i
Bar FBD: (Plane motion  Translation with D  Rotation about D.)
BD  BDk rB/D  (0.3 m)i  (0.1 m)j
vB  vD   BD  rB/D
 j  2i  (BD k )  ( 0.3i  0.1j)
 j  2i  0.3 BD j  0.1BD i
Bar AB:
(Rotation about A)
 AB   AB k rB/A  (0.42 m)j
vB   AB  rB/A  (AB k )  (0.42 j)  0.42AB i
Equating components of the two expressions for vB ,
(a)
j:
1  0.3BD  0
BD  3.3333 rad/s
i:
2  0.1BD  0.42AB
rF/D  CrB/D

2  (0.1)(3.3333)  0.42AB
AB  3.9683 rad/s
Bar FBD:
BD  3.33 rad/s
AB  3.97 rad/s
where C 
h  0.3
0.3
vF  vB   BD  rF/D
 j  2i  C (0.3BD j  0.1BD i )
With
(b)
 j  2i  C ( j  0.33333i )
0.8
h  500 mm  0.5 m, C 
 2.6667
0.3
vF  j  2i  2.6667 j  0.88889i
vF  (1.11111 m/s)i  (1.66667 m/s)j
vF  2.00 m/s
56.3 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.69
In the position shown, bar DE has a constant angular
velocity of 10 rad/s clockwise. Determine (a) the distance h
for which the velocity of Point F is vertical, (b) the
corresponding velocity of Point F.
SOLUTION
Bar DE:
(Rotation about E )
DE  10 rad/s
 (10 rad/s)k
rD/E  (0.1 m)i  (0.2 m)j
vD  DE  rD /E  (10k )  ( 0.1i  0.2 j)
 (1 m/s)j  (2 m/s)i
Bar FBD: (Plane motion  Translation with D  Rotation about D.)
BD  BDk rB/D  (0.3 m)i  (0.1 m)j
vB  vD   BD  rB/D
 j  2i  (BD k )  (0.3i  0.1j)
 j  2i  0.3BD j  0.1BD i
Bar AB:
(Rotation about A)
 AB   AB k rB/A  (0.42 m)j
vB   AB  rB/A  (AB k )  (0.42 j)  0.42AB i
Equating components of the two expressions for vB ,
(a)
j:
1  0.3BD  0
BD  3.3333 rad/s
i:
2  0.1BD  0.42AB
2  (0.1)(3.3333)  0.42AB
AB  3.9683 rad/s
Bar FBD:
rF/D  CrB/D
AB  3.97 rad/s
where C 
h  0.3
0.3
vF  vB   BD  rF/D
 j  2i  C (0.3BD j  0.1BD i )
 j  2i  C ( j  0.33333i )
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.69 (Continued)
But vF  vF j. Equating components of the two expressions for vF ,
i:
(a)
(b)
0  2  0.33333C
C6
h  0.3C  0.3  (0.3)(6)  0.3
j: vF  j  Cj  (1  6) j 
h  1.500 m 
vF  5.00 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.70
Both 6-in.-radius wheels roll without slipping on
the horizontal surface. Knowing that the distance
AD is 5 in., the distance BE is 4 in. and D has a
velocity of 6 in./s to the right, determine the
velocity of Point E.
SOLUTION
Disk D:
Velocity at the contact Point P with the ground is zero.
v0  6 in./s
D 
vD
6 in./s

 1 rad/s
rD/P
6 in.
D  1 rad/s
vA  rA/PD  (6 in.  5 in.)(1 rad/s)  11in./s
At Point A,
v A  11 in./s
Disk E:
Velocity at the contact Point Q with the ground is zero. ωE  E
 E k.
rB /Q  (4 in.)i  (6 in.) j
v B  v B /Q  ω E  rB /Q  E k  (4i  6 j)
v B  6E i  4E j
(1)
rB /A  ( 142  52 )i  5 j in inches.
Connecting rod AB:
v B /A  171i  5 j
ω AB   AB k
v B  v A  v B /A  v A   AB k  ( 171i  5 j)
 11i  5 AB i  171 AB j
(2)
Equating expressions (1) and (2) for vB gives
6E i  4E j  11i  5 AB i  171 AB j
Equating like components and transposing terms,
5AB  6E  11
i:
j:
(3)
171 AB  4E  0
(4)
Solving the simultaneous equations (3) and (4),
 AB  0.75265 rad/s,
Velocity of Point E.
E  2.4605 rad/s
v E  E k  rE /Q  2.4605k  6 j
v E  14.76 in./s i  14.76 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.71
The 80-mm-radius wheel shown rolls to the left with a velocity
of 900 mm/s. Knowing that the distance AD is 50 mm,
determine the velocity of the collar and the angular velocity of
rod AB when (a)   0, (b)   90.
SOLUTION
(a)   0.
vC  0,
Wheel AD.
 AD 
vD  45 in./s
vD 900

 11.25 rad/s
CD 80
CA  (CD )  ( DA)  80  50  30 mm
vA  (CA)AD  (30)(11.25)  337.5 mm/s
vB  vA  vB/ A
Rod AB.
]  [337.5
[ vB
]  [vB/ A
]
Wheel AD.
vC  0,
tan  
CA 
AD  11.25 rad/s
DA 50
 ,
DC 80
  32.005
DC
 94.34 mm
cos 
vA  (CA) AD  (94.34)(11.25)  1061.3 mm/s
v A  [1061.3 mm/s
Rod AB.
32.005]
v B  vB
sin  

AB  0 
vB/ A  0
(b)   90.
v B  338 mm/s
80
,   18.663
250
Plane motion  Translation with A  Rotation about A.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.71 (Continued)
vB  vA  vB/ A
]  [vA
[vB
 ]  [vB/A
]
Draw velocity vector diagram.
  180    (90   )
 90  32.005  18.663  39.332
Law of sines.
vB/ A
vB
vA


sin  sin  sin (90   )
vB 
v A sin 
(1061.3)sin 39.332

sin (90   )
sin 108.663
 710 mm/s
vB/ A 
vB  710 mm/s

 AB  2.37 rad/s

v A sin 
(1061.3) sin 32.005

sin (90   )
sin108.663
 593.8 mm/s
 AB 
vB/ A
AB

593.8
 2.37 rad/s
250
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.72*
For the gearing shown, derive an expression for the angular velocity C of
gear C and show that C is independent of the radius of gear B. Assume that
Point A is fixed and denote the angular velocities of rod ABC and gear A by
ABC and A respectively.
SOLUTION
Label the contact point between gears A and B as 1 and that between gears B and C as 2.
Rod ABC:
ABC  ABC
Assume
for sketch.
vA  0
vB  (rA  rB )ABC
vC  (rA  2rB  rC )ABC
Gear A:
Gear B:
A  0,
vA  0,
v1  0
v1  vB  rB B  0
(rA  rB )ABC  rB B  0
 rA  rB
 rB
B  

  ABC

v2  vB  rB B
 2(rA  rB )ABC
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PROBLEM 15.72* (Continued)
v2  vC  rC C
Gear C:
2(rA  rB ) ABC  (rA  2rB  rC ) ABC  rCC
C  (rA  rC ) ABC   rC C

C  1 

rA 
  ABC 
rC 
Note that the result is independent of rB .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.73
A juggling club is thrown vertically into the air. The center of
gravity G of the 20 in. club is located 12 in. from the knob.
Knowing that at the instant shown G has a velocity of 4 ft/s
upwards and the club has an angular velocity of 30 rad/s
counterclockwise, determine (a) the speeds of Point A and B,
(b) the location of the instantaneous center of rotation.
SOLUTION
Unit vectors:
Relative positions:
i 1
,
j 1 , k 1
8 
rA/G  (1 ft)i, rB/A   ft  i
 12 
Angular velocity:
ω  30 rad/s
Velocity at A:
vA  v G  v A /G  v G  ω  rA /G
 (30 rad/s)k
 (4 ft/s) j  (30 rad/s)k  ( 1 ft)i
 (4 ft/s) j  (30 ft/s) j  (26 ft/s) j
 26 ft/s
vA  26.0 ft/s 
Velocity at B:
vB  v G  v B/G  v G  ω  rB/G
 8 
 (4 ft/s) j  (30 rad/s)k   ft  i
 12 
 (4 ft/s) j  (20 ft/s) j  (24 ft/s) j
 24 ft/s
vB  24.0 ft/s 
Let rC/ G  x i bet the position of the instantaneous center C relative to G.
vC  vG  vC/G  vG  ω  ( x i )
 (4 ft/s) j  (30 rad/s)k  ( x i )
 (4 ft/s) j  (30 ft/s) x j  0
x
4 ft/s
4
  ft  1.6 in.
30 rad/s
30
Point C lies 1.6 in. to the left of G. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.74
At the instant shown during deceleration, the velocity of an
automobile is 40 ft/s to the right. Knowing that the velocity of the
contact point A of the wheel with the ground is 5 ft/s to the right,
determine (a) the instantaneous center of rotation of the wheel,
(b) the velocity of point B, (c) the velocity of point D.
SOLUTION
 
(a)
vO  vA
40  5

 35 rad/s
l AO
1
lCA 
vA


5
1
 ft
35 7
 1.714 in.
C lies 1.714 in. below A. 
(b)
lCB  2 
1 15
ft

7
7
vB  lCB 
15
 35  75 ft/s
7
v B  75.0 ft/s
(c)
lOD  1 ft, lCO  1 

1 8
 ft
7 7
2
8
lCD  12     1.5186 ft
7
vD  lCD  1.5186  35   53.2 ft/s
tan  
lOD
7
 ,
lCO
8
  41.2
v D  53.2 ft/s
41.2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.75
A helicopter moves horizontally in the x direction at a speed of
120 mi/h. Knowing that the main blades rotate clockwise with an
angular velocity of 180 rpm, determine the instantaneous axis of
rotation of the main blades.
SOLUTION
v0  120 mi/h  176 ft/s
  180 rpm 
(180)(2 )
 18.85 rad/s
60
Top view
v0  z
z
v0


176
 9.34 ft
18.85
Instantaneous axis is parallel to the y axis and passes through the point
x0 
z  9.34 ft 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.76
A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as
shown. One of the drums rolls without sliding on the surface shown, and a
cord is wound around the other drum. Knowing that end E of the cord is
pulled to the left with a velocity of 120 mm/s, determine (a) the angular
velocity of the drums, (b) the velocity of the center of the drums, (c) the
length of cord wound or unwound per second.
SOLUTION
Since the drum rolls without sliding, its instantaneous center lies at D.
v E  v B  120 mm/s
v A  v A /D  ,
(a)

vB  rB/D
vB
120

 3 rad/s
vB/D 100  60
ω  3.00 rad/s 
(b)
vA  (100)(3)  300 mm/s
v A  300 mm/s

Since vA is greater than vB , cord is being wound.
vA  vB  300  120  180 mm/s
(c)
Cord wound per second  180.0 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.77
A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as shown.
One of the drums rolls without sliding on the surface shown, and a cord is
wound around the other drum. Knowing that end E of the cord is pulled to
the left with a velocity of 120 mm/s, determine (a) the angular velocity of the
drums, (b) the velocity of the center of the drums, (c) the length of cord
wound or unwound per second.
SOLUTION
Since the drum rolls without sliding, its instantaneous center lies at B.
v E  v D  120 mm/s
vA  rA/B, vD  rD /B
(a)

vD
120

 3 rad/s
rD /B 100  60
ω  3.00 rad/s
(b)

vA  (60)(3.00)  180 mm/s
v A  180 mm/s

Since v A is to the right and v D is to the left, cord is being unwound.
vA  vE  180  120  300 mm/s
(c)
Cord unwound per second  300 mm 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.78
The spool of tape shown and its frame assembly are pulled upward at a
speed vA  750 mm/s. Knowing that the 80-mm-radius spool has an angular
velocity of 15 rad/s clockwise and that at the instant shown the total
thickness of the tape on the spool is 20 mm, determine (a) the instantaneous
center of rotation of the spool, (b) the velocities of Points B and D.
SOLUTION
v A  750 mm/s
  15 rad/s
x
(a)
vA


750
 50 mm
15
The instantaneous center lies 50 mm to the right of the axle. 
CB  80  20  50  50 mm
(b)
vB  (CB)  (50)(15)  750 mm/s
vB  750 mm/s 
CD  80  50  130 mm
vD  (CD)  (130)(15)  1950 mm/s
vD  1.950 m/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.79
The spool of tape shown and its frame assembly are pulled upward at a
speed vA  100 mm/s. Knowing that end B of the tape is pulled downward
with a velocity of 300 mm/s and that at the instant shown the total thickness
of the tape on the spool is 20 mm, determine (a) the instantaneous center of
rotation of the spool, (b) the velocity of Point D of the spool.
SOLUTION
vD  vA  100 mm/s
(a)
Since v0 and vB are parallel, instantaneous center C is located at intersection of BC and line joining
end points of vD and vB .
Similar triangles.
OC BC OC  BC


v0
vB
v0  vB
OC 
v0
(OC  BC )
v0  vB
OC 
100 mm/s
(100 mm)
(100  300) mm/s
 25 mm
(b)
v
vD
 0 ;
( DO)  (OC ) (OC )
vD
100 mm/s

(80  25) mm
25 mm

vD  420 mm/s 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.80
The arm ABC rotates with an angular velocity of 4 rad/s
counterclockwise. Knowing that the angular velocity of the
intermediate gear B is 8 rad/s counterclockwise, determine
(a) the instantaneous centers of rotation of gears A and C,
(b) the angular velocities of gears A and C.
SOLUTION
Contact points:
1
between gears A and B.
2
between gears B and C.
ABC  4 rad/s
Arm ABC:
vA  (0.300)(4)  1.2 m/s
vC  (0.300)(4)  1.2 m/s
B  8 rad/s
Gear B:
v1  (0.200)(8)  1.6 m/s
v2  (0.100)(8)  0.8 m/s
Gear A:
A 
v1  v A 1.6  1.2

0.100
0.100
A  4 rad/s
A 
vA
A

1.2
 0.3 m  300 mm
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.80 (Continued)
Gear C:
C 
vC  v2 1.2  0.8

0.200
0.2
C  2 rad/s
C 
vC
C

1.2
 0.6 m
2
(a) Instantaneous centers.
Gear A: 300 mm left of A 
Gear C: 600 mm left of C 
(b) Angular velocities.
 A  4.00 rad/s

C  2.00 rad/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.81
The double gear rolls on the stationary left rack R. Knowing that the rack on the
right has a constant velocity of 2 ft/s, determine (a) the angular velocity of the
gear, (b) the velocities of Points A and D.
SOLUTION
Since the rack R is stationary, Point C is the instantaneous center of the double gear.
v B  2 ft/s  24 in./s
Given:
Make a diagram showing the locations of Points A, B, C, and
D on the double gear.
vB  lCB

vB 24 in./s

 2.40 rad/s
10 in.
lCB
ω  2.40 rad/s
(a)
Angular velocity of the gear.
(b)
Velocity of Point A.
vA  l AC  (4 in.)(2.40 rad/s)
Geometry:
lCD  (4 in.)2  (6 in.) 2  52 in.
tan  
Velocity of Point D.
4 in.
6 in.

vA  9.60 in./s  0.800 ft/s 
  33.7
v D  lCD  52(2.40)  17.31 in./s
vD  1.442 ft/s
33.7° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.82
An overhead door is guided by wheels at A and B that roll in
horizontal and vertical tracks. Knowing that when   40 the
velocity of wheel B is 1.5 ft/s upward, determine (a) the angular
velocity of the door, (b) the velocity of end D of the door.
SOLUTION
Locate instantaneous center at intersection of lines drawn perpendicular
to vA and vB.
(a)
Angular velocity.
vB  ( BC )
1.5 ft/s  (3.214 ft)
  0.4667 rad/s
(b)
  0.467 rad/s

Velocity of D:
In CDE:
6.427
 59.2
3.83
6.427
CD 
 7.482 ft
sin 
vD  (CD)
  tan 1
 (7.482 ft)(0.4667 rad/s)
 3.49 ft/s
vD  3.49 ft/s
59.2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.83
Rod ABD is guided by wheels at A and B that roll in horizontal and
vertical tracks. Knowing that at the instant shown   60 and the
velocity of wheel B is 40 in./s downward, determine (a) the angular
velocity of the rod, (b) the velocity of Point D.
SOLUTION
Rod ABD:
We locate the instantaneous center by drawing lines perpendicular to vA and vD.
(a)
Angular velocity.
vB  ( BC )
40 in./s  (12.99 in.)
  3.079 rad/s
(b)
  3.08 rad/s

Velocity of D:
In CDE:
  tan 1
7.5
25.98
 16.1; CD 
 27.04 in.
25.98
cos 
vD  (CD)  (27.04 in.)(3.079 rad/s)  83.3 in./s
vD  83.3 in./s 16.1
vD  83.3 in./s
73.9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.84
Rod BDE is partially guided by a roller at D that moves in a
vertical track. Knowing that at the instant shown the angular
velocity of crank AB is 5 rad/s clockwise and that   25,
determine (a) the angular velocity of the rod, (b) the velocity of
Point E.
SOLUTION
Crank AB:
ω AB  5 rad/s
rB /A  120 mm
vB  AB rB /A  (5)(0.120)
v B  0.6 m/s
Rod BDE: Draw a diagram of the geometry of the rod and note that v B  0.6 m/s
and v D  v D .
Locate Point C, the instantaneous center, by noting that BC is perpendicular to v B and DC is perpendicular to
vD. Calculate lengths of BC and CD.
lBC  500sin 25  211.31 mm
lCD  500 cos 25  453.15.
(a)
Angular velocity of the rod.
BCD 
vB
0.6 m/s

 2.8394 rad/s
lBC 0.21131 m
BCD  2.84 rad/s



Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.84 (Continued)
(b)
Velocity of Point E.
Locate Point F on the diagram.
CF  700 cos 25 mm
FE  200sin 25
FE 200sin 25 2

 tan 25  0.13323
CF 700 cos 25 7
  7.6
  90    82.4
tan  
lCE  (CF ) 2  ( FE )2  640.02 mm  0.64002 m
vE  lCE   (0.64002)(2.8394)
v E  1.817 m/s
82.4° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.85
Rod BDE is partially guided by a roller at D which moves in a
vertical track. Knowing that at the instant shown   30, Point E
has a velocity of 2 m/s down and to the right, determine the
angular velocities of rod BDE and crank AB.
SOLUTION
Crank AB: When AB is vertical, the velocity v B at Point B is horizontal.
Rod BDE: Draw a diagram of the geometry of the rod and note that v B is horizontal and vD is vertical.
Locate Point C, the instantaneous center C, by noting that CB is vertical and CD is horizontal. From the
diagram, with Point F added,
CF  700cos 30 mm
FE  200sin 30 mm
CE  (CF ) 2  ( FE ) 2  614.41 mm  0.61441 m
Angular velocity of rod BDE
BDE 
vE
2 m/s

 3.2552 rad/s
(CE ) 0.61441 m
BDE  3.26 rad/s
Velocity of B.

CB  500sin 30 mm  250 mm  0.250 m
vB  (CB)BDE  (0.250)(3.2552)
v B  0.81379 m/s
Angular velocity of crank AB:
AB  120 mm  0.120 m
 AB 
vB
0.81379 m/s

( AB )
0.120 m
ω AB  6.78 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.86
A motor at O drives the windshield wiper
mechanism so that OA has a constant
angular velocity of 15 rpm. Knowing that
at the instant shown linkage OA is vertical,
= 30°, and = 15°, determine (a) the
angular velocity of bar AB, (b) the velocity
of the center of the center of bar AB.
SOLUTION
 =30,  =15, OA  15
Given:
Geometry:
rev min 2 rad 
 rad/s
min 60 s rev
2
Perpendiculars to vA and vB intersect at
the instantaneous center, C, of link AB.
lBE  100 cos   96.593 mm
lBE  lBC cos 30
lBC  111.536 mm
EAB  75
ECB  60
Therefore:
l AC  lBC cos 60  l AB cos 75
 81.650 mm
Using Law of Cosines for Triangle ACG
2
2
2
lCG
 l AC
 l AG
 2l AC l AG cos 75
 81.652  502  2  81.65  50  cos 75
lCG  83.984
(a) Velocities:
v A  lOAOA  l AC  AB
 AB 
lOAOA
  AB  0.2886 rad/s
l AC
 AB  0.289 rad/s
(b)
vG  lCG  AB  vG  24.236 mm/s
vG  24.236 mm/s

30 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.87
A motor at O drives the windshield wiper
mechanism so that point B has a speed of
20 mm/s. Knowing that at the instant
shown linkage OA is vertical, = 40°, and
= 15°, determine (a) the angular velocity
of bar OA, (b) the velocity of the center of
the center of bar AB.
SOLUTION
 =40,  =15
Given:
vB  20 mm/s
Geometry:
Perpendiculars to vA and vB intersect at
the instantaneous center, C, of link AB.
lBE  100 cos 
 96.593 mm
lBE  lBC cos 40
lBC  126.093 mm
Law of cosines:
2
2
2
lCG
 lBC
 lBG
 2lBC lBG cos 65
lCG  105.674 mm
(a) Velocities:
v B  lBC  AB
 AB 
vB
lBC
 0.159 rad/s
 AB  0.159 rad/s
(b)

vG  lCG  AB
 16.761 mm/s
vG  16.761 mm/s
25 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.88
Rod AB can slide freely along the floor and the inclined
plane. Denoting by vA the velocity of Point A, derive an
expression for (a) the angular velocity of the rod, (b) the
velocity of end B.
SOLUTION
Locate the instantaneous center at intersection of lines drawn perpendicular to vA and vB .
Law of sines.
AC
BC

sin[90  (    )] sin(90   )
l

sin 
AC
BC

cos(    ) cos 
l

sin 
cos(    )
AC  l
sin 
cos 
BC  l
sin 
(a)
Angular velocity:
vA  ( AC )  l
cos(    )

sin 

(b)
Velocity of B:
vB  ( BC )  l
cos 
sin 
vB  vA
v
sin  
  

 l cos(    ) 
vA
sin 


l cos(    )
cos 

cos(    )
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.89
Small wheels have been attached to the ends of bar AB
and roll freely along the surfaces shown. Knowing that
the velocity of wheel B is 7.5 ft/s to the right at the
instant shown, determine (a) the velocity of end A of the
bar, (b) the angular velocity of the bar, (c) the velocity of
the midpoint of the bar.
SOLUTION
Given:
v A  vA
v B  7.5 ft/s
45,
Locate the instantaneous center (point C) of rod AB by noting that velocity
directions at points A and B are known. Draw AC perpendicular to v A and
BC perpendicular to v B.
l  AB  24 in.  2 ft
Let
Law of sines for triangle ABC.
b
a
l


 2.8284 ft
sin 75 sin 60 sin 45
a  2.4495 ft,
 
(a)
b  2.73205 ft
vB
7.5

 2.7452 rad/s
2.73205
b
v A  a   2.4495 2.7452  6.724 ft/s
v A  6.72 ft/s
  2.75 rad/s 
(b)
(c)
45.0 
Let M be the midpoint of AB.
Law of cosines for triangle CMB.
2
l
l
m2  b 2     2b cos 60
 2
2
  2.73205  1   2 2.732051 cos 60
2
2
m  2.3942 ft
Law of sines.
sin 
l
2

sin 60
,
m
sin  
1sin 60
,
2.3942
  21.2
vM  m   2.3942 2.7452  6.573 ft/s,
v M  6.57 ft/s
21.2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.90
Two slots have been cut in plate FG and the plate has been
placed so that the slots fit two fixed pins A and B. Knowing
that at the instant shown the angular velocity of crank DE is
6 rad/s clockwise, determine (a) the velocity of Point F,
(b) the velocity of Point G.
SOLUTION
vE  ( DE )DE  (72 mm)(6 rad/s)
Crank DE:
vE  432 mm/s
(vF ) y  vE  432 mm/s
Rod EF:
Plate FG:
vA and vB are velocities of points on the plate next to the pins A and B. We draw lines perpendicular to
vA and vB to locate the instantaneous center C.
(a)
Velocity of Point F:
vF  (CF )
(vF ) y  [(CF ) ]cos   [(CF ) cos  ]
But
(vF ) y  432 mm/s and (CF ) cos   480 mm:
432 mm/s  (480 mm)
  0.9 rad/s
  0.9 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.90 (Continued)
CF  487.97 mm
  10.37
vF  (CF )
 (487.97 mm)(0.9 rad/s)
 439.18 mm/s
vF  439 mm/s
(b)
10.4°
v F  439 mm/s
79.6° 
vG  411 mm/s
20.5° 
Velocity of Point G:
CG  456.78 mm
  20.50
vG  (CG )
 (456.78 in.)(0.9 rad/s)
vG  411.11 mm/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.91
The disk is released from rest and rolls down the incline.
Knowing that the speed of A is 1.2 m/s when   0, determine
at that instant (a) the angular velocity of the rod, (b) the velocity
of B. Only portions of the two tracks are shown.
SOLUTION
Draw the slider, rod, and disk at   0.
Let Point P be the contact point between the disk and the incline. It is the instantaneous center of the disk.
vA is parallel to the incline. So that
vA  vA
30°
Constraint of slider:
vB  vB
To locate the instantaneous center C of the rod AB, extend the line AP to
meet the vertical line through B at Point C.
l AC  l AB / sin 30
lBC  l AB / tan 30
(a)
Angular velocity of rod AB.
 AB 
vA v A sin 30 (1.2 m/s)sin 30


l AC
l AB
0.6 m
ω AB  1.000 rad/s
(b)
Velocity of Point B.

vB  lBC AB
vB 
l AB v A sin 30
 v A cos 30  1.2 cos 30
tan 30
l AB
v B  1.039 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.92
The pin at B is attached to member ABD and can slide freely along the slot
cut in the fixed plate. Knowing that at the instant shown the angular velocity
of arm DE is 3 rad/s clockwise, determine (a) the angular velocity of
member ABD, (b) the velocity of point A.
SOLUTION
 DE  3 rad/s
vD   DE   DE  160  3
 480 mm/s
v D is perpendicular to DE.
v B  vB
Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are
known. Draw BC perpendicular to v B and DC perpendicular to v D .
BD  120 mm,
cos  
DK
120 cos30

,
ED
160
DK   BD  cos 30  120 cos 30
  49.495,
  180  30    100.505
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.92 (Continued)
Law of sines for triangle BCD.
CD
BC
BD


sin 30 sin 
sin 
CD 
BC 
 BD  sin 30
sin 
 BD  sin 
sin 
Law of sines.
(a)
120 sin 30
 78.911 mm
sin 
120 sin 
 155.177 mm
sin 
 AC 2   BC 2   AB 2  2  AB  BC  cos150
Law of cosines for triangle ABC.
 AC 2


 155.177 2  2002   2 155.177  200  cos150,
sin 
sin150

AB
AC
 ABD 
sin  
AC  343.27 mm
200 sin150
,
343.27
  16.9
vD
480

 6.0828 rad/s
CD
78.911
 ABD  6.08 rad/s
(b)

v A   AC   ABD   343.27  6.0828   2088 mm/s
v A  2.09 m/s
73.1 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.93
Two identical rods ABF and DBE are connected by a pin at B. Knowing
that at the instant shown the velocity of point D is 200 mm/s upward,
determine the velocity of (a) point E, (b) point F.
SOLUTION
vB   AB   ABF ,
v B  vB
75
v D  200 mm/s
Locate the instantaneous center (point C) of bar DBE by noting that the velocity directions at points B and D
are known. Draw BC perpendicular to v B and DC perpendicular to v D .
CD
BC
BD


sin150 sin15 sin15
Law of sines for triangle BCD.
BC  BD  180 mm
 DBE 
CD 
180sin150
 347.73 mm
sin15
vD
200

 0.57515 rad/s
CD
347.73
vB   BC   DBE  180  0.57515   103.528 mm/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.93 (Continued)
 ABF 
vB
103.528

 0.57515 rad/s
AB
180
vF   AF   ABF   300  0.57515   172.546 mm/s
 CE 2   CD 2   DE 2  2  CD  DE  cos15
Law of cosines for triangle DCE.
 CE 2
 347.732  3002   2  347.73 300  cos15,
CE  96.889 mm
EH  DE sin 15  300 sin15
cos  
(a)
(b)
EH
300 sin15

CE
96.889
  36.7
vE   CE   BCD   96.889  0.57515    55.7  mm/s,
v E  55.7 mm/s
36.7 
v F  172.5 mm/s
75.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.94
Arm ABD is connected by pins to a collar at B and to crank DE.
Knowing that the velocity of collar B is 16 in./s upward, determine
(a) the angular velocity of arm ABD, (b) the velocity of point A.
SOLUTION
v B  16 in./s
tan  
EF
5

DF 12
v D  vD

Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are
known. Draw BC perpendicular to v B and DC perpendicular to v D .
5
CJ   DJ  tan    6.4     2.6667 in.,
 12 
CB  JB  CJ  12.8  2.6667  10.1333 in.
(a)
 ABD 
vB
16

 1.57895 rad/s
CB 10.1333
 ABD  1.579 rad/s

CK  CB  BK  10.1333  7.2  17.3333 in.
tan  
KA
3.6
,   11.733,

CK 17.3333
AC 
(b)
90    78.3
CK
17.3333

 17.7032 in.
cos 
cos11.733
v A   AC   ABD  17.7032 1.57895   28.0 in./s,
v A  28.0 in./s
78.3 
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PROBLEM 15.95
Two 25-in. rods are pin-connected at D as shown. Knowing that B
moves to the left with a constant velocity of 24 in./s, determine at
the instant shown (a) the angular velocity of each rod, (b) the
velocity of E.
SOLUTION
Rod AB:
Draw lines perpendicular to vA and vB to locate instantaneous center CAB .
vB  ( BC AB ) AB
24 in./s  (20 in.) AB


Velocity of D:

 AB  1.200 rad/s
DC AB  12.5 in.
vD  ( DC AB ) AB
 (12.5 in.)(1.2 rad/s)
vD  15 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.95 (Continued)
Rod DE:
Draw lines perpendicular to vD and vE to locate instantaneous center CDE .
DCDE  (20) 2  (26.667) 2  33.333 in.
vD  ( DCDE ) DE
(a)
15 in./s  (33.333 in.)DE ; DE  0.45 rad/s
DE  0.450 rad/s

vE  ( ECDE )DE  (11.667 in.)(0.45 rad/s)
(b)
vE  5.25 in./s
vE  5.25 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.96
Two rods ABD and DE are connected to three collars as shown. Knowing
that the angular velocity of ABD is 5 rad/s clockwise, determine at the
instant shown (a) the angular velocity of DE, (b) the velocity of collar E.
SOLUTION
ABC  5 rad/s
v B  vB
v A  vA
v E  vE
Locate Point I, the instantaneous center of rod ABD by drawing IA perpendicular to vA and IB perpendicular
to vB.
200
  26.565
400
400
I ID 
 447.21 mm
cos 
vD   ABD lID  (5)(447.21 mm)
tan  
vD  2236.1 mm/s

Locate Point J, the instantaneous center of rod DE by drawing JD perpendicular to vD and JE perpendicular
to vE.
l JD 
DE 
400
 447.21 mm
cos 
vD 2236.1 mm/s

 5 rad/s
447.21
lJD
DE  5.00 rad/s 
(a)
lJE  200  lJD cos   600 mm
vE  lJE DE  (600)(5)  3000 mm/s
(b)
v E  3.00 m/s 
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PROBLEM 15.97
At the instant shown, the velocity of collar A is 0.4 m/s to the
right and the velocity of collar B is 1 m/s to the left. Determine
(a) the angular velocity of bar AD, (b) the angular velocity of
bar BD, (c) the velocity of point D.
SOLUTION
Method 1
Assume v D has the direction indicated by the angle  as shown.
Draw CDI perpendicular to v D. Then, point C is the instantaneous
center of rod AD and point I is the instantaneous center of rod BD.
 0.27 2   0.36 2
AD 
Geometry.
BD 
sin  
 0.182   0.1352
0.27
 0.6,
0.45
c
0.45
0.45


sin 
sin  90    cos 
c
 0.225 m
sin  
0.18
 0.8
0.225
d
0.225
0.225


sin 
sin  90    cos 
0.45sin 
cos 
d 
a  0.36  0.27 tan 
 0.45 m
0.225sin 
cos 
b  0.135  0.18 tan 
Kinematics. vA  0.4 m/s, vB  1 m/s
vA
v
v
v
 D,
 BD  B  D
a
c
b
d
cv
dv
vD  A  B
a
b
cv
0.45sin 
cos 
0.4
a  Ab 
b  0.6b


dvB
cos 
0.225sin  1
 AD 
0.36  0.27 tan   0.6  0.135  0.18 tan  
0.279  0.378 tan  ,
tan   0.73809,
  36.43
a  0.36   0.27  0.73809   0.1607 m
b  0.135   0.18  0.73809   0.2679 m
c 
 0.45  0.6 
cos 
 0.3356 m
d 
 0.225  0.8 
cos 
 0.2237 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.97 (Continued)
(a )
 AD 
vA
0.4

 2.4891 rad/s
a
0.1607
AD  2.49 rad/s

(b)
 BD 
vB
1

 3.733 rad/s
b
0.2679
BD  3.73 rad/s

(c)
vD  c AD   0.3356  2.4891  0.835 m/s,
v D  0.835 m/s
53.6 
Method 2
Consider the motion using a frame of reference that is translating with
collar A. For motion relative to this frame.
v A  0.4 m/s
v B  1 m/s
v A/ A  0,
v B/ A  1.4 m/s
0.27
  36.87
,
0.36
0.36
 0.45 m
AD 
cos

v D/ A  0.45 AD
tan  
Locate the instantaneous center (point C) for the relative motion of bar
BD by noting that the relative velocity directions at points B and D are
known. Draw BC perpendicular to v B/A and DC perpendicular to v D/A.
0.18
 0.3 m
sin 
BC   CD  cos  0.135  0.375 m
CD 
 BD 
vB/ A
CB

1.4
 3.73 rad/s
0.375
vD/ A   CD   BD   0.3 3.73  1.120 m/s
(a )
 AD 
vD/ A
AD

1.120
0.45
(b)
(c)
v D  v A  v D/ A   0.4 m/s
 0.835 m/s
53.6
AD  2.49 rad/s

BD  3.73 rad/s

  1.120
v D  0.835 m/s

53.6  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.98
Two rods AB and DE are connected as shown. Knowing that
Point D moves to the left with a velocity of 40 in./s, determine
(a) the angular velocity of each rod, (b) the velocity of Point A.
SOLUTION
We locate two instantaneous centers at intersections of lines drawn as follows:
C 1:
For rod DE, draw lines perpendicular to vD and vE.
C 2:
For rod AB, draw lines perpendicular to vA and vB.
Geometry:
BC1  (8 in.) 2  8 2 in.
DC1  16 in.
BC2  (9 in.  8 in.) 2  17 2 in.
AC2  25 in.
(a)
Rod DE:
vD  ( DC1 ) DE
40 in./s  (16 in.) DE
DE  2.5 rad/s
DE  2.5 rad/s
vB  ( BC ) DE
 (8 2 in.)(2.5 rad/s)
vB  20 2 in./s
45°
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.98 (Continued)
Rod AB:
vB  ( BC2 ) AB
20 2 in./s  (17 2 in.) AB
 AB 
20
rad/s  1.1765 rad/s
17
 AB  1.177 rad/s
(b)

vA  ( AC2 ) AB
 (25 in.)(1.1765 rad/s)
vA  29.41in./s
vA  29.4 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.99
Describe the space centrode and the body centrode of rod ABD of Problem
15.83. (Hint: The body centrode need not lie on a physical portion of the
rod.)
PROBLEM 15.83 Rod ABD is guided by wheels at A and B that roll in
horizontal and vertical tracks. Knowing that at the instant shown   60
and the velocity of wheel B is 40 in./s downward, determine (a) the
angular velocity of the rod, (b) the velocity of Point D.
SOLUTION
Draw x and y axes as shown with origin at the intersection of the two
slots. These axes are fixed in space.
vA  vA
vB  vB
,
Locate the space centrode (Point C ) by noting that velocity directions
at Points A and B are known. Draw AC perpendicular to vA and BC
perpendicular to vB.
 The coordinates of Point C are xC  l sin  and yC  l cos 

xC2  yC2  l 2  (15 in.)2

The space centrode is a quarter circle of 15 in. radius centered at O. 

Redraw the figure, but use axes x and y that move with the body. Place
origin at A.


xC  ( AC ) cos 

 l cos 2  

yC  ( AC )sin 
l
(1  cos 2 )
2
 l cos  sin  
2

l
sin 2 
2
2
l

l
2
2
2
2
 xC  2   yC   2   ( xC  7.5)  yC  7.5


 
The body centrode is a semicircle of 7.5 in. radius centered
midway between A and B.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.100
Describe the space centrode and the body centrode of the gear of
Sample Problem 15.6 as the gear rolls on the stationary horizontal
rack.
SOLUTION
Let Points, A, B, and C move to A, B, and C  as shown.
Since the instantaneous center always lies on the fixed lower rack, the space centrode is the lower rack.
space centrode: lower rack 
Since the point of contact of the gear with the lower rack is always a point on the circumference of the gear,
the body centrode is the circumference of the gear.
body centrode: circumference of gear 
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PROBLEM 15.101
Using the method of Section 15.3, solve Problem 15.60.
PROBLEM 15.60 In the eccentric shown, a disk of 2-in.-radius
revolves about shaft O that is located 0.5 in. from the center A of the
disk. The distance between the center A of the disk and the pin at B is
8 in. Knowing that the angular velocity of the disk is 900 rpm
clockwise, determine the velocity of the block when   30.
SOLUTION
(OA)  0.5 in. OA  900 rpm  30 rad/s
vA  (OA)OA  (0.5)(30 )  15 in./s
vA  vA
60,
vB  vA
Locate the instantaneous center (Point C ) of bar BD by noting that velocity directions at Point B and A are
known. Draw BC perpendicular to vB and AC perpendicular to vA.
sin  
(OA)sin 30 0.5sin 30

,   1.79
AB
8
OB  (OA) cos30  ( AB) cos   0.5cos30  8cos 
 8.4291 in.
AC 
OB
8.4291
 OA 
 0.5  9.2331 in.
cos30
cos30
BC  (OB ) tan 30  4.8665 in.
 AB 
vA
v
 B
AC BC
(4.8665)(15 )
 BC 
 24.84 in./s
vB  
 vA 
AC
9.2331


vB  24.8 in./s
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
PROBLEM 15.102
Using the method of Section 15.3, solve Problem 15.64.
PROBLEM 15.64 In the position shown, bar AB has an angular velocity
of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.
SOLUTION
ω AB  4 rad/s
Bar AB: (Rotation about A)
vB   AB ( AB )  (4)(175)
AB  175 mm
v B  700 mm/s
ωDE  DE
Bar DE: (Rotation about E )
DE  (275) 2  (75) 2  285.04 mm
v D  285.04DE
tan  

75 mm
 0.27273
275 mm
v B  700 mm/s ,
Bar BD:
v D  285.04DE

Locate the instantaneous center of bar BD by drawing line BC perpendicular to vB and line DC perpendicular
to vD.
BD  200 mm
CB 
BD
(200)(275)

 733.3 mm
tan 
75
CD 
BD
(200)(285.04)

 760.11 mm
sin 
75
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PROBLEM 15.102 (Continued)
BD 
vB
700 mm/s

 0.95455 rad/s
CB 733.33 mm
vD  BD (CD )  (0.95455 rad/s)(760.11 mm)  725.56 mm/s
DE 
Angular velocities:
vD
725.56

 2.5455 rad/s
285.04 285.04
BD  0.955 rad/s

DE  2.55 rad/s

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.103
Using the method of Section 15.3, solve
Problem 15.65.
PROBLEM 15.65 Linkage DBEF is part
of a windshield wiper mechanism, where
points O, F and D are fixed pinned
connections. At the position shown, =
60° and link EB is horizontal. Knowing
that link EF has a counterclockwise angular
velocity of 4 rad/s at the instant shown,
determine the angular velocity of links EB
and DB.
SOLUTION
Given:
 =60
EF  4 rad/s
Geometry:
30cos   20cos    =41.41
  90     =30.0
  90     =48.59
CEB  180    CEB  131.41
  180  131.41  30.0    18.590
Law of sines:
lBC
lEC
85 mm


sin131.41
sin 
sin 30.0
lBC  199.97 mm
lEC  133.31 mm
Velocity of E:
v E  lEF  EF  lEC  EB
 EB 
lEF  EF
lEC
 0.600 rad/s
EB  0.600 rad/s
Velocity of B:

v B  lBD BD  lBC  EB
 BD 
lBC  EB
lBD
 4.000 rad/s
BD  4.000 rad/s 
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PROBLEM 15.104
Using the method of section 15.3, solve Problem 15.38.
PROBLEM 15.38 An automobile travels to the right at a
constant speed of 48 mi/h. If the diameter of a wheel is 22 in.,
determine the velocities of Points B, C, D, and E on the rim of
the wheel.
SOLUTION
vA  48 mi/h  70.4 ft/s
vC  0 
d  22 in., r 
1
d  11 in.  0.91667 ft
2
Point C is the instantaneous center.
vA
70.4

 76.8 rad/s
r
9.1667
CB  2r  1.8333 ft

vB  (CB )  (1.8333)(76.8)  140.8 ft/s
vB  140.8 ft/s

1
2
  (30)  15
CD  2r cos15  (2)(0.91667) cos15  1.7709 ft
vD  (CD)  (1.7709)(76.8)  136.0 ft/s
vD  136.0 ft/s
15.0° 
CE  r 2  0.91667 2  1.2964 ft
vE  (CE )  (1.2964)(76.8)  99.56 ft/s
vE  99.6 ft/s
45.0° 
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PROBLEM 15.105
A 5-m steel beam is lowered by means of two cables
unwinding at the same speed from overhead cranes. As the
beam approaches the ground, the crane operators apply brakes
to slow the unwinding motion. At the instant considered the
deceleration of the cable attached at B is 2.5 m/s 2 , while that
of the cable attached at D is 1.5 m/s 2. Determine (a) the
angular acceleration of the beam, (b) the acceleration of points
A and E.
SOLUTION
 
  0,
,
a B  2.5 m/s 2 ,

a D  1.5 m/s 2
a D  aB  a D/B
1.5
(a )
(b)
  2.5
rD / B  2 m
t   aD/B n
   2    2  0 2
1.5  2.5  2

a A  a B  a A/B
  2.5


  0.500 rad/s 2

t   a A/B n
  1.5 0.5   0

 3.25 m/s2
a A  3.25 m/s 2

a E  a B  a E/ B
  2.5

t   a E/B n
   2  1.5 0.5
   0

 0.75 m/s2
a E  0.750 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.106
For a 5-m steel beam AE the acceleration of point A is 2 m/s 2
downward and the angular acceleration of the beam is
1.2 rad/s 2 counterclockwise. Knowing that at the instant
considered the angular velocity of the beam is zero, determine
the acceleration (a) of cable B, (b) of cable D.
SOLUTION
a A  2 m/s 2 ,
rB/ A  1.5 m
  1.2 rad/s 2
,
rD/ A  3.5 m
0
(a )

a B  a A  a B/ A
 2
t   a B/A n
  1.51.2 
2
  1.5  0 



 0.2 m/s2
a B  0.200 m/s 2
(b)

a D  a A  a D/ A
 2

t   a D/A n
   3.51.2     3.5 0 2


 2.2 m/s2
a D  2.20 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.107
A 900-mm rod rests on a horizontal table. A force P applied as
shown produces the following accelerations: aA  3.6 m/s 2 to
the right,   6 rad/s 2 counterclockwise as viewed from above.
Determine the acceleration (a) of Point G, (b) of Point B.
SOLUTION



]
aG  [3.6 m/s 2
]  [(0.45 m)(6 rad/s 2 )
aG  [3.6 m/s 2
]  [2.7 m/s 2
aG  0.9 m/s 2

a B  1.8 m/s 2

]
a B  [3.6 m/s 2
]  [(0.9 m)(6 rad/s 2 )
a B  [3.6 m/s 2
]  [5.4 m/s 2

]
]
]  [( AB )
a B  a A  a B/ A  [a A
(b)

]  [( AG )
a G  a A  a G/ A  [ a A
(a)
]
]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.108
In Problem 15.107, determine the point of the rod that (a) has
no acceleration, (b) has an acceleration of 2.4 m/s2 to the right.
SOLUTION
(a)
For aQ  0:
 ( AQ )
aQ  a A  aQ/A  a A
0  3.6 m/s2
y
 ( y )(6 rad/s 2 )
3.6 m/s 2
 0.6 m
6 rad/s
a  0 at 0.6 m from A 
(b)
For aQ  2.4 m/s 2
:
]  [( AQ )
]
]  [( y )(6 rad/s 2 )
]
aQ  a A  aQ /A  [aA
2.4 m/s2
 [3.6 m/s 2
1.2 m/s2
 ( y )(6 rad/s 2 )
y  0.2 m
a  2.4 m/s2
at 0.2 m from A 
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PROBLEM 15.109
Knowing that at the instant shown crank BC has a constant angular
velocity of 45 rpm clockwise, determine the acceleration (a) of Point A,
(b) of Point D.
SOLUTION
Geometry.
Let  be angle BAC.
sin  
Velocity analysis.
4 in.
  30
8 in.
BC  45 rpm
 4.7124 rad/s
vA  vA
vB  ( BC )BC  (4)(4.7124)  18.8496 in./s
vB  18.8496 in./s
vA and vB are parallel; hence, the instantaneous center of rotation of rod AD lies at infinity.
 AD  0
Acceleration analysis.
Crank BC:
vA  vB  18.8496 in./s
 BC  0
( aB )t  ( BC )  0
2
( aB ) n  ( BC ) BC
 (4)(4.7124) 2
a B  88.827 in./s 2
Rod ABD:
 AD   AD
a A  aA
a A  a B  (a A/B )t  (a A/B ) A
[a A ]  [88.827
]  [8 AD
2
30]  [8 AD
60]
Resolve into components.
:
(a)
0  88.827  8 AD cos 30  0
 AD  12.821 rad/s 2
: a A  8 AD sin 30  (8)( 12.821)sin 30  51.284 in./s 2
a A  51.3 in./s 2 


Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.109 (Continued)
(b)
a D  a B  (a D/B )t  (a D/B )t
 [88.827
]  [8 BD
 [88.827
]  [(8)(12.821)
 [88.827
]  [102.568
2
30]  [8BD
[
60]
30]  0
30]  [177.653
 51.284 ]
a D  184.9 in./s 2
16.1 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.110
End A of rod AB moves to the right with a constant velocity of 6 ft/s. For the
position shown, determine (a) the angular acceleration of rod AB, (b) the
acceleration of the midpoint G of rod AB.
SOLUTION
Use units of ft and seconds.
i 1
Geometry and unit vectors:
,
j 1 ,
k 1
rB /A  (10 cos 30)i  (10sin 30) j
rB/D  4 j
Velocity analysis.
v A  6 ft/s
Rod AB:
,
v B  vB
Since vA and vB are parallel, the instantaneous center lies at infinity, so AB  0 and vB  v A .
Acceleration analysis.
α AB   AB
Rod AB: a A  0 since vA is constant.
  AB k
2
a B  a A  a B /A  a A  a AB  rB /A   AB
rB /A
 0   AB k  (10cos 30i  5 j)  0
a B  (10 cos 30) AB j  5 AB j
Rod BD:
(1)
a D  0, α BD   BDk
2
a B  a D  a B /D  0  α BD  rB /D   BD
rB /D
  BD k  (4 j)  (1.5)2 (4 j)
 4 BD i  9 j
(2)
Equating the coefficients of j in the expressions (1) and (2) for a B .
(10cos30) AB  9
 AB  1.0392
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.110 (Continued)
(a)
Angular acceleration of rod AB:
(b)
Acceleration of midpoint G of rod AB.
α AB  1.039 rad/s 2

2
aG  a A  aG /A  a A  α AB k  rG /A   AB
rG /A
 0  1.0392k  ( 5cos 30i  5sin 30 j)
aG  (2.60 ft/s 2 )i  (4.50 ft/s 2 ) j  5.20 ft/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
60°
PROBLEM 15.111
An automobile travels to the left at a constant speed of 72 km/h.
Knowing that the diameter of the wheel is 560 mm, determine
the acceleration (a) of Point B, (b) of Point C, (c) of Point D.
SOLUTION
vA  72 km/h 
h
1000 m

 20 m/s
3600 s
km
Rolling with no sliding, instantaneous center is at C.
vA  ( AC ) ; 20 m/s  (0.28 m)
  71.429 rad/s
Acceleration.
Plane motion  Trans. with A  Rotation about A
aB /A  aC /A  aD /A  r 2  (0.280 m)(71.429 rad/s) 2  1428.6 m/s 2
(a)
a B  a A  a B /A  0  1428.6 m/s 2
a B  1430 m/s 2 
(b)
aC  a A  aC /A  0  1428.6 m/s 2
aC  1430 m/s 2 
(c)
a D  a A  a D /A  0  1428.6 m/s 2

60
a D  1430 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
60 
PROBLEM 15.112
The 18-in.-radius flywheel is rigidly attached to a 1.5-in. -radius
shaft that can roll along parallel rails. Knowing that at the instant
shown the center of the shaft has a velocity of 1.2 in./s and an
acceleration of 0.5 in./s2 , both directed down to the left, determine
the acceleration (a) of Point A, (b) of Point B.
SOLUTION
Velocity analysis.
Let Point G be the center of the shaft and Point C be the point of contact with the rails. Point C is the
instantaneous center of the wheel and shaft since that point does not slip on the rails.
vG  r ,  
vG 1.2

 0.8 rad/s
r 1.5
Acceleration analysis.
Since the shaft does not slip on the rails,
aC  aC
20
aG  [0.5 in./s 2
Also,
20]
aC  aG  (aC/G )t  (aC/G )n
[aC
Components
(a)
20]  [0.5 in./s 2
20:
0.5  1.5
20]  [1.5
20]  [1.5 2
20]
  0.33333 rad/s 2
Acceleration of Point A.
aA  aG  (aA/G )t  (aA/G )n
 [0.5
(b)
20]  [18
]  [18 2 ]
 [0.4698
]  [0.1710 ]  [6
 [6.4698
]  [11.670 ]
]  [11.52 ]
a A  13.35 in./s 2
61.0 
Acceleration of Point B.
a B  aG  (a B/G )t  (a B/G )n
 [0.5
20]  [18
]  [18 2 ]
 [0.4698
]  [0.1710 ]  [6
 [5.5302
]  [11.349 ]
]  [11.52 ]
a B  12.62 in./s 2
64.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.113
A 3-in.-radius drum is rigidly attached to a 5-in.-radius drum as shown.
One of the drums rolls without sliding on the surface shown, and a cord is
wound around the other drum. Knowing that at the instant shown end D of
the cord has a velocity of 8 in./s and an acceleration of 30 in./s2, both
directed to the left, determine the accelerations of Points A, B, and C of the
drums.
SOLUTION
Velocity analysis.
v D  vA  8 in./s
Instantaneous center is at Point B.
vA  ( AB), 8  (5  3)
  4 rad/s
Acceleration analysis.
a B  [aB ] for no slipping.
 
a A  [30 in./s 2
aG  [aG
]  [(aA )n ]
]
aB  a A  (aB/A )t  (aB/A )n
[aB ]  [30
Components
:
]  [(aA )n ]  [(5  3)
0  30  2
]  [5  3) 2 ]
  15 rad/s2
aB  aG  (a B /G )t  (a B / G )n
[aB ]  [aG
Components
:
:
]  [5
0   aG  5
]  [5 2 ]
aG  5  75 in./s 2
aB  (5)(4)2  80 in./s 2
a B  80.0 in./s 2 
a A  aG  (a A/G )t  (a A/G )n
 [75
]  [3
]  [3 2 ]
 [75
]  [45
]  [48 ]
 [30 in./s 2
]  [48 in./s 2 ]
a A  56.6 in./s 2
58.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.113 (Continued)
aC  aG  (aC/G )t  (aC/G )n
 [75
]  [5 ]  [5 2
 [75
]  [75 ]  [80
 [155 in./s 2

]
]
]  [75 in./s 2 ]
aC  172.2 in./s 2
25.8 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.114
A 3-in.-radius drum is rigidly attached to a 5-in.-radius drum as shown. One
of the drums rolls without sliding on the surface shown, and a cord is wound
around the other drum. Knowing that at the instant shown end D of the cord
has a velocity of 8 in./s and an acceleration of 30 in./s2 , both directed to the
left, determine the accelerations of Points A, B, and C of the drums.
SOLUTION
Velocity analysis.
vD  vB  8 in./s
Instantaneous center is at Point A.
vB  ( AB),
8  (5  3)
  4 rad/s
Acceleration analysis.
a A  [a A ] for no slipping.   
a B  [30 in./s 2
aG  [aG
]  [(aB )n ]
]
a A  a B  (a A/B )t  (a A/B )n
[a A ]  [30
Components
:
]  [(aB )n ]  [(5  3)
0  30  2
]  [(5  3)] 2 ]
  15 rad/s2
a A  aG  (a A/G )t  (a A/G )n
aA  [aG
Components
:
:
]  [3
0  aG  3
]  [3 2 ]
aG  3  45 in./s 2
a A  3 2  (3)(4) 2  48 in./s 2
a A  48.0 in./s 2 
a B  aG  (a B/G )t  (a B/G )n
 [45
]  [5
]  [5 2 ]
 [45
]  [75
]  [80 ]
 [30 in./s 2
]  [80 in./s 2 ]
a B  85.4 in./s 2
69.4 
PROBLEM 15.114 (Continued)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
aC  aG  (aC/G )t  (aC/G )n
 [45
]  [5 ]  [5 2
 [45
]  [75 ]  [80
 [35 in./s 2
]
]
]  [75 in./s 2 ]
aC  82.8 in./s 2
65.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.115
A heavy crate is being moved a short distance using three
identical cylinders as rollers. Knowing that at the instant shown
the crate has a velocity of 200 mm/s and an acceleration of
2
400 mm/s , both directed to the right, determine (a) the angular
acceleration of the center cylinder, (b) the acceleration of point
A on the center cylinder.
SOLUTION
Geometry.
Let point C be the center of the center cylinder, B its contact point with
the crate, and D its contact point with the ground. Let r be the radius of the
cylinder. r  100 mm.
Since the contacts at B and D are rolling contacts without slipping,
vB  200 mm/s
and v D  0. Point D is the instantaneous center of
rotation.
Velocity analysis.

vB 200 mm/s

 1 rad/s
2r
200 mm
Point C moves on a horizontal line.
Acceleration analysis.
aC  [aC
a D  aC  (a D/C )t  (a D/C ) n  [aC
Component
]  [r
]  [r 2 ]
  aC  r
:
a B  aC  (a B/C )t  (a B/C ) n  [aC
Component
]
]  [r
(1)
]  [r 2 ]
400 mm/s 2  aC  r
:
(2)
Solving (1) and (2) simultaneously,
aC  200 mm/s 2
r  200 mm/s 2
(a)

200 mm/s 2
100 mm
  2.00 rad/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.115 (Continued)
a A  aC  (a A/C )t  (a A/C ) n  [aC
(b)
]  [r ]  [r 2
]
 [200 mm/s 2
]  [200 mm/s 2 ] + [(100 mm) (1 rad/s) 2
 [100 mm/s 2
]  [200 mm/s 2 ]
aA  223.6 mm/s
]
a A  0.224 m/s 2
63.4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.116
A wheel rolls without slipping on a fixed cylinder. Knowing that at
the instant shown the angular velocity of the wheel is 10 rad/s
clockwise and its angular acceleration is 30 rad/ s 2 counterclockwise,
determine the acceleration of (a) Point A, (b) Point B, (c) Point C.
SOLUTION
r  0.04 m
Velocity analysis.
ω  10 rad/s
Point C is the instantaneous center of the wheel.
v A  [(r )
]  [(0.04)(10)
]  0.4 m/s
]
  30 rad/s2
Acceleration analysis.
  R  r  0.16  0.04  0.2 m.
Point A moves on a circle of radius
aC  aC
Since the wheel does not slip,
a C  a A  (a C / A ) t  (a C / A ) n
[aC ]  [(aA )t
 v2
] A


  [r


  [(0.04)(30)

 [(aA )t
 (0.4) 2
] 
 0.2
 [(aA )t
]  [0.8 ]  [1.2
Components.
]  [r 2 ]
]  [(0.04)(10) 2 ]
]  [4 ]
:  ( a A )t  1.2  0
(a A )t  1.2 m/s 2
: aC  0.8  4.0
aC  3.2 m/s 2
(a) Acceleration of Point A.
a A  [1.2 m/s 2
]  [0.8 m/s 2 ]
a A  1.442 m/s 2
33.7 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.116 (Continued)
a B  a A  (a B/ A )t  (a B/ A ) n
(b) Acceleration of Point B.
aB  [1.2
 [1.2
]  [0.8 ]  [r
]  [r 2
]
]  [0.8 ]  [(0.04)(30) ]  [(0.04)(10)2
 [2.8 m/s 2
]  [2 m/s 2 ]
(c) Acceleration of Point C.
aC  aC
]
a B  3.44 m/s 2
35.5 
aC  3.20 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.117
The 100 mm radius drum rolls without slipping on a portion of a
belt that moves downward to the left with a constant velocity of
120 mm/s. Knowing that at a given instant the velocity and
acceleration of the center A of the drum are as shown, determine
the acceleration of Point D.
SOLUTION
Velocity analysis.
v A  v B  v A/B
[180 mm/s
Components
]  [(100 mm)
]  [120 mm/s
]
:
180  120  100
  3 rad/s
Acceleration analysis.
Point A moves on a path parallel to the belt. The path is assumed to be straight.
a A  720 mm/s 2
30
Since the drum rolls without slipping on the belt, the component of acceleration of Point B on the drum
parallel to the belt is the same as the belt acceleration. Since the belt moves at constant velocity, this
component of acceleration is zero. Thus
a B  aB
Let the angular acceleration of the drum be 
60
.
a B  a A  (a B/ A )t  (a B/ A ) n
[aB
Components
:
]  [720
]  [r
]  [r 2
]
0  720  100
  7.2 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.117 (Continued)
Acceleration of Point D.
a D  a A  (a D/ A )t  (a D/ A ) n
 [a A
30]  [r
 [720
Components:
30:
Components:
60:
60]  [r 2
30]  [(100)(7.2)
30]
60]  [(100) (3) 2
30]
 720  900  180 mm/s 2
720 mm/s 2
aD  1802  7202  742.16 mm/s 2
tan  
720
180
  76.0
  30  46.0
a D  742 mm/s 2
46.0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.118
In the planetary gear system shown the radius of gears A, B, C, and D is
3 in. and the radius of the outer gear E is 9 in. Knowing that gear A has a
constant angular velocity of 150 rpm clockwise and that the outer gear E is
stationary, determine the magnitude of the acceleration of the tooth of gear
D that is in contact with (a) gear A, (b) gear E.
SOLUTION
Velocity.
T  Tooth of gear D in contact with gear A
vT  rA  (3 in.)A
Gears:
Since vE  0, E is instantaneous center of gear D.
vT  2rD
(3 in.)A  2(3 in.)D
1
2
D   A
1
vD  rD  (3 in.)  A  (1.5 in.)A
2
Spider:
vD  (6 in.)S
(1.5 in.)A  (6 in.)S
1
4
S   A
 A  150 rpm  15.708 rad/s
1
 D   A  7.854 rad/s
2
1
S   A  3.927 rad/s
4
Acceleration.
Spider:
S  3.927 rad/s
aD  ( AD )S2  (6 in.)(3.927 rad/s) 2
a D  92.53 in./s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.118 (Continued)
Gear D:
Plane motion
(a)
 Trans. with D

Rotation about D
Tooth T in contact with gear A.
aT  a D  aT/D  aD  ( DT )D2
 92.53 in./s2
 (3 in.)(7.854 rad/s) 2
 92.53 in./s2
 185.06 in./s2
aT  92.53 in./s 2
aT  92.5 in./s 2 
(b)
Tooth E in contact with gear E.
a E  a D  a E/D  a D  ( ED )D2

 92.53 in./s2
 (3 in.)(7.854 rad/s) 2
 92.53 in./s2
 185.06 in./s2
a E  277.6 in./s 2
aE  278 in./s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.119
The 200-mm-radius disk rolls without sliding on the
surface shown. Knowing that the distance BG is 160
mm and that at the instant shown the disk has an
angular velocity of 8 rad/s counterclockwise and an
angular acceleration of 2 rad/s2 clockwise, determine
the acceleration of A.
SOLUTION
m, m/s, m/s 2
Units:
Unit vectors: i  1
, j 1 , k 1
Geometric analysis.
Let P be the point where the disk contacts the flat surface.
rG /A  0.200 j
rB / G  0.16i
rA /B   0.600i  0.200 j
Velocity analysis.
ωG  (8 rad/s)k , v P  0, v A  vAi
v B  v P  v G /P  v P  ωG  rG /P
 0  8k  (0.160i  0.200 j)  1.6i  1.28 j
v A  v B  v A /B  v B  ω AB  rA /B
v A i  1.6i  1.28 j   AB k  ( 0.600i  0.200 j)
 1.6i  1.28 j  0.77460 AB j  0.2 AB i
Resolve into components and transpose terms.
j:
Acceleration analysis:
0.77460AB  1.28
 AB  1.6525 rad/s 
a A  a A j, aG  2 rad/s 2 k
a P  (G2 r ) j  (8)2 (0.2) j  (12.8 m/s 2 ) j
a B  a P  a B /P  a P  aG  rB /P  G2 rB /P
 12.8 j  (2k )  (0.160i  0.200 j)  (8) 2 (0.160i  0.200 j)
 12.8 j  0.32 j  0.4i  10.24i  12.8 j
 10.64i  0.32 j
2
a A  a B  a A /B  a B   AB k  rA /B   AB
rA /B
 10.64i  0.32 j   AB k  ( 0.600i  0.200 j)  (1.6525) 2 ( 0.600i  0.200 j)
 10.64i  0.32 j  0.77460 AB j  0.2 AB i  2.115i  0.54615 j
a A i  12.755i  0.86615 j  0.2 AB i  0.77460 AB j
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.119 (Continued)
Resolve into components and transpose terms.
j:
i:
0  0.86615  0.77460 AB
 AB  1.1182
aA  12.755  0.2 AB  12.755  (0.2)(1.1182)  12.98
a A  (12.98 m/s 2 )i  12.98 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.120
Knowing that crank AB rotates about Point A with a constant angular velocity of
900 rpm clockwise, determine the acceleration of the piston P when   60.
SOLUTION
sin  sin 60

0.05
0.15
Law of sines.
  16.779
AB  900 rpm  30 rad/s
Velocity analysis.
v B  0.05AB  1.5 m/s
v D  vD
60
BD  BD

v D/B  0.15  BD
v D  v B  v D/B
[vD ]  [1.5
Components
:
]
0  1.5 cos60  0.15BD cos 
 BD 
Acceleration analysis.
60]  [0.15BD
1.5 cos 60
 16.4065 rad/s
0.15 cos 
 AB  0
2
a B  0.05 AB
 (0.05)(30 ) 2  444.13 m/s 2
aD  aD
a D/B  [0.15 AB
 [0.15 BD
a D  a B  a D/B
30
 BD   BD
2
 ]  [0.15 BD
 ]  [40.376
]
]
Resolve into components.
PROBLEM 15.120 (Continued)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
:
0  444.13 cos 30  0.15 BD cos   40.376 sin 
 BD  2597.0 rad/s 2
:
aD  444.13 sin 30  (0.15)(2597.0)sin   40.376 cos 
 148.27 m/s 2
aP  aD
a P  148.3 m/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.121
Knowing that crank AB rotates about Point A with a constant angular velocity of
900 rpm clockwise, determine the acceleration of the piston P when   120.
SOLUTION
sin  sin120

,
0.05
0.15
Law of sines.
  16.779
AB  900 rpm  30 rad/s
Velocity analysis.
v B  0.05AB  1.5 m/s
v D  vD
60
BD  BD

v D/B  0.15BD
v D  v B  v D/B
60]  [0.15BD
[vD ]  [1.5
Components
:
0  1.5 cos 60  0.15BD cos 
BD  
Acceleration analysis.
]
1.5 cos 60
 16.4065 rad/s
0.15 cos 
 AB  0
2
a B  0.05 AB
 (0.05)(30 ) 2  444.13 m/s 2
aD  aD
30
 BD   BD
a D/B  [0.15 AB
 [6 BD
a D  a B  a D/B
]
2
 ]  [0.15 BD
 ]  [40.376
]
Resolve into components.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.121 (Continued)
:
0  444.13 cos 30  0.15 BD cos   40.376 sin 
 BD  2597.0 rad/s 2
:
aD  444.13 sin 30  (0.15)(2597.0)sin   40.376 cos 
 296 m/s 2
aP  aD
a P  296 m/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.122
In the two-cylinder air compressor shown, the connecting rods
BD and BE are each 190 mm long and crank AB rotates about the
fixed Point A with a constant angular velocity of 1500 rpm
clockwise. Determine the acceleration of each piston when   0.
SOLUTION
Crank AB.
vA  0, a A  0, AB  1500 rpm  157.08 rad/s ,  AB  0
vB  vA  vB/ A  0  [0.05 AB
45°]  [7.854 m/s
45]
a B  aA  (a B/ A )t  (a B/ A ) n
 0  [0.05AB
2
45°]  [0.05 AB
 [(0.05)(157.08) 2
v D  vD
Rod BD.
45°
45°]
45°]  1233.7 m/s
2
45°
BD  BD
v D  v B  v B/D
vD
Components
45]  [0.19BD
45  [7.854
0  7.854  0.19BD
45:
aD  aD
45]
BD  41.337 rad/s
45
a D  a B  (a D/B )t  (a D/B ) n
[aD
45]  [1233.7
45]  [0.19 BD
2
45]  [0.19 BD
45]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.122 (Continued)
Components
aD  1233.7  (0.19)(41.337)2  1558.4 m/s 2
45:
a D  1558 m/s 2
sin  
Rod BE.
45° 
0.05
,   15.258,   45    29.742
0.19
v E  vE
45°
BE  0.
Since v E is parallel to v B ,
a E  aE
2
0
45° (aB/E )n  0.19BE
(a E /B )t  (aE /B )t

aE  aB  (aB /E )t
Draw vector addition diagram.
  45  
 15.258
aE  aB tan 
 1233.7 tan 
 336.52 m/s 2
a E  337 m/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
45° 
PROBLEM 15.123
The disk shown has a constant angular velocity of 500 rpm counter-clockwise.
Knowing that rod BD is 10 in. long, determine the acceleration of collar D
when (a)   90, (b)   180.
SOLUTION
 A  500 rpm
Disk A.
 52.36 rad/s
 A  0, ( AB)  2 in.
vB  ( AB) A  (2)(52.36)  104.72 in./s
aB  ( AB) A2  (2)(52.36) 2  5483.1 in./s 2
(a)
  90.
v B  104.72 m/s , v D  vD
sin  
2 in.
 0.4
5 in.
  23.58
vD and vB are parallel.
BD  0
a B  5483.1 in./s 2
a D/B  [( BD) BD
 [10  BD
aD  a B  aD/B
:
:
,
aD  aD ,
2
 ]  [( BD ) BD
 BD   BD
]
] 0
Resolve into components.
0  5483.1  (10 cos  ) BD
 BD  598.26 rad/s 2
aD  0  (10 sin )( 598.26)  0  2393.0 in./s 2
a D  199.4 ft/s 2 







Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.123 (Continued)
(b)
  180.
vB  104.72 in./s
sin  
6 in.
 0.6
10 in.
vB  104.72 in./s
,
vD  vD
  36.87
,
vD  vD
Instantaneous center of bar BD lies at Point C.
BD 
vB
104.72

 13.09 rad/s
( BD ) 10 cos 
a B  5483.1 in./s 2 , a D  aD , BD   BD
a D/B  [( BD) BD
 [10BD
a D  a B  a D/B
:
2
 ]  [( BD ) BD
 ]  [1713.5
]
]
Resolve in components.
0  0  (10 cos  ) BD  1713.5 sin
 BD  128.51 rad/s 2
:
aD  5483.1  (10 sin  )( 128.51)  1713.5cos 
 7625.0 in./s2
a D  635 ft/s 2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.124
Arm AB has a constant angular velocity of 16 rad/s
counterclockwise. At the instant when   90, determine
the acceleration (a) of collar D, (b) of the midpoint G of bar
BD.
SOLUTION
2
aB  ( AB) AB
Rod AB:
 (3 in.)(16 rad/s)2
a B  768 in./s 2
Rod BD: instantaneous center is at OD; BD  0
sin   (3 in.)/(10 in.)  0.3;   17.46
Acceleration.
Plane motion

Trans. with B

Rotation about B
a D  a B  a D/B  aB  (a D/B )t  (a D/B ) n
(a)
 [aB ]  [( BD )
2
 ]  [( BD ) BD
 [768 in./s 2 ]  [(10 in.) 
a D   [768 in./s ]  [(10 in.)
]
 ]  [(10 in.)(0) 2
]
]
Vector diagram:
a D  (768 in./s 2 ) tan17.46
 241.62 in./s 2
a D  242 in./s 2
(10 in.)  (768 in./s 2 )/cos17.46
(10 in.)  805.08 in./s 2
α  80.5 rad/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.124 (Continued)
aG  a B  aG /B  a B  (aG /B )t  (aG /B )n
(b)
 [aB
]  [( BG )
 [768 in./s 2
aG  [768 in./s 2
components:
2
 ]  [( BG ) BD
]  [(5 in.)(80.5 rad/s 2 )
]  [402.5 in./s 2
]
 ]  [( BG )(0) 2 ]
17.46]
(aG ) x  (402.5 in./s 2 )sin 17.46
(aG ) x  120.77 in./s 2
components:
(aG ) y  768 in./s 2  (402.5 in./s 2 ) cos 17.46
 768 in./s 2  384 in./s 2
(aG ) y  384 in./s2
aG  403 in./s 2
72.5° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.125
Arm AB has a constant angular velocity of 16 rad/s
counterclockwise. At the instant when   60, determine
the acceleration of collar D.
SOLUTION
  sin 1
3.403 in.
10 in.
  19.89
Velocity.
v B  ( AB)AB  (3 in.)(16 rad/s)  48 in./s
30°
Rod BD:
Plane motion

Trans. with B

Rotation about B
v D  v B  v D/B  v B  [( BD ) BD
v D   [48 in./s
components:
]
30]  [(10 in.)BD
19.89]
(48 in./s)sin 30  (10 in.)BD cos19.89
ωBD 
(48 in./s)sin 30
 2.552 rad/s
(10 in.) cos19.890
Acceleration.
Rod AB:
2
a B  [( AB ) AB
60]  [(3 in.)(16 rad/s)2
a B  768 in./s 2
60°
60]
Rod BD:
Plane motion

Trans. with B

Rotation about B
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.125 (Continued)
a D  a B  a B/D  a B  (a D/B )t  (a D/B ) n
aD  [aB
60]  [( BD) BD
 [768 in./s 2
aD   [768 in./s 2
2
 ]  [( BD ) DB
60]  [(10 in.) BD
]  [(10 in.) BD
]
 ]  [(10 in.)(2.552 rad/s 2 )
19.89]  [65.14 in./s 2
]
19.89]
Vector diagram.
y components:
768sin 60  65.14sin19.89  10 BD cos19.89  0
:
 BD  73.09 rad/s 2
x components:
:
aD  768cos60  65.14cos19.89  (10)(73.09)sin19.89
aD  693.9 in./s 2
a D  694 in./s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.126
A straight rack rests on a gear of radius r  3 in. and is attached to
a block B as shown. Knowing that at the instant shown   20°,
the angular velocity of gear D is 3 rad/s clockwise, and it is
speeding up at a rate of 2 rad/s2, determine (a) the angular
acceleration of AB, (b) the acceleration of block B.
SOLUTION
Let Point P on the gear and Point Q on the rack be located at the contact point between them.
Units: inches, in./s, in./s2
i 1
Unit vectors:
, j 1 , k 1
rP /D  3(sin  i  cos  j)
Geometry:
3
(cos  i  sin  j)
tan 
  20
rB /Q 
αgear  2 rad/s2
ωgear  3 rad/s
Gear D:
v P  9 in./s
vP  gear r  (3)(3)  9 in./s

2
(aP )n  gear
r  (3)2 (3)  27 in./s2

(a P ) n  27 in./s 2
(aP )t   gear r  (2)(3)  6 in./s2
(a P )t  6 in./s 2



Velocity analysis.
Gear to rack contact:
Rack AQB:
v Q  v P  9 in./s
ω AB  AB ,
v B  vB
,

α AB   AB
αB  B
v B  vQ  v B /Q  vQ   AB k  rB /Q
vB i  9(cos  i  sin  j)   AB k  (7.74535i  2.81908 j)
 (9 cos   2.81907 AB )i  (9sin   2.81907 AB ) j
Equating like components,
j:
0  9sin   7.74535AB
 AB  0.39742
 AB  0.39742 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.126 (Continued)
Acceleration analysis.
Gear to rack contact:
(aQ )t  (a P )t  6 in./s2
2
(aQ )n  (a P )n  rrd


ωrd  ω AB  ωD  3.39742 rad/s
where
(aQ )n  27 in./s2
20  (3)(3.39742) 2
 7.6274 in./s 2
20°
20°
aQ  6(cos  i  sin  j)  7.6274(sin  i  cos  j)
Then,
 (8.2469 in./s 2 )i  (5.1153 in./s2 ) j
2
a B  aQ  a B /Q  aQ   AB k  rB /Q   AB
rB /Q
aB i  8.2469i  5.1153j   AB k  (7.74535i  2.81908 j)
 (0.39742) 2 (7.74535i  2.81908 j)
 (8.2469  2.81908 AB  1.22332)i
 (5.1153  7.74535 AB  0.44526) j
Equating like components of aB ,
j:
0  5.1153  7.74535 AB  0.44526
 AB  0.71792 rad/s 2
i:
aB  8.2469  (2.81908)(0.71792)  1.22332
aB  5.00 in./s 2
(a)
Angular acceleration of AB:
(b)
Acceleration of block B:
α AB  0.718 rad/s 2
a B  5.00 in./s 2
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

PROBLEM 15.127
The elliptical exercise machine has fixed axes of rotation at points A
and E. Knowing that at the instant shown the flywheel AB has a
constant angular velocity of 6 rad/s clockwise, determine the
acceleration of point D.
SOLUTION
Given:
 AB  6k rad/s,  AB  0 rad/s 2
Geometry:
rB / A  0.2 j m
rD / B  1.2i  0.2 j m
rD / E  0.12i  0.8 j m
Relative Velocity:
v B  v A   AB k  rB / A
(1)
v D  v B   BD k  rD / B
(2)
v D  v E   DEF k  rD / E
(3)
Sub (1) into (2) = (3):
 AB k  rB / A   BD k  rD / B   DEF k  rD / E
Put in known values:
6k  0.2 j  BD k  1.2i  0.2 j =DEF k   0.12i  0.8 j
Equate components:
i:
1.2  0.2 BD  0.8 DEF
j:
1.2 BD  0.12 DEF
Solve simultaneously:
 BD  0.1463 rad/s,  DEF  1.463 rad/s
Relative acceleration:
2
a B  a A   AB k  rB / A   AB
rB / A
(4)
2
a D  a B   BD k  rD / B   BD
rD / B
(5)
a D  a E   DEF k
2
 rD / E   DEF
rD / E
(6)
Sub (4) into (5) = (6):
2
2
2
 AB
rB / A   BD k  rD / B  BD
rD / B   DEF k  rD / E  DEF
rD / E
Put in known values:
  6   0.2 j   BD k  1.2i  0.2 j   0.1463  1.2i  0.2 j =
2
2
 DEF k   0.12i  0.8 j  1.463  0.12i  0.8 j
2
Equate components:
Solve simultaneously:
i:
0.2 BD  0.02570  0.8 DEF  0.2570
j:
7.2  1.2 BD  0.004283  0.12 DEF  1.7132
 BD  7.278 rad/s2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.127 (Continued)
Relative acceleration:
2
a D  a E   DEF k  rD/ E   DEF
rD/ E
 1.466k   0.12i  0.8 j  1.463   0.12i  0.8 j
2
a D  1.430i  1.537 j m/s 2
a F  2.099 m/s2
47.08 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.128
The elliptical exercise machine has fixed axes of rotation at points A
and E. Knowing that at the instant shown the flywheel AB has a
constant angular velocity of 6 rad/s clockwise, determine (a) the
angular acceleration of bar DEF, (b) the acceleration of point F.
SOLUTION
Given:
 AB  6k rad/s,  AB  0 rad/s 2
Geometry:
rB / A  0.2 j m
rD / B  1.2i  0.2 j m
rD / E  0.12i  0.8 j m
rF / E  0.09i  0.6 j m
Relative Velocity:
v B  v A   AB k  rB / A
(1)
v D  v B   BD k  rD / B
(2)
v D  v E   DEF k  rD / E
(3)
Sub (1) into (2) = (3):
 AB k  rB / A   BD k  rD / B   DEF k  rD / E
Put in known values:
6k  0.2 j   BD k  1.2i  0.2 j = DEF k   0.12i  0.8 j
Equate components:
i:
1.2  0.2 BD  0.8 DEF
j:
1.2BD  0.12DEF
Solve simultaneously:
 BD  0.1463 rad/s,  DEF  1.463 rad/s
Relative acceleration:
2
a B  a A   AB k  rB / A   AB
rB / A
(4)
2
a D  a B   BD k  rD / B   BD
rD / B
(5)
a D  a E   DEF k
2
 rD / E   DEF
rD / E
(6)
Sub (4) into (5) = (6):
2
2
2
 AB
rB / A   BD k  rD / B   BD
rD / B   DEF k  rD / E   DEF
rD / E
Put in known values:
  6   0.2 j   BD k  1.2i  0.2 j   0.1463  1.2i  0.2 j =
2
2
 DEF k   0.12i  0.8 j  1.463  0.12i  0.8 j
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.128 (Continued)
Equate components:
i:
0.2 BD  0.02570  0.8 DEF  0.2570
j:
7.2  1.2 BD  0.004283  0.12 DEF  1.7132
(a) Solve simultaneously:
 DEF  1.466 rad/s 2
(b) Relative acceleration:
2
a F  a E   DEF k  rF/ E   DEF
rF/ E
α DEF  1.466 rad/s 2

 1.466k   0.09i  0.6 j  1.463  0.09i  0.6 j
2
a F  1.0724i  1.153 j m/s 2
a F  1.575 m/s 2
47.08 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.129
Knowing that at the instant shown bar AB has a constant angular
velocity of 19 rad/s clockwise, determine (a) the angular acceleration
of bar BGD, (b) the angular acceleration of bar DE.
SOLUTION
Velocity analysis.
 AB  19 rad/s
vB  ( AB ) AB  (8)(19)  152 in./s
v B  vB
, v D  vD
Instantaneous center of bar BD lies at C.
 BD 
vB
152

 19 rad/s
8
BC
vD  (CD ) BD  (19.2)(19)  364.8 in./s
 DE 
Acceleration analysis.
vD
364.8

 24 rad/s 2
15.2
DE
 AB  0.
2
a B  [( AB) AB
]  [(8)(19)2 ]  2888 in./s
a D  [( DE ) DE
 [15.2 DE
2
]  [( DE ) DE
]  [8755.2 in./s 2
(a D/B )t  [19.2 BD ]  [8 DB
2
(a D/B )n  [19.2BD
 [6931.2 in./s 2
(b)
]
]  [2888 in./s 2 ]
Resolve into components.
8755.2  0  8 BD  6931.2
(a)
:
]
2
]
]  [8BD
a D  a B  (a D/B )t  (aD/B ) n
:
]
 BD  228 rad/s 2

 DE  92.0 rad/s 2

15.2 DE  2888  (19.2)(228)  2888
 DE  92 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.130
Knowing that at the instant shown bar DE has a constant angular velocity
of 18 rad/s clockwise, determine (a) the acceleration of Point B, (b) the
acceleration of Point G.
SOLUTION
Velocity analysis.
 DE  18 rad/s
v D  ( DE ) DE  (15.2)(18)  273.6 in./s
v D  vD , v B  v B
Point C is the instantaneous center of bar BD.
 BD 
vD
273.6

 14.25 rad/s
19.2
CD
vB  (CB ) BD  (8)(14.25)  114 in./s
 AB 
Acceleration analysis.
vB
114

 14.25 rad/s
8
AB
 DE  0
2
a D  [( DE ) DE
]  [(15.2)(18) 2
a B  [( AB) AB
2
]
]  [( AB) AB
 [8 AB
]
]  [1624.5 in./s 2 ]
(a D/B )t  [19.2 BD ]  [8 BD
2
(a D/B )n  [19.2BD
]  [4924.8 in./s 2
]
2
]
]  [8BD
a D  a B  (a D/B )t  (a D/B ) n
Resolve into components.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.130 (Continued)
: 0  1624.5  19.2 BD  1624.5,
 BD  169.21875 rad/s 2
: 4924.8  8 AB  (8)(169.21875)  3898.8
 AB  40.96875 rad/s 2
(a)
 [327.75 in./s 2
(b)
]  [1624.5 in./s 2 ]
a B  [(8)(40.96875)
a G  a B  a G/ B  a B 
 aB 
]  [1624.5 in./s 2 ],
a B  138.1 ft/s 2
78.6 
aG  203 ft/s 2
19.5 
1
a D/B
2
1
1
(a D  a B )  (a B  a D )
2
2
 327.75  4924.8
 
2

 [2298.5 in./s 2 ]
 1624.5
 2
 



 [812.25 in./s 2 ]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.131
Knowing that at the instant shown bar AB has a constant
angular velocity of 4 rad/s clockwise, determine the angular
acceleration (a) of bar BD, (b) of bar DE.
SOLUTION
rB /A  (20 in.)i  (40 in.) j
Relative position vectors.
rD /B  (40 in.)i
rD /E  (20 in.)i  (25 in.) j
Velocity analysis.
Bar AB (Rotation about A):
ω AB  4 rad/s
 (4 rad/s)k
rB /A  (20 in.)i  (40 in.) j
v B  ω AB  rB /A  ( 4k )  ( 20i  40 j)
v B   (160 in./s)i  (80 in./s) j
(Plane motion  Translation with B + Rotation about B):
Bar BD
ω BD  BD k
rD /B  (40 in.)i
v D  v B  ω BD  rD /B  v B  ( BD k )  (40i )
v D  (160 in/s)i  (40BD  80 in./s) j
Bar DE (Rotation about E):
ω DE  DE k
rD /E  (20 in.)i  (25 in.) j
v D  ω DE  rD /E  (DE k )  (20i  25 j)
v D  20DE j  25DE i
Equating components of the two expression for vD,
i:
160  25 DE
 DE  6.4 rad/s
j:
40 BE  80  20 DE
40 BD  80  20( 6.4)
 BD  5.2 rad/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.131 (Continued)
Summary of angular velocities:
ω AB  4 rad/s
Acceleration analysis.
 AB  0,  BD   BD k ,  DE   DE k
Bar AB (Rotation about A):
ω DE  6.4 rad/s
ω BD  5.2 rad/s
2
a B   AB  rB /A   AB
rB /A
 0  (4)2 (20i  40 j)
 (320 in./s 2 )i  (640 in./s 2 ) j
Bar BD (Translation with B  Rotation about B):
2
a D  a B   BD  rD /B   BD
rD /B
 320i  640 j   BD k  (40i )  (5.2) 2 (40)i
 320i  640 j  40 BD j  1081.6i
 761.60i  (640  40 BD ) j
(1)
Bar DE (Rotation about E):
2
a D   DE  rD /E  DE
rD /E
  DE k  (20i  25 j)  (6.4) 2 (20i  25 j)
 20 DE j  25 DE i  819.20i  1024 j
 (25 DE  819.20)i  (20 DE  1024) j
(2)
Equate like components of aD expressed by Eqs. (1) and (2).
i:
761.60  25 DE  819.20
 DE  2.3040 rad/s 2
j:
640  40 BD  (20)(2.304)  1024
 BD  10.752 rad/s 2
(a)
Angular acceleration of bar BD.
α BD  10.75 rad/s 2

(b)
Angular acceleration of bar DE.
α DE  2.30 rad/s 2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.132
Knowing that at the instant shown bar AB has a constant angular velocity of
4 rad/s clockwise, determine the angular acceleration (a) of bar BD, (b) of bar
DE.
SOLUTION
Velocity analysis.
Bar AB (Rotation about A):
 AB  4 rad/s
 (4 rad/s)k
rB /A  (175 mm)i
vB   AB  rB /A  ( 4k )  (175i )
vB  (700 mm/s)j
Bar BD (Plane motion  Translation with B  Rotation about B):
 BD  BD k
rD /B  (200 mm)j
vD  vB   BD  rD /B  700 j  ( BD k )  (200 j)
vD  700 j  200 BD i
Bar DE (Rotation about E ):
 DE  DE k
rD /E  (275 mm)i  (75 mm)j
vD   DE  rD /E  (DE k )  (275i  75 j)
vD  275DE j  75DE i
Equating components of the two expressions for vD ,
j:
700  275 DE
i : 200 BD  75 BD
 DE  2.5455 rad/s
3
8
 DE   BD
3
 BD     (2.5455)  0.95455 rad/s
8
Acceleration analysis.
Bar AB:
Bar BD:
 DE  2.55 rad/s
 BD  0.955 rad/s
 AB  0
2
a B   AB
rB /A  (4) 2 (175i )  (2800 mm/s 2 )i
 BD   BD k
2
a D  a B   BD  rD /B  BD
rD /B
 2800i   BD k  (200 j)  (0.95455) 2 (200 j)
 (2800  200  BD )i  182.23 j
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 15.132 (Continued)
 DE   DE k
Bar DE:
2
a D   DE  rD /E  DE
rD /E
  DE k  (275i  75 j)  (2.5455) 2 (275i  75 j)
 275 DE j  75 DE i  1781.8i  485.95 j
 ( 75 DE  1781.8)i  (275 DE  485.95) j
(2)
Equate like components of aD expressed by Eqs. (1) and (2).
j:
i:
182.23   (275 DE  485.95)
(2800  200 BD )  [ (75)( 2.4298)  1781.8]
 DE  2.4298 rad/s 2
 BD  4.1795 rad/s 2
(a)
Angular acceleration of bar BD.
 BD  4.18 rad/s 2

(b)
Angular acceleration of bar DE.
 DE  2.43 rad/s 2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.133
Knowing that at the instant shown bar AB has an angular
velocity of 4 rad/s and an angular acceleration of 2 rad/s2, both
clockwise, determine the angular acceleration (a) of bar BD,
(b) of bar DE by using the vector approach as is done in Sample
Problem 15.16.
SOLUTION
Relative position vectors.
rB /A  (20 in.)i  (40 in.) j
rD /B  (40 in.)i
rD /E  (20 in.)i  (25 in.) j
Velocity analysis.
Bar AB (Rotation about A):
ω AB  4 rad/s
 (4 rad/s)k
rB /A  (20 in.)i  (40 in.) j
v B  ω AB  rB /A  ( 4k )  ( 20i  40 j)
v B   (160 in./s)i  (80 in./s) j
(Plane motion  Translation with B  Rotation about B):
Bar BD
ω BD  BD k
rD /B  (40 in.)i
v D  v B  ω BD  rD /B  v B  ( BD k )  (40i )
v D  (160 in/s)i  (40BD  80 in./s) j
Bar DE (Rotation about E):
ω DE  DE k
rD /E  (20 in.)i  (25in.) j
v D  ω DE  rD /E  (DE k )  (20i  25 j)
v D  20DE j  25DE i
Equating components of the two expression for vD,
i:
160  25 DE
 DE  6.4 rad/s
j:
40 BE  80  20 DE
40 BD  80  20( 6.4)
 BD  5.2 rad/s
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PROBLEM 15.133 (Continued)
Summary of angular velocities:
ω AB  4 rad/s
Acceleration analysis.
 AB  (2 rad/s 2 )k ,  BD   BD k ,  DE   DE k
Bar AB (Rotation about A)
ω DE  6.4 rad/s
ω BD  5.2 rad/s
2
a B   AB  rB /A   AB
rB /A
 (2k )  (20i  40 j)  (4)2 (20i  40 j)
 (80 in./s 2 )i  (40 in./s 2 ) j  (320 in./s 2 )i  (640 in./s 2 ) j
 (240 in./s 2 )i  (680 in./s 2 ) j
Bar BD (Translation with B  Rotation about B):
2
a D  a B   BD  rD /B  BD
rD /B
 240i  680 j   BD k  (40i )  (5.2)2 (40)i
 240i  680 j  40 BD j  1081.6i
 841.60i  (680  40 BD ) j
Bar DE
(1)
(Rotation about E):
2
a D   DE  rD /E  DE
rD /E
  DE k  (20i  25 j)  (6.4) 2 (20i  25 j)
 20 DE j  25 DE i  819.20i  1024 j
 (25 DE  819.20)i  (20 DE  1024) j
(2)
Equate like components of aD expressed by Eqs. (1) and (2).
i:
841.60  25 DE  819.20
j:
680  40 BD  (20)( 0.896)  1024
 DE  0.896 rad/s 2
 BD  8.152 rad/s 2
(a)
Angular acceleration of bar BD.
α BD  8.15 rad/s 2

(b)
Angular acceleration of bar DE.
α DE  0.896 rad/s 2

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PROBLEM 15.134
Knowing that at the instant shown bar AB has an angular velocity of 4 rad/s
and an angular acceleration of 2 rad/s2, both clockwise, determine the angular
acceleration (a) of bar BD, (b) of bar DE by using the vector approach as is
done in Sample Problem 15.16.
SOLUTION
Velocity analysis.
 (4 rad/s)k
rB /A  (175 mm)i
vB   AB  rB /A  (4k )  (175i )
vB  (700 mm/s)j
 AB  4 rad/s
Bar AB (Rotation about A):
Bar BD (Plane motion  Translation with B  Rotation about B ):
 BD  BD k
rD /B  (200 mm)j
vD  vB   BD  rD /B  700 j  ( BD k )  (200 j)
vD  700 j  200 BD i
 DE
rD /E
vD
vD
Bar DE (Rotation about E ):
  DE k
 (275 mm)i  (75 mm)j
  DE  rD /E  ( DE k )  ( 275i  75 j)
 275 DE j  75 DE i
Equating components of the two expressions for vD ,
j:
700  275 DE
i : 200 BD  75 DE
 DE  2.5455 rad/s
3
BD   DE
8
3
 BD     (2.5455)  0.95455 rad/s
8
Acceleration analysis.
Bar AB:
Bar BD:
 DE  2.55 rad/s
 BD  0.955 rad/s
 AB  2 rad/s 2  (2 rad/s 2 )k
2
a B   AB  rB /A   AB
rB /A
 BD
 ( 2k )  ( 175i )  (4) 2 (175i )  2800 mm/s 2 i  350 mm/s 2 j
  BD k
2
a D  a B   BD  rD /B  BD
rD /B
 2800i  350 j   BD k  (200 j)  (0.95455) 2 (200 j)
 (2800  200  BD )i  532.23 j
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(1)
PROBLEM 15.134 (Continued)
 DE   DE k
Bar DE:
2
a D   DE  rD /E  DE
rD /E
  DE k  (275i  75 j)  (2.5455) 2 (275i  75 j)
 275 DE j  75 DE i  1781.8i  485.95 j
 ( 75 DE  1781.8)i  (275 DE  485.95) j
(2)
Equate like components of a D expressed by Eqs. (1) and (2).
j:
i:
532.23   (275 DE  485.95)
(2800  200 BD )  [ (75)( 3.7025)  1781.8]
 DE  3.7025 rad/s 2
 BD  3.7025 rad/s 2
(a)
Angular acceleration of bar BD.
 BD  3.70 rad/s 2

(b)
Angular acceleration of bar DE.
 DE  3.70 rad/s 2

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PROBLEM 15.135
Robert’s linkage is named after Richard Robert (1789–1864) and can be used
to draw a close approximation to a straight line by locating a pen at Point F.
The distance AB is the same as BF, DF and DE. Knowing that at the instant
shown bar AB has a constant angular velocity of 4 rad/s clockwise, determine
(a) the angular acceleration of bar DE, (b) the acceleration of Point F.
SOLUTION
inches, in./s, in./s2
Units:
Unit vectors:
i 1
, j 1 , k 1 .
Geometry:
rB /A  3i  122  32 j  3i  135 j
rD /B  6i
rF /B  3i  135 j
rD /E  3i  135 j
 AB  4 rad/s 2
Velocity analysis:
 4 k
Bar AB:
vB   AB  rB /A  4k  (3i  135 j)  4 135i  12 j
Object BDF:
vD  v B  v D /B  v B  ω BD  rD /B
 4 135i  12 j  BD k  6i
 4 135i  12 j  6 BD j
(1)
v D   DE  rD /E   DE k  ( 3i  135 j)
Bar DE:
  135 DE i  3 DE j
(2)
Equating like components of v D from Eqs. (1) and (2),
i:
j:
4 135   135 DE
(3)
12  6 BD  3 DE
From Eq. (3),
 DE  4
From Eq. (4),
 BD  (12  3 DE )  5
1
6
(4)
 DE  5 rad/s
 BD  4 rad/s
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PROBLEM 15.135 (Continued)
α AB  0
Acceleration Analysis:
2
a B  α AB  rB /A   AB
rB /A
Bar AB:
 0  (4)2 (3i  135 j)
 48i  16 135 j
2
a D  a B  a D /B  a B  α BD  rD /B   BD
rD /B
Object BDF:
a D  48i  16 135 j   BD k  (6i )  (4) 2 (6i )
 144i  16 135 j  6 BD j
(5)
2
a D  α DE  rD /E   DE
rD /E
Bar DE:
a D   DE k  (3i  135 j)  (4) 2 (3i  135 j)
  135 DE i  3 DE j  48i  16 135 j
(6)
Equating like components of aD from Eqs. (5) and (6),
i :  144   135 DE  48
(7)
j:  16 135  6 BD  3 DE  16 135
(8)
192
From Eq. (7),
 DE 
From Eq. (8),
 BD    BD  
135
1
2
(a)
Angular acceleration of bar DE:
(b)
Acceleration of Point F:
96
135
α DE  16.53 rad/s 2

2
a F  a B  a F /B  a B   BD  rF /B  BD
rF /B
 96 
 48i  16 135 j   
k   (3i  135 j)  (4) 2 (3i  135 j)
 135 
288
 48i  16 135 j 
j  96i  48i  16 135 j
135
288
 192i 
j
135
a F  (192.0 in./s 2 )i  (24.8 in./s 2 ) j  193.6 in./s 2
7.36° 
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PROBLEM 15.136
For the oil pump rig shown, link AB causes the beam
BCE to oscillate as the crank OA revolves. Knowing
that OA has a radius of 0.6 m and a constant
clockwise angular velocity of 20 rpm, determine the
velocity and acceleration of Point D at the instant
shown.
SOLUTION
Units: meters, m/s, m/s2
Unit vectors:
i 1
Crank OA:
rOA  0.6 m, ω OA  20 rpm
 2.0944 rad/s
v A  OA rOA  (2.0944)(0.6)
v A  1.25664 m/s
,
k 1 .
j 1 ,
α OA  0
( a A )t  0
2
(a A ) n  OA
rOA  (2.0944)2 (0.6)  2.6319 m/s 2
a A  2.6319 m/s 2
v B  vA ,
Rod AB:
Since v B and v A are parallel, v A  v B and  AB  0.
v B  1.25664 m/s
2
a B  a A  a B /A  a A   AB k  rB /A   AB
rB /A
 2.6319i   AB k  (0.6i  2 j)  0
 (2.6319  2 AB )i  0.6 AB j
(1)
Beam BCE: Point C is a pivot.
vB   BCE rBC
 BCE 
vB 1.25664

 0.41888
rBC
3
vE   BCE rCE  (0.41888)(3.3)  1.38230
ω BCE  0.41888 rad/s
v E  1.38230 m/s
2
a B  α BCE  rB /C  BCE
rB /C
  BCE k  (3i )  (0.41888)2 (3i )
 0.52638i  3 BCE j
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(2)
PROBLEM 15.136 (Continued)
Equating like components of α B expressed by Eqs. (1) and (2),
i:
2.6319  2 AB  0.52638
 AB  1.05276 rad/s
j:
0.6 AB  3 BCE
 BCE  0.21055 rad/s 2
α AB  1.055276 rad/s
α BCE  0.21055 rad/s 2
a E  (a E /C )t  (a E /C ) n
(a E /C )t  α BCE  rE /C  (0.21055k )  (3.3i )
 (0.69482 m/s 2 ) j
String ED:
vD  vE
a D  (a E /C )t  (0.69482 m/s 2 ) j
v D  1.382 m/s 
a D  0.695 m/s 2 
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PROBLEM 15.137
Denoting by rA the position vector of Point A of a rigid slab that is in plane
motion, show that (a) the position vector rC of the instantaneous center of rotation is
rC  rA 
ω  vA
2
where  is the angular velocity of the slab and vA is the velocity of Point A,
(b) the acceleration of the instantaneous center of rotation is zero if, and only if,
aA 

v  ω  vA
 A
where    k is the angular acceleration of the slab.
SOLUTION
(a)
At the instantaneous center C,
vC  0
vA  vC    rA /C    rA /C
  vA    (  rA /C )   2rA /C
rA /C  
rC  rA  
  vA

2
 rC/ A
or
rC/ A 
  vA

  vA
2
rC  rA 
2
  vA
2

aA  aC    rA /C    vA/ C
(b)
 aC   k 
 vA
2
   ( vA  vC )

k  (k  vA )    vA
2

 aC  vA    vA

 aC 
Set a C  0.
aA 

v    vA 
 A
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PROBLEM 15.138*
The drive disk of the scotch crosshead mechanism shown has an angular
velocity  and an angular acceleration , both directed counterclockwise.
Using the method of Section 15.4 B, derive expressions for the velocity and
acceleration of Point B.
SOLUTION
Origin at A.
yB  l  yP  l  b sin 
v  y  b cos   b cos  
B
B
vB  b cos  
aB  
yB
d
vB
dt
d
 (b cos )
dt
aB  b sin  2  b cos 

aB  b cos   b 2 sin  
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PROBLEM 15.139*
The wheels attached to the ends of rod AB roll along the
surfaces shown. Using the method of Section 15.4 B, derive an
expression for the angular velocity of the rod in terms of
vB ,  , l , and  .
SOLUTION
Law of sines.
dB
l

sin  sin 
l
dB 
sin 
sin 
d
(d B )
dt
l
d

cos 
dt
sin 
l

cos  
sin 
vB 

vB sin 

l cos 
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PROBLEM 15.140*
The wheels attached to the ends of rod AB roll along the
surfaces shown. Using the method of Section 15.4 B and
knowing that the acceleration of wheel B is zero, derive an
expression for the angular acceleration of the rod in terms of
vB ,  , l , and .
SOLUTION
Law of sines.
dB
l

sin  sin 
l
dB 
sin 
sin 
d
(d B )
dt
l
d

cos 
sin 
dt
l

cos  
sin 
v sin 
 B
l cos 
vB 
Note that
dvB
 0.
dt
d  vB sin  sin  d




dt
l
cos 2 dt
aB 

vB sin  sin  vB sin 

l cos 
l cos 2
2
 vB sin   sin 
 cos3 
l


 
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PROBLEM 15.141*
A disk of radius r rolls to the right with a constant velocity v. Denoting by P the point of the rim in contact
with the ground at t  0, derive expressions for the horizontal and vertical components of the velocity of P at
any time t.
SOLUTION
x A  r ,
yA  r
xP  x A  r sin 
 r  r sin 
yP  y A  r cos 
 r  r cos 
x
 A
r
v
y A  0,  
r
vt
x A  vt ,  
r
x A  v,
vt  v

xP  vx  r  r cos   r  1  cos 
r r

vt  v

y P  v y  r sin   r  sin 
r r

vt 

vx  v 1  cos  
r 

v y  v sin
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
vt

r
PROBLEM 15.142*
Rod AB moves over a small wheel at C while end A moves to the
right with a constant velocity vA. Using the method of Section
15.4 B, derive expressions for the angular velocity and angular
acceleration of the rod.
SOLUTION
tan  
b
xA
cot  
xA
u
b
  cot 1 u
u
1  u2
(2uu )u
  
 
But
(1  u )
   and   
u
Then
u
1  u2

2 2
xA
x
v
v
, u  A   A , u   A  0
b
b
b
b
vA
b

1
 
xA
b
  



1    
2
xA
b
vA
b
xA
b
bvA

2
2
2
b 
x A2

,
bvA
b  x A2
2

2
2
0

2bxAv A2
b 2  x A2

2
,


2bxA v A2
b 2  x A2

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2

PROBLEM 15.143*
Rod AB moves over a small wheel at C while end A moves to
the right with a constant velocity vA. Using the method of
Section 15.4 B, derive expressions for the horizontal and
vertical components of the velocity of Point B.
SOLUTION
sin  
b
b
2

x A2

1/ 2
cos 
,
b
xA
2
 x A2

1/ 2
xB  l cos   x A


lx A
b 2  x A2
yB  l sin  
xB 


lx A
b 2  x A2
b
y B  
But

1/ 2
lb
b
2

1/ 2
lb 2 x A
2
b

x A2
 xA

3/ 2
 x A2


lx A x A x A

1/ 2
b 2  x A2

3/ 2
 x A
 x A
lbx A x A
2
x A  v A ,
 x A2

3/ 2
x B  (vB ) x ,
y B  (vB ) y
(vB ) x  v A 
b
(vB ) y 
lb 2v A
2

x A2



3/ 2
lbx Av A
2
b  x A2

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3/ 2

PROBLEM 15.144
Crank AB rotates with a constant clockwise angular velocity ω.
Using the method of Section 15.4 B, derive expressions for the
angular velocity of rod BD and the velocity of the point on the rod
coinciding with Point E in terms of  ,  , b, and l.
SOLUTION
Law of cosines for triangle ABE.
u 2  l 2  b2  2bl cos(180   )
 l 2  b2  2bl cos 
l  b cos 
u
b sin 
tan  
l  b cos 
cos  
d
(l  b cos  )(b cos  )  (b sin  )(b cos  )
(tan  )  sec2  
dt
(l  b cos  )2
 

But,
   ,
(cos 2  )[bl cos   b 2 (cos 2   sin 2  )]
(l  b cos  ) 2
bl cos   b 2 
b(b  l cos  ) 
 2

2
u
l  b 2  2bl cos 
  BD ,
and
vE  u
BD 
Hence,
b(b  l cos  )

l  b 2  2bl cos 
2

Differentiate the expression for u 2 .
2uu  2bl sin 
vE  u 
bl sin 

l 2  b 2  2bl cos 
vE 
bl sin 

l  b 2  2bl cos 
2
 b sin  
tan 1 
 
 l  b cos  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.145
Crank AB rotates with a constant clockwise angular velocity ω.
Using the method of Section 15.4 B, derive an expression for the
angular acceleration of rod BD in terms of  ,  , b, and l.
SOLUTION
Law of cosines for triangle ABE.
u 2  l 2  b2  2bl cos(180   )
 l 2  b2  2bl cos 
l  b cos 
u
b sin 
tan  
l  b cos 
cos  
d
(l  b cos  )(b cos  )  (b sin  )(b cos  )
(tan  )  sec2  
dt
(l  b cos  )2
 
(cos 2  )[bl cos   b 2 (cos 2   sin 2  )]
(l  b cos  )2
bl cos   b 2 
b(b  l cos  ) 
 2

2
u
l  b 2  2bl cos 
b(b  l cos  ) 
  2

l  b 2  2bl cos 
(l 2  b 2  2bl cos  )(bl sin  )  b(b  l cos  )(2bl sin  )  2


(l 2  b 2  2bl cos  ) 2


But,
   ,
b(b  l cos  )
l 2  b 2  2bl cos 
    0,
 
bl (l 2  b 2 )sin 
 2
(l 2  b 2  2bl cos  )2
   BD
 BD 
bl (l 2  b 2 ) sin 
2
l 2  b 2  2bl cos 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.146
Solve the engine system from Sample Prob. 15.15 using the methods of
Section 15.4B. Hint: Define the angle between the horizontal and the
crank AB as  and derive the motion in terms of this parameter.
SOLUTION
r  3 in, l  8 in,   40
Given:
  2000 rpm  209.4 rad/s
=0
Law of sines:
sin  sin 

l
r
At   40
  sin 1 
Differentiate (1) with respect to t:
cos  cos 

l
r
At   40
 
Differentiate (2) with respect to t:
2




r   sin   cos  cos   cos  sin  



l
cos 2 


At   40
  9940.2 rad/s 2
(1)
 r sin 
 l

  13.948

r cos 
  
l cos 
3cos 40  209.4 
8cos13.948
(2)
 61.99 rad/s or   61.99 rad/s


α BD  9940.2 rad/s 2
Write an equation for d:
d  r cos   l cos 
Differentiate with respect to t:
d  r sin   l sin 
Differentiate again:
d   r cos  2  r sin    l cos  2  l sin 
At   40

2
2
d  3cos 40  209.4   8 cos13.95  61.99   8sin13.95  9940.2 
 111, 477.5 in/s
a D  9289.8 ft/s 2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.147*
The position of rod AB is controlled by a disk of radius r that is attached to yoke CD.
Knowing that the yoke moves vertically upward with a constant velocity v0, derive
expression for the angular velocity and angular acceleration of rod AB.
SOLUTION
From geometry,
y
r
sin 
dy
r cos  d

dt
sin 2 dt
dy
 v0
dt
But,
and
d

dt
r cos 

sin 2 
v sin 2 
 0
r cos 
v0 
From geometry,
But,

v0 sin 2 
r cos 
r
sin 
dy
r cos  d

dt
sin 2  dt
y
dy
  v0
dt
and
d

dt
r cos 

sin 2 
v sin 2 
 0
r cos 
v0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.147* (Continued)
Angular acceleration.


d  d  d d 



dt
d dt d
v0 (2cos 2  sin   sin 3  )  v0 sin 2 

r
cos 2 
 r cos 



2
 v  (1  cos 2  )sin 3 
 0 
cos3 
 r 
2
v 
   0  (1  cos 2  ) tan 3 
 r 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 15.148*
A wheel of radius r rolls without slipping along the inside of a fixed cylinder
of radius R with a constant angular velocity ω. Denoting by P the point of the
wheel in contact with the cylinder at t  0, derive expressions for the
horizontal and vertical components of the velocity of P at any time t. (The
curve described by Point P is a hypocycloid.)
SOLUTION
Define angles  and  as shown.
   ,
  t
Since the wheel rolls without slipping, the arc OC is equal to arc PC.
r (   )  R

 
r
Rr
r

r
Rr
Rr
r t

Rr
xP  ( R  r )sin   r sin 
(vP ) x  xP
 ( R  r ) cos   r cos 
r t  r 

 ( R  r )  cos
  r (cos t )( )
R  r 

 R  r 
r t


(vP ) x  r  cos
 cos  t  
R
r



yP  R  ( R  r ) cos   r cos 
(vP ) y  y P
 ( R  r )sin   r sin 
r t  r 

 ( R  r )  sin

  r (sin  t)( )
R
 r  R  r 

r t


 sin  t  
(vP ) y  r  sin
R
r



Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.149*
In Problem 15.148, show that the path of P is a vertical straight line when
r  R/2. Derive expressions for the corresponding velocity and acceleration of
P at any time t.
SOLUTION
Define angles  and  as shown.
   ,
   t,
  0
Since the wheel rolls without slipping, the arc OC is equal to arc PC.
r (   )  R
 2r
 0
    
    0
xP  ( R  r )sin   r sin 
 r sin   r sin 
0
The path is the y axis. 
yP  R  ( R  r ) cos   r cos 
 R  r cos   r cos 
 R(1  cos  )
v  y P  R sin 
v  ( R sin  t ) j 
a  v
 ( R cos  2  sin )
 R 2 cos 
a  ( R 2 cos  t ) j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.150
Pin P is attached to the collar shown; the motion of the pin is guided by a
slot cut in rod BD and by the collar that slides on rod AE. Knowing that
at the instant considered the rods rotate clockwise with constant angular
velocities, determine for the given data the velocity of pin P.
 AE  8 rad/s, BD  3 rad/s
SOLUTION
AB  500 mm  0.5 m, AP  0.5 tan 30, BP 
ω AE  8 rad/s
,
0.5
cos 30
ωBD  3 rad/s
Let P be the coinciding point on AE and u1 be the outward velocity of the collar along the rod AE.
v P  v P  v P/ AE  [( AP) AE ]  [u1
]
Let P be the coinciding point on BD and u2 be the outward speed along the slot in rod BD.
30]  [u2
v P  v P  v P/BD  [( BP ) BD
60]
Equate the two expressions for v P and resolve into components.
:
 0.5 
u1  
 (3)(cos30)  u2 cos 60
 cos30 
u1  1.5  0.5u2
or
:
u2 
From (1),
(1)
 0.5 
(0.5 tan 30)(8)   
 (3)sin 30  u2 sin 60
 cos30 
1
[1.5 tan 30  4 tan 30]  1.66667 m/s
sin 60
u1  1.5  (0.5)(1.66667)  0.66667 m/s
v P  [(0.5tan 30)(8) ]  [0.66667
]  [2.3094 m/s ]  [0.66667 m/s
]
vP   2.30942  0.66667 2  2.4037 m/s
tan 
2.3094
0.66667
  73.9
v P  2.40 m/s
73.9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.151
Pin P is attached to the collar shown; the motion of the pin is guided by
a slot cut in rod BD and by the collar that slides on rod AE. Knowing that
at the instant considered the rods rotate clockwise with constant angular
velocities, determine for the given data the velocity of pin P.
AE  7 rad/s, BD  4.8 rad/s
SOLUTION
AB  500 mm  0.5 m, AP  0.5 tan 30, BP 
ω AE  7 rad/s
0.5
cos 30
ωBD  4.8 rad/s
,
Let P be the coinciding point on AE and u1 be the outward velocity of the collar along the rod AE.
v P  v P  v P/ AE  [( AP ) AE ]  [u1
]
Let P be the coinciding point on BD and u2 be the outward speed along the slot in rod BD.
v P  v P  v P/BD  [( BP ) BD
30]  [u2
60]
Equate the two expressions for v P and resolve into components.
 0.5 
u1  
 (4.8)(cos30)  u2 cos60
 cos30 
:
u1  2.4  0.5u2
or
 0.5 
(0.5 tan 30)(7)   
 (4.8)sin 30  u2 sin 60
 cos30 
:
u2 
From (1),
(1)
1
[2.4 tan 30  3.5 tan 30]  0.73333 m/s
sin 60
u1  2.4  (0.5)(0.73333)  2.0333 m/s
v P  [(0.5tan30)(7)
]  [2.0333
]  [2.0207 m/s ]  [2.0333 m/s
]
vP  (2.0333) 2  (2.0207) 2  2.87 m/s
tan  
2.0207
,
2.0333
  44.8
v P  2.87 m/s
44.8  
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.152
Two rotating rods are connected by slider block P. The rod
attached at A rotates with a constant angular velocity A . For the
given data, determine for the position shown (a) the angular
velocity of the rod attached at B, (b) the relative velocity of slider
block P with respect to the rod on which it slides.
b  8 in.,
A  6 rad/s.
SOLUTION
Dimensions:
Law of sines.
8 in.
AP
BP


sin 20 sin120 sin 40
AP  4.2567 in.
BP  10.7784 in.
 AP  6 rad/s
Velocities.
Note: P  Point of BE coinciding with P.
vP  ( AP) AP
 (4.2567 in.)(6 rad/s)
 25.540 in./s
30°
vP  vP  vP/BE
[25.540
(a)
30]  [vP
30]
vP  (25.54) cos 40
BE
(b)
70]  [vP/BE
 19.565 in./s
v
 P
BP
19.565 in./s

10.7784 in.
 1.8152 rad/s
BE  1.815 rad/s

vP/BE  (25.54)sin 40
 16.417 in./s
v P/BE  16.42 in./s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
20° 
PROBLEM 15.153
Two rotating rods are connected by slider block P. The rod
attached at A rotates with a constant angular velocity A . For
the given data, determine for the position shown (a) the
angular velocity of the rod attached at B, (b) the relative
velocity of slider block P with respect to the rod on which it
slides.
A  10 rad/s.
b  300 mm,
SOLUTION
Dimensions:
Law of sines.
AP
BP
300 mm


sin 20 sin120 sin 40
AP  159.63 mm
BP  404.19 mm
 AD  10 rad/s
Velocities.
Note: P  Point of AD coinciding with P.
vP  ( AP) AD
 (159.63 mm)(10 rad/s)
 1596.3 mm/s
30°
vP  vP  vP/AD
[ vP
(a)
70]  [1596.3
60]
vP  (1596.3)/cos 40°
 BP
(b)
30]  [vP/AD
 2083.8 mm/s
v
 P
BP
2083.8 mm/s

404.19 mm
 5.155 rad/s
vP /AD  (1596.3)tan 40  1339.5 mm/s
BD  5.16 rad/s
v P/AD  1.339 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

60° 
PROBLEM 15.154
Pin P is attached to the wheel shown and slides in a slot cut in bar
BD. The wheel rolls to the right without slipping with a constant
angular velocity of 20 rad/s. Knowing that x  480 mm when
  0, determine the angular velocity of the bar and the relative
velocity of pin P with respect to the rod for the given data.
(a)   0, (b)   90.
SOLUTION
Coordinates.
x A  ( x A )0  r ,
xB  0,
xC  x A ,
yA  r
yB  r
yC  0
xP  x A  e sin 
yP  r  e cos 
( xA )0  480 mm  0.48 m
Data:
r  200 mm  0.20 m
e  140 mm  0.14 m
Velocity analysis.
 AC  AC
,
BD  BD ,
v P  v A  v P/ A  [r AC
]  [eAC
v P  [ xPBD ]  [(e cos )]BD
v P/F  [u cos 
]
]
]  [u sin  ]
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.154 (Continued)
Use v P  v P  v P/F and resolve into components.
(a)
  0.
:
(r  e cos )AC  (e cos )BD  (cos  )u
(1)
:
(e sin  )AC  xP BD  (sin  )u
(2)
xA  0.48 m,
tan 
xP  0.48 m,
e cos 
0.14
,

xP
0.48
AC  20 rad/s
  16.26°
Substituting into Eqs. (1) and (2),
(0.20  0.14)(20)  0.14BD  (cos16.26)u
(1)
0  0.48BD  (sin16.26)u
(2)
Solving simultaneously,
 BD  3.81 rad/s,
v P/F  6.53 m/s
u  6.53 m/s,
(b)   90.
ωBD  3.81 rad/s

16.26 
 
xP  0.48  (0.20)    0.14  0.93416 m
2
 0
Substituting into Eqs. (1) and (2),
(0.20)(20)  u
(1)
u  4 m/s
(0.14)(20)  0.93416BD
BD  2.9973 rad/s,
(2)
BD  3.00 rad/s
v P/F  4.00 m/s
Copyright © McGraw-Hill Education. Permission required for reproduction or display.


PROBLEM 15.155
Knowing that at the instant shown the angular velocity of bar
AB is 15 rad/s clockwise and the angular velocity of bar EF is
10 rad/s clockwise, determine (a) the angular velocity of rod
DE, (b) the relative velocity of collar B with respect to rod DE.
SOLUTION
 EF  10 rad/s
v E   rE/F  EF
,
v B  v E  v B/E  150 in./s
  1510  150 in./s
   20DE 
v B/ED  u
v B  v B  v B/ED  150 in./s
v B   AB   AB
   20DE   u
 15 
45  
 15 
 cos 45 


45   225 in./s

   225 in./s

Equate the two expressions for v B and resolve into components.
(a)
(b)
:
150  u  225,
:
20DE  225,
u  75 in./s
 DE  11.25 rad/s
 DE  11.25 rad/s
v B/DE  75.0 in./s
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

PROBLEM 15.156
Knowing that at the instant shown the angular velocity of rod
DE is 10 rad/s clockwise and the angular velocity of bar EF is
15 rad/s counterclockwise, determine(a) the angular velocity of
bar AB, (b) the relative velocity of collar B with respect to rod
DE.
SOLUTION
 EF  15 rad/s
v E   rE/F EF
v B  v E  v B/ED   225 in./s
  15 15   225 in./s
   20DE    225 in./s
   200 in./s 
v B/ED  u
   200 in./s   u
v B  v B  v B/ED   225 in./s
 15
 AB
45   
 cos 45
v B   AB   AB

45  15 AB


  15 AB

Equate the two expressions for v B and resolve into components.
:
225  u  15 AB
:
200  15 AB
 AB 
(1)
200
 13.333 rad/s
15
 AB  13.33 rad/s
(a)
From (1),
(b)

u  225  1513.333  25 in./s
v B/ED  25.0 in./s
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
PROBLEM 15.157
The motion of pin P is guided by slots cut in rods AD and BE.
Knowing that bar AD has a constant angular velocity of 4 rad/s
clockwise and bar BE has an angular velocity of 5 rad/s
counterclockwise and is slowing down at a rate of 2 rad/s2,
determine the velocity of P for the position shown.
SOLUTION
Units: meters, m/s, m/s2
i 1
Unit vectors:
,
j 1 ,
k 1 .
Geometry: Slope angle  of rod BE.
tan  
0.15
 0.5
0.3
  26.565

rP /A  0.1i  0.15j
rP /B  0.3i  0.15j
Angular velocities:
ω AD  (4 rad/s)k
ωBE  (5 rad/s)k
Angular accelerations:
α AD  0
α BE  (2 rad/s2 )k
Velocity of Point P on rod AD coinciding with the pin:
v P  ω AD  rP /A  (4k )  (0.1i  0.15j)  0.6i  0.4 j
Velocity of the pin relative to rod AD:
v P /AD  u1  u1 j
v P  v P  v P /AD  0.6i  0.4 j  u1 j
Velocity of P:
Velocity of Point P on rod BE coinciding with the pin:
v P  ωBE  rP /B  5k  (0.3i  0.15j)  0.75i  1.5j
Velocity of the pin relative to rod BE:
v P /BE  u2
Velocity of P:
  u2 cos i  u2 sin  j
v P  v P  v P /BE
 0.75i  1.5 j  u2 cos  i  u2 sin  j
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.157 (Continued)
Equating the two expressions for vP and resolving into components,
i:
0.6  0.75  u2 cos 
u2  
j:
1.35
 1.50965
cos 26.565
0.4  u1  1.5  u2 sin 
u1  1.1  (1.50935)sin 26.535  1.77500
Velocity of P:
v P  0.6i  0.4 j  1.775j  0.6i  2.175j
v P  2.26 m/s
74.6° 
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PROBLEM 15.158
Four pins slide in four separate slots cut in a circular plate as shown. When
the plate is at rest, each pin has a velocity directed as shown and of the
same constant magnitude u. If each pin maintains the same velocity relative
to the plate when the plate rotates about O with a constant
counterclockwise angular velocity  , determine the acceleration of each
pin.
SOLUTION
For each pin:
aP  aP  aP/F  aC
Acceleration of the coinciding Point P of the plate.
2
For each pin a P  r towards the center O.
Acceleration of the pin relative to the plate.
For pins P1 , P2 and P4 ,
a P/F  0
For pin P3 ,
a P/F 
u2

r
Coriolis acceleration aC .
For each pin aC  2u with aC in a direction obtained by rotating u through 90 in the sense of , i.e., .
Then
a1  [r 2 ]  [2u ]
a1  r 2i  2uj 
a2  [r 2 ]  [2u ]
a2  2ui  r 2 j 
 2 
a3  [ r 2 ]   u    [2 u ]
 r


a4  [r 2 ]  [2u ]


u2
a3    r 2 
 2 u  i 
r


a4  (r 2  2u) j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.159
Solve Problem 15.158, assuming that the plate rotates about O with a
constant clockwise angular velocity  .
PROBLEM 15.158 Four pins slide in four separate slots cut in a circular
plate as shown. When the plate is at rest, each pin has a velocity directed as
shown and of the same constant magnitude u. If each pin maintains the
same velocity relative to the plate when the plate rotates about O with a
constant counterclockwise angular velocity  , determine the acceleration
of each pin.
SOLUTION
For each pin:
aP  aP  aP/F  aC
Acceleration of the coinciding Point P of the plate.
2
For each pin a P  r towards the center O.
Acceleration of the pin relative to the plate.
For pins P1 , P2 and P4 ,
aP/F  0
For pin P3 ,
a P/F 
u2

r
Coriolis acceleration aC .
For each pin aC  2u with aC in a direction obtained by rotating u through 90 in the sense of .
Then
a1  [r 2 ]  [2u ]
a1  r 2i  2uj 
a2  [r 2 ]  [2u ]
a2  2ui  r 2 j 
 2 
a3  [r 2 ]   u    [2 u ]
 r


a4  [r 2 ]  [2u ]

u2 
a3   2u  r 2   i 
r 

a4  (r 2  2u) j 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.160
The cage of a mine elevator moves downward at a constant speed of 12.2 m/s. Determine the magnitude and
direction of the Coriolis acceleration of the cage if the elevator is located (a) at the equator, (b) at latitude 40
north, (c) at latitude 40 south.
SOLUTION
Earth makes one revolution  2 radians  in 23.933 h (86160 s).

2
j  72.926  106 rad/s j
86160


Velocity relative to the Earth at latitude angle  .
v P/earth  12.2   cos  i  sin  j
Coriolis acceleration ac .
ac  2  v P/earth


cos   k
  2  72.926  106 j  12.2   cos  i  sin  j 

 1.7794  103
(a)   0, cos   1.000
ac  1.779  103 m/s2 west 
(b)   40, cos   0.76604
ac  1.363  103 m/s2 west 
(c)   40, cos   0.76604
ac  1.363  103 m/s2 west 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 15.161
Pin P is attached to the collar shown; the motion of the pin is guided by
a slot cut in bar BD and by the collar that slides on rod AE. Rod AE
rotates with a constant angular velocity of 6 rad/s clockwise and the
distance from A to P increases at a constant rate of 8 ft/s. Determine at
the instant shown (a) the angular acceleration of bar BD, (b) the relative
acceleration of pin P with respect to bar BD.
SOLUTION
AP  16 in.  1.3333 ft,
AE  6 rad/s
Given:
,
 AE  0,
 BD
Find:
BP  16 2 in.  1.3333 2 ft
and
v P/AE  8 ft/s ,
aP/ AE  0.
aP/BD.
Velocity of coinciding point P on rod AE.
v P   AP   AE  1.3333 6  8 ft/s
=  8 ft/s  i
v P/ AE   8 ft/s  j
Velocity of P relative to rod AE.
v P  v P  v P/ AE   8 ft/s  i   8 ft/s  j
Velocity of point P.
Velocity of coinciding point P on rod BD.
v P   BP  BD
45  1.3333 2 BD
45  1.3333BDi  1.3333 BD j
Velocity of P relative to rod BD.
v P/BD   cos 45  ui   sin 45  uj
Velocity of point P.
v P  v P  v P/BD
v P  1.3333BDi  1.3333BD j   cos 45 ui   sin 45 uj
Equating the two expressions for v P and resolving into components.
i : 8  1.3333BD   cos 45 u
j:
Solving (1) and (2),
(1)
8  1.3333BD   sin 45 u
 BD  0,
u  8 2 ft/s,
(2)
v P/BD   8 ft/s  i   8 ft/s  j
Acceleration of coinciding point P on rod AE.
2
a P   AP  AE i   AP   AE
j  0  1.3333 6  j    48 ft/s 2  j
2
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PROBLEM 15.161 (Continued)
aP/AE  0
Acceleration of P relative to rod AE.
2 AE  v P/ AE   2  6k   8 j   96 ft/s 2  i
Coriolis acceleration.
Acceleration of point P.
a P  a P  a P/ AE  2 AE  v P/ AE   96 ft/s 2  i   48 ft/s 2  j
Acceleration of coinciding point P on rod BD.
2
a P   BDk  rP/B   BD
rP/B  1.3333 BD i  1.3333 BD j  0
a P/BD   cos 45  ar i   sin 45  ar j
Acceleration of P relative to rod BD.
2BD  v P/BD  0
Coriolis acceleration.
Acceleration of point P.
aP  aP  aP/ AE  2BD  v P/BD
 1.3333 BDi  1.3333 BD j   cos 45 ar i   sin 45 ar j
Equating the two expressions for a P and resolving into components.
Solving (3) and (4),
(a)
(b)
i:
96  1.3333 BD   cos 45  ar
(3)
j:
 48  1.3333 BD   sin 45  ar
(4)
 BD  54 rad/s 2 ,
ar  24 2 ft/s
α BD  54 rad/s 2
a P/BD  33.9 ft/s 2
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
45 
PROBLEM 15.162
A rocket sled is tested on a straight track that is built along a meridian. Knowing that the track is located at
latitude 40 north, determine the Coriolis acceleration of the sled when it is moving north at a speed of 900 km/h.
SOLUTION
Earth makes one revolution (2 radians) in 23.933 h  86,160 s.

2
86,160
 (72.926  10 6 rad/s)j
u  900 km/h
 250 m/s
Speed of sled.
Velocity of sled relative to the Earth.
vP/earth  250(sin  i  cos j)
Coriolis acceleration.
aC  2  v P/earth
aC  (2)(72.926  106 j)  [250(sin  i  cos  j)]
 0.036463sin  k
At latitude   40,
aC  0.036463 sin 40°k
 (0.0234 m/s 2 )k

aC  0.0234 m/s2 west 
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PROBLEM 15.163
Solve the Geneva mechanism Sample Prob. 15.20 using vector
algebra.
SAMPLE PROBLEM 15.20 In the Geneva mechanism of Sample
Prob. 15.9, disk D rotates with a constant counterclockwise angular
velocity  D of 10 rad/s. At the