SOLUTION
MANUAL
CHAPTER 11
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) 
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3
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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4
PROBLEM 11.1
The motion of a particle is defined by the relation x = t 4 − 10t 2 + 8t + 12 , where x and t are expressed in
inches and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle
when t = 1 s.
SOLUTION
x = t 4 − 10t 2 + 8t + 12
At t = 1 s,
v=
dx
= 4t 3 − 20t + 8
dt
a=
dv
= 12t 2 − 20
dt
x = 1 − 10 + 8 + 12 = 11
x = 11.00 in. 
v = 4 − 20 + 8 = −8
v = −8.00 in./s 
a = 12 − 20 = −8 
a = −8.00 in./s 2 
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5
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=
a=
dx
= 6t 2 − 18t + 12 = 6(t 2 − 3t + 2)
dt
= 6(t − 2)(t − 1)
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
t = 1.000 s 
a1 = 12 − 18 = −6
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 
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6
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
= 20 cos 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
x1 = 102.9 mm 
v1 = 20cos 2 − 30sin 2 = −35.6
v1 = −35.6 mm/s 
a1 = −40sin 2 − 60 cos 2 = −11.40
a1 = −11.40 mm/s 2 
(b) Maximum velocity occurs when a = 0.
−40sin 2t − 60cos 2t = 0
tan 2t = −
60
= −1.5
40
2t = tan −1 (−1.5) = −0.9828 and −0.9828 + π
Reject the negative value. 2t = 2.1588
t = 1.0794 s
t = 1.0794 s for vmax
so
vmax = 20cos(2.1588) − 30sin(2.1588)
vmax = −36.1 mm/s 
= −36.056
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/s 2
amax = 72.1 mm/s 2 
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7
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
x0 = 0 mm 
v0 = 960 mm/s
a0 = −9216 mm/s 2
(b) At t = 0.3 s,
v0 = 960 mm/s

a0 = 9220 mm/s 2

x0.3 = 14.16 mm

v0.3 = 87.9 mm/s

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
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/s 2

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8
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
t = 0.667 s 
or
x2/3 = 0.259 m 
2
2
2
2
v2/3 = 24   − 6   − 24   + 3
3
3
3
or v2/3 = −8.56 m/s 
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9
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. 
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10
PROBLEM 11.7
The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x is expressed in meters
and t in seconds. 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
v=
dt
dv
a=
= 12t − 30
dt
(a)
v = 0 when
6t 2 − 30t + 24 = 0
6(t − 1)(t − 4) = 0
(b)
a = 0 when
t = 1.000 s and t = 4.00 s 
12t − 30 = 0
t = 2.5 s
x2.5 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4
For t = 2.5 s:
x2.5 = +1.500 m 
To find total distance traveled, we note that
x1 = 2(1)3 − 15(1)2 + 24(1) + 4
v = 0 when t = 1 s:
x1 = +15 m
For t = 0,
x0 = +4 m
Distance traveled
From t = 0 to t = 1 s:
x1 − x0 = 15 − 4 = 11 m
From t = 1 s to t = 2.5 s:
x2.5 − x1 = 1.5 − 15 = 13.5 m
Total distance traveled = 11 m + 13.5 m
Total distance = 24.5 m 
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11
PROBLEM 11.8
The motion of a particle is defined by the relation x = t 3 − 6t 2 − 36t − 40, where x and t are expressed in feet
and seconds, respectively. Determine (a) when the velocity is zero, (b) the velocity, the acceleration, and the
total distance traveled when x = 0.
SOLUTION
We have
x = t 3 − 6t 2 − 36t − 40
Then
v=
dx
= 3t 2 − 12t − 36
dt
and
a=
dv
= 6t − 12
dt
(a)
When v = 0:
3t 2 − 12t − 36 = 3(t 2 − 4t − 12) = 0
(t + 2)(t − 6) = 0
or
t = −2 s (Reject) and t = 6 s
or
(b)
When x = 0:
t 3 − 6t 2 − 36t − 40 = 0
Factoring
(t − 10)(t + 2)(t + 2) = 0 or t = 10 s
Now observe that
0 ≤ t < 6 s:
v<0
6 s < t ≤ 10 s:
v>0
and at t = 0:
t = 6 s:
t = 6.00 s 
x0 = −40 ft
x6 = (6)3 − 6(6)2 − 36(6) − 40
= −256 ft
t = 10 s:
Then
v10 = 3(10) 2 − 12(10) − 36
or v10 = 144.0 ft/s 
a10 = 6(10) − 12
or
a10 = 48.0 ft/s 2 
| x6 − x0 | = | − 256 − (−40)| = 216 ft
x10 − x6 = 0 − (−256) = 256 ft
Total distance traveled = (216 + 256) ft
Total distance = 472 ft 
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12
PROBLEM 11.9
The brakes of a car are applied, causing it to slow down at a
rate of 10 m/s2. Knowing that the car stops in 100 m,
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
dv
= a = −10
dx
0
xf
 vdv = − (−10)dx
0
v0
0−
(b)
v02
= −10 x f = −(10)(300)
2
v02 = 6000
v0 = 77.5 ft/s 2 
Time to stop.
dv
= a = −10
dx

0
v0
dv = −
tf
 −10dt
0
0 − v0 = −10t f
tf =
v0 77.5
=
10
10
t f = 7.75 s 
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13
PROBLEM 11.10
The acceleration of a particle is directly proportional to the time t. At t = 0, the velocity of the particle
is v = 16 in./s. Knowing that v = 15 in./s and that x = 20 in. when t = 1 s, determine the velocity, the position,
and the total distance traveled when t = 7 s.
SOLUTION
We have
a = kt
Now
dv
= a = kt
dt
v
At t = 0, v = 16 in./s:

or
v − 16 =
dv =
16
t
 kt dt
0
1 2
kt
2
v = 16 +
or
At t = 1 s, v = 15 in./s:
k = constant
1 2
kt (in./s)
2
1
k (1 s) 2
2
15 in./s = 16 in./s +
or
k = −2 in./s3
Also
dx
= v = 16 − t 2
dt
x
t
20
1
and v = 16 − t 2
At t = 1 s, x = 20 in.:
 dx =  (16 − t )dt
or
1 

x − 20 = 16t − t 3 
3 1

2
t
or
1
13
x = − t 3 + 16t + (in.)
3
3
Then
At t = 7 s:
When v = 0:
v7 = 16 − (7) 2
or v7 = −33.0 in./s 
1
13
x7 = − (7)3 + 16(7) +
3
3
or
x7 = 2.00 in. 
16 − t 2 = 0 or t = 4 s
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14
PROBLEM 11.10 (Continued)
At t = 0:
x0 =
13
3
1
13
x4 = − (4)3 + 16(4) + = 47 in.
3
3
t = 4 s:
Now observe that
Then
0 ≤ t < 4 s:
v>0
4 s < t ≤ 7 s:
v<0
13
= 42.67 in.
3
| x7 − x4 | = |2 − 47| = 45 in.
x4 − x0 = 47 −
Total distance traveled = (42.67 + 45) in.
Total distance = 87.7 in. 
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15
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
We have
a = kt 2
Now
dv
= a = kt 2
dt
v
t
k = constant
At t = 6 s, v = 18 m/s:

or
1
v − 18 = k (t 3 − 216)
3
18
dv =
 kt dt
2
6
or
1
v = 18 + k (t 3 − 216)(m/s)
3
Also
dx
1
= v = 18 + k (t 3 − 216)
dt
3
x
t
1

At t = 0, x = 24 m:

or
1 1

x − 24 = 18t + k  t 4 − 216t 
3 4

24
dx =
 18 + 3 k (t − 216) dt
3
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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16
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
8
2
−8
0
 dv =  kt dt
(a)
2
1
8 − ( −8) = k (2)3
3
(b)
k = 6.00 m/s 4 
Substituting k = 6 m/s 4 into (1)
dv
= a = 6t 2
dt
t = 0, v = −8 m/s:

v
−8
dv =
t
a = 6t 2 
 6t dt
2
0
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
t
(2t 3 − 8) dt ; x =
1 4
t − 8t
2
2
1
 1

x =  t 4 − 8t  −  (2) 4 − 8(2) 
2
 2

1
x = t 4 − 8t − 8 + 16
2
x=
1 4
t − 8t + 8 
2
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17
PROBLEM 11.13
The acceleration of Point A is defined by the relation a = −1.8sin kt , where a and t are
expressed in m/s 2 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
Given:
a = −1.8sin kt m/s 2 ,
v0 = 0.6 m/s, x0 = 0,
t
t
v − v0 =  0 a dt = −1.8  0 sin kt dt =
v − 0.6 =
t
1.8
cos kt
0
k
1.8
(cos kt − 1) = 0.6cos kt − 0.6
3
v = 0.6cos kt m/s
Velocity:
t
t
x − x0 =  0 v dt = 0.6  0 cos kt dt =
x−0=
t
0.6
sin kt
0
k
0.6
(sin kt − 0) = 0.2sin kt
3
x = 0.2sin kt m
Position:
When t = 0.5 s,
k = 3 rad/s
kt = (3)(0.5) = 1.5 rad
v = 0.6cos1.5 = 0.0424 m/s
v = 42.4 mm/s 
x = 0.2sin1.5 = 0.1995 m
x = 199.5 mm 
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18
PROBLEM 11.14
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
Given:
a = −1.08sin kt − 1.44 cos kt m/s 2 ,
x0 = 0.16 m,
k = 3 rad/s
v0 = 0.36 m/s
t
t
t
v − v0 =  0 a dt = −1.08  0 sin kt dt − 1.44  0 cos kt dt
v − 0.36 =
=
t
t
1.08
1.44
cos kt −
sin kt
0
0
k
k
1.08
1.44
(cos kt − 1) −
(sin kt − 0)
3
3
= 0.36 cos kt − 0.36 − 0.48sin kt
Velocity:
v = 0.36 cos kt − 0.48sin kt m/s
t
t
t
x − x0 =  0 v dt = 0.36  0 cos kt dt − 0.48  0 sin kt dt
x − 0.16 =
t
t
0.36
0.48
sin kt +
cos kt
0
0
k
k
0.36
0.48
(sin kt − 0) +
(cos kt − 1)
3
3
= 0.12sin kt + 0.16cos kt − 0.16
=
Position:
When t = 0.5 s,
x = 0.12sin kt + 0.16cos kt m
kt = (3)(0.5) = 1.5 rad
v = 0.36 cos1.5 − 0.48sin1.5 = −0.453 m/s
v = −453 mm/s 
x = 0.12sin1.5 + 0.16 cos1.5 = 0.1310 m
x = 131.0 mm 
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19
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
xf
v0
0
 vdv = − 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.
1
1
0 − (4) 2 = − k (0.02) 2
2
2
k = 40,000 s −2
Maximum acceleration.
amax = −kxmax : ( −40,000)(0.02) = −800 m/s 2
a = 800 m/s 2

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20
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/s 2 
or
(b)
dx
= v = v0 − kx
dt
We have
At t = 0, x = 0:
dx
=
0 v − kx
0

x
t
 dt
0
or
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


1
ln

2700
1/s
3.9
1
1
ft
−
) 
2700 s 
900 ft/s ( 12
t = 1.366 × 10−3 s 
or
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21
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
v

x
dx
0.6 x
x
1 2
v
= − k ln x
2 15
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 /s 2
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 
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22
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 /s 2 . Determine the maximum velocity and
acceleration of B.
SOLUTION
The maximum veolocity occurs when a = 0.
xm2 =
0 = −9.81 +
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 +
k dx
x2
v
x
x dx
0
x0
x0
 vdv = −9.81 dx + k  x
2
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 /s2
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/s 2

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23
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
dv
x 

= a = −  0.1 + sin
dx
0.8 

v
x
1
0

x 
 vdv =  −  0.1 + sin 0.8  dx
x
1 2
x 

(v − 1) = − 0.1x − 0.8 cos
2
0.8  0

or
x
1 2
v = −0.1x + 0.8 cos
− 0.3
2
0.8
or
(a)
When x = −1 m:
1 2
−1
v = −0.1(−1) + 0.8 cos
− 0.3
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
vmax = 1.004 m/s 
or
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24
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 + x 2 

Separate variables and integrate:

vf
v0
vdv = −100

0

lx
x−
 dx
x0 
l 2 + x 2 

 x2

1 2 1 2
− l l 2 + x 2 
v f − v0 = −100 
2
2
 2

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 ) 
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25
PROBLEM 11.21
The acceleration of a particle is defined by the relation a = −0.8v where a is expressed in m/s2 and
v in m/s. Knowing that at t = 0 the velocity is 1 m/s, determine (a) the distance the particle will travel
before coming to rest, (b) the time required for the particle’s velocity to be reduced by 50 percent of its
initial value.
SOLUTION
(a)
Determine relationship between x and v.
a=
vdv
= −0.8v
dx
dv = −0.8dx
Separate and integrate with v = 1 m/s when x = 0.
v
x
1
0
 dv = −0.8 dx
v − 1 = −0.8 x
Distance traveled.
For v = 0,
(b)
x=
−1

−0.8
x = 1.25 m 
Determine realtionship between v and t.
a=
v dv
 v
1
dv
= 0.8v
dt
=−
x
 0.8dt
0
v
ln   = −0.8t
1
1
t = 1.25ln  
v
For v = 0.5(1 m/s) = 0.5 m/s,
 1 
t = 1.25ln 

 0.5 
t = 0.866 s 
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26
PROBLEM 11.22
Starting from x = 0 with no initial velocity, a particle is given an acceleration a = 0.1 v 2 + 16,
where a and v are expressed in ft/s2 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
vdv
0
v 2 + 16
=
x
 0.1 dx
0
v
(v 2 + 16)1/2 = 0.1 x
0
2
1/2
(v + 16)
(a)
(b)
− 4 = 0.1 x
x = 10[(v 2 + 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
From (1),
v 2 = 3.36 ft 2 /s 2
v = 1.833 ft/s 
a = 0.1(1.8332 + 16)1/2
a = 0.440 ft/s 2 
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27
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
 dt
0
v
−
1
ln(10 − 0.8v) = t
0.8
v0
 10 − 0.8v 
ln 
 = −0.8t
 10 − 0.8v0 
10 − 0.8v = (10 − 0.8v0 )e −0.8t
or
0.8v = 10 − (10 − 0.8v0 )e−0.8t
v = 12.5 − (12.5 − v0 )e −0.8t
With v0 = 16.5ft/s
v = 12.5 + 4e−0.8t
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28
PROBLEM 11.23 (Continued)
Integrate to determine x as a function of t.
dx
= 12.5 + 4e−0.8t
v=
dt
x
t
0
0
 dx =  (12.5 + 4e
−0.8t
x = 12.5t − 5e −0.8t
(a)
t
0
)dt
= 12.5t − 5e−0.8t + 5
At t = 35 s,
x = 12.5(3) − 5e −2.4 + 5 = 42.046 ft
(b)
x = 42.0 ft 
v = 12.5 + 4e −2.4 = 12.863 ft/s
v = 12.86 ft/s 
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29
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
x
When x = 0, v = 81 m/s:

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:
2
(363/2 − 729) = − k (18)
3
v dv =
81
 −k dx
0
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:

or
v
2[ v ]81
= −19t
or
2( v − 9) = −19t
When v = 0:
v dv
81
v
=
v = 29.3 m/s 
t
 −19dt
0
2(−9) = −19t
t = 0.947 s 
or
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30
PROBLEM 11.25
A particle is projected to the right from the position x = 0 with an initial velocity of 9 m/s. If the acceleration
of the particle is defined by the relation a = −0.6v3/ 2 , where a and v are expressed in m/s2 and m/s,
respectively, determine (a) the distance the particle will have traveled when its velocity is 4 m/s, (b) the time
when v = 1 m/s, (c) the time required for the particle to travel 6 m.
SOLUTION
(a)
We have
When x = 0, v = 9 m/s:
v

v
9
dv
= a = −0.6v3/2
dx
v − (1/2) dv =
x
 −0.6dx
0
or
2[v1/2 ]9v = −0.6 x
or
x=
1
(3 − v1/ 2 )
0.3
When v = 4 m/s:
x=
1
(3 − 41/ 2 )
0.3
x = 3.33 m 
or
(b)
dv
= a = −0.6v3/2
dt
We have
v
t
When t = 0, v = 9 m/s:
v
or
−2[v − (1/2) ]9v = −0.6t
or
When v = 1 m/s:
− (3/2)
9
1
v
1
1
dv =
 −0.6dt
0
−
1
= 0.3t
3
−
1
= 0.3t
3
t = 2.22 s 
or
(c)
We have
(1)
1
v
−
1
= 0.3t
3
2
or
9
 3 
v=
 =
1
0.9
t
+
(1 + 0.9t ) 2


Now
dx
9
=v=
dt
(1 + 0.9t ) 2
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31
PROBLEM 11.25 (Continued)
x
t
0
0
9
 dx =  (1 + 0.9t ) dt
At t = 0, x = 0:
2
t
1 
 1
x = 9 −

 0.9 1 + 0.9t  0
or
1 

= 10 1 −

 1 + 0.9t 
9t
=
1 + 0.9t
When x = 6 m:
6=
9t
1 + 0.9t
t = 1.667 s 
or
An alternative solution is to begin with Eq. (1).
x=
1
(3 − v1/ 2 )
0.3
dx
= v = (3 − 0.3x) 2
dt
Then
Now
At t = 0, x = 0:

dx
=
0 (3 − 0.3 x ) 2
x
t
 dt
0
x
or
1  1 
x
=
t=


0.3  3 − 0.3x  0 9 − 0.9 x
which leads to the same equation as above.
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32
PROBLEM 11.26
The acceleration of a particle is defined by the relation a = 0.4(1 − kv), where k is a constant. Knowing that
at t = 0 the particle starts from rest at x = 4 m and that when t = 15 s, v = 4 m/s, determine (a) the constant k,
(b) the position of the particle when v = 6 m/s, (c) the maximum velocity of the particle.
SOLUTION
(a)
dv
= a = 0.4(1 − kv)
dt
We have
dv
v
t
 1 − k v =  0.4dt
At t = 0, v = 0:
0
0
1
− [ln(1 − k v)]v0 = 0.4t
k
or
or
ln(1 − k v) = −0.4 kt
At t = 15 s, v = 4 m/s:
ln(1 − 4k ) = −0.4k (15)
(1)
= −6k
k = 0.145703 s/m
Solving yields
k = 0.1457 s/m 
or
(b)
We have
v
v
dv
= a = 0.4(1 − kv)
dx
vdv
x
 1 − k v =  0.4dx
When x = 4 m, v = 0:
0
4
v
1
1/k
=− +
1− kv
k 1− kv
Now
v
1
1

x
Then
 − k + k (1 − k v)  dv =  0.4dx
or
 v 1

x
 − k − k 2 ln(1 − k v)  = 0.4[ x]4

0
or
v 1

−  + 2 ln(1 − kv)  = 0.4( x − 4)
k
k


When v = 6 m/s:


6
1
ln(1 − 0.145 703 × 6)  = 0.4( x − 4)
−
+
2
 0.145 703 (0.145 703)

0
4
v
or
0.4( x − 4) = 56.4778
x = 145.2 m 
or
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33
PROBLEM 11.26 (Continued)
(c)
The maximum velocity occurs when a = 0.
a = 0: 0.4(1 − k vmax ) = 0
vmax =
or
1
0.145 703
vmax = 6.86 m/s 
or
An alternative solution is to begin with Eq. (1).
ln(1 − k v) = −0.4 kt
v=
Then
Thus, vmax is attained as t
1
(1 − k −0.4 kt )
k
∞
vmax =
1
k
as above.
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34
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/s 2 
or
(b)
0.18v0
dx
=v=
dt
x
We have
3
t3
From x = 1 m to x = 3 m:

or
1 2 
 2 x  = 0.18v0 (t3 − t1 )

1
or
(t3 − t1 ) =
1
xdx =
 0.18v dt
0
t1
3
1 (9 − 1)
2
0.18(3.6)
t3 − t1 = 6.17 s 
or
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35
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
x
dx
 (1 − 0.04 x)
At t = 0, x = 0:
0
0.3
=
t
 7.5dt
0
or
1 
1 
[(1 − 0.04 x)0.7 ]0x = 7.5t
−

0.7  0.04 
or
1 − (1 − 0.04 x)0.7 = 0.21t
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)
2
From Eq. (1)
t=
When x = 6 mi:
t=
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 
or
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36
PROBLEM 11.29
The acceleration due to gravity at an altitude y above the surface of the earth can be
expressed as
a=
−32.2
[1 + ( y/20.9 × 106 )]2
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
32.2
(1 +
y
20.9 × 106
When
y = 0,
provided that v does reduce to zero,
y = ymax , v = 0
2
v = v0
0
ymax
v0
0
 v dv = 
Then
)
−32.2
(
1 + 20.9 y× 106
)
2
dy
y
or

 max
1 2
1
6

− v0 = −32.2  −20.9 × 10
y


2
1+
6
20.9 × 10  0

or


1

v02 = 1345.96 × 106 1 −
 1 + ymax 
20.9 × 106 

ymax =
or
(a)
v0 = 1800 ft/s:
ymax =
v02
64.4 −
v02
20.9 × 106
(1800)2
2
64.4 − (1800) 6
20.9 × 10
ymax = 50.4 × 103 ft 
or
(b)
v0 = 3000 ft/s:
ymax =
(3000) 2
2
64.4 − (3000) 6
20.9 × 10
ymax = 140.7 × 103 ft 
or
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37
PROBLEM 11.29 (Continued)
(c)
v0 = 36,700 ft/s:
ymax =
(36, 700) 2
2
64.4 − (36,700)6
20.9 × 10
= −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
∞ 
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38
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
0
gR 2
dr
r2

or
1
1 
− ve2 = gR 2  
2
 r R
or
ve = 2 gR
ve
vdv =

∞
then
R
−
∞
1/2
5280 ft 

=  2 × 32.2 ft/s 2 × 3960 mi ×

1 mi 

ve = 36.7 × 103 ft/s 
or
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39
PROBLEM 11.31
The velocity of a particle is v = v0 [1 − sin (π 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
 π t 
= v = v0 1 − sin   
dt
 T 

We have
At t = 0, x = 0:
x

t
 π t 
 dx =  v 1 − sin  T  dt
0
0
0
t
At t = 3T :
 T
 T
 π t 
 πt  T 
x = v0 t + cos    = v0 t + cos   − 
 T 0
T  π
 π
 π
(1)

T
2T 
 π × 3T  T 

−  = v0  3T −
x3T = v0 3T + cos 

π
π 
 T  π


x3T = 2.36 v0T 
a=
At t = 3T :
(b)
π
πt
dv d  
 π t   
= v0 1 − sin     = −v0 cos
dt dt  
T
T
 T   
a3T = −v0
π
T
cos
π × 3T
a3T =
T
π v0
T

Using Eq. (1)
At t = 0:
T
T

x0 = v0 0 + cos(0) −  = 0
π
π


At t = T :

T
2T 
 πT  T 

−  = v0  T −
= 0.363v0T
xT = v0 T + cos 

π
π 
 T  π


Now
vave =
xT − x0 0.363v0T − 0
=
Δt
T −0
vave = 0.363v0 
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40
PROBLEM 11.32
The velocity of a slider is defined by the relation v = v′ sin (ωn t + φ ). Denoting the velocity and the position
of the slider at t = 0 by v0 and x0 , respectively, and knowing that the maximum displacement of the slider
is 2 x0 , show that (a) v′ = (v02 + x02ωn2 )/2 x0ωn , (b) the maximum value of the velocity occurs when
x = x0 [3 − (v0 /x0ωn ) 2 ]/2.
SOLUTION
(a)
v0 = v′ sin (0 + φ ) = v′ sin φ
At t = 0, v = v0 :
Then
2
cos φ = v′ − v02 v′
dx
= v = v′ sin (ωn t + φ )
dt
Now
x
t
At t = 0, x = x0 :

or
 1

x − x0 = v′  −
cos (ωn t + φ ) 
 ωn
0
x0
dx =
 v′ sin (ω t + φ )dt
n
0
t
or
x = x0 +
v′
ωn
[cos φ − cos (ωn t + φ )]
Now observe that xmax occurs when cos (ωn t + φ ) = −1. Then
xmax = 2 x0 = x0 +
ωn
[cos φ − ( −1)]
2
2

v′  v′ − v0

x0 =
+
1

ωn 
v1


Substituting for cos φ
or
v′
x0ωn − v′ = v′2 − v02
Squaring both sides of this equation
x02ωn2 − 2 x0ωn + v′2 = v′2 − v02
or
v′ =
v02 + x02ωn2
2 x0ωn
Q. E. D.
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41
PROBLEM 11.32 (Continued)
(b)
First observe that vmax occurs when ωn t + φ = π2 . The corresponding value of x is
v′ 
 π 
cos φ − cos   
ωn 
 2 
v′
= x0 +
cos φ
xvmax = x0 +
ωn
Substituting first for cos φ and then for v′
xvmax = x0 +
2
v′ − v02
v′
v′
ωn
1/2
2


1  v02 + x02ωn2 
2
= x0 +

 − v0 

ωn  2 x0ωn 


1/2
1
v04 + 2v02 x02ωn2 + x04ωn4 − 4 x02ωn2 v02
= x0 +
2
2 x0ωn
(
= x0 +
= x0 +
)
1
(
)
1/ 2
 x 2ω 2 − v 2 2 
0
2  0 n

2 x0ωn 
x02ωn2 − v02
2 x0ωn2
2
x0   v0  

=
3−
 
2   x0ωn  


Q. E. D.
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42
PROBLEM 11.33
A stone is thrown vertically upward from a point on a bridge located 40 m above the water. Knowing that it
strikes the water 4 s after release, determine (a) the speed with which the stone was thrown upward, (b) the
speed with which the stone strikes the water.
SOLUTION
Uniformly accelerated motion. Origin at water.
1
y = y0 + v0t + a t 2
2
v = v0 + a t
where y0 = 40 m and a = −9.81 m/s2 .
(a)
Initial speed.
y = 0 when t = 4 s.
1
0 = 40 + v0 (4) − (9.81)(4)2
2
v0 = 9.62 m/s
(b)
v0 = 9.62 m/s 
Speed when striking the water. (v at t = 4 s)
v = 9.62 − (9.81)(4) = −29.62 m/s
v = 29.6 m/s 
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43
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 v0 = 54 km/h = 15 m/s
(a)
x = x0 + v0t +
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/s 2 
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 
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44
PROBLEM 11.35
A motorist enters a freeway at 30 mi/h and accelerates uniformly
to 60 mi/h. From the odometer in the car, the motorist knows
that she traveled 550 ft while accelerating. Determine (a) the
acceleration of the car, (b) the time required to reach 60 mi/h.
SOLUTION
(a)
Acceleration of the car.
v12 = v02 + 2a( x1 − x0 )
a=
Data:
v12 − v02
2( x1 − x0 )
v0 = 30 mi/h = 44 ft/s
v1 = 60 mi/h = 88 ft/s
x0 = 0
x1 = 550 ft
a=
(b)
(88) 2 − (44) 2
(2)(55 − 0)
a = 5.28 ft/s 2 
Time to reach 60 mi/h.
v1 = v0 + a (t1 − t0 )
v −v
t1 − t0 = 1 0
a
88 − 44
=
5.28
= 8.333 s
t1 − t0 = 8.33 s 
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45
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 = ymax , v = 0
Then
0 = (252 ft/s) 2 + 2( −32.2 ft/s 2 )( ymax − 89.6) ft
ymax = 1076 ft 
or
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46
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/s 2 )(3 − 0) m
Then, at B
= 28.8 m 2 /s 2
(vBC = 5.3666 m/s)
2
vD2 = vBC
+ 2aCD ( xD − xC )
and at D
d = xD − 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/s 2 )t AB
t AB = 1.11804 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 = xB + vBC t BC
or
3 m = (5.3666 m/s)t BC
or
t BC = 0.55901 s
Finally,
t D = t AB + t BC + tCD = (1.11804 + 0.55901 + 0.38196) s
t D = 2.06 s 
or
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47
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
Given:
0 ≤ x ≤ 35 m, a = constant
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:
vmax = 12.96 m/s 
x = x1 + v0 (t − t1 )
for
35 m < x ≤ 100 m
100 m = 35 m + (12.9628 m/s)(t2 − 5.4) s
t2 = 10.41 s 
or
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48
PROBLEM 11.39
As relay runner A enters the 20-m-long exchange zone with a
speed of 12.9 m/s, he begins to slow down. He hands the baton to
runner B 1.82 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
(a)
For runner A:
At t = 1.82 s:
x A = 0 + (v A )0 t +
1
a At 2
2
20 m = (12.9 m/s)(1.82 s) +
1
a A (1.82 s) 2
2
a A = −2.10 m/s 2 
or
Also
At t = 1.82 s:
v A = (v A ) 0 + a A t
(v A )1.82 = (12.9 m/s) + ( −2.10 m/s 2 )(1.82 s)
= 9.078 m/s
For runner B:
vB2 = 0 + 2aB [ xB − 0]
When
xB = 20 m, vB = v A : (9.078 m/s) 2 = 2aB (20 m)
or
aB = 2.0603 m/s 2
aB = 2.06 m/s 2 
(b)
For runner B:
vB = 0 + aB (t − t B )
where t B is the time at which he begins to run.
At t = 1.82 s:
or
9.078 m/s = (2.0603 m/s 2 )(1.82 − t B ) s
t B = −2.59 s
Runner B should start to run 2.59 s before A reaches the exchange zone. 
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49
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 A t
(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/s 2 
or
(b)
We have
x A = ( x A ) 0 + (v A )0 t +
1
1
a At 2 = 50 m + (50 m/s)(8 s) + (1.5625 m/s 2 )(8 s) 2 = 500 m
2
2
and
xB = 0 + (vB )0 t +
At t = 8 s:
x A = xB
1
aB t 2
2
500 m = (50 m/s)(8 s) +
(vB )0 = 50 m/s
1
aB (8 s) 2
2
aB = 3.13 m/s 2 
or
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50
PROBLEM 11.41
A police officer in a patrol car parked in a 45 mi/h speed zone observes a passing automobile traveling at a
slow, constant speed. Believing that the driver of the automobile might be intoxicated, the officer starts his
car, accelerates uniformly to 60 mi/h in 8 s, and, maintaining a constant velocity of 60 mi/h, overtakes the
motorist 42 s after the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing
the motorist, determine (a) the distance the officer traveled before overtaking the motorist, (b) the motorist’s
speed.
SOLUTION
(vP )18 = 0
(a)
(vP )26 = 60 mi/h = 88 ft/s
(vP ) 42 = 90 mi/h = 88 ft/s
Patrol car:
For 18 s < t ≤ 26 s:
At t = 26 s:
vP = 0 + aP (t − 18)
88 ft/s = aP (26 − 18) s
or
aP = 11 ft/s 2
Also,
xP = 0 + 0(t − 18) −
At t = 26 s:
For 26 s < t ≤ 42 s:
At t = 42 s:
(xP ) 26 =
1
aP (t − 18) 2
2
1
(11 ft/s 2 )(26 − 18) 2 = 352 ft
2
xP = ( xP )26 + (vP ) 26 (t − 26)
(xP ) 42 = 352 m + (88 ft/s)(42 − 26) s
= 1760 ft
( xP ) 42 = 1760 ft 
(b)
For the motorist’s car:
xM = 0 + vM t
At t = 42 s, xM = xP :
1760 ft = vM (42 s)
or
vM = 41.9048 ft/s
vM = 28.6 mi/h 
or
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51
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 At = 35.2 + 1.8t
x A = ( x A ) 0 + (v A ) 0 t +
1
1
a At 2 = 0 + 35.2t + (1.8)t 2
2
2
(1)
(2)
Motion of auto B:
vB = (vB )0 + aB t = 52.8 − 1.2t
x B = ( xB ) 0 + ( vB ) 0 t +
(a)
1
1
aB t 2 = 75 + 52.8t + (−1.2)t 2
2
2
(3)
(4)
A overtakes B at t = t1 .
x A = xB : 35.2t + 0.9t12 = 75 + 52.8t1 − 0.6t12
1.5t12 − 17.6t1 − 75 = 0
Eq. (2):
t1 = −3.22 s and t1 = 15.0546
t1 = 15.05 s 
x A = 35.2(15.05) + 0.9(15.05) 2
x A = 734 ft 
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52
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

vB = 23.7 mi/h

vB = 52.8 − 1.2(15.05)
vB = 34.74 ft/s
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53
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 vB = 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
1
aB (42 s) 2
2
aB = 0.83447 ft/s 2
When the cars pass each other
Then (95.333 ft/s)t AB +
(c)
a A = −0.767 ft/s 2 
(vB )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/s 2 
x A + xB = 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
Solving
t = 20.685 s and t = −4563 s
We have
vB = ( vB ) 0 + a B t
At t = t AB :
vB = 58.667 ft/s + (0.83447 ft/s 2 )(20.685 s)
= 75.927 ft/s
t >0
t AB = 20.7 s 
vB = 51.8 mi/h 
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54
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.
yB = ( yB )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 ( yE )0 = −10 m and vE = 4 m/s.
(a)
Set yB = yE
Time of impact.
3t − 4.905t 2 = −10 + 4t
4.905t 2 + t − 10 = 0
t = 1.3295 and −1.5334
(b)
t = 1.330 s 
Location of impact.
yB = (3)(1.3295) − (4.905)(1.3295)2 = −4.68 m
yE = −10 + (4)(1.3295) = −4.68 m
(checks)
4.68 m below the man 
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55
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 2
gt = 100t − 4.905t 2
2
(1)
(2)
vB = (vB )0 − g (t − t1 ) = 100 − 9.81(t − t1 )
(3)
1
g (t − t1 ) 2
2
= 100(t − t1 ) − 4.905(t − t1 )2
(4)
yB = ( yB )0 + (vB )0 (t − t1 ) −
Time of explosion of rockets A and B. y A = yB = 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 and 3.655 s
Since rocket A is falling,
t = 16.732 s
Since rocket B is rising,
t − t1 = 3.655 s
(a)
Time t1:
(b)
Relative velocity at explosion.
t1 = t − (t − t1 )
From (1),
v A = 100 − (9.81)(16.732) = −64.15 m/s
From (3),
vB = 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 
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56
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 = vB , 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 (vB )0 = 60 mi/h = 88 ft/s
At t = 5 s:
xB = (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
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
a t2
2 A AB
=
aA
(t AB − 5)
6
1
2
a At AB
2
88
294
2
44t AB
− 343t AB + 245 = 0
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57
PROBLEM 11.46 (Continued)
Solving
(a)
With t AB > 5 s,
t AB = 0.795 s and t AB = 7.00 s
294 ft =
1
a A (7.00 s) 2
2
a A = 12.00 ft/s 2 
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

d = 906 ft 
or
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58
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 yE = constant
vC + 2vE = 0
vE = 4 m/s
But,
or
(b)
vC = −2vE = −8 m/s
vC = 8.00 m/s 
Velocity of counterweight W.
yW + yE = constant
vW + vE = 0 vW = −vE = −4 m/s
(c)
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 
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59
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/s 2
(a)
Accelerations of E and C.
Since
Thus:
Also,
Thus:
(b)
aW = 2.4 ft/s 2
yW + yE = constant vW + vE = 0, and aW + aE = 0
aE = − aW = −(2.4 ft/s 2 ),
a E = 2.40 ft/s 2 
yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0
aC = −2aE = −2(−2.4 ft/s 2 ) = +4.8 ft/s 2 ,
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 
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60
PROBLEM 11.49
Slider block A moves to the left with a constant velocity of 6 m/s.
Determine (a) the velocity of block B, (b) the velocity of portion D of
the cable, (c) the relative velocity of portion C of the cable with respect
to portion D.
SOLUTION
From the diagram, we have
x A + 3 yB = constant
Then
v A + 3vB = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Substituting into Eq. (1)
6 m/s + 3vB = 0
v B = 2.00 m/s 
or
(b)
From the diagram
yB + yD = constant
Then
vB + vD = 0

v D = 2.00 m/s 
(c)
From the diagram
x A + yC = constant
Then
v A + vC = 0
Now
vC/D = vC − vD = (−6 m/s) − (2 m/s) = −8 m/s
vC = −6 m/s
vC/D = 8.00 m/s 
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61
PROBLEM 11.50
Block B starts from rest and moves downward with a constant
acceleration. Knowing that after slider block A has moved 9 in. its
velocity is 6 ft/s, determine (a) the accelerations of A and B, (b) the
velocity and the change in position of B after 2 s.
SOLUTION
From the diagram, we have
x A + 3 yB = constant
Then
v A + 3vB = 0
(1)
and
a A + 3aB = 0
(2)
(a)
Eq. (2): a A + 3aB = 0 and a B is constant and
positive  a A is constant and negative
Also, Eq. (1) and (vB )0 = 0  (v A )0 = 0
v A2 = 0 + 2a A [ x A − ( x A )0 ]
Then
When |Δ x A | = 0.4 m:
(6 ft/s) 2 = 2a A (9/12 ft)
a A = 24.0 ft/s 2
or

Then, substituting into Eq. (2):
−24 ft/s 2 + 3aB = 0
aB =
or
24
ft/s 2
3
a B = 8.00 ft/s 2 
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62
PROBLEM 11.50 (Continued)
(b)
We have
vB = 0 + a B t
At t = 2 s:
 24

vB = 
ft/s 2  (2 s)
 3

v B = 16.00 ft/s 
or
yB = ( yB )0 + 0 +
Also
At t = 2 s:
or
y B − ( y B )0 =
1
aB t 2
2
1  24

ft/s 2  (2 s) 2 
2  3

y B − (y B )0 = 16.00 ft 

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63
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
From the diagram
xB + ( xB − x A ) − 2 x A = constant
Then
2vB − 3v A = 0
(1)
and
2aB − 3a A = 0
(2)
Also, we have
− xD − x A = constant
Then
vD + v A = 0
(a)
Substituting into Eq. (1)
2(300 mm/s) − 3v A = 0
or
(b)
From the diagram
Then
Substituting
(3)
v A = 200 mm/s

vC = 600 mm/s

xB + ( xB − xC ) = constant
2vB − vC = 0
2(300 mm/s) − vC = 0
or
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64
PROBLEM 11.51 (Continued)
(c)
From the diagram
Then
Substituting
( xC − x A ) + ( xD − x A ) = constant
vC − 2v A + vD = 0
600 mm/s − 2(200 mm/s) + vD = 0
or
(d)
We have
v D = 200 mm/s

vC/A = 400 mm/s

vC/A = vC − v A
= 600 mm/s − 200 mm/s
or
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65
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
xB + ( xB − x A ) − 2 x A = constant
From the diagram
Then
2vB − 3v A = 0
(1)
and
2aB − 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 aB = constant  a A = constant
v A2 = (v A )02 + 2a A [ x A − ( x A )0 ]
Then
When x A − ( x A )0 = 240 mm:
(60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm)
or
aA = −
40
mm/s 2
3
a A = 13.33 mm/s 2
or

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66
PROBLEM 11.52 (Continued)
Then, substituting into Eq. (2)
 40

2aB − 3  −
mm/s 2  = 0
 3

aB = −20 mm/s 2
or
(b)
a B = 20.0 mm/s 2

a D = 13.33 mm/s 2

v B = 70.0 mm/s

From the diagram, − xD − x A = constant
vD + v A = 0
Then
Substituting
aD + a A = 0
 40

aD +  −
mm/s 2  = 0
3


or
(c)
We have
vB = ( vB ) 0 + a B t
At t = 4 s:
vB = 150 mm/s + ( −20.0 mm/s 2 )(4 s)
or
Also
At t = 4 s:
x B = ( xB ) 0 + ( vB ) 0 t +
1
aB t 2
2
xB − ( xB )0 = (150 mm/s)(4 s)
+
1
(−20.0 mm/s 2 )(4 s) 2
2
x B − (x B )0 = 440 mm
or

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67
PROBLEM 11.53
Collar A starts from rest and moves upward with a constant acceleration. Knowing
that after 8 s the relative velocity of collar B with respect to collar A is 24 in./s,
determine (a) the accelerations of A and B, (b) the velocity and the change in position
of B after 6 s.
SOLUTION
From the diagram
2 y A + yB + ( yB − y A ) = constant
Then
v A + 2 vB = 0
(1)
and
a A + 2aB = 0
(2)
(a)
Eq. (1) and (v A )0 = 0  (vB )0 = 0
Also, Eq. (2) and a A is constant and negative  a B is constant and
positive.
Then
v A = 0 + a At
vB = 0 + a B t
Now
vB/A = vB − v A = (aB − a A )t
From Eq. (2)
1
aB = − a A
2
So that
3
vB/A = − a At
2
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68
PROBLEM 11.53 (Continued)
At t = 8 s:
3
24 in./s = − a A (8 s)
2
a A = 2.00 in./s 2 
or
and then
1
aB = − (−2 in./s 2 )
2
a B = 1.000 in./s 2 
or
(b)
At t = 6 s:
vB = (1 in./s 2 )(6 s)
v B = 6.00 in./s 
or

Now
At t = 6 s:
yB = ( yB )0 + 0 +
y B − ( y B )0 =
1
aB t 2
2
1
(1 in./s 2 )(6 s) 2
2
y B − (y B )0 = 18.00 in. 
or
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69
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
2v L + v B = 0
vM = 100 mm/s
with
v
vB = − M = −33.333 m/s
3
vB
vL =
= −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 
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70
PROBLEM 11.55
Block C starts from rest at t = 0 and moves downward with a constant
acceleration of 4 in./s2. Knowing that block B has a constant velocity of 3 in./s
upward, determine (a) the time when the velocity of block A is zero, (b) the
time when the velocity of block A is equal to the velocity of block D, (c) the
change in position of block A after 5 s.
SOLUTION
From the diagram:
Cord 1:
2 y A + 2 y B + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
( y D − y A ) + ( y D − y B ) = constant
Then
2vD − v A − v B = 0
and
2aD − a A − aB = 0
(1)
(2)
Use units of inches and seconds.
Motion of block C:
vC = vC 0 + aC t
where aC = −4 in./s 2
= 0 + 4t
Motion of block B:
vB = −3 in./s;
Motion of block A:
From (1) and (2),
aB = 0
1
1
v A = −vB − vC = 3 − (4t ) = 3 − 2t in./s
2
2
1
1
a A = −aB − aC = 0 − (4) = −2 in./s 2
2
2
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71
PROBLEM 11.55 (Continued)
(a)
Time when vB is zero.
3 − 2t = 0
Motion of block D:
From (3),
vD =
(b)
t = 1.500 s 
1
1
1
1
v A + vB = (3 − 2t ) − (3) = −1t
2
2
2
2
Time when vA is equal to v0.
3 − 2t = −t
(c)
t = 3.00 s 
Change in position of block A (t = 5 s).
1
a At 2
2
1
= (3)(5) + (−2)(5)2 = −10 in.
2
Δy A = ( v A ) 0 t +
Change in position = 10.00 in. 
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72
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 + 2vB = 0
(1)
aC + a A + 2aB = 0
(2)
Use units of inches and seconds.
Motion of block A:
v A = a At + 6t
Δy A =
Motion of block B:
1
1
a At 2 = (6)t 2 = 3t 2
2
2
v B = 3 in./s
vB = −3 in./s
ΔyB = vB t = −3t
Motion of block C:
From (1),
vC = −v A − 2vB = −6t − 2(−3) = 6 − 6t
ΔyC =
(a)
Time when vC is zero.
(b)
Corresponding position.
t
 v dt = 6t − 3t
0
C
2
6 − 6t = 0
t = 1.000 s 
ΔyC = (6)(1) − (3)(1)2 = 3 in.
ΔyC = 3.00 in. 
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73
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 yB + xC = constant
Then
3v A + 4vB + vC = 0
(1)
and
3a A + 4aB + aC = 0
(2)
(vB ) = 0,
Given:
a A = constent
(aC ) = 75 mm/s 2
At t = 2 s,
v B = 480 mm/s
vC = 280 mm/s
(a)
Eq. (2) and a A = constant and aC = constant  aB = constant
vB = 0 + a B t
Then
At t = 2 s:
480 mm/s = aB (2 s)
aB = 240 mm/s 2

or
a B = 240 mm/s 2 
or
a A = 345 mm/s 2 
Substituting into Eq. (2)
3a A + 4(240 mm/s 2 ) + (75 mm/s 2 ) = 0
a A = −345 mm/s


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74
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
( vC )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
xC − (xC )0 = 728 mm

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75
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 yB + xC = constant
Then
3v A + 4vB + vC = 0
(1)
and
3a A + 4aB + 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)
( vC )0 = 10.00 mm/s
or
1
aC t 2
2
We have
xC = ( xC )0 + (vC )0 t +
At t = 3 s:
57 mm = (10 mm/s)(3 s) +
1
aC (3 s)2
2
aC = 6 mm/s 2

Now

or
aC = 6.00 mm/s 2

v B = constant → aB = 0
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76
PROBLEM 11.58 (Continued)
Then, substituting into Eq. (2)
3a A + 4(0) + (6 mm/s 2 ) = 0
a A = −2 mm/s 2
(c)
We have
At t = 5 s:
y A = ( y A )0 + (v A )0 t +
or
a A = 2.00 mm/s 2 
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
y A − (y A )0 = 175.0 mm 
or
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77
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
Cable 1:
2 y A + 2 yB + yC = constant
Then
2v A + 2vB + vC = 0
(1)
and
2a A + 2aB + aC = 0
(2)
Cable 2:
( yD − y A ) + ( yD − yB ) = constant
Then
− v A − v B + 2 vD = 0
(3)
and
− a A − aB + 2aD = 0
(4)
Given: At t = 0, v = 0; all accelerations constant;
aC/B = 60 mm/s 2 , aD /A = 110 mm/s 2
(a)
We have
aC/B = aC − aB = −60 or aB = aC + 60
and
aD/A = aD − a A = 110 or a A = aD − 110
Substituting into Eqs. (2) and (4)
Eq. (2):
2(aD − 110) + 2( aC + 60) + aC = 0
3aC + 2aD = 100
or
Eq. (4):
−(aD − 110) − ( aC + 60) + 2aD = 0
− aC + aD = −50
or
(5)
(6)
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78
PROBLEM 11.59 (Continued)
Solving Eqs. (5) and (6) for aC and aD
aC = 40 mm/s 2
aD = −10 mm/s 2
Now
vC = 0 + aC t
At t = 3 s:
vC = (40 mm/s 2 )(3 s)
vC = 120.0 mm/s 
or
(b)
We have
At t = 5 s:
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 
or
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79
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/s 2 , (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 yB + yC = constant
Then
2v A + 2vB + vC = 0
(1)
and
2a A + 2aB + aC = 0
(2)
Cable 2: ( yD − y A ) + ( yD − yB ) = constant
Then
− v A − vB − 2 vD = 0
(3)
and
− a A − aB + 2aD = 0
(4)
t =0
Given: At
v=0
( y A )0 = ( yB )0 = ( yC )0
All accelerations constant.
At t = 2 s
yC /A = 280 mm
vB /A = 80 mm/s
When
y A − ( y A )0 = 160 mm
yB − ( yB )0 = 320 mm
aB > 10 mm/s 2
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80
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)
2a A + 2aB + (a A − 140) = 0
or
1
a A = (140 − 2aB )
3
Now
vB = 0 + a B t
(6)
v A = 0 + a At
vB/A = vB − v A = (aB − a A )t

Also
yB = ( yB )0 + (0)t +
1
aB t 2
2
When
v B /A = 80 mm/s :
80 = (aB − a A )t
Δy A = 160 mm :
160 =
1
a At 2
2
ΔyB = 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/s 2 
and
320 =
1
aB (4) 2
2
or
a B = 40.0 mm/s 2 
Note that Eq. (6) is not used; thus, the problem is over-determined.
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81
PROBLEM 11.60* (Continued)
(b)
Substituting into Eq. (5)
aC = 20 − 140 = −120 mm/s 2
and into Eq. (4)
−(20 mm/s 2 ) − (40 mm/s 2 ) + 2aD = 0
or
aD = 30 mm/s 2
Now
vC = 0 + aC t
When vC = −600 mm/s:
−600 mm/s = ( −120 mm/s 2 )t
or
t =5s
Also
yD = ( yD )0 + (0)t +
At t = 5 s:
yD − ( yD )0 =
1
aD t 2
2
1
(30 mm/s 2 )(5 s)2
2
y D − (y D )0 = 375 mm 
or
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82
PROBLEM 11.61
A particle moves in a straight line with the acceleration shown in the figure.
Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t
curves for 0 < t < 15 s and determine (a) the maximum value of the velocity
of the particle, (b) the maximum value of its position coordinate.
SOLUTION
v0 = −14 ft/s
Change in v = area under a−t curve.
t = 0 to t = 2 s:
v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s
v2 = −8 ft/s
t = 2 s to t = 5 s:
v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s
v5 = +16 ft/s
t = 5 s to t = 8 s:
v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s
v8 = +25 ft/s
t = 8 s to t = 10 s:
v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s
v10 = +15 ft/s
t = 10 s to t = 15 s:
v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s
v15 = −10 ft/s
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83
PROBLEM 11.61 (Continued)
Plot v−t curve. Then by similar triangles Δ’s find t for v = 0.
x0 = 0
Change in x = area under v−t curve
t = 0 to t = 2 s:
1
x2 − x0 = (−14 − 8)(2) = −22 ft
2
x2 = −22 ft
t = 2 s to t = 3 s:
1
x3 − x2 = (−8)(1) = −4 ft
2
x8 = −26 ft
t = 3 s to t = 5 s:
1
x5 − x3 = (+16)(2) = +16 ft
2
x5 = −10 ft
t = 5 s to t = 8 s:
1
x8 − x5 = (+16 + 25)(3) = +61.5 ft
2
x8 = +51.6 ft
t = 8 s to t = 10 s:
1
x10 − x8 = (+25 + 15)(2) = + 40 ft
2
x10 = +91.6 ft
t = 10 s to t = 13 s:
1
x13 − x10 = (+15)(3) = +22.5 ft
2
x13 = +114 ft
t = 13 s to t = 15 s:
1
x15 − x13 = (−10)(2) = −10 ft
2
x15 = +94 ft
(a)
Maximum velocity: When t = 8 s,
vm = 25.0 ft/s 
(b)
Maximum x: When t = 13 s,
xm = 114.0 ft 
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84
PROBLEM 11.62
For the particle and motion of Problem 11.61, plot the v−t and x−t curves for
0 < t < 15 s and determine the velocity of the particle, its position, and the
total distance traveled after 10 s.
PROBLEM 11.61 A particle moves in a straight line with the acceleration
shown in the figure. Knowing that it starts from the origin with
v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine
(a) the maximum value of the velocity of the particle, (b) the maximum
value of its position coordinate.
SOLUTION
v0 = −14 ft/s
Change in v = area under a−t curve.
t = 0 to t = 2 s:
v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s
v2 = −8 ft/s
t = 2 s to t = 5 s:
v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s
v5 = +16 ft/s
t = 5 s to t = 8 s:
v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s
v8 = +25 ft/s
t = 8 s to t = 10 s:
v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s
v10 = +15 ft/s
t = 10 s to t = 15 s:
v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s
v15 = −10 ft/s
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85
PROBLEM 11.62 (Continued)
Plot v−t curve. Then by similar triangles Δ’s find t for v = 0.
x0 = 0
Change in x = area under v−t curve
t = 0 to t = 2 s:
1
x2 − x0 = (−14 − 8)(2) = −22 ft
2
x2 = −22 ft
t = 2 s to t = 3 s:
1
x3 − x2 = (−8)(1) = −4 ft
2
x8 = −26 ft
t = 3 s to t = 5 s:
1
x5 − x3 = (+16)(2) = +16 ft
2
x5 = −10 ft
t = 5 s to t = 8 s:
1
x8 − x5 = (+16 + 25)(3) = +61.5 ft
2
x8 = +51.6 ft
t = 8 s to t = 10 s:
1
x10 − x8 = (+25 + 15)(2) = + 40 ft
2
x10 = +91.6 ft
t = 10 s to t = 13 s:
1
x13 − x10 = (+15)(3) = +22.5 ft
2
x13 = +114 ft
t = 13 s to t = 15 s:
1
x15 − x13 = (−10)(2) = −10 ft
2
x15 = +94 ft
when t = 10 s:
v10 = +15 ft/s 
x10 = +91.5 ft/s 
Distance traveled: t = 0 to t = 105
t = 0 to t = 3 s:
Distance traveled = 26 ft
t = 3 s to t = 10 s
Distance traveled = 26 ft + 91.5 ft = 117.5 ft
Total distance traveled = 26 + 117.5 = 143.5 ft 
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86
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 t2 )
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
tv = 0 = 22 s
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
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87
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 
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88
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
v = constant  a = 0
From t = 0 to t = 10 s:
−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=
−5 − (−20)
= 3 m/s 2
46 − 41
v = constant  a = 0
t = 46 s:
x2 = x1 + (area under v −t curve from t1 to t2 )
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
tv = 0 = 22 s
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
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89
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
100 = 420 −
1
(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
t2 = 40.0 s 
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90
PROBLEM 11.65
During a finishing operation the bed of an industrial planer moves alternately 30 in. to the right and 30 in. to
the left. The velocity of the bed is limited to a maximum value of 6 in./s to the right and 12 in./s to the left; the
acceleration is successively equal to 6 in./s2 to the right, zero 6 in./s2 to the left, zero, etc. Determine the time
required for the bed to complete a full cycle, and draw the v−t and x−t curves.
SOLUTION
We choose positive to the right, thus the range of permissible velocities is −12 in./s < v < 6 in./s since
acceleration is −6 in./s 2 , 0, or + 6 in./s 2 . The slope the v−t curve must also be −6 in./s2, 0, or +6 in./s2.
Planer moves = 30 in. to right: +30 in. = 3 + 6t1 + 3
t1 = 4.00 s
Planer moves = 30 in. to left: −30 in. = −12 − 12t2 + 12
t2 = 0.50 s
Total time = 1 s + 4 s + 1 s + 2 s + 0.5 s + 2 s = 10.5 s
ttotal = 10.50 s 
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91
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
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92
PROBLEM 11.66 (Continued)
(a)
Now Δ 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
ttotal = (0.4032 + 40.0346 + 4.3203) s
(b)
We have
ainitial =
=
ttotal = 44.8 s 
Δ vinitial
t1
[−13.888 − (−55.555)] m/s
0.4032 s
= 103.3 m/s 2
ainitial = 103.3 m/s 2 
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93
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
1h
 20 + v2 
(7.5 − 5) min × 
mi/h ×
= 0.5 mi

60 min
 2 
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/s 2 
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94
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
or
t1 = 7 s
and
A2 = 60 in.
or
{[(t2 − 7) − 1.5] s}(30 in./s) = 60 in.
or
t2 = 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 = 5 s:
1
(2) (12) = 12 in.
2
x5 = 12 + (5 − 2)(12)
t = 7 s:
= 48 in.
x7 = 60 in.
t = 2 s:
t = 8.5 s:
t = 9 s:
t = 10.5 s:
x2 =
1
x8.5 = 60 − (1.5)(30)
2
= 37.5 in.
x9 = 37.5 − (0.5)(30)
= 22.5 in.
x10.5 = 0
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95
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/s 2 ;
at
t = 0 a = −12 m/s 2 ;
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)
vt1 = v0 + A1
We have
 12 + 2 
1.8 m/s = 6 m/s − (t1 s) 
m/s2

 2 
t1 = 0.6 s 
(b)
v1.4 = vt1 + A2
We have
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,
−2 − (−12)
50
t − 12 = t − 12
0.6
3
for 0 < t < t1:
a=
Now
dv
50
= a = t − 12
dt
3
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96
PROBLEM 11.69 (Continued)
At t = 0, v = 6 m/s:
v
t  50
6
0

 dv =   3 t − 12  dt
25 2
t − 12t
3
or
v=6+
We have
dx
25
= v = 6 − 12t + t 2
dt
3
Then
A3 =
xt1
0.6
0
0
25 2
 dx =  (6 − 12t + 3 t )dt
0.6
25 

= 6t − 6t 2 + t 3  = 2.04 m
9 0

Also
 1.8 + 0.2 
A4 = (1.4 − 0.6) 
 = 0.8 m
2


Then
x1.4 = (2.04 + 0.8) m
x1.4 = 2.84 m 
or
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97
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:
= Δx = 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 
Δt = 8 s
Δ x = (60 m/s)(14 s − 6 s) − (shaded area)
= (60 m/s)(8 s) − 6 m = 480 m − 6 m = 474 m
vaverage =
Δ x 474 m
=
Δt
8s
vaverage = 59.25 m/s 
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98
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 vB 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, t2 = 25 − 4 = 21 s
A1 =
1
(4)(v A )max = 2(vA ) max
2
A2 = 21(v A )max
A1 + A2 = Δ x = 200 m
23(v A ) max = 200
Runner B:
or
t1 = 5 s,
A1 =
(v A ) max = 8.6957 m/s
t2 = 25.2 − 5 = 20.2 s
1
(5)(vB ) max = 2.5(vB ) max
2
A2 = 20.2(vB ) max
A1 + A2 = Δ x = 200 m
22.7(vB ) max = 200
Sketch v −t curve for second 200 m.
A3 = vmaxt3 −
t3 =
Runner A:
or
(vB ) max = 8.8106 m/s
Δv = | a |t3 = 0.1t3
1
Δvt3 = 200
2
or
0.05t32 − vmaxt3 + 200 = 0
(
vmax ± (vmax ) 2 − (4)(0.05)(200)
= 10 vmax ± (vmax )2 − 40
(2)(0.05)
(vmax ) A = 8.6957,
Reject the larger root. Then total time
(t3 ) A = 146.64 s
and
)
27.279 s
t A = 25 + 27.279 = 52.279 s
t A = 52.2 s 
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99
PROBLEM 11.71 (Continued)
Runner B: (vmax ) B = 8.8106, (t3 ) B = 149.45 s
and
26.765 s
t B = 25.2 + 26.765 = 51.965 s
Reject the larger root. Then total time
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 Δ x
Δ x = vave (Δt ) =
1
(5.999 + 5.968)(52.279 − 51.965)
2
Δ x = 1.879 m 
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100
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/s2 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:
Δ x = 16 + 40 + 50 + 40 = 146 ft
Allowable increase in speed:
Δvm = 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: Δ x = A1 + A2 + A3 :
146 ft = 48.4 + (22)t2 + 12.1
ttotal = t1 + t2 + t3 = 4.4 s + 3.89 s + 1.1 s = 9.39 s
t2 = 3.89 s
t B = 9.39 s 
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101
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/s2 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:
Δ x = 16 + 40 + 50 + 40 = 146 ft
Δvm = 5t1 = 20t2 ;
Δ x = A1 + A2 :
t2 =
146 ft =
1
(Δvm )(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
ttotal = t1 + t2 = 6.835 + 1.709
t B = 8.54 s 
Δvm = 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 
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102
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 (vB )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
Given:
(v A )0 = 60 mi/h, (vB )0 = 1.5 mi/h; at t = 0,
( 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,
aB = 0
First note
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, xB = 200 ft:
 22 +(vB )5 
(5 s) 
 ft/s = 200 ft
2


(vB )5 = 58 ft/s
or
Then, using similar triangles, we have
(88 − 22) ft/s (58 − 22) ft/s
=
( = aB )
t1
5s
or
t1 = 9.16 67 s
Finally, at t = t1


 22 + 88 
ft/s 
xB/A = xB − x A =  (9.1667 s) 

 2 


− [−380 ft + (9.1667 s) (88 ft/s)]
xB/A = 77.5 ft 
or
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103
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 vE = 0; For 0 < vE ≤ 7.8 m/s, aE = 1.2 m/s 2 ↑;
For vE = 7.8 m/s, aE = 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/s2 , while the slope of the curve for the ball is − g (−9.81 m/s 2 ).
The time t1 is the time when vE reaches 7.8 m/s.
Thus,
vE = (0) + aE t
or
7.8 m/s = (1.2 m/s 2 )t1
or
t1 = 6.5 s
The time ttop is the time at which the ball reaches the top of its trajectory.
Thus,
vB = (vB )0 − g (t − 2)
or
0 = 20 m/s − (9.81 m/s 2 ) (t top − 2) s
or
ttop = 4.0387 s
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104
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
y E = −12 m +
At
t = [2 + 2(4.0387 − 2)] s = 6.0774 s, yB = 0
and at t = t1 ,
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
1
(6.5 s) (7.8 m/s) = 13.35 m
2
The ball hits the elevator ( yB = yE ) when ttop ≤ t ≤ t1.
For t ≥ ttop :
1

yB = 20.387 m −  g (t − ttop )2  m
2


Then,
when
yB = yE
1
20.387 m − (9.81 m/s 2 ) (t − 4.0387)2
2
1
= −12 m + (1.2 m/s 2 ) (t s)2
2
or
Solving
5.505t 2 − 39.6196t + 47.619 = 0
t = 1.525 s and t = 5.67 s
t = 5.67 s 
Since 1.525 s is less than 2 s,
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105
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 = (tB )1:
Car B enters the speed zone.
t = tA:
Car A reaches its final speed.
t = tmin :
v A = vB
t = (t B )2 :
Car B reaches its final speed.
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106
PROBLEM 11.76 (Continued)
(a)
(v A )final − (v A )0
tA
aA =
We have
(44 − 58.667) ft/s
tA
−16 ft/s 2 =
or
or
t A = 0.91669 s
Also
60 ft = (tB )1 (vB )0
or
60 ft = (tB )1 (66 ft/s)
and
aB =
or
−20 ft/s 2 =
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 > v A . Therefore, ( x A/B ) min will occur when v A = vB ,
which occurs for
(tB )1 < tmin < (tB ) 2
For this time interval
v A = 44 ft/s
vB = (vB )0 + aB [t − (tB )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 )tmin − ( xB )tmin
  (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 
= (0.91669 s) 
ft/s + (2.00909 − 0.91669) s × (44 ft/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 ( x A/B ) min = 34.6 ft 
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107
PROBLEM 11.76 (Continued)
(b)
Since ( x A/B ) ≤ 60 ft for t ≤ tmin , it follows that x A/B = 70 ft for t > (tB ) 2
[Note (tB ) 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 − (t B )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




t = 14.14 s 
or
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108
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
2
(0.4)(0.1) = 0.0267 ft,
3
x0.3 = 0.227 ft 
x0.2 = 0.8 − ( A5 + A6 ) − A7
At t = 0.2 s,
With A5 + A6 =
and
0.4
x = 0.8 −  t vdt
x0.3 = 0.8 − A5 − (5.47)(0.1)
At t = 0.3 s,
With A5 =
or
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 
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109
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
t1 = 18.00 s 
or
(b)
Using the above values of the velocities, the v −t curve is drawn as shown.
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110
PROBLEM 11.78 (Continued)
Now
x at t = 18 s
x18 = 0 + Σ (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
vave =
Δ x (178.75 − 15) m
=
= 9.6324 m/s
Δt
(18 − 1) s
vave = 34.7 km/h 
or
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111
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/s 2
Given:
 da 
= 0.8 ft/s2 /s; vmax = 20 mi/h
 
 dt max
First note
(a)
20 mi/h = 29.333 ft/s
1.6 mi = 8448 ft
To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the
time for which v = vmax . The time Δ t required for the train to have an acceleration of 4 ft/s2 is found
from
a
 da 
= max
 dt 
Δt
 max
or
or
4 ft/s2
0.8 ft/s2 /s
Δt = 5 s
Δt =
Now,
da


 since dt = constant 


after 5 s, the speed of the train is
v5 =
1
(Δt )(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.
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112
PROBLEM 11.79 (Continued)
Now
at t = (10 + Δt1 ) s, v = vmax :
1

2  (5 s)(4 ft/s2 )  + (Δt1 )(4 ft/s2 ) = 29.333 ft/s
2

or
Δt1 = 2.3333 s
Then
at t = 5 s:
t = 7.3333 s:
1
(5)(4) = 10 ft/s
2
v = 10 + (2.3333)(4) = 19.3332 ft/s
v =0+
1
t = 12.3333 s: v = 19.3332 + (5)(4) = 29.3332 ft/s
2
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 + Δt2 ) s × (29.3332 ft/s) = 8448 ft
or
Δt2 = 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
vave =
Δx
1.6 mi 3600 s
=
×
Δt 300.34 s
1h
vave = 19.18 mi/h 
or
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113
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 = Δt :
t = 2Δt :
1
1
v = 0 + (Δt )(amax ) = amax Δt
2
2
vmax =
1
1
amax Δt + ( Δt )( amax ) = amax Δt
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
(2Δt )(vmax ) = xmax
vmax = amax Δt  2amax Δt 2 = xmax
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114
PROBLEM 11.80 (Continued)
Now
amax
= 4.8 ft/s2 /s so that
Δt
2(4.8Δt ft/s3 )Δt 2 = 1.2 ft
or
Δt = 0.5 s
Then
tmin = 4Δt
tmin = 2.00 s 
or
(b)
We have
vmax = amax Δt
= (4.8 ft/s 2 /s × Δt)Δt
= 4.8 ft/s 2 /s × (0.5 s) 2
vmax = 1.2 ft/s 
or
Also
vave =
Δx
1.2 ft
=
Δttotal 2.00 s
vave = 0.6 ft/s 
or
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115
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 Δt = 0.25 s. The area
(Δt)(aave) of each rectangle is approximately equal to the change in velocity Δv for the specified
interval of time. Thus,
Δv ≅ aave Δt
where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table.
2.
v(2) = v0 +
Now
and approximating the area
2
 a dt = 0
0
2
 a dt under the a−t curve by Σa Δt ≈ ΣΔv, the initial velocity is then
ave
0
equal to
v0 = −ΣΔv
Finally, using
v2 = v1 + Δv12
where Δv12 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
(Δt )(vave ) of each rectangle is approximately equal to the change in position Δ x for the specified
interval of time. Thus
Δ x ≈ vave Δt
where vave and Δ x are given in columns 4 and 5, respectively, of the table.
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116
PROBLEM 11.81 (Continued)
4.
With x0 = 0 and noting that
x2 = x1 + Δ x12
where Δ 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 
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117
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 Δt = 2 s. The
area (Δt )(aave ) of each rectangle is approximately equal to the change in velocity Δv for the
specified interval of time. Thus,
Δv ≅ aave Δt
where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table.
2.
Noting that v0 = 0 and that
v2 = v1 + Δv12
where Δv12 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 Δt = 2 s. The
area (Δt )(vave ) of each rectangle is approximately equal to the change in position Δx for the
specified interval of time.
Thus,
Δx ≅ vave Δt
where vave and Δx are given in columns 4 and 5, respectively, of the table.
4.
With x0 = 0 and noting that
x2 = x1 + Δ x12
where Δ 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.
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118
PROBLEM 11.82 (Continued)
(a)
At t = 8 s, v = 32.58 m/s
(b)
At t = 20 s
v = 117.3 km/h 
or
x = 660 m 
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119
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
Δt = 2 1
a
Δx =
or
For the five regions shown above, we have
Region
v1 , ft/s
v2 , ft/s
a, ft/s 2
Δx, ft
Δt , 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
Σ
(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 
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120
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
. Now
First note that the slope of the above curve is dv
dx
a=v
(a)
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./s 2 
or
(b)
When v = 80 in./s, we have
 40 in./s 
a = 80 in./s 

 28 in. 
a = 114.3 in./s 2 
or
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, Δv and Δ x were measured directly, so different scales could be used.
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121
PROBLEM 11.85
Using the method of Section 11.8, derive the formula x = x0 + v0t + 12 at 2 for the position coordinate of a
particle in uniformly accelerated rectilinear motion.
SOLUTION
The a −t curve for uniformly accelerated motion is as shown.
Using Eq. (11.13), we have
x = x0 + v0t + (area under a −t curve) (t − t )
 1 
= x0 + v0t + (t × a)  t − t 
 2 
1
= x0 + v0 t + at 2
Q.E.D.
2

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122
PROBLEM 11.86
Using the method of Section 11.8 determine the position of the
particle of Problem 11.61 when t = 8 s.
PROBLEM 11.61 A particle moves in a straight line with the
acceleration shown in the figure. Knowing that it starts from the
origin with v0 = −14 ft/s, plot the v−t and x−t curves for
0 < t < 15 s and determine (a) the maximum value of the velocity
of the particle, (b) the maximum value of its position coordinate.
SOLUTION
x0 = 0
v0 = −14 ft/s
when t = 8s:
x = x0 + v0t + Σ A(t1 − t )
= 0 − (14 ft/s)(8 s) + [(3 ft/s2 )(2 s)](7 s) + [(8 ft/s2 )(3 s)](4.5 s) + [(3 ft/s)(3 s)](1.5 s)
x8 = −112 ft + 42 ft + 108 ft + 13.5 ft
x8 = 51.5 ft 
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123
PROBLEM 11.87
The acceleration of an object subjected to the pressure wave of a
large explosion is defined approximately by the curve shown. The
object is initially at rest and is again at rest at time t1. Using the
method of section 11.8, determine (a) the time t1, (b) the distance
through which the object is moved by the pressure wave.
SOLUTION
(a)
Since v = 0 when t = 0 and when t = t1 the change in v between t = 0 and t = t1 is zero.
Thus, area under a−t curve is zero
A1 + A2 + A3 = 0
1
1
1
(30)(0.6) + ( −10)(0.2) + ( −10)(t1 − 0.8) = 0
2
2
2
9 − 1 − 5t1 + 4 = 0
(b)
t1 = 2.40 s 
Position when t = t1 = 2.4 s
2
x = x0 + v0t1 + A1 (t1 − 0.2) + A2 (t1 − 0.733) + A3   (t1 − 0.8)
3
1
2
= 0 + 0 + (9)(2.4 − 0.2) + ( −1)(2.4 − 0.733) +  ( −10)(2.4 − 0.8)  (2.4 − 0.8)
2
3
= 19.8 m − 1.667 m − 8.533 m
x = 9.60 m 
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124
PROBLEM 11.88
For the particle of Problem 11.63, draw the
a −t curve and determine, using the method
of Section 11.8, (a) the position of the
particle when t = 52 s, (b) the maximum
value of its position coordinate.
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=
We have
dv
dt
where dv is the slope of the v −t curve. Then
dt
t=0
from
to t = 10 s:
v = constant  a = 0
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:
−20 − 60
= −5 m/s 2
26 − 10
−5 − (−20)
= 3 m/s 2
46 − 41
v = constant  a = 0
The a −t curve is then drawn as shown.
(a)
From the discussion following Eq. (11.13),
we have
x = x0 + v0t1 + ΣA(t1 − t )
where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s
x = −540 m + (60 m/s)(52 s) + {−[(16 s)(5 m/s 2 )](52 − 18) s
+ [(5 s)(3 m/s 2 )](52 − 43.5)s}
= [−540 + (3120) + (−2720 + 127.5)] m
x = −12.50 m 
or
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125
PROBLEM 11.88 (Continued)
(b)
Noting that xmax occurs when v = 0 ( dx
= 0 ), it is seen from the v–t curve that xmax occurs for
dt
10 s < t < 26 s. Although similar triangles could be used to determine the time at which x = xmax
(see the solution to Problem 11.63), the following method will be used.
For
10 s < t1 < 26 s, we have
x = −540 + 60t1
1

− [(t1 − 10)(5)]  (t1 − 10)  m
2

5
= −540 + 60t1 − (t1 − 10) 2
2
When x = xmax :
or
Then
dx
= 60 − 5(t1 − 10) = 0
dt
(t1 ) xmax = 22 s
5
xmax = −540 + 60(22) − (22 − 10)2
2
xmax = 420 m 
or
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126
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) 
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127
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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128
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) 
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129
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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130
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 α of the acceleration of block B?
(a)
(b)
(c)
(d)
(e)
SOLUTION



Since aB = a A + aB/A we get
Answer: (d ) 
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131
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 y2 = 52 + 3.82 = 6.28 m/s
tan θ =
(b)
vy
vx
=
−3.8
5
Horizontal distance:
θ = −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 
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132
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/s 2
69.4° 
v = 8.29 mm/s
36.2° 
a = 336 mm/s 2
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
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133
PROBLEM 11.91
The motion of a vibrating particle is defined by the position vector
r = (4sin π t )i − (cos 2π 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 π t )i − (cos 2π t ) j
v = (4π cos π t )i + (2π sin 2π t ) j
a = −(4π 2 sin π t )i + (4π 2 cos 2π t ) j
(a)
(b)
When t = 1 s:
v = (4π cos π )i + (2π sin 2π ) j
v = −(4π in/s)i 
a = −(4π 2 sin π )i − (4π 2 cos π ) j
a = −(4π 2 in/s 2 ) j 
Path of particle:
Since r = xi + y j ;
x = 4sin π t ,
y = − cos 2π t
Recall that cos 2θ = 1 − 2sin 2 θ and write
y = − cos 2π t = −(1 − 2sin 2 π t )
But since x = 4sin π t or sin π t =
(1)
1
x, Eq.(1) yields
4
2

1  
y = − 1 − 2  x  
 4  

y=
1 2
x − 1 (Parabola) 
8
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134
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 ≤ 2π ,
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
π
2
π
3
π
2
2π
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135
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 y2 = (10 − 5cos t )2 + 25sin 2 t = 125 − 100cos t
d (v ) 2
= 100sin t = 0 t = 0. ± π . ± 2π  ± N π
dt
(b)
When t = 2 N π .
cos t = 1.
and
v2 is minimum.
When t = (2 N + 1)π .
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
vmax = 15 ft/s 
When v = vmin .
When N = 0,1, 2,
x = 10(2π N ) − 5sin(2π N )
x = 20π N ft 
y = 10 − 5cos(2π N )
y = 5 ft 
vx = 10 − 5cos(2π N )
vx = 5 ft/s 
v y = 5sin(2π N )
tan θ =
vy
vx
vy = 0 
θ =0 
= 0,
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136
PROBLEM 11.92 (Continued)
t = (2 N + 1)π s 
When v = vmax .
x = 10[2π ( N − 1)] − 5sin[2π ( N + 1)]
x = 20π ( N + 1) ft 
y = 10 − 5cos[2π ( N + 1)]
y = 15 ft 
vx = 10 − 5cos[2π ( N + 1)]
vx = 15 ft/s 
v y = 5sin[2π ( N + 1)]
tan θ =
vy
vx
vy = 0 
θ =0 
= 0,
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137
PROBLEM 11.93
The damped motion of a vibrating particle is defined by
the position vector r = x1[1 − 1/(t + 1)]i + ( y1e−π t/2 cos 2π 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 

−π t/2
r = 30 1 −
cos 2π t ) j
 i + 20(e
 t +1
Then
v=
= 30
1
 π

i + 20  − e −π t/2 cos 2π t − 2π e−π t/2 sin 2π t  j
2
(t + 1)
 2

= 30

1
1

i − 20π e −π t/ 2  cos 2π t + 2 sin 2π t   j
2
2
(t + 1)



a=
and
dr
dt
dv
dt
 π

2
1

i − 20π  − e−π t/2  cos 2π t + 2 sin 2π t  + e−π t/2 (−π sin 2π t + 4 cos 2π t )  j
3
(t + 1)
2

 2

60
=−
i + 10π 2 e−π t/2 (4 sin 2π t − 7.5 cos 2π t ) j
(t + 1)3
= −30
(a)
At t = 0:
 1
r = 30 1 −  i + 20(1) j
 1
r = 20 mm 
or
 1
1

v = 30   i − 20π (1)  + 0   j
1

 2
or
a=−
v = 43.4 mm/s
46.3° 
a = 743 mm/s 2
85.4° 
60
i + 10π 2 (1)(0 − 7.5) j
(1)
or
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138
PROBLEM 11.93 (Continued)
(b)
At t = 1.5 s:
1 

−0.75π
r = 30 1 −
(cos 3π ) 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/s 2
86.9° 
30
1

i − 20π e−0.75π  cos 3π + 0  j
2
(2.5) 2


= (4.80 mm/s)i + (2.9778 mm/s) j
v=
or
a=−
60
i + 10π 2 e−0.75π (0 − 7.5 cos 3π ) j
(2.5)3
= (−3.84 mm/s 2 )i + (70.1582 mm/s 2 ) j
or
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139
PROBLEM 11.94
The motion of a particle is defined by the position vector
r = A(cos t + t sin t )i + A(sin t − t cos t ) j, where t is expressed in
seconds. Determine the values of t for which the position vector
and the acceleration are (a) perpendicular, (b) parallel.
SOLUTION
We have
r = A(cos t + t sin t )i + A(sin t − t cos t ) j
Then
v=
and
a=
(a)
dr
= A( − sin t + sin t + t cos t )i
dt
+ A(cos t − cos t + t sin t ) j
= A(t cos t )i + A(t sin t ) j
dv
= A(cos t − t sin t )i + A(sin t + t cos t ) j
dt
When r and a are perpendicular, r ⋅ a = 0
A[(cos t + t sin t )i + (sin t − t cos t ) j] ⋅ A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0
(b)
or
(cos t + t sin t )(cos t − t sin t ) + (sin t − t cos t )(sin t + t cos t ) = 0
or
(cos 2 t − t 2 sin 2 t ) + (sin 2 t − t 2 cos 2 t ) = 0
or
1− t2 = 0
or
t =1 s 
or
t=0 
When r and a are parallel, r × a = 0
A[(cos t + t sin t )i + (sin t − t cos t ) j] × A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0
or
Expanding
[(cos t + t sin t )(sin t + t cos t ) − (sin t − t cos t )(cos t − t sin t )]k = 0
(sin t cos t + t + t 2 sin t cos t ) − (sin t cos t − t + t 2 sin t cos t ) = 0
2t = 0
or
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140
PROBLEM 11.95
The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin
ωnt)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 ωn t )i + ctj + ( Rt sin ωn t )k
Then
v=
and
a=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k
dt
dv
dt
= R(−ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t )i
+ R(ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t )k
= R(−2ωn sin ωn t − ωn2 t cos ωn t )i + R(2ωn cos ωn t − ωn2 t sin ωn t )k
Now
v 2 = vx2 + v 2y + vz2
= [ R(cos ωn t − ωn t sin ωn t )]2 + (c)2 + [ R(sin ωn t + ωn t cos ωn t )]2
(
)
= R 2  cos 2 ωn t − 2ωn t sin ωn t cos ωn t + ωn2 t 2 sin 2 ωn t

+ sin 2 ωn t + 2ωn t sin ωn t cos ωn t + ωn2 t 2 cos 2 ωn t  + c 2

(
(
)
)
= R 2 1 + ωn2 t 2 + c 2
(
Also,
)
v = R 2 1 + ωn2 t 2 + c 2 
or
a 2 = ax2 + a 2y + az2
(
)
+  R ( 2ω cos ω t − ω t sin ω t ) 


= R ( 4ω sin ω t + 4ω t sin ω t cos ω t + ω t cos ω t )

+ ( 4ω cos ω t − 4ω t sin ω t cos ω t + ω t sin ω t ) 

= R ( 4ω + ω t )
2
=  R −2ωn sin ωn t − ωn2 t cos ωn t  + (0) 2


n
2
2
n
2
n
2
n
2
2
2
n
n
3
n
n
n
2
2
n
3
n
n
n
n
n
4 2
n
4 2
n
2
2
n
n
4 2
n
a = Rωn 4 + ωn2 t 2 
or
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141
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
x
At
y = A t 2 + 1 z = Bt sin t
sin t =
2
 y
t2 =   −1
 A
z
Bt
Then
cos t =
Now
 x   z 
cos 2 t + sin 2 t = 1    +   = 1
 At   Bt 
2
x z
t2 =   +  
 A  B
2
or
2
2
2
2
Then
 y
x z
 A  −1 =  A  +  B 
 
   
or
 y  x  z
 A  −  A  −  B  =1
     
2
(a)
2
2
2

Q.E.D.
With A = 3 and B = 1, we have
v=
dr
t
j + (sin t + t cos t )k
= 3(cos t − t sin t )i + 3
2
dt
t +1
( )
t
and
t 2 + 1 − t t 2 +1
dv
a=
j
= 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
1
j + (2 cos t − t sin t )k
(t + 1)3/2
2
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142
PROBLEM 11.96 (Continued)
At t = 0:
v = 3(1 − 0)i + (0) j + (0)k
v = vx2 + v y2 + vz2
v = 3 ft/s 
or
and
a = −3(0)i + 3(1) j + (2 − 0)k
Then
a 2 = (0) 2 + (3) 2 + (2) 2 = 13
a = 3.61 ft/s 2 
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 cos 2 t − 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0
10 + 8 cos 2 t − 8t sin t cos t = 0
or (with t ≠ 0)
7 + 2 cos 2t − 2t sin 2t = 0
or
Using “trial and error” or numerical methods, the smallest root is
t = 3.82 s 
Note: The next root is t = 4.38 s. 
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143
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
At B:
1
−300 ft = − (32.2 ft/s 2 )t 2
2
or
t B = 4.31666 s
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
At B:
d = (264 ft/s)(4.31666 s)
d = 1140 ft 
or
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144
PROBLEM 11.98
A helicopter is flying with a constant horizontal velocity of
180 km/h and is directly above Point A when a loose part
begins to fall. The part lands 6.5 s later at Point B on an
inclined surface. Determine (a) the distance d between
Points A and B, (b) the initial height h.
SOLUTION
Place origin of coordinates at Point A.
Horizontal motion:
(vx )0 = 180 km/h = 50 m/s
x = x0 + (vx )0 t = 0 + 50t m
At Point B where t B = 6.5 s,
(a)
Distance AB.
325
cos10°
From geometry
d=
Vertical motion:
y = y0 + ( v y ) 0 t −
At Point B
(b)
xB = (50)(6.5) = 325 m
d = 330 m 
1 2
gt
2
1
− xB tan10° = h + 0 − (9.81)(6.5) 2
2
h = 149.9 m 
Initial height.
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145
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 α corresponding
to h = 788 mm and h = 1068 mm.
SOLUTION
(a)
y0 = 1.5 m, (v y )0 = 0
Vertical motion:
or
t =
2( y0 − y )
g
y =h
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
x0 = 0, (vx )0 = v0 ,
x = v0t
or
v0 =
x
x
= B
t
tB
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146
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 α = −
115.3 km/h ≤ v0 ≤ 148.0 km/h 
(v y ) B
dy
gt
=−
= B
dx
(vx ) B
v0
For h = 0.788 m,
tan α =
(9.81)(0.3810)
= 0.11673,
32.02
α = 6.66° 
For h = 1.068 m,
tan α =
(9.81)(0.2968)
= 0.07082,
41.11
α = 4.05° 
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147
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 = v0 t
At B:
y:
t B = 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 
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148
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
1
−8.8 ft = (−0.64705 ft/s) t − (32.2 ft/s 2 ) t 2
2
or
t BC = 0.719491 s
(the other root is negative)
Horizontal motion. (Uniform)
x = 0 + (v x ) 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
0 < d < 1.737 ft 
Since the trough is 2 ft wide, it then follows that
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149
PROBLEM 11.102
Milk is poured into a glass of height 140 mm and inside
diameter 66 mm. If the initial velocity of the milk is 1.2 m/s at
an angle of 40° with the horizontal, determine the range of
values of the height h for which the milk will enter the glass.
SOLUTION
First note
(vx )0 = (1.2 m/s) cos 40° = 0.91925 m/s
(v y )0 = −(1.2 m/s) sin 40° = −0.77135 m/s
Horizontal motion. (Uniform)
x = 0 + (v x ) 0 t
Vertical motion. (Uniformly accelerated motion)
y = y0 + ( v y ) 0 t −
1 2
gt
2
Milk enters glass at B.
x:
0.08 m = (0.91925 m/s) t or tB = 0.087028 s
y : 0.140 m = hB + (−0.77135 m/s)(0.087028 s)
1
− (9.81 m/s 2 )(0.087028 s) 2
2
or
hB = 0.244 m
Milk enters glass at C.
x : 0.146 m = (0.91925 m/s) t or tC = 0.158825 s
y : 0.140 m = hC + (−0.77135 m/s)(0.158825 s)
1
− (9.81 m/s 2 )(0.158825 s) 2
2
or
hC = 0.386 m
0.244 m < h < 0.386 m 

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150
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 + (v x ) 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:
1
0 = 2.1 m + (4.5831 m/s)t − (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
b = (16.01 − 9.00) m = 7.01 m from the net 
The ball lands
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151
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°
and at B
Now
yB = − d sin 5°
Horizontal motion. (Uniform)
x = 0 + (v x ) 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 t B
 cos 5° 
1  cos 5°  2
− d sin 5° = (160 sin 25°) 
d − g 
 d
2  160 cos 25° 
 160 cos 25° 
2
or
2
(160 cos 25°) 2 (tan 5° + tan 25°)
32.2 cos 5°
= 726.06 ft
d=
d = 242 yd 
or
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152
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 + (v x ) 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 t B

 1 

14
14
1.5 = (v0 sin 40°) 
− g

 v0 cos 40°  2  v0 cos 40° 
or
2
1
(32.2)(196)/ cos 2 40°
2
2
v0 =
−1.5 + 14 tan 40°
v0 = 22.9 ft/s 
or
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153
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 2
gt
2
1
= 2 + (v0 sin 40°) = − (9.81)t 2
2
v0 t =
Then
y = 2 + x tan 40° − 4.905t 2
t2 =
(2)
x
cos 40°
From (1),
Point B:
(1)
(3)
2 + x tan 40° − y
4.905
(4)
x = 8 + 10 cos 35° = 16.1915 m
y = 1.5 + 10sin 35° = 7.2358 m
16.1915
v0 t =
= 21.1365 m
cos 40°
2 + 16.1915 tan 40° − 7.2358
t2 =
4.905
21.1365
v0 =
1.3048
t = 1.3048 s
v0 = 16.199 m/s
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154
PROBLEM 11.106 (Continued)
Point C:
x = 8 + (10 + 7) cos 35° = 21.9256 m
y = 1.5 + (10 + 7)sin 35° = 11.2508 m
v0 t =
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 
Range of values of v0.
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155
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 + (v x ) 0 t
(16 − d ) = (v0 cos 30°) t or t B =
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  13 (16 − d ) − 3.2 


2(32.2) (16 − 129 )
2
2(32.2) (16 − 17
12 )
2
3  13 (16 − 129 ) − 3.2


17
3  13 (16 − 12
) − 3.2

v0 = 29.8 ft/s 
v0 = 29.6 ft/s 
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156
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 cos 5° and (v y )0 = v0 sin 5°.
Horizontal motion:
x = x0 + (vx )0 t = (v0 cos 5°)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
x
cos 5°
From (1),
v0 t =
Then
y = 2.5 − x tan 5° − 4.905t 2
t2 =
(2)
(3)
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
v0 t =
t = 0.32517 s
v0 = 37.66 m/s
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157
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
v0 t =
= 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
37.7 m/s < v0 < 44.3 m/s 
Range for v0.
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158
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°
y = (0.175 m) cos 10°
At Point B:
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
gtB
2
Substituting for t B
 0.050388  1
 0.050388 
−0.032659 = (−v0 sin 6°) 
 − (9.81) 

 v0 cos 6°  2
 v0 cos 6° 
2
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159
PROBLEM 11.109 (Continued)
v02 =
or
1
(9.81)(0.050388)2
2
cos 2 6°(0.032659 − 0.050388 tan 6°)
(v0 ) B = 0.678 m/s
or
x = (0.175 m) cos 30°
y = (0.175 m) sin 30°
At Point C:
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 =
2
1
(9.81)(0.171554) 2
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 

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160
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
(vx )0 = v0 cos 65°
(v y )0 = v0 sin 65°
Horizontal motion. (Uniform)
x = 0 + (v x ) 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 = tB ,C

 1 

5
5
yB ,C = (v0 sin 65°) 
− g

 v0 cos 65°  2  v0 cos 65° 
2
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161
PROBLEM 11.110 (Continued)
or
v02 =
1 (9.81)(25)
2
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 

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162
PROBLEM 11.111
The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle α with the
horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle α, (b) the angle θ 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 α = (20 m/s) cos α
and
(v y )0 = v0 sin α = (20 m/s) sin α
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (20 cos α ) t
At Point B:
14 = (20 cos α )t or t B =
7
10 cos α
Vertical motion. (Uniformly accelerated motion)
y = 0 + (v y )0 t −
At Point B:
1 2
1
gt = (20 sin α )t − gt 2
2
2
0.08 = (20 sin α )t B −
( g = 9.81 m/s 2 )
1 2
gt B
2
Substituting for t B

 1 

7
7
0.08 = (20 sin α ) 
− g

 10 cos α  2  10 cos α 
2
1
49
g
2 cos 2 α
or
8 = 1400 tan α −
Now
1
= sec2 α = 1 + tan 2 α
2
cos α
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163
PROBLEM 11.111 (Continued)
Then
or
Solving
8 = 1400 tan α − 24.5 g (1 + tan 2 α )
240.345 tan 2 α − 1400 tan α + 248.345 = 0
α = 10.3786° and α = 79.949°
Rejecting the second root because it is not physically reasonable, we have
α = 10.38° 
(b)
We have
vx = (vx )0 = 20 cos α
and
v y = (v y )0 − gt = 20 sin α − gt
At Point B:
(v y ) B = 20 sin α − gt B
= 20 sin α −
7g
10 cos α
Noting that at Point B, v y < 0, we have
tan θ =
=
=
|(v y ) B |
vx
7g
− 20 sin α
10 cos α
20 cos α
7
9.81
− sin 10.3786°
200 cos 10.3786°
cos 10.3786°
θ = 9.74° 
or
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164
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 α 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,
Horizontal motion:
x = v0 t sin α
Vertical motion:
y = v0t cos α −
cos α =
sin 2 α + cos 2 α =
y0 = 0
sin α =
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
(v0 t )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)
)
(3)
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
t = 15.8916 s
and 1.6458 s 2
and 1.2829 s
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165
PROBLEM 11.112 (Continued)
Restrictions on α :
0 < α < 120°
tan α =
x
y + 12 gt 2
=
100 cos 30°
= 0.0729
−50 + (4.905)(15.8916) 2
α = 4.1669°
100 cos 30°
= −2.0655
−50 + (4.905)(1.2829)2
α = 115.8331°
and
Use α = 4.1669° corresponding to the steeper possible trajectory.
(a)
Angle α .
(b)
Maximum height.
α = 4.17° 
v y = 0 at
y = ymax
v y = v0 cos α − gt = 0
t=
v0 cos α
g
ymax = v0 t cos α −
=
(c)
Duration of the flight.
v 2 cos 2 α
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 

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166
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 α 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 α = (154 ft/s) cos α
and
(v y )0 = v0 sin α = (154 ft/s) sin α
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (154 cos α )t
At the front of the net,
x = 16 ft
Then
16 = (154 cos α )t
or
tenter =
8
77 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (154 sin α ) 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 α ) tenter −
1 2
gtenter
2

 1 

8
8
= (154 sin α ) 
− g

77
cos
2
77
cos
α
α




32 g
= 16 tan α −
5929 cos 2 α
2
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167
PROBLEM 11.113 (Continued)
Now
1
= sec2 α = 1 + tan 2 α
cos 2 α
Then
yfront = 16 tan α −
or
tan 2 α −
32 g
(1 + tan 2 α )
5929
 5929

5929
tan α + 1 +
yfront  = 0
2g
32 g


(


5929
±  − 5929
2g
2g
) − 4 (1 +
2
1/2
5929
y
32 g front
)
Then
tan α =
or
 5929 2 
5929
5929

±  −
tan α =
yfront  
 − 1 +
4 × 32.2  4 × 32.2   32 × 32.2
 
or
tan α = 46.0326 ± [(46.0326) 2 − (1 + 5.7541 yfront )]1/2
2
1/2
Now 0 < yfront < 4 ft so that the positive root will yield values of α > 45° for all values of yfront.
When the negative root is selected, α increases as yfront is increased. Therefore, for α max , set
yfront = yC = 4 ft
(b)
Then
tan α = 46.0326 − [(46.0326) 2 − (1 + 5.7541 + 4)]1/ 2
or
α max = 14.6604°
We had found
tenter =
α max = 14.66° 
8
77 cos α
8
=
77 cos 14.6604°
tenter = 0.1074 s 
or
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168
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 α.
SOLUTION
First note
(vx )0 = v0 cos α = (11.5 m/s) cos α
(v y )0 = v0 sin α = (11.5 m/s) sin α
By observation, d max occurs when
ymax = 1.1 m.
Vertical motion. (Uniformly accelerated motion)
v y = (v y )0 − gt
= (11.5 sin α ) − gt
y = ymax
When
Then
at
1 2
gt
2
1
= (11.5 sin α ) t − gt 2
2
y = 0 + (v y ) 0 t −
B , (v y ) B = 0
(v y ) B = 0 = (11.5 sin α ) − gt
11.5 sin α
g
( g = 9.81 m/s 2 )
or
tB =
and
yB = (11.5 sin α ) t B −
1 2
gt B
2
Substituting for t B and noting yB = 1.1 m
 11.5 sin α  1  11.5 sin α 
1.1 = (11.5 sin α ) 
− g

g
g

 2 

1
=
(11.5) 2 sin 2 α
2g
or
sin 2 α =
2.2 × 9.81
11.52
2
α = 23.8265°
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169
PROBLEM 11.114 (Continued)
(a)
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (11.5 cos α ) t
At Point B:
x = d max
where
tB =
Then
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 
or
(b)
α = 23.8° 
From above
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170
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 α 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 α = (8 m/s) cos α
(v y )0 = v0 sin α = (8 m/s) sin α
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (8 cos α ) t
At Point B:
or
x = d : d = (8 cos α ) t
d
tB =
8 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (8 sin α ) t − gt 2 ( g = 9.81 m/s 2 )
2
1
0 = (8 sin α ) t B − gt B2
2
y = 0 + (v y )0 t −
At Point B:
Simplifying and substituting for t B
0 = 8 sin α −
d=
or
(a)
1  d 
g

2  8 cos α 
64
sin 2α
g
(1)
When h = 0, the water can follow any physically possible trajectory. It then follows from
Eq. (1) that d is maximum when 2α = 90°
α = 45° 
or
Then
d=
64
sin (2 × 45°)
9.81
d max = 6.52 m 
or
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171
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 α increases
in value from 0 to 45° and then d decreases as α is further increased. Thus, d max occurs for the value
of α 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 α
Also, with ycorn = h, we have
h = (8 sin α ) tcorn −
1 2
gtcorn
2
Substituting for tcorn and noting h = 1.8 m,
 1.5  1  1.5 
1.8 = (8 sin α ) 
− g

 8 cos α  2  8 cos α 
or
Now
Then
or
1.8 = 1.5 tan α −
2
2.25 g
128 cos 2 α
1
= sec2 α = 1 + tan 2 α
cos 2 α
1.8 = 1.5 tan α −
2.25(9.81)
(1 + tan 2 α )
128
0.172441 tan 2 α − 1.5 tan α + 1.972441 = 0
α = 58.229° and α = 81.965°
Solving
From the above discussion, it follows that d = d max when
α = 58.2° 
Finally, using Eq. (1)
d=
64
sin (2 × 58.229°)
9.81
d max = 5.84 m 
or
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172
PROBLEM 11.116*
A mountain climber plans to jump from A to B over a
crevasse. Determine the smallest value of the climber’s initial
velocity v0 and the corresponding value of angle α so that he
lands at B.
SOLUTION
First note
(vx )0 = v0 cos α
(v y )0 = v0 sin α
Horizontal motion. (Uniform)
x = 0 + (vx )0 t = (v0 cos α ) t
At Point B:
1.8 = (v0 cos α )t
or
tB =
1.8
v0 cos α
Vertical motion. (Uniformly accelerated motion)
1 2
gt
2
1
= (v0 sin α ) t − gt 2
2
y = 0 + (v y )0 t −
At Point B:
−1.4 = (v0 sin α ) t B −
( g = 9.81 m/s 2 )
1 2
gt B
2
Substituting for t B
 1.8  1  1.8 
−1.4 = (v0 sin α ) 
− g

 v0 cos α  2  v0 cos α 
or
v02 =
1.62 g
cos α (1.8 tan α + 1.4)
=
1.62 g
0.9 sin 2α + 1.4 cos 2 α
2
2
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173
PROBLEM 11.116* (Continued)
Now minimize v02 with respect to α.
We have
dv02
−(1.8 cos 2α − 2.8 cos α sin α )
= 1.62 g
=0
dα
(0.9 sin 2α + 1.4 cos 2 α ) 2
1.8 cos 2α − 1.4 sin 2α = 0
or
tan 2α =
or
18
14
α = 26.0625° and α = 206.06°
or
Rejecting the second value because it is not physically possible, we have
α = 26.1° 
Finally,
v02 =
1.62 × 9.81
cos 26.0625°(1.8 tan 26.0625° + 1.4)
2
(v0 )min = 2.94 m/s 
or
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174
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
30
17.80450
=
sin α
sin 15°
or
α = 25.8554°
α + 25° = 50.8554°
vA /B = 17.8 ft/s
50.9° 
Alternative solution.
vA /B = v A − 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°
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175
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
v A + vD = 0
Cable 2:
( yB − yD ) + ( yC − yD ) = constant
Then
vB + vC − 2vD = 0
(1)
(2)
Combining Eqs. (1) and (2) to eliminate vD ,
2v A + vB + vC = 0
(3)
Now
v A/C = v A − vC = −300 mm/s
(4)
and
vB/A = vB − v A = 200 mm/s
(5)
Then
(3) + (4) − (5) 
(2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200)
v A = 125 mm/s 
or
vB − (−125) = 200
and using Eq. (5)
v B = 75 mm/s 
or
−125 − vC = −300
Eq. (4)
vC = 175 mm/s 
or
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176
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 β sin110°
=
66
90.99
β = 42.97°
α = 90° − β = 90° − 42.97° = 47.03°
(a)
Relative velocity:
(b)
Change in position for Δt = 4 s.
ΔrB/A = vB/A Δt = (91.0 ft/s)(4 s)
(c)
v B/A = 91.0 ft/s
47.0° 
rB/A = 364 ft
47.0° 
Distance between autos 2 seconds after auto A has passed intersection.
Auto A travels for 2 s.
v A = 66 ft/s
20°
rA = v At = (66 ft/s)(2 s) = 132 ft
rA = 132 ft
20°
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177
PROBLEM 11.119 (Continued)
v B = 44 ft/s
Auto B
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 
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178
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
18
3.1974
=
sin α sin 10°
or
α = 77.84°
Noting that
v R = 3.20 km/h
17.8° 
Alternatively one could use vector algebra.
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179
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
v C = v A + v C/ A
and
v C = v B + v C/ B
Then
v A + vC/A = v B + v C/B
or
v B − v A = v C/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
400 405.175
=
sin α
sin 65°
or
α = 63.474°
75° − α = 11.526°
v B/A = 405 km/h
11.53° 
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180
PROBLEM 11.121 (Continued)
(b)
We have
v C = v A + v C/ 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 + v C t
rC/B = rC − rB = [(rC )0 − (rB )0 ] + ( vC − v B )t
= [(rC )0 − (rB )0 ] + v C/B t
Then
ΔrC/B = (rC/B )t2 − (rC/B )t1 = v C/B (t2 − t1 ) = vC/B Δt
For Δt = 15 min:
1 
ΔrC/B = (400 km/h)  h  = 100 km
4 
ΔrC/B = 100 km
40° 
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181
PROBLEM 11.122
Pin P moves at a constant speed of 150 mm/s in a counterclockwise sense along a
circular slot which has been milled in the slider block A shown. Knowing that the
block moves downward at a constant speed 100 mm/s determine the velocity of
pin P when (a) θ = 30°, (b) θ = 120°.
SOLUTION
v P = v A + v P/A
v P = 100 ms (− j) + 150(cos θ i + sin θ j)mm/s
(a)
For θ = 30°
v P = −100 mm/s ( j) + 150( − cos(30°)i + sin(30°) j)mm/s
v P = (−75i + 29.9038 j)mm/s
v P = 80.7 mm/s
(b)
For θ = 120°
21.7° 
v P = −100 mm/s ( j) + 150(− cos(120°)i + sin(120°) j)mm/s
v P = (−129.9038i + −175 j) mm/s
v P = 218 mm/s
53.4° 
vP = 80.7 mm/s
21.7° 
Alternative Solution
(a)
For θ = 30°, vP /A = 7.5 in./s
30°
vP = v A + vP/A
Law of cosines
vP2 = (150) 2 + (100) 2 − 2(100)(150) cos 30°
vP = 80.7418 mm/s
Law of sines
sin β
sin 30°
=
150 80.7418
β = 111.7°
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182
PROBLEM 11.122 (Continued)
(b)
For θ = 120°, vP/A = 150 mm/s
30°
Law of cosines
vP2 = (150) 2 + (100) 2 − 2(100)(150) cos120°
vP = 217.9449 mm/s
Law of sines
sin β
sin120°
=
β = 36.6°
150
217.9449
α = 90 − β = 90° − 36.6 = 53.4°
vP = 218 mm/s
53.4° 
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183
PROBLEM 11.123
Knowing that at the instant shown assembly A has a velocity of 9 in./s and an
acceleration of 15 in./s2 both directed downward, determine (a) the velocity
of block B, (b) the acceleration of block B.
SOLUTION
Length of cable = constant
L = x A + 2 xB/A = constant
v A + 2vB/A = 0
(1)
a A + 2aB/A = 0
(2)
a A = 15 in./s 2
Data:
v A = 9 in./s
Eqs. (1) and (2)
a A = −2aB/A
v A = −2vB/A
15 = −2aB/A
9 = −2vB/A
aB/A = −7.5 in./s 2
vB/A = −4.5 in./s
a B/A = 7.5 in./s 2
(a)
40°
v B/A = −4.5 in./s
Velocity of B.
v B = v A + v B/A
Law of cosines:
vB2 = (9) 2 + (4.5) 2 − 2(9)(4.5) cos 50°
40°
vB = 7.013 in./s
Law of sines:
sin β sin 50°
=
4.5
7.013
β = 29.44°
α = 90° − β = 90° − 29.44° = 60.56°
v B = 7.01 in./s
60.6° 
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184
PROBLEM 11.123 (Continued)
(b)
Acceleration of B. a B may be found by using analysis similar to that used above for vB . An alternate
method is
a B = a A + a B/A
a B = 15 in./s 2 ↓ +7.5 in./s 2
40°
= −15 j − (7.5 cos 40°)i + (7.5 sin 40°) j
= −15 j − 5.745i + 4.821j
a B = −5.745i − 10.179 j
a B = 11.69 in./s 2
60.6° 
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185
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
2v A + vB/A = 0
| vB/A | = 16 in./s
or
2a A + aB/A = 0
and
| aB/A | = 12 in./s 2
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
8
8.5278
=
sin α sin 15°
or
α = 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
a B = 6.40 in./s 2
or
54.1° 
* Note the orientation of the coordinate axes on the sketch of the system.
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186
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
a Dt 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° 
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187
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
a C = a B + a C/ B
The graphical representation of this equation is then as shown.
First note
α = 180° − (20° + 105°)
= 55°
Then
aC
2
=
sin 20° sin 55°
aC = 0.83506 mm/s 2
(b)
aC = 0.835 mm/s 2
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
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188
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
y-direction.
vs y = 2 gh = (2)(32.2 ft/s 2 )(3 ft) = 13.90 ft/s ↓
Relative velocity.
v S/B = v S − v B
(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° 
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189
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)
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190
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
v B = v A + v B/A
or
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
v B/A = 10.54 ft/s
or
81.3° 
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191
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
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192
PROBLEM 11.129 (Continued)
3.706
21.294
β = 7.24°
21.294
= 21.47 km/h = 5.96 m/s
vR =
cos 7.24°
tan β =
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 − cos 30° 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 − cos 30° 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 cos 30° = −vR / T − 2 cos 45°
j:
vR / T −1 cos 30° + 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 
θ = tan −1 
v R = 21.4654 km/hr = 5.96 m/s
v R = 5.96 m/s
82.8° 
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193
PROBLEM 11.130
As observed from a ship moving due east at 9 km/h, the wind appears to blow from the south. After the ship
has changed course and speed, and as it is moving north at 6 km/h, the wind appears to blow from the
southwest. Assuming that the wind velocity is constant during the period of observation, determine the
magnitude and direction of the true wind velocity.
SOLUTION
v wind = v ship + v wind/ship
v w = v s + v w/s
Case 
v s = 9 km/h →; v w /s
Case 
v s = 6 km/h ↑ ; v w /s
15
= 1.6667
9
α = 59.0°
tan α =
vw = 92 + 152 = 17.49 km/h
v w = 17.49 km/h
59.0° 
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194
PROBLEM 11.131
When a small boat travels north at 5 km/h, a flag mounted on its
stern forms an angle θ = 50° with the centerline of the boat as
shown. A short time later, when the boat travels east at 20 km/h,
angle θ 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
θ = 180° − (50° + 90° + α )
= 40° − α
We have
vW
5
=
sin 50° sin (40° − α )
φ = 180° − (50° + α )
= 130° − α
vW
20
=
sin 50° sin (130° − α )
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195
PROBLEM 11.131 (Continued)
Therefore
or
or
5
20
=
sin (40° − α ) sin (130° − α )
sin 130° cos α − cos 130° sin α = 4(sin 40° cos α − cos 40° sin α )
tan α =
sin 130° − 4 sin 40°
cos 130° − 4 cos 40°
or
α = 25.964°
Then
vW =
5 sin 50°
= 15.79 km/h
sin (40° − 25.964°)
vW = 15.79 km/h
26.0° 
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196
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° + α )
Equating the expressions for
vD
3
=
sin 7° sin (23° − α )
vD
3
sin 8°
sin 7°
=
sin (22° + α ) sin (23° − α )
or
or
sin 8° (sin 23° cos α − cos 23° sin α )
= sin 7° (sin 22° cos α + cos 22° sin α )
tan α =
sin 8° sin 23° − sin 7° sin 22°
sin 8° cos 23° + sin 7° cos 22°
or
α = 2.0728°
Then
vD =
3 sin 8°
= 1.024 ft/s
sin (22° + 2.0728°)
v D = 1.024 ft/s
2.07° 
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197
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
α = −(vD cos α )i + (vD sin α ) 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° 
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198
PROBLEM 11.CQ8
The Ferris wheel is rotating with a constant angular velocity ω. 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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199
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) 
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200
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 ω 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 − rθ 2 )er + (rθ + 2rθ)eθ
(1)
r = 0, θ = 0, r = 0, θ = ω, r = -u. When we substitute
From the information given, we know 
these values into (1), we will only have a term in the −θ direction.
Answer: (d ) 
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201
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/s 2 .
SOLUTION
an =
v2
an = 0.8 m/s 2
ρ
v = 72 km/h = 20 m/s
0.8 m/s 2 =
(20 m/s) 2
ρ
ρ = 500 m  
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202
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 ρ is 25 m
and the normal component of their acceleration cannot exceed 3 g.
SOLUTION
We have
an =
v2
ρ
Then
(vmax ) 2AB = (3 × 9.81 m/s 2 )(25 m)
or
(vmax ) AB = 27.124 m/s
(vmax ) AB = 97.6 km/h 
or
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203
PROBLEM 11.135
A bull-roarer is a piece of wood that produces a roaring sound when attached to the
end of a string and whirled around in a circle. Determine the magnitude of the
normal acceleration of a bull-roarer when it is spun in a circle of radius 0.9 m at
a speed of 20 m/s.
SOLUTION
an =
v2
ρ
=
(20 m/s) 2
= 444.4 m/s 2
0.9 m
an = 444 m/s 2 

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204
PROBLEM 11.136
To test its performance, an automobile is driven around a circular test track of diameter d. Determine (a) the
value of d if when the speed of the automobile is 45 mi/h, the normal component of the acceleration
is 11 ft/s 2 , (b) the speed of the automobile if d = 600 ft and the normal component of the acceleration is
measured to be 0.6 g.
SOLUTION
(a)
First note
v = 45 mi/h = 66 ft/s
Now
an =
ρ=
v2
ρ
(66 ft/s) 2
= 396 ft
11 ft/s 2
d = 2ρ
(b)
d = 792 ft 
v2
We have
an =
Then
1

v 2 = (0.6 × 32.2 ft/s 2 )  × 600 ft 
2

ρ
v = 76.131 ft/s
v = 51.9 mi/h 
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205
PROBLEM 11.137
An outdoor track is 420 ft in diameter. A runner increases her speed at a constant
rate from 14 to 24 ft/s over a distance of 95 ft. Determine the magnitude of the total
acceleration of the runner 2 s after she begins to increase her speed.
SOLUTION
We have uniformly accelerated motion
v22 = v12 + 2at Δs12
Substituting
(24 ft/s)2 = (14 ft/s)2 + 2at (95 ft)
or
at = 2 ft/s 2
Also
v = v1 + at t
At t = 2 s:
v = 14 ft/s + (2 ft/s 2 )(2 s) = 18 ft/s
Now
an =
At t = 2 s:
an =
Finally
a 2 = at2 + an2
At t = 2 s:
a 2 = 22 + 1.542862
v2
ρ
(18 ft/s) 2
= 1.54286 ft/s 2
210 ft
a = 2.53 ft/s 2 
or
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206
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/s 2
Speed:
v = at t
Normal acceleration:
an =
where
ρ = 0.8 m = 800 mm
(a)
When t = 4 s
ρ
=
at2 t 2
ρ
v = (10)(4) = 40 mm/s
an =
Acceleration:
v2
(40)2
= 2 mm/s 2
800
a = at2 + an2 = (10) 2 + (2) 2
a = 10.20 mm/s 2 
(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 
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207
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 ρ = 400 m,
an =
v2
ρ
=
(20)2
= 1.000 m/s 2
400
a = an2 + at2
But
at = a 2 − an2 = (1.5) 2 − (1.000) 2 = ± 1.11803 m/s 2
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/s 2 

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208
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:
π
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 =
π
2
(150) = 235.6 m
vB2 = 0 + (2)(1.1495)(235.6) = 541.69 m 2 /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
ρ = 150 m
Since v < vB , the car is still on the curve.
(b)
Normal component of acceleration:
an =
Magnitude of total acceleration:
|a| =
v2
ρ
=
(22.99)2
= 3.524 m/s 2
150
at2 + an2 =
(1.1495) 2 + (3.524) 2
| a | = 3.71 m/s 2 
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209
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°) ← + 55.56sin 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/s 2
(aB )t = 6 m/s 2
an =
Normal accelerations:
ρ
( ρ = ∞)
Car B:
( ρ = 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/s 2
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
= (6cos 50° + 10.288cos 40° + 10)
+ (6sin 50° − 10.288sin 40°)
= 21.738 → + 2.017
a B/A = 21.8 m/s 2
5.3° 
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210
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 α sin 60°
α = 68.9°
v B/A = 501 km/h
(b)
First note
Now
a A = 8 m/s 2
( aB ) n =
v 2B
ρB
=
Then
60°
(150 m/s) 2
300 m
(a B ) n = 75 m/s 2

(a B )t = 3 m/s 2
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/s 2
25.1° 
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211
PROBLEM 11.143
From a photograph of a homeowner using a snowblower, it
is determined that the radius of curvature of the trajectory of
the snow was 30 ft as the snow left the discharge chute at A.
Determine (a) the discharge velocity v A of the snow, (b) the
radius of curvature of the trajectory at its maximum height.
SOLUTION
(a) The acceleration vector is 32.2 ft/s .
At Point A, tangential and normal components of a are as shown
in the sketch.
an = a cos 40° = 32.2 cos 40° = 24.67 ft/s 2
v A2 = ρ A (a A ) n = (30)(24.67) = 740.0 ft 2 /s 2
v A = 27.2 ft/s
40° 
vx = 27.20 cos 40° = 20.84 ft/s
(b)
At maximum height,
v = vx = 20.84 ft/s
an = g = 32.2 ft/s 2 ,
ρ =
v2
(20.84) 2
=
32.2
an
ρ = 13.48 ft 
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212
PROBLEM 11.144
A basketball is bounced on the ground at Point A and rebounds with a velocity v A of
magnitude 2.5 m/s as shown. 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)
(a A ) n =
We have
ρA =
or
v 2A
ρA
(2.5 m/s) 2
(9.81 m/s 2 ) sin 15°
ρ A = 2.46 m 
or
(b)
( aB ) n =
We have
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A sin 15°

ρB =
Then
[(2.5 m/s) sin 15°]2
= 0.0427 m
9.81 m/s 2
ρ B = 42.7 mm 
or
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213
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 =
ρA =
v 2A
ρA
(50 m/s)2
(9.81 m/s 2 ) cos 25°
ρ A = 281 m 
or
(b)
We have
( aB ) n =
vB2
ρB
where Point B is the highest point of the trajectory, so that
vB = (v A ) x = v A cos 25°
Then
ρB =
[(50 m/s) cos 25°]2
9.81 m/s 2
ρ B = 209 m 
or
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214
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 = v0 t
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
Speed:
v 2 = vx2 + v 2y = v02 + 2 gh
Direction of velocity.
cos θ =
v0
v
Direction of normal acceleration.
an = g cos θ =
gv0 v 2
=
v
ρ
v3
gv0
Radius of curvature:
ρ=
At Point B,
hB = 1 m; xB = 7 m
tB =
(2)(1 m)
= 0.45152 s
9.81 m/s 2
xB = v0t 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
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215
PROBLEM 11.146 (Continued)
(a)
Radius of curvature at Point B.
ρB =
At Point C
(259.97 m 2 /s 2 )3/ 2
(9.81 m/s 2 )(15.504 m/s)
ρ B = 27.6 m 
hC = 1 m + 2 m = 3 m
vC2 = (15.504) 2 + (2)(9.81)(3) = 299.23 m 2 /s 2
(b)
Radius of curvature at Point C.
ρC =
(299.23 m 2 /s 2 )3/2
(9.81 m/s 2 )(15.504 m/s)
ρC = 34.0 m 
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216
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/s 2
(a) At Point A.
Sketch tangential and normal components of acceleration at A.
(a A ) n = g cos 50°
ρA =
vA2
(2) 2
=
(a A ) n
9.81cos 50°
ρ A = 0.634 m 
(b) At Point B, 1 meter below Point A.
Horizontal motion: (vB ) x = (v A ) x = 2 cos 50° = 1.286 m/s
Vertical motion:
(vB ) 2y = (v A ) 2y + 2a y ( yB − y A )
= (2 cos 40°)2 + (2)(−9.81)(−1)
= 21.97 m 2 /s 2
(vB ) y = 4.687 m/s
tan θ =
(vB ) y
(vB ) x
=
4.687
,
1.286
or
θ = 74.6°
aB = g cos 74.6°
2
2
(v ) + (vB ) y
v 2
ρB = B = B x
(aB )n
g cos 74.6°
=
(1.286) 2 + 21.97
9.81cos 74.6°
ρ B = 9.07 m 
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217
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 2A
ρA
or
4

v A2 =  (9.81 m/s 2 )  (25 m)
5


or
v A = 14.0071 m/s
vA = 14.01 m/s
(b)
We have
( aB ) n =
vB2
ρB
4
vA
5
Where
vB = ( v A ) x =
Then
( 4 × 14.0071 m/s )
ρ = 5
B
36.9° 
2
9.81 m/s 2
ρ B = 12.80 m 
or
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218
PROBLEM 11.149
A child throws a ball from Point A with an initial velocity v A of
20 m/s at an angle of 25° with the horizontal. Determine the
velocity of the ball at the points of the trajectory described by the
ball where the radius of curvature is equal to three-quarters of its
value at A.
SOLUTION
Assume that Points B and C are the points of interest, where yB = yC and vB = vC .
v 2A
Now
(a A ) n =
or
ρA =
v 2A
g cos 25°
Then
ρB =
v A2
3
3
ρA =
4
4 g cos 25°
ρA
vB2
We have
( aB ) n =
where
(aB ) n = g cos θ
so that
v A2
vB2
3
=
4 g cos 25° g cos θ
or
vB2 =
ρB
3 cos θ 2
vA
4 cos 25°
(1)
Noting that the horizontal motion is uniform, we have
( v A ) x = ( vB ) x
where
(v A ) x = v A cos 25°
Then
v A cos 25° = vB cos θ
or
cos θ =
(vB ) x = vB cos θ
vA
cos 25°
vB
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219
PROBLEM 11.149 (Continued)
Substituting for cos θ in Eq. (1), we have
or
vB2 =
 v A2
3  vA
 cos 25° 
4  vB
 cos 25°
vB3 =
3 3
vA
4
3
vB = 3 v A = 18.17 m/s
4
4
cos 25°
3
θ = ± 4.04°
cos θ = 3
and
v B = 18.17 m/s
4.04° 
v B = 18.17 m/s
4.04° 
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220
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 θ 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 ρ = ρ min /cos3 θ .
SOLUTION
For the arbitrary Point C, we have
(aC ) n =
ρC =
or
vC2
ρC
vC2
g cos θ
Noting that the horizontal motion is uniform, we have
(v A ) x = (vC ) x
where
(v A ) x = v0 cos α
(vC ) x = vC cos θ
Then
v0 cos α = vC cos θ
or
vC =
so that
v 2 cos 2 α
1  cos α 
ρC =
v0  = 0

g cos θ  cos θ 
g cos3 θ
cos α
v0
cos θ
2
(a)
In the expression for ρC , v0 , α , and g are constants, so that ρC is minimum where cos θ is
maximum. By observation, this occurs at Point B where θ = 0.
ρ min = ρ B =
(b)
v02 cos 2 α
g
ρC =
1  v02 cos 2 α 


g
cos3 θ 

ρC =
ρ min
cos3 θ
Q.E.D.

Q.E.D.

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221
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 ωnt)i + ctj + (Rt sin ωnt)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 ωn t − ωn t sin ωn t )i + cj + R (sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
= R − ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t i
dt
(
)
(
)
+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k
or
a = ωn R [−(2 sin ωn t + ωn t cos ωn t )i + (2 cos ωn t − ωn t sin ωn t ) k ]
Now
v 2 = R 2 (cos ωn t − ωn t sin ωn t )2 + c 2 + R 2 (sin ωn t + ωn t cos ωn t )2
(
)
= R 2 1 + ωn2 t 2 + c 2
(
and
Now
At t = 0:
)
1/2
v =  R 2 1 + ωn2 t 2 + c 2 


Then
R 2ωn2 t
dv
=
1/ 2
dt  2
R 1 + ωn2 t 2 + c 2 


(
a
2
)
2
 v2 
+  
 dt   ρ 
 dv 
= at2 + an2 =  
2
dv
=0
dt
a = ωn R(2 k ) or a = 2ωn R
v2 = R2 + c2
Then, with
dv
= 0,
dt
we have
a=
or
2ωn R =
v2
ρ
R 2 + c2
ρ=
ρ
R 2 + c2

2ωn R
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222
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.
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

 t 
2
dv
 t2 + 1   j

t
t
+
1
−
a=
= 3(− sin t − sin t − t cos t )i + 3



dt
2
t
+
1


)
(
 3t 
dr
= 3(cos t − t sin t )i +  2
j + (sin t + t cos t )k
 t + 1 
dt


+ (cos t + cos t − t sin t )k
= − 3(2sin t + t cos t )i + 3
1
j
2
t
+
(
1)1/2
+ (2 cos t − t sin t )k
t2
v 2 = 9 (cos t − t sin t )2 + 9 2
+ (sin t + t cos t )2
t +1
Then
Expanding and simplifying yields
v 2 = t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t
Then
v = [t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t ]1/ 2
and
dv 4t 3 + 38t + 8(−2 cos t sin t + 4t 3sin 2 t + 2t 4 sin 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
Now
2
2
 dv   v 
a 2 = at2 + an2 =   +  
 dt   ρ 
2
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223
PROBLEM 11.152* (Continued)
At t = 0:
a = 3j + 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
ρ=
v2
ρ
9 ft 2 /s 2
ρ = 2.50 ft 
13 ft/s 2
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224
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/s 2 ,
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/s 2 ,
For the sun,
R=
and
Given that an =
1
1
D =   (1.39 × 109 ) = 0.695 × 109 m
2
2
gR 2
v2
and
that
for
a
circular
orbit
=
a
n
r
r2
r =
Eliminating an and solving for r,
v = 107 × 106 m/h = 29.72 × 103 m/s
For the planet Earth,
Then
gR 2
v2
r =
(274)(0.695 × 109 ) 2
= 149.8 × 109 m
(29.72 × 103 ) 2
r = 149.8 Gm 
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225
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/s 2
For the sun,
R=
and
Given that an =
1
1
D =   (1.39 × 109 ) = 0.695 × 109 m
2
2
gR 2
v2
.and
that
for
a
circular
orbit:
=
a
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 
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226
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/s 2 , R = 3761 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
Eliminating an and solving for v,
v= R
For Venus,
g
r
g = 29.20 ft/s 2
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 
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227
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/s 2 , R = 2102 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
Eliminating an and solving for v,
v= R
For Mars,
g
r
g = 12.17 ft/s 2
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
11.6266 × 106
v = 7740 mi/h 
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228
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/s 2 , R = 44, 432 mi.
SOLUTION
From Problems 11.153 and 11.154,
an =
gR 2
r2
For a circular orbit,
an =
v2
r
Eliminating an and solving for v,
v= R
For Jupiter,
g
r
g = 75.35 ft/s 2
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
6
235.13 × 10
v = 90600 mi/h 
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229
PROBLEM 11.158
A satellite is traveling in a circular orbit around Mars at an altitude of 300 km. After the altitude of the satellite
is adjusted, it is found that the time of one orbit has increased by 10 percent. Knowing that the radius of Mars
is 3382 km, determine the new altitude of the satellite. (See information given in Problems 11.153–11.155.)
SOLUTION
an = g
We have
Then
g
R2
r2
and an =
v2
r
where
r =R+h
R2 v2
=
r
r2
v=R
g
r
The circumference s of a circular orbit is equal to
s = 2π r
Assuming that the speed of the satellite in each orbit is constant, we have
s = vtorbit
Substituting for s and v
2π r = R
torbit =
=
Now
g
torbit
r
2π r 3/2
R g
2π ( R + h)3/2
R
g
(torbit ) 2 = 1.1(torbit )1
2π ( R + h2 )3/2
2π ( R + h1 )3/2
= 1.1
R
R
g
g
h2 = (1.1)2/3 ( R + h1 ) − R
= (1.1)2/3 (3382 + 300) km − (3382 km)
h2 = 542 km 
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230
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
where
r =R+h
R2 v2
=
r
r2
v=R
g
r
The circumference s of the circular orbit is equal to
s = 2π r
Assuming that the speed of the telescope is constant, we have
s = vtorbit
Substituting for s and v
2π r = R
or
torbit =
=
g
torbit
r
2π r 3/2
R g
2π
[(6370 + 590) km]3/2
1h
×
−3
2 1/2
6370 km [9.81 × 10 km/s ]
3600 s
torbit = 1.606 h 
or
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231
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 = 2π r
Assuming that the speeds of the satellites are constant, we have
s = vT
Substituting for s and v
2π r = R
g
T
r
2π r 3/ 2 2π ( R + h)3/2
=
R g
R
g
or
T=
Now
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
 2π ( R + hB )3/2 
 2π ( R + hA )3/2 
(
1)
N
=
N
+



g
g
 R

 R

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232
PROBLEM 11.160 (Continued)
N=
or
=
Then
( R + hA )3/2
( R + hB )
3/ 2
− ( R + hA )
1
(
3960 + 200
3960 +120
) −1
3/2
TC = N (T ) B = N
3/2
=
1
(
R + hB
R + hA
) −1
3/2
= 33.835 orbits
2π ( R + hB )3/2
R
g
[(3960 + 200) mi] × 1 h
2π
3960 mi 32.2 ft/s 2 × 1 mi 1/2 3600s
3/2
= 33.835
(
5280 ft
)
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 θ A and θ B
swept out by radial lines drawn to the satellites must differ by 2π . That is,
θ A = θ B + 2π
For a circular orbit
s = rθ
From above
s = vt and v = R
Then
θ=
At time TC :
or
R g
( R + hA )
3/2
TC =
TC =
=
g
r
R g
R g
s vt 1 
g
= = R
t
 t = 3/2 t =

r r r
r 
r
( R + h)3/2
R g
( R + hB )3/ 2
TC + 2π
2π

1
1
R g
−
 ( R + hA )3/ 2 ( R + hB )3/ 2 
2π
(
1 mi
(3960 mi) 32.2 ft/s 2 × 5280
ft
×
×
)
1/2
1
1
1
−
[(3960 + 120) mi ]3/ 2 [(3960 + 200) mi ]3/ 2
1h
3600 s
TC = 51.2 h 
or
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233
PROBLEM 11.161
The oscillation of rod OA about O is defined by the relation θ = (3/π )(sin π t ),
where θ 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.
θ=
r = 12e −2t in/s
θ = 3cosπ t rad/s
r = −24e−2t in/s 2
θ = −3π sin π t rad/s2
π
sin π t rad
At t = 1 s,
(a)
3
r = 6 − 6e −2 = 5.1880 in.
θ=
r = 12e −2 = 1.6240 in/s
θ = 3cos π = −3 rad/s

r = −24e−2 = −3.2480 in/s 2
θ = −3π sin π = 0
π
sin π = 0
Velocity of the collar.
v = rer + rθeθ = 1.6240 e r + (5.1880)(−3)eθ
v = (1.624 in/s)er + (15.56 in/s)eθ 
(b)
Acceleration of the collar.
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
= [ −3.2480 − (5.1880)( −3) 2 ]er + (5.1880)(0) + (2)(1.6240)(−3)]eθ
(−49.9 in/s 2 )er + (−9.74 in/s 2 )eθ 
(c)
Acceleration of the collar relative to the rod.
a B /OA = (−3.25 in/s 2 )er 
a B /OA = 
re r
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234
PROBLEM 11.162
The rotation of rod OA about O is defined by the relation θ = t 3 − 4t , where
θ and t are expressed in radians and seconds, respectively. Collar B slides along
the rod so that its distance from O is r = 2.5t 3 − 5t 2 , 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 radius of curvature of the path of
the collar.
SOLUTION
Calculate the derivatives with respect to time.
r = 2.5t 3 − 5t 2
θ = t 3 − 4t
r = 7.5t 2 − 10t
θ = 3t 2 − 4

r = 15t − 10
θ = 6t
At t = 1 s,
(a)
r = 2.5 − 5 = −2.5 in.
θ = 1 − 4 = −3 rad
r = 7.5 − 10 = −2.5 in./s
θ = 3 − 4 = −1 rad/s
r = 15 − 10 = 5 in./s 2
θ = 6 rad/s 2
Velocity of the collar.
v = rer + rθeθ = −2.5er + (−2.5)( −1)eθ
v = ( −2.50 in./s)er + (2.50 in./s)eθ 
v = (2.50) 2 + (2.50) 2 = 3.5355 in./s
Unit vector tangent to the path.
et =
(b)
v
= −0.70711e r + 0.70711eθ
v
Acceleration of the collar.
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
= [5 − (−2.5)(−1)2 ]er + [(−2.5)(6) + (2)(−2.5)(−1)]eθ
a = (7.50 in/s 2 )er + (−10.00 in/s 2 )eθ 
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235
PROBLEM 11.162 (Continued)
Magnitude:
a = (7.50) 2 + (10.00) 2 = 12.50 in./s 2
at = aet
Tangential component:
at = (7.50)(−0.70711) + (−10.00)(0.70711) = −12.374 in./s 2
Normal component:
(c)
an = a 2 − at2 = 1.7674 in./s 2
Radius of curvature of path.
an =
ρ=
v2
ρ
v 2 (3.5355 in./s) 2
=
an
1.7674 in./s 2
ρ = 7.07 in. 
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236
PROBLEM 11.163
The path of particle P is the ellipse defined by the relations
r = 2/(2 − cos π t ) and θ = π t , where r is expressed in meters, t is
in seconds, and θ is in radians. Determine the velocity and the
acceleration of the particle when (a) t = 0, (b) t = 0.5 s.
SOLUTION
We have
r=
2
2 − cos π t
θ = πt
Then
r =
−2π sin π t
(2 − cos π t )2
θ = π
and
r = −2π
(a)
At t = 0:
Now
π cos π t (2 − cos π t ) − sin π t (2π sin π t )
(2 − cos π t )3
= −2π 2
2cos π t − 1 − sin 2 π t
(2 − cos π t )3
r=2m
θ =0
r = 0
θ = π rad/s
r = −2π 2 m/s 2
θ = 0
v = rer + rθeθ = (2)(π )eθ
v = (2π m/s)eθ 
or
and
θ = 0
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
= [ −2π 2 − (2)(π )2 ]er
a = −(4π 2 m/s 2 )er 
or
(b)
At t = 0.5 s:
θ=
r =1 m
r =
−2π
π
= − m/s
2
2
(2)
r = −2π 2
−1 − 1 π 2
=
m/s 2
3
2
(2)
π
rad
2
θ = π rad/s
θ = 0
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237
PROBLEM 11.163 (Continued)
Now
 π
v = rer + rθeθ =  −  e r + (1)(π )eθ
 2
π

v = −  m/s  er + (π m/s)eθ 
2


or
and
a = (r − rθ 2 )e r + (rθ + 2rθ)eθ
π 2

  π 
=
− (1)(π ) 2  e r +  2  −  (π )  eθ
  2 
 2

π2

a = − 
m/s 2  er − (π 2 m/s 2 )eθ 
 2

or
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238
PROBLEM 11.164
The two-dimensional motion of a particle is defined by the relations r = 2a cos θ and θ = bt 2 /2, where a and b
are constants. Determine (a) the magnitudes of the velocity and acceleration at any instant, (b) the radius of
curvature of the path. What conclusion can you draw regarding the path of the particle?
SOLUTION
(a)
We have
r = 2a cos θ
1
θ = bt 2
2
Then
r = −2aθ sin θ
θ = bt
and

r = −2a (θ sin θ + θ 2 cos θ )
θ = b
Substituting for θ and θ
r = −2abt sin θ

r = −2ab(sin θ + bt 2 cos θ )
vθ = rθ = 2abt cos θ
Now
vr = r = −2abt sin θ
Then
v = vr2 + vθ2 = 2abt[( − sin θ ) 2 + (cos θ ) 2 ]1/2
v = 2abt 
or
Also
ar = r − rθ 2 = −2ab(sin θ + bt 2 cos θ ) − 2ab 2 t 2 cos θ
= −2ab(sin θ + 2bt 2 cos θ )
and
aθ = rθ + 2rθ = 2ab cos θ − 4ab 2 t 2 sin θ
= −2ab(cos θ − 2bt 2 sin θ )
Then
a = ar2 + aB2 = 2ab[(sin θ + 2bt 2 cos θ ) 2
+ (cos θ − 2bt 2 sin θ ) 2 ]1/2
a = 2ab 1 + 4b 2 t 4 
or
(b)
2
2
 v2 
+  
 dt   ρ 
 dv 
= at2 + an2 =  
Now
a
Then
dv d
= (2abt ) = 2ab
dt dt
2
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239
PROBLEM 11.164 (Continued)
so that
or
(
2ab 1 + 4b 2 t 4
) = (2ab) + a
2
2
2
n
4a 2b 2 (1 + 4b 2t 4 ) = 4a 2 b 2 + an2
or
an = 4ab 2 t 2
Finally
an =
v2
(2abt ) 2
ρ=
ρ
4ab 2 t 2
ρ =a 
or
Since the radius of curvature is a constant, the path is a circle of radius a.

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240
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 θ ) and that θ = kt , determine the
velocity and acceleration of P when (a) θ = 0, (b) θ = 90°.
SOLUTION
2b
θ = kt
1 + cos kt
2bk sin kt
θ = k θ = 0
r =
(1 + cos kt ) 2
2bk
[(1 + cos kt )2 k cos kt + (sin kt )2(1 + cos kt )( k sin kt )]
r =
(1 + cos kt ) 4
r=
(a)
When θ = kt = 0:
2bk
1
[(2)2 k (1) + 0] = bk 2
4
2
(2)
r =b
r = 0

r=
θ =0
θ = k
θ = 0
vθ = rθ = bk
vr = r = 0
v = bk eθ 
1
1

ar = 
r − rθ 2 = bk 2 − bk 2 = − bk 2 
2
2


aθ = rθ + 2rθ = b(0) + 2(0) = 0

(b)
1
a = − bk 2 er 
2
When θ = kt = 90°:
r = 2b
r = 2bk
θ = 90°
θ = k
2bk
[0 + 2k ] = 4bk 2
19
θ = 0
r =
vθ = rθ = 2bk
v = 2bk er + 2bk eθ 
ar = r − rθ 2 = 4bk 2 − 2bk 2 = 2bk 2
a = rθ + 2rθ = 2b(0) + 2(2bk )k = 4bk 2
a = 2bk 2 e r + 4bk 2 eθ 
vr = r = 2bk
θ
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241
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 θ,
(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 θ
r = −2b sin θ θ
Since θ = constant, θ = 0
r = −2b cos θ θ 2
ar = r − rθ 2 = −2b cos θ θ 2 − (2b cos θ )θ 2
a = −4b cos θ θ 2
r
aθ = rθ + 2rθ = (2b cos θ )(0) + 2(−2b sin θ )θ 2
a = −4b sin θ θ 2
θ
a = ar2 + aθ2 = 4bθ 2 (− cos θ ) 2 + (− sin θ )2
a = 4bθ 2
Since both b and θ are constant, we find that
a = constant 
a
ar
 −4b sin θ θ 2 
 2 
 −4b cos θ θ 
γ = tan −1 θ = tan −1 
γ = tan −1 (tan θ )
γ =θ
Thus, a is directed toward A 
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242
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, θ , and θ, (b) the magnitude of the acceleration
in terms of b, θ , θ, and θ.
SOLUTION
(a)
We have
r=
b
cos θ
Then
r =
bθ sin θ
cos 2 θ
We have
v 2 = vr2 + vθ2 = (r)2 + (rθ) 2
2
or
 bθ sin θ   bθ 2
= 
 + 

2
 cos θ   cos θ 
 b2θ 2
b2θ 2  sin 2θ
=
+
1
 =
2 
2
4
cos θ  cos θ
 cos θ
bθ
v=±
cos 2θ
For the position of the car shown, θ is decreasing; thus, the negative root is chosen.
v=−
bθ

cos 2θ
v=−
bθ

cos 2θ
Alternative solution.
From the diagram
or
r = −v sin θ
bθ sin θ
= −v sin θ
cos 2θ
or



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243
PROBLEM 11.167 (Continued)
(b)
For rectilinear motion
a=
dv
dt
Using the answer from Part a
v=−
Then
a=
bθ
cos 2θ
d 
bθ 
 −

dt  cos 2θ 
= −b
θ cos 2θ − θ(−2θ cos θ sin θ )
cos 4θ
a=−
or
b
(θ + 2θ 2 tan θ ) 
cos 2θ
Alternative solution
Then
b
bθ sin θ
r =
cos θ
cos 2 θ
(θ sin θ + θ 2 cos θ )(cos 2θ ) − (θ sin θ )(−2θ cos θ sin θ )
r = b
cos 4θ
 θ sin θ θ 2 (1 + sin 2 θ ) 
= b
+

2
cos3θ
 cos θ

Now
a 2 = ar2 + aθ2
From above
where
and
r=
 θ sin θ θ 2 (1 + sin 2 θ )  bθ 2
+
ar = r − rθ 2 = b 
−
2
cos 2θ
 cos θ
 cos θ
b  
2θ 2 sin 2 θ 
=
+
θ
sin
θ


cos θ 
cos 2θ 
b sin θ 
ar =
(θ + 2θ 2 tan θ )
2
cos θ
aθ = rθ + 2rθ =
=
Then
bθ
bθ 2 sin θ
+2
cos θ
cos 2θ
b cos θ 
(θ + 2θ tan θ )
cos 2θ
a=±
b
(θ + 2θ 2 tan θ )[(sin θ ) 2 + (cos θ )2 ]1/ 2
2
cos θ
For the position of the car shown, θ is negative; for a to be positive, the negative root is chosen.
b
a=−
(θ + 2θ 2 tan θ ) 
cos 2 θ
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244
PROBLEM 11.168
After taking off, a helicopter climbs in
a straight line at a constant angle β . Its
flight is tracked by radar from Point A.
Determine the speed of the helicopter
in terms of d, β , θ , and θ.
SOLUTION
From the diagram
r
d
=
sin (180° − β ) sin ( β − θ )
or
d sin β = r (sin β cos θ − cos β sin θ )
tan β
tan β cos θ − sin θ
or
r=d
Then
r = d tan β
−(− tan β sin θ − cos θ ) 
θ
(tan β cos θ − sin θ )2
tan β sin θ + cos θ
= dθ tan β
(tan β cos θ − sin θ ) 2
From the diagram
vr = v cos ( β − θ )
where
vr = r
Then
dθ tan β
tan β sin θ + cos θ
= v(cos β cos θ + sin β sin θ )
(tan β cos θ − sin θ )2
= v cos β (tan β sin θ + cos θ )
v=
or
dθ tan β sec β

(tan β cosθ − sin θ )2
Alternative solution.
We have
v 2 = vr2 + vθ2 = (r) 2 + ( rθ) 2
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245
PROBLEM 11.168 (Continued)
Using the expressions for r and r from above

tan β sin θ + cos θ 
v =  dθ tan β

(tan β cos θ − sin θ ) 2 

2
1/ 2
or
 (tan β sin θ + cos θ )2

dθ tan β
v=±
+ 1

2
(tan β cos θ − sin θ )  (tan β cos θ − sin θ )

1/ 2


tan 2 β + 1
dθ tan β
=±

2
(tan β cos θ − sin θ )  (tan β cos θ − sin θ ) 
Note that as θ increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.
v=
dθ tan β sec β

(tan β cos θ − sin θ )2
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246
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. What are
r , θ and θ for this
the recorded values of r, 
instant?
SOLUTION
Geometry. The polar coordinates are
r = (2400) 2 + (1800)2 = 3000 ft
Velocity Analysis.
 1800 
 = 36.87°
 2400 
θ = tan −1 
v = 315 mi/h = 462 ft/s
vr = 462 cos θ = 369.6 ft/s
vθ = −462sin θ = −277.2 ft/s
vr = r
v
r
vθ = rθ
θ = θ = −
r = 370 ft/s 
277.2
3000
θ = −0.0924 rad/s 
Acceleration analysis.
at = 10 ft/s 2
an =
v2
ρ
=
(462) 2
= 40.425 ft/s 2
5280
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247
PROBLEM 11.169 (Continued)
ar = at cos θ + an sin θ = 10 cos 36.87° + 40.425 sin 36.87° = 32.255 ft/s 2
aθ = − at sin θ + an cos θ = −10 sin 36.87° + 40.425 cos 36.87° = 26.34 ft/s 2
a = 
r − rθ 2 
r = a + rθ 2
r
r

r = 32.255 + (3000)( −0.0924) 2

r = 57.9 ft/s 2 
aθ = rθ + 2rθ
a
2rθ
θ = θ −
r
r
26.34 (2)(369.6)(−0.0924)
=
−
3000
3000
θ = 0.0315 rad/s 2 
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248
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 ω. At the instant when β = 60°,
r and θ. Express your answers in terms
determine (a) r and θ, (b) 
of d and ω.
SOLUTION
Looking at d and β as polar coordinates with d = 0,
vβ = d β = d ω,
vd = d = 0
aβ = d β + 2d β = 0,
r = d 3 for angles shown.
Geometry analysis:
(a)
Velocity analysis:
ad = d − d β 2 = −d ω 2
Sketch the directions of v, er and eθ.
vr = r = v ⋅ er = d ω cos120°
1
r = − d ω 
2
vθ = rθ = v ⋅ eθ = d ω cos 30°
θ =
(b)
Acceleration analysis:
3
dω cos 30° dω 2
=
r
d 3
θ =
1
ω
2
Sketch the directions of a, er and eθ.
ar = a ⋅ e r = a cos150° = −
3
dω 2
2
3

r − rθ2 = −
d ω2
2

r =−
3
3
1 
dω 2 + rθ 2 = −
dω 2 + d 3  ω 
2
2
2 
2
r = −
3
dω 2 
4
1  1
 1
 1  
2
 − d ω − (2)  − d ω  ω  
2
2
3d 

 2  
θ = 0 
1
aθ = a ⋅ eθ = d ω 2 cos120° = − d ω 2
2
a = rθ + 2rθ
θ
θ =
1
(aθ − 2rθ) =
r
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249
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 θ = 60° to
the position θ = 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, θ , and
θ, (b) the magnitude of the acceleration in terms of b, θ ,
θ, and θ.
SOLUTION
From the diagram:
Δr12 = 25 tan 60° − 25 tan 35°
= 25.796 m
Now
vave =
=
Δr12
Δt12
25.796 m
0.5 s
= 51.592 m/s
vave = 185.7 km/h 
or
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250
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 θ = 20°, respectively. Four seconds later, the radar
station sighted the helicopter at r = 3320 ft and θ = 23.1°. Determine the average speed and the angle of
climb β 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 β . Its flight is
tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ.
SOLUTION
We have
r0 = 3000 ft
r4 = 3320 ft
θ0 = 20°
θ 4 = 23.1°
From the diagram:
Δr 2 = 30002 + 33202
− 2(3000)(3320) cos (23.1° − 20°)
or
Δr = 362.70 ft
Now
vave =
Δr
Δt
362.70 ft
=
4s
= 90.675 ft/s
vave = 61.8 mi/h 
or
Also,
or
Δr cos β = r4 cos θ 4 − r0 cos θ0
cos β =
3320 cos 23.1° − 3000 cos 20°
362.70
β = 49.7° 
or
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251
PROBLEM 11.173
A particle moves along the spiral shown; determine the magnitude of the velocity
of the particle in terms of b, θ , and θ.
SOLUTION
Hyperbolic spiral.
r=
b
θ
dr
b dθ
b
=− 2
= − 2 θ
dt
θ dt
θ
b 
b
vr = r = − 2 θ
vθ = rθ = θ
r =
θ
θ
2
 1  1
v = vr2 + vθ2 = bθ  − 2  +  
 θ  θ 
bθ
= 2 1+θ 2
2
θ
v=
b
θ2
1 + θ 2 θ 
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252
PROBLEM 11.174
A particle moves along the spiral shown; determine the magnitude of the velocity
of the particle in terms of b, θ , and θ.
SOLUTION
Logarithmic spiral.
r = ebθ
r =
dr
dθ
= bebθ
= bebθ θ
dt
dt
vr = r = bebθ θ vθ = rθ = ebθ θ
v=
vr2 + vθ2 = ebθ θ
2
b +1
v = ebθ 1 + b 2 θ 
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253
PROBLEM 11.175
A particle moves along the spiral shown. Knowing that θ is constant and
denoting this constant by ω , determine the magnitude of the acceleration of
the particle in terms of b, θ , and ω.
SOLUTION
b
Hyperbolic spiral.
r=
From Problem 11.173
r = −
θ

r =−
b
θ2
θ
b  2b  2
θ + 3θ
2
θ
θ
b
2b
b
ar = 
r − rθ 2 = − 2 θ + 3 θ 2 − θ 2
θ
θ
θ
b
b
b
 b 
aθ = rθ + 2rθ = θ + 2  − 2 θ θ = θ − 2 2 θ 2
θ
θ
θ
 θ 
Since θ = ω = constant, θ = 0, and we write:
b 2 bω 2
2
−
ω
ω = 3 (2 − θ 2 )
θ
θ3
θ
2
b
bω
aθ = −2 2 ω 2 = − 3 (2θ )
θ
θ
ar = +
2b
a = ar2 + aθ2 =
bω 2
θ
3
(2 − θ 2 ) 2 + (2θ ) 2 =
bω 2
θ
3
4 − 4θ 2 + θ 4 + 4θ 2
a=
bω 2
θ3
4 +θ 4 
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254
PROBLEM 11.176
A particle moves along the spiral shown. Knowing that θ is constant and
denoting this constant by ω , determine the magnitude of the acceleration of
the particle in terms of b, θ , and ω.
SOLUTION
Logarithmic spiral.
r = ebθ
dr
= bebθ θ
dt
r = bebθ θ + b 2 ebθ θ 2 = bebθ (θ + bθ 2 )
r =
ar = r − rθ 2 = bebθ (θ + bθ 2 ) − ebθ θ 2
a = rθ + 2rθ = ebθ θ + 2(bebθ θ)θ
θ
Since θ = ω = constant, θ = θ , and we write
ar = bebθ (bω 2 ) − ebθ ω 2 = ebθ (b 2 − 1)ω 2
aθ = 2bebθ ω 2
a = ar2 + aθ2 = ebθ ω 2 (b2 − 1)2 + (2b) 2
= ebθ ω 2 b4 − 2b 2 + 1 + 4b 2 = ebθ ω 2 b 4 + 2b 2 + 1
= ebθ ω 2 (b2 + 1) 2 = ebθ ω 2 (b2 + 1)
a = (1 + b 2 )ω 2 ebθ 
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255
PROBLEM 11.177
The motion of a particle on the surface of a right circular cylinder is
defined by the relations R = A, θ = 2π t , and z = B sin 2π 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
θ = 2π t
z = B sin 2π nt
θ = 2π
z = 2π n B cos 2π nt
 = 0
R
θ = 0

z = −4π 2 n 2 B sin 2π nt
Velocity (Eq. 11.49)
v = R e R + Rθeθ + zk
v=
+ A(2π )eθ + 2π n B cos 2π nt k
v = 2π A2 + n 2 B 2 cos 2 2π nt 
Acceleration (Eq. 11.50)
 - Rθ 2 )e + ( Rθ + 2 Rθ)e + 
a = (R
zk
R
θ
a = −4π 2 Aek − 4π 2 n 2 B sin 2π nt k
a = 4π 2 A2 + n4 B 2 sin 2 2π nt 
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256
PROBLEM 11.178
Show that r = hφ sin θ knowing that at the instant shown, step AB of
the step exerciser is rotating counterclockwise at a constant rate φ.
SOLUTION
From the diagram
r 2 = d 2 + h 2 − 2dh cos φ
Then
2rr = 2dhφ sin φ
Now
r
d
=
sin φ sin θ
or
r=
d sin φ
sin θ
Substituting for r in the expression for r
 d sin φ 

 sin θ  r = dhφ sin φ


or
r = hφ sin θ
Q.E.D.

Alternative solution.
First note
α = 180° − (φ + θ )
Now
v = v r + vθ = re r + rθeθ
With B as the origin
vP = dφ
( d = constant  d = 0)
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257
PROBLEM 11.178 (Continued)
With O as the origin
(vP )r = r
where
(vP )r = vP sin α
Then
r = dφ sin α
Now
h
d
=
sin α sin θ
or
d sin α = h sin θ
substituting
r = hφ sin θ
Q.E.D.
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258
PROBLEM 11.179
The three-dimensional motion of a particle is defined by the relations R = A(1 − e −t ), θ = 2π 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
θ = 2π t
z = B (1 − e−t )
θ = 2π
z = Be−t
 = − Ae−t
R
θ = 0

z = − Be−t
Velocity (Eq. 11.49)
v = R e R + Rθeθ + zk
v = Ae −t e R + 2π A(1 − e −t )eθ + Be−t k
(a)
When
t = 0: e−t = e0 = 1;
v = Ae R + Bk
v = A2 + B 2 
(b)
When
t = ∞ : e− t = e −∞ = 0
v = 2π Aeθ
v = 2π A 
Acceleration (Eq. 11.50)
 − Rθ1 )e + ( Rθ + 2 Rθ)e + 
a = (R
zk
R
θ
= [ − Ae−t − A(1 − e −t )4π 2 ]e R + [0 + 2 Ae−t (2π )]eθ − Be −t k
(a)
When
t = 0: e −t = e0 = 1
a = − Ae R + 4π Aeθ − B k
a = A2 + (4π A) 2 + B 2
(b)
When
a = (1 + 16π 2 ) A2 + B 2 
t = ∞ : e −t = e−∞ = 0
a = −4π 2 Ae R
a = 4π 2 A 
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259
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 ωnt)i + ctj + (Rt sin ωnt)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 ωn t )i + ctj + ( Rt sin ωn t )k
Then
v=
dr
= R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k
dt
and
a=
dv
dt
(
)
= R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i
(
+ R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t
)k
= ωn R[−(2sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t )k ]
It then follows that the vector ( v × a) is perpendicular to the osculating plane.
i
j
k
( v × a) = ωn R R(cos ωn t − ωn t sin ωn t ) c R(sin ωn t + ωn t cos ωn t )
−(2sin ωn t + ωn t cos ωn t ) 0 (2cos ωn t − ωn t sin ωn t )
= ωn R{c(2 cos ωn t − ωn t sin ωn t )i + R[ −(sin ωn t + ωn t cos ωn t )(2sin ωn t + ωn t cos ωn t )
− (cos ωn t − ωn t sin ωn t )(2 cos ωn t − ωn t sin ωn t )] j + c(2sin ωn t + ωn t cos ωn t )k
(
)
= ωn R c(2 cos ωn t − ωn t sin ωn t )i − R 2 + ωn2 t 2 j + c(2sin ωn t + ωn t cos ωn t )k 


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260
PROBLEM 11.180* (Continued)
The angle α formed by the vector ( v × a) and the y axis is found from
cos α =
Where
( v × a) ⋅ j
| ( v × a) || j |
|j| =1
(
( v × a) ⋅ j = −ωn R 2 2 + ωn2 t 2
)
(
)
2
|( v × a) | = ωn R c 2 (2 cos ωn t − ωn t sin ωn t ) 2 + R 2 2 + ωn2 t 2

1/ 2
+ c 2 (2sin ωn t + ωn t cos ωn t )2 
(
)
(
1/2
)
2
= ωn R c 2 4 + ωn2 t 2 + R 2 2 + ωn2t 2 


Then
(
)
ω R c ( 4 + ω t ) + R ( 2 + ω t ) 


−R ( 2 + ω t )
=
c 4 + ω t + R 2 + ω t 
) (
) 
 (
−ωn R 2 2 + ωn2t 2
cos α =
2
n
2 2
n
2
2 2
n
2
1/2
2 2
n
2
2 2
n
2
2 2
n
2 1/2
The angle β that the osculating plane forms with y axis (see the above diagram) is equal to
β = α − 90°
Then
cos α = cos ( β + 90°) = −sin β
− sin β =
(
)
c 4 + ω t + R 2 + ω t 
) (
) 
 (
− R 2 + ωn2t 2
2
Then
tan β =
2 2
n
(
R 2 + ωn2 t 2
c
2
2 2
n
2
1/ 2
)
4 + ωn2 t 2
(
)
 R 2 + ωn2t 2 
 
β = tan
 c 4 + ω 2t 2 
n


−1 
or
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261
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 = π /2 s.
SOLUTION
Given:
)
(
r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k
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 +
j + (sin t + t cos t )k
(
t
)
2
dv
t 2 +1
a=
= 3( − sin t − sin t − t cos t )i + 3 t + 12− t
j
dt
t +1
+ (cos t + cos t − t sin t )k
3
= −3(2sin t + t cos t )i + 2
j + (2 cos t − t sin t )k
(t + 1)3/2
and
(a)
t2 +1
At t = 0:
v = (3 ft/s)i
a = (3 ft/s 2 ) j + (2 ft/s 2 )k
Then
and
Then
v × a = 3i × (3j + 2k )
= 3(−2 j + 3k )
| v × a | = 3 (−2) 2 + (3) 2 = 3 13
eb =
3( −2 j + 3k )
3 13
cos θ x = 0
or
θ x = 90°
=
cos θ y = −
1
13
2
(−2 j + 3k )
13
θ y = 123.7°
cos θ z =
3
13
θ z = 33.7°

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262
PROBLEM 11.181* (Continued)
(b)
At t =
π
2
s:

 3π
  3π
v = −
ft/s  i + 
ft/s  j + (1 ft/s)k


 2
  π2 +4


 π
24

a = −(6 ft/s 2 )i +  2
ft/s 2  j −  ft/s 2  k
3/2

 (π + 4)
 2
Then
i
3π
v×a = −
2
−6
j
3π
2
(π + 4)1/ 2
24
(π 2 + 4)3/ 2
k
1
−
π
2

 
3π 2
24
3π 2 
6
i
= −
+
−
+

j

2
1/2
4 
(π 2 + 4)3/ 2  
 2(π + 4)


36π
18π
k
+ − 2
+ 2
3/2
1/2 
(π + 4) 
 (π + 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 θ x = −
cos θ y = −
cos θ z =
19.18829
19.18829
19.18829
eb =
θ x = 103.4°
θ y = 134.3°
θ z = 47.4°

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263
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 = 1 s, 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 
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264
PROBLEM 11.183
A particle starting from rest at x = 1 m is accelerated so that its velocity doubles in magnitude between
x = 2 m and x = 8 m. Knowing that the acceleration of the particle is defined by the relation a = k[ x − (A/x)],
determine the values of the constants A and k if the particle has a velocity of 29 m/s when x = 16 m.
SOLUTION
We have
v
dv
A

= a = kx− 
dx
x

v
x
0
1

A
 vdv =  k  x − x  dx
When x = 1 ft, v = 0:
x
1 2
1

v = k  x 2 − A ln x 
2
2
1
or
1
1
= k  x 2 − A ln x − 
2
2
At x = 2 ft:
1 2
1
1
3

v2 = k  (2) 2 − A ln 2 −  = k  − A ln 2 
2
2
2
2




x = 8 ft:
1 2
1
1
v8 = k  (8) 2 − A ln 8 −  = k (31.5 − A ln 8)
2
2
2
Now
1 2
v
k (31.5 − A ln 8)
2 8
= (2) 2 =
1 2
v
k 32 − A ln 2
2 2
v8
= 2:
v2
(
)
6 − 4 A ln 2 = 31.5 − A ln 8
1
25.5 = A(ln 8 − 4 ln 2) = A(ln 8 − ln 24 ) = A ln  
2
A=
When x = 16 m, v = 29 m/s:
25.5
ln 12
A = −36.8 m 2 
1
1
25.5
1
(29) 2 = k  (16) 2 −
ln(16) − 
1
2
2
2
ln ( 2 )


1

420.5k = k 128 + 102 −  = 230.5k
2

= 230.5k
k = 1.832 s −2 
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265
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 
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266
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)
We have
v B = v A + v B/A
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 α sin 155°
or
α = 10.50°
v B/A = 111.4 km/h
(b)
10.50° 
First note that
at t = 3 min, A is (66 km/h) ( 603 ) = 3.3 km west of the crossing.
7
at t = 3 min, B is (48 km/h) ( 60
) = 5.6 km southwest of the crossing.
Now
rB = rA + rB/A
Then at t = 3 min, we have
rB2/A = 3.32 + 5.62 − 2(3.3)(5.6) cos 25°
rB/A = 2.96 km 
or
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267
PROBLEM 11.186
Slider block B starts from rest and moves to the right with a constant
acceleration of 1 ft/s 2. Determine (a) the relative acceleration of
portion C of the cable with respect to slider block A, (b) the velocity
of portion C of the cable after 2 s.
SOLUTION
Let d be the distance between the left and right supports.
Constraint of entire cable: xB + ( xB − x A ) + 2(d − x A ) = constant
2vB − 3v A = 0
aA =
and
2
2
aB = (1) = 0.667 ft/s 2
3
3
Constraint of Point C:
2aB − 3a A = 0
a A = 0.667 ft/s 2
or
2(d − x A ) + yC/ A = constant
−2v A + vC/ A = 0
− 2a A + aC/ A = 0
and
aC/ A = 2a A = 2(0.667) = 1.333 ft/s 2
(a)
aC/ A = 1.333 ft/s 2

Velocity vectors after 2s: v A = (0.667)(2) = 1.333 ft/s
vC/ A = (1.333)(2) = 2.666 ft/s
v C = v A + v C/ A
Sketch the vector addition.
vC2 = v A2 + vC2 / A = (1.333) 2 + (2.666) 2 = 8.8889(ft/s) 2
vC = 2.981 ft/s
tan θ =
vC/ A
vA
=
2.666
= 2,
1.333
θ = 63.4°
vC = 2.98 ft/s
(b)
63.4° 
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268
PROBLEM 11.187
Collar A starts from rest at t = 0 and moves downward with a constant
acceleration of 7 in./s 2 . 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
−2v A − 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


At t = 2 s
(a)
y − (yB )0 = 20 in.
y B = ( y B ) 0 + ( vB ) 0 t +
We have
At t = 2 s:
−20 in. = (−8 in./s)(2 s) +
aB = −4 in./s 2
1
aB t 2
2
1
aB (2 s) 2
2
or
a B = 2 in./s 2 
Then, substituting into Eq. (2)
−2(7 in./s 2 ) − (−2 in./s 2 ) + 4aC = 0
aC = 3 in./s 2
or
aC = 3 in./s 2 
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269
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 
(c)
Now
vC = (vC )0 + aC t
When vC = 0:
0 = (−2 in./s) + (3 in./s 2 )t
or
t=
We have
At t = 23 s:
2
s
3
yC = ( yC )0 + (vC )0 t +
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. 
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270
PROBLEM 11.188
A golfer hits a ball with an initial velocity of magnitude v0 at an
angle α 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 α = 31°, (b) a five-iron with α = 27°.
SOLUTION
The horizontal and vertical motions are
x
t cos α
1 2
1
y = (v0 sin α )t − gt = x tan α − gt 2
2
2
x = (v0 cos α )t
v0 =
or
(1)
2( x tan α − y)
g
or
t2 =
At the landing Point C:
yC = 0,
And
xC = (v0 cos α )t =
(2)
t =
2v0 sin α
g
2v02 sin α cos α
g
(3)
(a) α = 31°
To clear tree A:
x A = 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
t A = 1.1084 s
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,
v0 = 38.06 m/s
From (3),
xC =
t B = 3.0652 s
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 
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271
PROBLEM 11.188 (Continued)
(b)
α = 27°
By a similar calculation,
t A = 0.81846 s,
(v0 ) A = 41.138 m/s,
t B = 2.7447 s,
(v0 ) B = 40.890 m/s,
v0 = 41.138 m/s
v0 = 41.1 m/s 
xC = 139.56 m,
d = 29.6 m 
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272
PROBLEM 11.189
As the truck shown begins to back up with a constant acceleration
of 4 ft/s 2 , the outer section B of its boom starts to retract with a
constant acceleration of 1.6 ft/s 2 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/s 2
For the boom,
a B/ A = 1.6 ft/s 2
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) cos 50°
aB = 3.214 ft/s 2
Law of sines:
sin α =
aB/ A sin 50°
aB
α = 22.4°,
=
1.6 sin 50°
= 0.38131
3.214
a B = 3.21 ft/s 2
22.4° 
v B = (vB )0 + aBt = 0 + (3.214)(2)
(b)
v B = 6.43 ft/s 2
22.4° 
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273
PROBLEM 11.190
A motorist traveling along a straight portion of a
highway is decreasing the speed of his automobile at
a constant rate before exiting from the highway onto a
circular exit ramp with a radius of 560-ft. He continues
to decelerate at the same constant rate so that 10 s after
entering the ramp, his speed has decreased to 20 mi/h,
a speed which he then maintains. Knowing that at this
constant speed the total acceleration of the automobile
is equal to one-quarter of its value prior to entering the
ramp, determine the maximum value of the total
acceleration of the automobile.
SOLUTION
First note
v10 = 20 mi/h =
88
ft/s
3
While the car is on the straight portion of the highway.
a = astraight = at
and for the circular exit ramp
a = at2 + an2
where
an =
v2
ρ
By observation, amax occurs when v is maximum, which is at t = 0 when the car first enters the ramp.
For uniformly decelerated motion
v = v0 + at t
and at t = 10 s:
v2
v = constant  a = an = 10
ρ
a=
Then
or
1
ast.
4
88
v 2 ( 3 ft/s )
1
astraight = at  at = 10 =
ρ
4
560 ft
2
at = −6.1460 ft/s 2
(The car is decelerating; hence the minus sign.)
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274
PROBLEM 11.190 (Continued)
Then at t = 10 s:
or
Then at t = 0:
88
ft/s = v0 + (−6.1460 ft/s 2 )(10 s)
3
v0 = 90.793 ft/s
amax =
 v2 
at2 +  0 
 
2
ρ 
1/2
2

 (90.793 ft/s) 2  

2 2
= (−6.1460 ft/s ) + 
 
560 ft

 

amax = 15.95 ft/s 2 
or
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275
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 α = 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 = v0 t cos 25°
(2)
v y = v0 sin 25° − gt
(3)
y = v0t sin 25° −
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
tB = 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 θ
−v y
vx
= 1.66632
θ = 59.03°
v = 41.36 ft/s
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276
PROBLEM 11.191 (Continued)
Components of acceleration.
a = 32.2 ft/s 2
at = 32.2sin θ
an = 32.2cos θ = 32.2 cos 59.03° = 16.57 ft/s 2
(b)
Radius of curvature at B.
an =
ρ=
v2
ρ
v 2 (41.36) 2
=
an
16.57
ρ = 103.2 ft 
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277
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
acceleration of 
r = −1.0 m/s 2 and lower it with a constant
angular acceleration θ = −0.5 rad/s 2 . 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/s 2
Radial motion.
1 2

rt = 5 + 0 − 0.5t 2
2
r = r0 + 
rt = 0 − 1.0t
r = r0 + r0 t +
At t = 2 s,
r = 5 − (0.5)(2)2 = 3 m
r = (−1.0)(2) = −2 m/s
Angular motion.
θ 0 = 60° =
π
3
rad, θ0 = 0, θ = −0.5 rad/s 2
1
π
+ 0 − 0.25t 2
2
3
θ = θ0 + θt = 0 − 0.5t
θ = θ0 + θ0 + θt 2 =
θ=
At t = 2 s,
π
+ 0 − (0.25)(2) 2 = 0.047198 rad = 2.70°
3

θ = −(0.5)(2) = −1.0 rad/s
Unit vectors er and eθ .
(a)
Velocity of Point B at t = 2 s.
v B = re r + rθeθ
= −(2 m/s)e r + (3 m)(−1.0 rad/s)eθ


v B = (−2.00 m/s)er + ( −3.00 m/s)eθ 

v
−3.0
tan α = θ =
= 1.5
vr −2.0
v=
vr2 + vθ2 =
α = 56.31°
2

2
(−2) + (−3) = 3.6055 m/s
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278
PROBLEM 11.192 (Continued)
Direction of velocity.
et =
v −2er − 3eθ
=
= −0.55470er − 0.83205eθ
3.6055
v
θ + α = 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 − rθ 2 )er + ( rθ + 2rθ)eθ
= [−1.0 − (3)(−1) 2 ]e r + [(3)(−0.5) + (2)(−1.0)(−0.5)]eθ
a B = ( −4.00 m/s 2 )er + (2.50 m/s 2 )eθ 
tan β =
aθ
2.50
=
= −0.625
ar −4.00
β = −32.00°
a = ar2 + aθ2 = ( −4) 2 + (2.5) 2 = 4.7170 m/s 2
θ + β = 2.70° − 32.00° = −29.30°
a B = 4.72 m/s 2
Tangential component:
29.3° 
at = (a ⋅ et )et
at = (−4er + 2.5eθ ) ⋅ (−0.55470er − 0.83205eθ )et
= [(−4)(−0.55470) + (2.5)( −0.83205)]et
= (0.138675 m/s 2 )et = 0.1389 m/s 2
Normal component:
59.0°
a n = a − at
a n = −4er + 2.5eθ − (0.138675)(−0.55470e r − 0.83205eθ )
= (−3.9231 m/s 2 )er + (2.6154 m/s 2 )eθ
an = (3.9231) 2 + (2.6154) 2 = 4.7149 m/s 2
(c)
Radius of curvature of the path.
an =
ρ=
v2
ρ
v 2 (3.6055 m/s)2
=
an
4.7149 m/s
ρ = 2.76 m 
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279
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,

r = −10 ft/s 2 , θ = 25°, θ = 0.07 rad/s, θ = 0.06 rad/s 2 .
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, θ = 25°)
v = vr e r + vθ eθ = re r + rθeθ
= ( −105 ft/s)e r + (500 ft)(0.07 rad/s)eθ
v = ( −105 ft/s)e r + (35 ft/s)eθ 
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
α = 6.565°
tan α =
=
vx −109.95
v = (105) 2 + (35)2 = 110.68 ft/s
v = 110.7 ft/s
6.57° 
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280
PROBLEM 11.193 (Continued)
(b)
Acceleration of the skier.
a = ar er + aθ eθ = (
r − rθ 2 )er + (rθ + 2rθ)eθ
ar = −10 − (500)(0.07)2 = −12.45 ft/s 2
aθ = (500)(0.06) + (2)(−105)(0.07) = 15.30 ft/s 2
a = (−12.45 ft/s 2 )er + (15.30 ft/s 2 )eθ 
a = (−12.45)(i cos 25° + j sin 25°) + (15.30)(−i cos 65° + j sin 65°)
= (−17.750 ft/s 2 )i + (8.6049 ft/s 2 ) j
ay
8.6049
tan β =
=
β = −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)


x0 = 0
x0 = 109.95 ft/s


x = x0 + x0 t = 109.95t 
Vertical motion:




(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 0 t +
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)
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281
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 
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282
CHAPTER 12
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
Answer: (c) The tension will be greater than 1200 lb and the normal force will be greater than 1000 lb.
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285
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. After it
leaves, it will travel in a straight line.
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286
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 larger acceleration than block A
(c) Block A will have a larger acceleration than block C
(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 is harder to accelerate than the system on the right. 
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287
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.
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288
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
experiences by the person at Point C.
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289
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 μk. Draw the FBD and KD for A immediately after it contacts
the belt.
SOLUTION
Answer:
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290
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 μk, 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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291
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 μk, 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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292
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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293
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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294
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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295
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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296
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 B, draw the FBD and KD of the collar at this point.
SOLUTION
where x = 7/12 ft and r = 5/12 ft.
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297
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 ⋅ s 2 /ft
2
g moon 5.30 ft/s
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,
w = 844 lb 
m = 26.2 slugs 
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298
PROBLEM 12.2
The value of g at any latitude φ may be obtained from the formula
g = 32.09(1 + 0.0053 sin 2φ ) 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. Determine to four significant figures (a) the weight in pounds, (b) the mass in pounds, (c) the mass
in lb ⋅ s 2 /ft, at the latitudes of 0°, 45°, and 60°, of a silver bar, the mass of which has been officially designated
as 5 lb.
SOLUTION
g = 32.09(1 + 0.0053 sin 2φ ) ft/s 2
(a)
Weight:
(b)
Mass: At all latitudes:
(c)
or
φ = 0° :
g = 32.09 ft/s 2
φ = 45°:
g = 32.175 ft/s 2
φ = 90°:
g = 32.26 ft/s 2
W = mg
φ = 0° :
W = (0.1554 lb ⋅ s 2 /ft)(32.09 ft/s 2 ) = 4.987 lb

φ = 45°:
W = (0.1554 lb ⋅ s 2 /ft)(32.175 ft/s 2 ) = 5.000 lb

φ = 90°:
W = (0.1554 lb ⋅ s 2 /ft)(32.26 ft/s 2 ) = 5.013 lb

m = 5.000 lb 
m=
5.00 lb
32.175 ft/s 2
m = 0.1554 lb ⋅ s 2 /ft 
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299
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 
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300
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/s 2 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
ΣF = ma : Fs − W = −
W
a
g

a
W 1 −  = Fs
g


or
W=
Fs
a
1−
g
=
60
1
1−
9.81
W = 66.8 N 
(b)
ΣF = ma : Fs − W =
W
a
g

a
Fs = W 1 + 
g

1 

= 66.811 +

9.81


Fs = 73.6 N 
For the balance system B,
ΣM 0 = 0: bFw − bFp = 0
Fw = Fp
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301
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
mw = 6.81 kg 
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302
PROBLEM 12.5
In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3 ft/s 2 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
ΣFx = ma: P =
W
alevel
g
Now consider when the bus is on the upgrade.
We have
Substituting for P
ΣFx = 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
xupgrade = 1280.16 ft
xupgrade = 0.242 mi 
or
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303
PROBLEM 12.6
A hockey player hits a puck so that it comes to rest 10 s after sliding 100 ft on the ice. Determine (a) the
initial velocity of the puck, (b) the coefficient of friction between the puck and the ice.
SOLUTION
(a)
Assume uniformly decelerated motion.
Then
v = v0 + at
At t = 10 s:
0 = v0 + a(10)
v
a=− 0
10
Also
v 2 = v02 + 2a( x − 0)
At t = 10 s:
0 = v02 + 2a(100)
Substituting for a
 v 
0 = v02 + 2  − 0  (100) = 0
 10 
v0 = 20.0 ft/s
a=−
and
Alternate solution to part (a)
or v0 = 20.0 ft/s 
20
= −2 ft/s 2
10
1 2
at
2
1 v 
d = v0 t +  − 0  t 2
2 t 
d = d0 + v0 t +
1
v0 t
2
2d
v0 =
t
d=
(b)
We have
+ ΣFy = 0: N − W = 0
Sliding:
N = W = mg
F = μ k N = μ k mg
− μk mg = ma
ΣFx = ma : −F = ma
μk = −
a
−2.0 ft/s 2
=−
g
32.2 ft/s 2
μk = 0.0621 
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304
PROBLEM 12.7
The acceleration of a package sliding at Point A is 3 m/s2.
Assuming that the coefficient of kinetic friction is the same
for each section, determine the acceleration of the package
at Point B.
SOLUTION
For any angle θ .
Use x and y coordinates as shown.
ay = 0
ΣFy = ma y : N − mg cos θ = 0
N = mg cos θ
ΣFx = max : mg sin θ − μ k N = max
ax = g (sin θ − μ k cos θ )
At Point A.
θ = 30°, ax = 3 m/s 2
μk =
g sin 30° − ax
g cos 30°
9.81 sin 30° − 3
9.81 cos 30°
= 0.22423
=
At Point B.
θ = 15°, μk = 0.22423
ax = 9.81(sin 15° − 0.22423 cos 15°)
= 0.414 m/s
a = 0.414 m/s 2
15° 
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305
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
ΣFy = 0: N1 + N 2 − W = 0
F1 + F2 = μ N1 + μ N 2
= μ ( N1 + N 2 ) = μ w
ΣFx = ma : F1 + F2 = ma
μ w = ma
μW
mg
= μ g = 0.80(9.81)
m
m
a = 7.848 m/s 2
a=
=μ
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
μ (0.6 W ) = ma
0.6μW 0.6μ mg
=
m
m
= 0.6 μ 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
v = 85.6 km/h 
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306
PROBLEM 12.8 (Continued)
(c)
Rear-wheel drive
F1 = ma
μ (0.4 W ) = ma
0.4μW 0.4μ mg
=
m
m
= 0.4 μ 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
v = 69.9 km/h 
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307
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.
ΣF = ma
W
− Fb − W sin 5° = a
g
F + W sin 5°
a=− b
g
W
= −(0.70789 + sin 5°)(9.81)
= −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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308
PROBLEM 12.9 (Continued)
(b)
Going down a 3 percent incline.
3
β = 1.71835°
100
W
− Fb + W sin β = a
g
a = −(0.70789 − sin β )(9.81)
= −6.65028 m/s
0 − (25) 2
x f = x0 =

(2)(−6.65028)
tan β =

x f − x0 = 47.0 m 
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309
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.
ΣF = ma :
x-direction:
mg sin 5° − μ k N − T cos15° = ma
y-direction:
N − mg cos5° + 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,
μk N
T cos15°
m
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° −
−
= −0.45128 m/s 2
(in x-direction.)
v0 = 7.2 km/h = 2 m/s
vf =
x0 = 0
1
v0 = 1 m/s
2
v f = v0 + at
t=
v f − v0
a
=
−1 m/s
= 2.2159 s
−0.45128 m/s 2
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310
PROBLEM 12.10 (Continued)
t = 2.22 s 
(a)
Time elapsed.
(b)
Corresponding distance.
x = x0 + v0t +
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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311
PROBLEM 12.11
The coefficients of friction between the load and the flat-bed
trailer shown are μs = 0.40 and μk = 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
= μs N
Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max .
ΣFy = 0: N − W = 0 N = W
Fm = μ s N = 0.40 W
ΣFx = 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
x = 51.0 m 
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312
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:
ΣFx = Σma: 60 × 103 N = (45 × 103 kg)a
a = 1.333 m/s 2
v 2 = v01 + 2ax: 0 = (25) 2 + 2(−1.333) x 

x = 234 m 
Stopping distance:
(b)
Car A:


ΣFx = 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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313
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
ΣFx = m A a A : T − (200 lb) sin 30° =
200
aA
32.2
(1)
Block B
350  1

aA 

32.2  2

ΣFy = mB aB : 350 lb − 2T =
(a)
(2)
Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:
−2(200) sin 30° + 350 = 2
200
350  1 
aA +
aA
32.2
32.2  2 
150 =
575
aA
32.2
1
1
a A = (8.40 ft/s 2 ),
2
2


aB =
(b)
From Eq. (1), T − (200) sin 30° =
200
(8.40)
32.2
a A = 8.40 ft/s 2
30° 
a B = 4.20 ft/s 2 
T = 152.2 lb 
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314
PROBLEM 12.14
Solve Problem 12.13, assuming that the coefficients of
friction between block A and the incline are μs = 0.25 and
μk = 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,
ΣFx = 0: 175 − 200sin 30° − Freq = 0
Freq = 75.0 lb
ΣFy = 0: N − 200 cos 30° = 0 N = 173.2 lb
FM = μs 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 = μk N = (0.20)(173.2) = 34.64 lb.
ΣFx = m A 0 A : − 200sin 30° − 34.64 + T =
200
aA
32.2
(1)
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315
PROBLEM 12.14 (Continued)
Block B
ΣFy = mB aB : 350 lb − 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 
200
350  1 
aA +
aA
32.2
32.2  2 
575
aA
32.2
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
T = 162.9 lb 
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316
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: y A + yB = constant
a A + aB = 0
For blocks A and B,
aB = − a A
or
ΣF = 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,
T = WA −
or
aA =
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 ]
v A − (v A )0 = a A t
t=
(2)
with (v A )0 = 0
vA
aA
(3)
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317
PROBLEM 12.15 (Continued)
(a)
Acceleration of block A.
System (1):
(a A )1 =
System (2):
WA = 200 lb, WB = 0, P = 50 lb
System (3):
By formula (1),
(c)
200 − 100
(32.2)
200 + 100
By formula (1),
By formula (1),
(b)
WA = 200 lb, WB = 100 lb, P = 0
(a A ) 2 =
200 − 100
(32.2)
200
(a A )1 = 10.73 ft/s 2 
(a A ) 2 = 16.10 ft/s 2 
WA = 2200 lb, WB = 2100 lb, P = 0
( a A )3 =
2200 − 2100
(32.2)
2200 + 2100
(a A )3 = 0.749 ft/s 2 
v A at y A − ( y A )0 = 10 ft. Use formula (2).
System (1):
(v A )1 = (2)(10.73)(10)
(v A )1 = 14.65 ft/s 
System (2):
(v A ) 2 = (2)(16.10)(10)
(v A ) 2 = 17.94 ft/s 
System (3):
(v A )3 = (2)(0.749)(10)
(v A )3 = 3.87 ft/s 
Time at v A = 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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318
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 ( μk ) A = 0.30 and ( μk ) B = 0.32, determine the
initial acceleration of each box.
SOLUTION
Assume that aB > a A so that the normal force NAB between the boxes is zero.
A:
ΣFy = 0: NA − WA cos 15° = 0
or
NA = WA cos 15°
Slipping:
FA = ( μk ) A NA
A:
= 0.3WA cos 15°
ΣFx = m A a A : FA − WA sin 15° = m A a A
0.3WA cos 15° − WA sin 15° =
or
WA
aA
g
a A = (32.2 ft/s 2 )(0.3 cos 15° − sin 15°)
or
= 0.997 ft/s 2
B:
B:
ΣFy = 0: N B − WB cos 15° = 0
or
N B = WB cos 15°
Slipping:
FB = ( μk ) B N B
= 0.32WB cos 15°
ΣFx = 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 > a A  assumption is correct
a A = 0.997 ft/s 2
15° 
a B = 1.619 ft/s 2
15° 
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319
PROBLEM 12.16 (Continued)
Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then assuming NAB to be compression,
a A = aB
and find (ΣFx = 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.
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320
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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321
PROBLEM 12.17 (Continued)
Horizontal components
(a)
: F − T = mT aT
Horizontal force between lines 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 aB
FAB = mB ( g + a) = 31.056(32.2 + 0.5)
(b)
FAB = 1016 lb 
Contact force:
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322
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 μs = 0.20 and μk = 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 x A + xB/A = constant
Then
2v A + vB/A = 0
and
2a A + aB/A = 0
Now
a B = a A + a B/A
Then
aB = a A + ( −2a A )
or
aB = − a A
(1)
First we determine if the blocks will move for the given value of θ . Thus, we seek the value of θ for
which the blocks are in impending motion, with the impending motion of A down the incline.
B:
ΣFy = 0: N AB − WB cos θ = 0
or
N AB = mB g cos θ
Now
FAB = μ s N AB
B:
= 0.2mB g cos θ
ΣFx = 0: − T + FAB + WB sin θ = 0
T = mB g (0.2cos θ + sin θ )
or
A:
A:
ΣFy = 0: N A − N AB − WA cos θ = 0
or
N A = (m A + mB ) g cos θ
Now
FA = μ s N A
= 0.2( mA + mB ) g cos θ
ΣFx = 0: − T − FA − FAB + WA sin θ = 0
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323
PROBLEM 12.18 (Continued)
T = m A g sin θ − 0.2(m A + mB ) g cos θ − 0.2mB g cos θ
or
= g[ mA sin θ − 0.2(m A + 2mB ) cos θ ]
Equating the two expressions for T
mB g (0.2cos θ + sin θ ) = g[m A sin θ − 0.2(mA + 2mB ) cos θ ]
8(0.2 + tan θ ) = [40 tan θ − 0.2(40 + 2 × 8)]
or
tan θ = 0.4
or
or θ = 21.8° for impending motion. Since θ < 25°, the blocks will move. Now consider the
motion of the blocks.
ΣFy = 0: N AB − WB cos 25° = 0
(a)
B:
or
N AB = mB g cos 25°
Sliding:
FAB = μk N AB = 0.15mB g cos 25°
ΣFx = 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)
ΣFy = 0: N A − N AB − WA cos 25° = 0
or
N A = (m A + mB ) g cos 25°
Sliding:
FA = μk N A = 0.15(m A + mB ) g cos 25°
ΣFx = m A a A : − T − FA − FAB + WA sin 25° = m A a A
Substituting and using Eq. (1)
T = m A g sin 25° − 0.15( mA + mB ) g cos 25°
− 0.15mB g cos 25° − m A (− aB )
= g[ mA sin 25° − 0.15(m A + 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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324
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
μ s = 0.20 and μk = 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 x A + xB/A = constant
Then
2v A + vB/A = 0
and
2a A + aB/A = 0
Now
a B = a A + a B/A
Then
aB = a A + ( −2a A )
or
aB = − a A
(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:
ΣFy = 0: N AB − WB cos 25° = 0
or
N AB = mB g cos 25°
Now
FAB = μ s N AB
B:
= 0.2 mB g cos 25°
ΣFx = 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
ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0
N A = (m A + mB ) g cos 25° − P sin 25°
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325
PROBLEM 12.19 (Continued)
Now
FA = μ s N A
or
FA = 0.2[(mA + mB ) g cos 25° − P sin 25°]
ΣFx = 0: − T − FA − FAB + WA sin 25° + P cos 25° = 0
or
−T − 0.2[(m A + mB ) g cos 25° − P sin 25°] − 0.2mB g cos 25° + mA g sin 25° + P cos 25° = 0
or
P(0.2 sin 25° + cos 25°) = T + 0.2[(m A + 2mB ) g cos 25°] − m A 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.
ΣFy = 0: N AB − WB cos 25° = 0
(a)
B:
or
N AB = mB g cos 25°
Sliding:
FAB = μk N AB
= 0.15 mB g cos 25°
ΣFx = 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:
ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0
or
N A = (m A + mB ) g cos 25° − P sin 25°
Sliding:
FA = μk N A
= 0.15[(m A + mB ) g cos 25° − P sin 25°]
ΣFx = m A a A : − T − FA − FAB + WA sin 25° + P cos 25° = m A a A
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326
PROBLEM 12.19 (Continued)
Substituting and using Eq. (1)
T = m A g sin 25° − 0.15[( mA + mB ) g cos 25° − P sin 25°]
− 0.15 mB g cos 25° + P cos 25° − m A (− aB )
= g[ mA sin 25° − 0.15(m A + 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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327
PROBLEM 12.20
A package is at rest on a conveyor belt which is initially at rest. The belt is
started and moves to the right for 1.3 s with a constant acceleration of
2 m/s2. The belt then moves with a constant deceleration a2 and comes to a
stop after a total displacement of 2.2 m. Knowing that the coefficients of
friction between the package and the belt are μ s = 0.35 and μk = 0.25,
determine (a) the deceleration a2 of the belt, (b) the displacement of the
package relative to the belt as the belt comes to a stop.
SOLUTION
(a)
Kinematics of the belt. vo = 0
1. Acceleration phase with a1 = 2 m/s 2
v1 = vo + a1t1 = 0 + (2)(1.3) = 2.6 m/s
x1 = xo + vo t1 +
1 2
1
a1t1 = 0 + 0 + (2)(1.3) 2 = 1.69 m
2
2
2. Deceleration phase: v2 = 0 since the belt stops.
v22 − v12 = 2a2 ( x2 − x1 )
a2 =
t2 − t1 =
(b)
v22 − v12
0 − (2.6)2
=
= −6.63
2( x2 − x1 ) 2(2.2 − 1.69)
a 2 = 6.63 m/s 2

v2 − v1 0 − 2.6
=
= 0.3923 s
−6.63
a2
Motion of the package.
1. Acceleration phase. Assume no slip. (a p )1 = 2 m/s 2
ΣFy = 0: N − W = 0 or N = W = mg
ΣFx = ma : F f = m(a p )1
The required friction force is Ff.
The available friction force is μ s N = 0.35W = 0.35mg
Ff
m
= (a p )1 , <
μs N
= μ s g = (0.35)(9.81) = 3.43 m/s 2
m
Since 2.0 m/s 2 < 3.43 m/s 2, the package does not slip.
(v p )1 = v1 = 2.6 m/s and (x p )1 = 1.69 m.
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328
PROBLEM 12.20 (Continued)
2. Deceleration phase. Assume no slip. (a p ) 2 = −11.52 m/s 2
ΣFx = ma : − F f = m( a p )2
Ff
m
μs N
m
=
μs mg
m
= (a p ) 2 = −6.63 m/s 2
= μ s g = 3.43 m/s 2 < 6.63 m/s 2
Since the available friction force μ s N is less than the required friction force Ff for no slip, the
package does slip.
(a p ) 2 < 6.63 m/s 2 ,
F f = μk N
ΣFx = m( a p )2 : − μk N = m(a p ) 2
μk N
= − μk g
m
= −(0.25)(9.81)
(a p ) 2 = −
= −2.4525 m/s 2
(v p ) 2 = (v p )1 + (a p ) 2 (t2 − t1 )
= 2.6 + (−2.4525)(0.3923)
= 1.638 m/s 2
1
( a p )2 (t2 − t1 )2
2
1
= 1.69 + (2.6)(0.3923) + ( −2.4525)(0.3923) 2
2
( x p ) 2 = ( x p )1 + (v p )1 (t2 − t1 ) +
= 2.521 m
Position of package relative to the belt
( x p )2 − x2 = 2.521 − 2.2 = 0.321
x p/belt = 0.321 m

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329
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.
ΣFy = ma y : − (mB + m A ) g cos 20° + N = 0
N = (m A + mB ) g cos 30° = (30)(9.81) cos 20° = 276.55 N
μ B N = (0.3)(276.55) = 82.965 N
ΣFx = max : μ B N + ( mA + mB ) g sin 20° = (m A + mB )a
a = g sin 20° +
μB N
m A mB
= 9.81sin 20° −
a = 0.58972 m/s 2
82.965
30
a = 0.58972 m/s 2
20°
Apply Newton’s second law to bag A alone.
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330
PROBLEM 12.21 (Continued)
ΣFy = 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 = m A a
FAB = m A ( g sin 20° − a) = (10)(9.81sin 20° − 0.58972)
= 27.655 N
Since bag A does not slide on suitcase B,
μs >
FAB 27.655
=
= 0.300
N AB 92.184
μs > 0.300 
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331
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 μs = 0.40 and μk = 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 a P/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 = μs N1 = 0.40 N1
ΣFy = mP a y : N1 − WP cos 20° = − mP aT sin 20°
N1 = mP ( g cos 20° − aT sin 20°)
ΣFx = 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)
= 0.309
aT =
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/s 2
(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°)
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332
PROBLEM 12.22 (Continued)
ΣFx = Σmax : 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 = μk N 2 = 0.30 N 2
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

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333
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.
ΣFy = m A a y :
F1 = μ s N1 = 0.30 N1
N1 − WA = mA a1 sin 65°
N1 = WA + mA a1 sin 65°
= m A ( g + a1 sin 65°)
ΣFx = m A ax : F1 = m A a1 cos 65°
F1 = μ s N
or
m A a1 cos 65° = 0.30m A ( 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
65° 
F2 = μ s N 2 = 0.30 N 2
ΣFy = ma y : N1 − WA = − mA a2 sin 65°
N1 = WA − mA a2 sin 65°
ΣFx = max :
F2 = mA a2 cos 65°
F2 = μ s N 2
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334
PROBLEM 12.23 (Continued)
or
m A a2 cos 65° = 0.30 m A ( g − a2 cos 65°)
0.30 g
cos 65° + 0.30 sin 65°
= (0.432)(9.81)
a2 =
= 4.24 m/s 2
a 2 = 4.24 m/s 2
65° 
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335
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

dv
dv
: 40 × 103 − 2.25 v 2 = (25 × 103 ) v
dx
dx
(25 × 103 )vdv
0
0 40 × 103 − 2.25v 2
25 × 103
x1 = −
[ln(40 × 103 − 2.25v 2 )]v01
2(2.25)
x1
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
= 5.556ln1.333
3
4.5
40 × 10 − 2.25(66.67) 2
= 1.5982 × 103 m
x1 = 1.598 km 
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336
PROBLEM 12.25
The propellers of a ship of weight W can produce a propulsive force F0; they produce a force of the same
magnitude but of opposite direction when the engines are reversed. Knowing that the ship was proceeding
forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship
travels before coming to a stop. Assume that the frictional resistance of the water varies directly with the
square of the velocity.
SOLUTION
At maximum speed a = 0.
F0 = kv02 = 0
k=
F0
v02
When the propellers are reversed, F0 is reversed.
ΣFx = ma : − F0 − kv 2 = ma
− F0 − F0
v2
= ma
v02
dx =

x
0
a−
(v + v )
F0
2
0
mv02
mv02 vdv
vdv
=
a
F0 v02 + v 2
dx = −
x=−
=−
(
mv02
F0
0
vdv
v0
2
0
 v +v
2
)
2
0
mv02 1
ln v02 + v 2
v0
F0 2
(
)
mv02  2
mv 2
ln v0 − ln 2v02  = 0 ln 2
 2 F0
2 F0 
( )
x = 0.347
m0 v02

F0
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337
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 kv 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
ΣF = 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


t
t
t
Pt
P k

−  − e− kt/m 
k 0 k m
0
x=

=
Pt P − kt/m
Pt P
+ (e
− 1) =
− (1 − e− kt/m )
k m
k m
x=
Pt kv
− , which is linear.
k
m
0
v dt =

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338
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 + x 2
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 − )
x
cos θ =
By geometry,
 + x2
2
ΣFx = max : − Fs cos θ = ma
− k (  2 + x 2 − )
a=−

v
0
v dv =
x
 + x2
2
= ma
k
x 
x−

m 
2 + x 2 
0
 0 a dx
x
v
1 2
k
v =−
m
2 0

0 
0
x 
k 1 2
2
2 
−
=
−
−
+
x
dx
x
x





x0 
m  2
x
2 + x 2 

0
k
1 2
1

v = −  0 −  2 − x02 +   2 + x02 
m
2
2

k
v2 =
2 2 + x02 − 2  2 + x02
m
k
=   2 + x02 − 2  2 + x02 +  2 

m 
)
(
(
)
answer: v =
k
m
(  + x − ) 
2
2
0
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339
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:
v A + vD = 0,
a A + aD = 0
( yB − yD ) + ( yC − yD ) = constant
Constraint of cord BC:
vB + vC − 2vD = 0,
aB + aC − 2aD = 0
2a A + aB + aC = 0
Eliminate aD .
(1)
We have uniformly accelerated motion because all of the forces are
constant.
y B = ( y B ) 0 + ( vB ) 0 t +
1
a B t 2 , ( vB ) 0 = 0
2
2[ yB − ( yB )0 ]
(2)(3)
= 1.5 m/s 2
2
(2)
aB =
t
2
=
ΣFy = 0: 2TBC − TAD = 0
Pulley D:
TAD = 2TBC
Block A:
ΣFy = ma y : WA − TAD = m A a A
or
aA =
WA − TAD WA − 2TBC
=
mA
mA
(2)
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340
PROBLEM 12.28 (Continued)
Block C:
ΣFy = ma y : WC − TBC = mC aC
or
aC =
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

 mC g − TBC 
 m g − 2TBC 
2 A
=0
 + aB + 
mA
mC




 4
1 
+

 TBC = 3g + aB
 mA mC 
 4 1
 10 + 5  TBC = 3(9.81) + 1.5 or TBC = 51.55 N


ΣFy = ma y : P + WB − TBC = mB aB
Block B:
(a)
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)
TAD = 103.1 N 
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341
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:
T −F =
ΣFx = ma :
40
a
g
(1)
Counterweight A: Its acceleration has two components
a A = a P + a A /P = a → + a
ΣFx = max : F =
25
a
g
ΣFg = 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/s 2

Substituting for a into (3):
25 − T =
25  25 
g
g  90 
T = 25 −
625
90
T = 18.06 lb 
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342
PROBLEM 12.29 (Continued)
(b)
Panel:
ΣFy = ma :
T=
40
a
g
(1)
25 − T =
25
a
g
(2)
Counterweight A:
ΣFy = 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:
ΣFx = ma :
T=
40
a
g
(1)
Counterweight A: Same free body as in Part (b):
ΣFy = 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/s 2
; T = 15.38 lb 
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343
PROBLEM 12.30
The coefficients of friction between blocks A and C and the
horizontal surfaces are μ s = 0.24 and μk = 0.20. Knowing
that m A = 5 kg, mB = 10 kg, and mC = 10 kg, determine
(a) the tension in the cord, (b) the acceleration of each
block.
SOLUTION
We first check that static equilibrium is not maintained:
( FA ) m + ( FC )m = μs ( mA + mC ) g
= 0.24(5 + 10) g
= 3.6 g
Since WB = mB g = 10g > 3.6g, equilibrium is not maintained.
ΣFy : N A = m A g
Block A:
FA = μk N A = 0.2m A g
ΣFλ = mA a A : T − 0.2mA g = mA a A
(1)
ΣFy : NC = mC g
Block C:
FC = μk NC = 0.2mC g
Σ Fx = mC aC : T − 0.2mC g = mC aC
ΣFy = mB aB
Block B:
mB g − 2T = mB aB
From kinematics:
(a)
(2)
aB =
(3)
1
(a A + aC )
2
Tension in cord. Given data:
(4)
m A = 5 kg
mB = mC = 10 kg
Eq. (1): T − 0.2(5) g = 5a A
a A = 0.2T − 0.2 g
(5)
Eq. (2): T − 0.2(10) g = 10aC
aC = 0.1T − 0.2 g
(6)
Eq. (3): 10 g − 2T = 10aB
aB = g − 0.2T
(7)
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344
PROBLEM 12.30 (Continued)
Substitute into (4):
1
(0.2T − 0.2 g + 0.1T − 0.2 g )
2
24
24
1.2 g = 0.35T
T=
g=
(9.81 m/s 2 )
7
7
g − 0.2T =
(b)
T = 33.6 N 
Substitute for T into (5), (7), and (6):
 24 
a A = 0.2  g  − 0.2 g = 0.4857(9.81 m/s 2 )
 7 
 24 
aB = g − 0.2  g  = 0.3143(9.81 m/s 2 )
 7 
 24 
aC = 0.1 g  − 0.2 g = 0.14286(9.81 m/s 2 )
 7 
a A = 4.76 m/s 2

a B = 3.08 m/s 2 
aC = 1.401 m/s 2

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345
PROBLEM 12.31
A 10-lb block B rests as shown on a 20-lb bracket A. The
coefficients of friction are μs = 0.30 and μk = 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 x A 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 − x A ) + ( xP − x A ) + yC + constant
Differentiating and noting that x P = 0,
vB − 2v A + vC = 0
−2a A + aB + aC = 0
(1)
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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346
PROBLEM 12.31 (Continued)
Block B:
Block C:
WA
aA
g
(2)
WB
aB
g
(3)
Bracket A:
ΣFx = max : 2T − FAB =
Block B:
ΣFx = max : FAB − T =
+ ΣFy = ma y : N AB − WB = 0
N AB = WB
or
Block C:
ΣFy = ma y : mC − T =
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 = a A
From Eq. (1),
a A = aB = aC = a
From Eq. (5),
a=
For impending slip,
(6)
WC g
WA + WB + WC
FAB = μs N AB = μsWB
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347
PROBLEM 12.31 (Continued)
Substituting into Eq. (6),
(WB − WC )(WC g )
= μ sWB − WC
WA + WB + WC
Solving for WC ,
WC =
=
μsWB (WA + WB )
WA + 2WB − μ 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 = μk N AB = μkWB
Eq. (6) becomes
W
WB
aB − C aC = μ kWB − WC
g
g
or
10aB − 2.6757 aC = [(0.25)(10) − 2.6757](32.2)
(7)
With numerical data, Eq. (5) becomes
20a A + 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 
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348
PROBLEM 12.32
The masses of blocks A, B, C and D are 9 kg, 9 kg, 6 kg and 7 kg,
respectively. Knowing that a downward force of magnitude 120 N is
applied to block D, determine (a) the acceleration of each block, (b) the
tension in cord ABC. Neglect the weights of the pulleys and the effect of
friction.
SOLUTION
Note: As shown, the system is in equilibrium.
From the diagram:
A:
Cord 1:
2 y A + 2 yB + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
( yD − y A ) + ( yD − yB ) = constant
Then
2 vD − v A − vB = 0
and
2aD − a A − aB = 0
(a)
ΣFy = mA a A : mA g − 2T1 + T2 = m A a A
9(9.81) − 2T1 + T2 = 9a A
or
(1)
(2)
(3)
ΣFy = mB aB : mB g − 2T1 + T2 = mB aB
B:
9(9.81) − 2T1 + T2 = 9aB
or
(4)
Note: Eqs. (3) and (4)  a A = a B
Then
Eq. (1)  aC = −4a A
Eq. (2)  aD = a A
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349
PROBLEM 12.32 (Continued)
ΣFy = mC aC : mC g − T1 = mC aC
C:
T1 = mC ( g − aC ) = 6( g + 4a A )
or
(5)
ΣFy = mD aD : mD g − 2T2 + ( FD )ext = mD aD
1
1
T2 = [mD ( g − aD ) + 120] = 94.335 − (7a A )
2
2
or
(6)
Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)
D:
1
9(9.81) − 2 × 6( g + 4a A ) + 94.335 − (7a A ) = 9a A
2
aA =
or
9(9.81) − 2 × 6(9.81) + 94.335
= 1.0728 m/s 2
48 + 3.5 + 9
a A = a B = a D = 1.073 m/s 2 
aC = −4(1.0728 m/s 2 )
and
(b)
or aC = 4.29 m/s 2 
Substituting into Eq. (5)
T1 = 6 ( 9.81 + 4(1.0728) )
or T1 = 84.6 N 
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350
PROBLEM 12.33
The masses of blocks A, B, C and D are 9 kg, 9 kg, 6 kg and 7 kg,
respectively. Knowing that a downward force of magnitude 50 N is applied
to block B and that the system starts from rest, determine at t = 3 s the
velocity (a) of D relative to A, (b) of C relative to D. Neglect the weights of
the pulleys and the effect of friction.
SOLUTION
Note: As shown, the system is in equilibrium.
From the constraint of the two cords,
Cord 1:
2 y A + 2 yB + yC = constant
Then
2v A + 2vB + vC = 0
and
2a A + 2aB + aC = 0
Cord 2:
( yD − y A ) + ( yD − yB ) = constant
Then
2 vD − v A − vB = 0
and
2aD − a A − aB = 0
(1)
(2)
We determine the accelerations of blocks A, C, and D using the
blocks as free bodies.
WA
aA
g
m A g − 2T1 + T2 = m A a A
ΣFy = m A a A : WA − 2T1 + T2 =
A:
or
WB
aB
g
mB g − 2T1 + T2 + ( FB )ext = mB aB
(3)
ΣFy = mB aB : WB − 2T1 + T2 + ( FB )ext =
B:
or
(4)
(3) − (4)  −( FB )ext = 9( a A − aB )
(F )
aB = a A + B ext
mB
Forming
or
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351
PROBLEM 12.33 (Continued)
Then

(F ) 
2a A + 2  a A + B ext  + aC = 0

mg 

Eq. (1):
aC = −4a A −
or
2( FB )ext
mB

(F ) 
2aD − a A −  a A + B ext  = 0
mB 

Eq. (2):
aD = a A +
or
C:
( FB )ext
2mB

2( FB )ext 
ΣFy = mC aC : WC − T1 = mC aC = mC  −4a A −

mB 

2mC ( FB )ext
mB
(5)

(F ) 
1
× mD  g − a A − B ext 
2
2mB 

(6)
T1 = mC g + 4mC a A +
or
D:
ΣFy = mD aD : WD − 2T2 = mD aD
T2 =
or
Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3)


2mC ( FB )ext  1
( FB )ext 
m A g − 2  mC g + 4mC a A +
 + × mD  g − a A −
 = mAa A
mB
2mB 

 2

4 mC ( FB )ext
m (F )
m g
− D4 mB ext + D2
mB
B
m
mA + 8mC + 2D
m A g − 2mC g −
or
aA =
Then
aC = −4( −2.2835 m/s 2 ) −
aD = −2.2835 m/s 2 +
= −2.2835 m/s 2
2(50)
= −1.9771 m/s 2
9
(50)
= 0.4943 m/s 2
2(9)
Note: We have uniformly accelerated motion, so that
v = 0 + at
(a)
We have
v D/ A = v D − v A
or
v D /A = aD t − a A t = [0.4943 − (−2.2835)] m/s 2 × 3 s
v D /A = 8.33 m/s 
or
(b)
And
v C/ D = v C = v D
or
vC /D = aC t − aD t = (−1.9771 − 0.4943) m/s 2 × 3 s
v C /D = 7.41m/s 
or
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352
PROBLEM 12.34
The 15-kg block B is supported by the 25-kg block A and is
attached to a cord to which a 225-N horizontal force is applied
as shown. Neglecting friction, determine (a) the acceleration of
block A, (b) the acceleration of block B relative to A.
SOLUTION
(a)
First we note a B = a A + a B/A , where a B/A is directed along the inclined surface of A.
B:
ΣFx = mB ax : P − WB sin 25° = mB a A cos 25° + mB aB/A
or
225 − 15 g sin 25° = 15( a A cos 25° + aB/A )
or
15 − g sin 25° = a A cos 25° + aB/A
B:
(1)
ΣFy = mB a y : N AB − WB cos 25° = −mB a A sin 25°
N AB = 15( g cos 25° − a A sin 25°)
or
A:
ΣFx′ = mA a A : P − P cos 25° + N AB sin 25° = m A a A
or
N AB = [25a A − 225(1 − cos 25°)] / sin 25°
A:
Equating the two expressions for N AB
15( g cos 25° − a A sin 25°) =
or
25a A − 225(1 − cos 25°)
sin 25°
3(9.81) cos 25° sin 25° + 45(1 − cos 25°)
5 + 3sin 2 25°
= 2.7979 m/s 2
aA =
a A = 2.80 m/s 2
(b)

From Eq. (1)
aB/A = 15 − (9.81)sin 25° − 2.7979cos 25°
a B/A = 8.32 m/s 2
or
25° 
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353
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:
ΣFx = m A a A : WA sin 30° + N AB cos 40° = m A a A
(a)
NAB =
or
22 ( a A − 12 g )
cos 40°
Now we note: a B = a A + a B /A , where a B/A is directed along the top surface of A.
B:
ΣFy ′ = mB a y′ : NAB − WB cos 20° = −mB a A sin 50°
NAB = 10 ( g cos 20° − a A 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°
ΣFx′ = mB ax′ : WB sin 20° = mB aB/A − mB a A cos 50°
or
aB/A = g sin 20° + a A cos 50°
= (9.81sin 20° + 6.4061cos 50°) m/s 2
= 7.4730 m/s 2
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354
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
7.4730 5.9447
=
sin α
sin 50°
or
α = 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 = a B/At
At t = 0.5 s:
vB/A = 7.4730 m/s 2 × 0.5 s
v B/A = 3.74 m/s
or
20° 
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355
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 θ that the
cord forms with pole BC, (b) the tension in the cord.
SOLUTION
a A = an =
First we note
v 2A
ρ
ρ = l AB sin θ
where
ΣFy = 0: TAB cos θ − WA = 0
(a)
TAB =
or
mA g
cos θ
ΣFx = m A a A : TAB sin θ = m A
v A2
ρ
Substituting for TAB and ρ
mA g
v A2
sin θ = m A
cos θ
l AB sin θ
1 − cos 2 θ =
or
sin 2 θ = 1 − cos 2 θ
(4 m/s) 2
cos θ
1.8 m × 9.81 m/s 2
cos 2 θ + 0.906105cos θ − 1 = 0
cos θ = 0.64479
Solving
θ = 49.9° 
or
(b)
From above
TAB =
m A g 0.450 kg × 9.81 m/s 2
=
cos θ
0.64479
TAB = 6.85 N 
or
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356
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 ρ = 0.93 m and θ = 60°, determine (a) the tension in
wire BC, (b) the speed of the hammer’s head.
SOLUTION
First we note
a A = an =
v 2A
ρ
ΣFy = 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 
ΣFx = m A a A : TBC cos 60° = mA
(b)
or
v A2 =
v A2
ρ
(80.426 N) cos 60° × 0.93 m
7.1 kg
v A = 2.30 m/s 
or
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357
PROBLEM 12.38
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
ΣFx = ma: T (sin 30° + sin 45°) =
mv 2
ρ
(1)
ΣFy = 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
=
cos 30° + cos 45° ρ g
v 2 = 0.76733ρ g = 0.76733 (1.6 m)(9.81 m/s 2 ) = 12.044 m 2 /s 2
v = 3.47 m/s 
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358
PROBLEM 12.39
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 values of v for which both wires remain taut.
SOLUTION
ΣFx = ma: TAC sin 30° + TBC sin 45° =
mv 2
(1)
ρ
ΣFy = 0: TAC cos 30° + TBC cos 45° − mg = 0
Divide Eq (1) by Eq. (2):
TAC cos 30° + TBC cos 45° = mg
(2)
TAC sin 30° + TBC sin 45° v 2
=
TAC cos 30° + TBC cos 45° ρ g
(3)
When AC is slack, TAC = 0.
Eq. (3) yields
v12 = ρ g tan 45° = (1.6 m) (9.81 m/s2 ) tan 45° = 15.696 m 2 /s2
Wire AC will remain taut if v ≤ v1 , that is, if
v1 = 3.96 m/s
v ≤ 3.96 m/s 
When BC is slack, TBC = 0.
Eq. (3) yields
v22 = ρ g tan 30° = (1.6 m)(9.81 m/s 2 ) tan 30° = 9.0621 m 2 /s 2
Wire BC will remain taut if v ≥ v2 , that is, if
v2 = 3.01 m/s
v ≥ 3.01 m/s 
Combining the results obtained, we conclude that both wires remain taut for
3.01 m/s ≤ v ≤ 3.96 m/s 
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359
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
ρ
(1)
TAC cos 30° + TBC cos 45° = mg
(2)
Subtract Eq. (1) from Eq. (2). Since sin 45° = cos 45°, we obtain
TAC (cos 30° − sin 30°) = mg −
mv 2
(3)
ρ
Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract:
TBC (sin 45° cos 30° − cos 45° sin 30°) =
TBC sin15° =
mv 2
ρ
mv 2
ρ
cos 30° − mg sin 30°
cos 30° − mg sin 30°
(4)
Making TAC = 60 N, m = 5 kg, ρ = 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(cos 30° − 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
v1 = 2.94 m/s
v ≥ 2.94 m/s 
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360
PROBLEM 12.40 (Continued)
Making TBC = 60 N, m = 5 kg, ρ = 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
v ≤ 3.85 m/s 
Combining the results obtained, we conclude that the range of allowable value is
3.01 m/s ≤ v ≤ 3.85 m/s 
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361
PROBLEM 12.41
A 100-g sphere D is at rest relative to drum ABC which rotates
at a constant rate. Neglecting friction, determine the range of
the allowable values of the velocity v of the sphere if neither of
the normal forces exerted by the sphere on the inclined
surfaces of the drum is to exceed 1.1 N.
SOLUTION
vD2
First we note
aD = an =
where
ρ = 0.2 m
ρ
ΣFx = mD aD : N1 cos 60° + N 2 cos 20° = mD
vD2
ρ
(1)
ΣFy = 0: N1 sin 60° + N 2 sin 20° − WD = 0
N1 sin 60° + N 2 sin 20° = mD g
or
(2)
Case 1: N1 is maximum.
N1 = 1.1 N
Let
Eq. (2)
(1.1 N) sin 60° + N 2 sin 20° = (0.1 kg) (9.81 m/s 2 )
N 2 = 0.082954 N
or
( N 2 )( N1 )max < 1.1 N
OK
0.2 m
(1.1cos 60° + 0.082954 cos 20°) N
0.1 kg
Eq. (1)
(vD2 )( N1 )max =
or
(vD )( N1 )max = 1.121 m/s
Now we form
(sin 20°) × [Eq. (1)] − (cos 20°) × [Eq. (2)]
N1 cos 60° sin 20° − N1 sin 60° cos 20° = mD
or
− N1 sin 40° = mD
vD2
ρ
vD2
ρ
sin 20° − mD g cos 20°
sin 20° − mD g cos 20°
(vD )min occurs when N1 = ( N1 ) max
(vD )min = 1.121 m/s
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362
PROBLEM 12.41 (Continued)
Case 2: N 2 is maximum.
N 2 = 1.1 N
Let
Eq. (2)
N1 sin 60° + (1.1 N)sin 20° = (0.1 kg)(9.81 m/s 2 )
N1 = 0.69834 N
or
( N1 )( N 2 )max ≤ 1.1 N
OK
0.2 m
(0.69834 cos 60° + 1.1cos 20°) N
0.1 kg
Eq. (1)
(vD2 )( N 2 )max =
or
(vD )( N 2 )max = 1.663 m/s
(sin 60°) × [Eq. (1)] − (cos 60°) × [Eq. (2)]
Now we form
N 2 cos 20° sin 60° − N 2 sin 20° cos 60° = mD
or
N 2 cos 40° = mD
vD2
ρ
vD2
ρ
sin 60° − mD g cos 60°
sin 60° − mD g cos 60°
(vD ) max occurs when N 2 = ( N 2 )max
(vD ) max = 1.663 m/s
For N1 ≤ N 2 < 1.1 N
1.121 m/s < vD < 1.663 m/s 
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363
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
First note
aC = an =
where
ρ = 3 ft
vC2
ρ
ΣFx = mC aC : TCA sin 40° + TCB sin15° =
WC vC2
g ρ
ΣFy = 0: TCA cos 40° − TCB cos15° − WC = 0
(1)
(2)
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]
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364
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 ρ
TCA sin 55° =
WC vC2
cos15° + WC sin15°
g ρ
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).
Eq. (2)
TCA cos 40° − (12 lb) = 0
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 
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365
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
ρ = 0.5 ft
ρ
W v2
g ρ
(1)
ΣFy = 0: TDA cos 20° − TDE cos 30° − W = 0
(2)
ΣFx = 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 cos 30° − (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) cos 30° − (1.2 lb) = 0
TDA = 16.9443 lb OK ( ≤ 17 lb)
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366
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
or
v(TDA )max = 13.85 ft/s
Now form
(cos 30°) × [Eq. (1)] + (sin 30°) × [Eq. (2)]
TDA sin 20° cos 30° + TDA cos 20° sin 30° =
W v2
cos 30° + W sin 30°
g ρ
TDA sin 50° =
W v2
cos 30° + W sin 30°
g ρ
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 =
or
(32.2 ft/s 2 ) (0.5 ft)
(1.27701 lb) sin 20°
1.2 lb
vmin = 2.42 ft/s
0 < TAB , TBC , TAD , TDE < 17 lb
2.42 ft/s < v < 13.85 ft/s 
when
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367
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
ΣFn = 0: TBA − WB cos 20° = 0
TBA = (130 lb) × cos 20°
or
TBA = 122.2 lb 
or
ΣFn = man : TBA − WB = mB
(b)
or
(vB ) 2D
LAB
 130 lb   (13.2 ft/s) 2  
TBA = (130 lb) + 
 
2 

 32.2 ft/s   45 ft  
TBA = 145.6 lb 
or
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368
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 ρ of the vertical profile of the road at A. (b) Using
the value of ρ 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
ΣFn = man : Wcar =
or
Wcar v A2
g ρ
(146.667 ft/s) 2
32.2 ft/s 2
= 668.05 ft
ρ=
ρ = 668 ft 
or
(b)
Note: v is constant  at = 0; 50 mi/h = 73.333 ft/s
ΣFn = man : W − N =
or
W v 2A
g ρ


(73.333 ft/s) 2
N = (160 lb) 1 −

2
 (32.2 ft/s )(668.05 ft) 
N = 120.0 lb 
or
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369
PROBLEM 12.46
A child having a mass of 22 kg sits on a swing and is held in the
position shown by a second child. Neglecting the mass of the swing,
determine the tension in rope AB (a) while the second child holds
the swing with his arms outstretched horizontally, (b) immediately
after the swing is released.
SOLUTION
Note: The factors of “ 12 ” are included in the following free-body diagrams because there are two ropes and
only one is considered.
(a)
For the swing at rest
1
ΣFy = 0: TBA cos 35° − W = 0
2
TBA =
or
22 kg × 9.81 m/s 2
2 cos 35°
TBA = 131.7 N 
or
(b)
At t = 0, v = 0, so that
an =
v2
ρ
=0
1
ΣFn = 0: TBA − W cos 35° = 0
2
or
TBA =
1
(22 kg)(9.81 m/s 2 ) cos 35°
2
TBA = 88.4 N 
or
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370
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 ( μk = 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
ΣFn = man : N − mg = m
(a)
v2
ρ

v2 
N = m  g + 
ρ


v2 
F = μk N = μk m  g + 
ρ

ΣFt = mat : F = mat
at =
Given data:

F
v2 
= μ k  g + 
ρ
m

μk = 0.20, v = 72 km/h = 20 m/s
ρ = 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 
ΣFn = man = 0: N − mg = 0
N = mg
F = μ k N = μk mg
ΣFt = mat : F = mat
at =
F
= μ k g = 0.20(9.81)
m
at = 1.962 m/s 2 
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371
PROBLEM 12.47 (Continued)
(c)
ΣFn = man : mg − N =
mv 2
ρ

v2 
N = m  g − 
ρ


v2 
F = μk N = μk m  g − 
ρ

ΣFt = mat : F = mat
at =


F
v2 
(20) 2 
= μk  g −  = 0.20 9.81 −

m
45 
ρ


at = 0.1842 m/s 2 
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372
PROBLEM 12.48
A 250-g block fits inside a small cavity cut in arm OA, which rotates
in the vertical plane at a constant rate such that v = 3 m/s. Knowing
that the spring exerts on block B a force of magnitude P = 1.5 N and
neglecting the effect of friction, determine the range of values of θ
for which block B is in contact with the face of the cavity closest to
the axis of rotation O.
SOLUTION
ΣFn = man : P + mg sin θ − Q = m
v2
ρ
To have contact with the specified surface, we need Q ≥ 0,
or
Q = P + mg sin θ −
sin θ >
Data:
mv 2
ρ
>0
1  v2 P 
 − 
g  ρ m 
(1)
m = 0.250 kg, v = 3 m/s, P = 1.5 N, ρ = 0.9 m
Substituting into (1):
sin θ >
1  (3) 2 1.5 
−


9.81  0.9 0.25 
sin θ > 0.40775
24.1° < θ < 155.9° 
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373
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 θ defining the Point B where the packages first
slip relative to the belt.
SOLUTION
Assume package does not slip.
at = 0, F f ≤ μ s N
On the curved portion of the belt
an =
v2
ρ
=
(1 m/s) 2
= 4 m/s 2
0.250 m
For any angle θ
ΣFy = ma y : N − mg cos θ = − man = −
mv 2
N = mg cos θ −
ρ
mv 2
ρ
(1)
ΣFx = max : −F f + mg sin θ = mat = 0
F f = mg sin θ
(a)
At Point A,
θ = 0°
N = (0.5)(9.81)(1.000) − (0.5)(4)
(b)
At Point B,
(2)
N = 2.905 N 
F f = μs N
mg sin θ = μs ( mg cos θ − man )

a 
4 

sin θ = μs  cos θ − n  = 0.40 cos θ −
g 
9.81 


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374
PROBLEM 12.49 (Continued)
Squaring and using trigonometic identities,
1 − cos 2 θ = 0.16 cos 2 θ − 0.130479cos θ + 0.026601
1.16 cos 2 θ − 0.130479cos θ − 0.97340 = 0
cos θ = 0.97402
θ = 13.09° 
Check that package does not separate from the belt.
N=
Ff
μs
=
mg sin θ
μs
N > 0. 
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375
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:
v2
ΣFn = man : N A − W = m A
ρ
 1680 N

v A2 = (1200 m) 
− 9.81 m/s 2 
 54 kg

or
= 25,561.3 m 2 /s 2
At C:
v2
ΣFn = man : N C + W = m C
or
ρ
 350 N

vC2 = (1200 m) 
+ 9.81 m/s 2 
54
kg


= 19,549.8 m 2 /s 2
Since at = constant, we have from A to C
vC2 = v 2A + 2at Δs AC
or
or
19,549.8 m 2/s 2 = 25,561.3 m 2 /s 2 + 2at (π × 1200 m)
at = −0.79730 m/s 2
Then from A to B
vB2 = v 2A + 2at Δs AB
π

= 25,561.3 m 2/s 2 + 2(−0.79730 m/s 2 )  × 1200 m 
2

= 22,555 m 2/s 2
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376
PROBLEM 12.50 (Continued)
At B:
v2
ΣFn = man : N B = m B
ρ
22,555 m 2 /s 2
1200 m
or
N B = 54 kg
or
N B = 1014.98 N
ΣFt = 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
(Fpilot ) B = 1126 N
or
25.6° 
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377
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
ΣF = ma:
Components up the incline,
70°:
− m A 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
N = 2290 N 
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378
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 12.6 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 θ , (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
Weight
W = mg
Acceleration
a=
v2
ρ
ΣFx = max : F + W sin θ = ma cos θ
F=
mv 2
ρ
cos θ − mg sin θ
(1)
ΣFy = ma y : N − W cos θ = ma sin θ
N=
(a)
mv 2
ρ
sin θ + mg cos θ
(2)
Banking angle. Rated speed v = 120 mi/h = 176 ft/s. F = 0 at rated speed.
0=
mv 2
ρ
cos θ − mg sin θ
(176)2
v2
=
= 0.96199
ρ g (1000) (32.2)
θ = 43.89°
tan θ =
(b)
Slipping outward.
θ = 43.9° 
v = 180 mi/h = 264 ft/s
F = μN
μ=
F v 2 cos θ − ρ g sin θ
=
N v 2 sin θ + ρ g cos θ
(264) 2 cos 43.89° − (1000) (32.2)sin 43.89°
(264) 2 sin 43.89° + (1000) (32.2) cos 43.89°
= 0.39009
μ=
μ = 0.390 
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379
PROBLEM 12.52 (Continued)
(c)
Minimum speed.
F = −μ N
v 2 cos θ − ρ g sin θ
v 2 sin θ + ρ g cos θ
ρ g (sin θ − μ cos θ )
v2 =
cos θ + μ sin θ
−μ =
=
(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
v = 78.8 mi/h 
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380
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 θ = 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 (φ = 0), (b) the
required angle of tilt φ if the passenger is to feel no side force.
(See Sample Problem 12.6 for the definition of rated speed.)
SOLUTION
Rated speed:
vR = 60 mi/h = 88 ft/s, 100 mi/h = 146.67 ft/s
From Sample Problem 12.6,
vR2 = g ρ tan θ
ρ=
or
vR2
(88) 2
=
= 2288 ft
g tan θ 32.2 tan 6°
Let the x-axis be parallel to the floor of the car.
ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ )
=
(a)
mv 2
ρ
cos (θ + φ )
φ = 0.
 v2

cos (θ + φ ) − sin (θ + φ ) 
Fs = W 
 gρ

 (146.67) 2

=W 
cos 6° − sin 6° 
 (32.2)(2288)

= 0.1858W
Fs = 0.1858W 
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381
PROBLEM 12.53 (Continued)
(b)
For Fs = 0,
v2
cos (θ + φ ) − sin (θ + φ ) = 0
gρ
v2
(146.67)2
=
= 0.29199
g ρ (32.2)(2288)
θ + φ = 16.28°
φ = 16.28° − 6°
tan (θ + φ ) =
φ = 10.28° 
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382
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 φ 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 ρ tan θ
or
ρ=
vR2
(88) 2
=
= 2288 ft
g tan θ 32.2 tan 6°
Let the x-axis be parallel to the floor of the car.
ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ )
=
mv 2
ρ
cos (θ + φ )
Solving for Fs,
 v2

cos (θ + φ ) − sin (θ + φ ) 
Fs = W 
 gρ

Now
v2
(146.67) 2
=
= 0.29199 and Fs = 0.10W
g ρ (32.2)(2288)
So that
Let
Then
0.10W = W [0.29199 cos (θ + φ ) − sin (θ + φ )]
u = sin (θ + φ )
cos (θ + φ ) = 1 − u 2
0.10 = 0.29199 1 − u 2 − u or 0.29199 1 − u 2 = 0.10 + u
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383
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,
θ + φ = sin −1 u = 10.77°
φ = 10.77° − 6°
φ = 4.77° 
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384
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 β be the slope angle of the dish. tan β =
At r = 2 m, tan β = 1
or
dy
1
= r
dr
2
β = 45°
Draw free body sketches of the sphere.
ΣFy = 0: N cos β − μS N sin β − mg = 0
N =
mg
cos β − μ S sin β
ΣFn = man: N sin β + μ S N cos β =
mv 2
ρ
mg (sin β + μS N cos β ) mv 2
=
cos β − μ S sin β
ρ
v2 = ρ g
sin β + μS cos β
sin 45° + 0.5cos 45°
= (2)(9.81)
= 58.86 m 2 /s 2
cos β − μ S sin β
cos 45° − 0.5sin 45°
v = 7.67 m/s 
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385
PROBLEM 12.56
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. If the polisher is
started so that the fleece along the circumference undergoes a
constant tangential acceleration of 4 m/s 2 , 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)
ΣFt = mat : Ft = mat
or
Ft = (1.6 × 10−6 kg)(4 m/s 2 )
= 6.4 × 10−6 N
ΣFn = man : Fn = m
At t = 3 s:
Fn = (1.6 × 10−6 kg)
v2
ρ
(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
Ftuft = 2.05 × 10−3 N 
or
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386
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 + ( a B ) t t
At t = 10 s:
vB = (0.24 m/s 2 )(10 s) = 2.4 m/s
In the plane of the turntable
ΣF = mB a B : F = mB (a B )t + mB (a B ) n
Then
F = mB ( aB )t2 + (aB ) n2
= mB (aB )t2 +
( )
vB2
2
ρ
+ ΣFy = 0: N − W = 0
or
N = mB g
At t = 10 s:
F = μ s N = μ s mB g
Then
 2



2
μs mB g = mB ( aB )t2 +  vρB 
1/2
or
2

 (2.4 m/s)2  
1

2 2
μs =
(0.24 m/s ) + 
 
2.5 m  
9.81 m/s 2 



μs = 0.236 
or
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387
PROBLEM 12.58
A small, 300-g collar D can slide on portion AB of a rod which is bent as shown.
Knowing that α = 40° and that the rod rotates about the vertical AC at a constant rate of
5 rad/s, determine the value of r for which the collar will not slide on the rod if the effect
of friction between the rod and the collar is neglected.
SOLUTION
First note
vD = rθABC
+ ΣFy = 0: N sin 40° − W = 0
or
N=
mg
sin 40°
v2
ΣFn = man : N cos 40° = m D
r
or
( rθABC ) 2
mg
cos 40° = m
sin 40°
r
or
r=
g
2

θ
1
ABC tan 40°
9.81 m/s 2
1
2
(5 rad/s) tan 40°
= 0.468 m
=
r = 468 mm 
or
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388
PROBLEM 12.59
A small, 200-g collar D can slide on portion AB of a rod which is bent as shown.
Knowing that the rod rotates about the vertical AC at a constant rate and that α = 30°
and r = 600 mm, determine the range of values of the speed v for which the collar will not
slide on the rod if the coefficient of static friction between the rod and the collar is 0.30.
SOLUTION
Case 1: v = vmin , impending motion downward
ΣFx = max : N − W sin 30° = m
or
v2
cos 30°
r


v2
N = m  g sin 30° + cos 30° 
r


ΣFy = ma y : F − W cos 30° = −m
v2
sin 30°
r
or


v2
F = m  g cos 30° − sin 30° 
r


Now
F = μs N
Then




v2
v2
m  g cos 30° − sin 30°  = μs × m  g sin 30° + cos 30° 
r
r




or
v 2 = gr
1 − μ s tan 30°
μ s + tan 30°
= (9.81 m/s 2 )(0.6 m)
or
1 − 0.3tan 30°
0.3 + tan 30°
vmin = 2.36 m/s
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389
PROBLEM 12.59 (Continued)
Case 2: v = vmax , impending motion upward
ΣFx = max : N − W sin 30° = m
v2
cos 30°
r


v2
N = m  g sin 30° + cos 30° 
r


or
ΣFy = ma y : F + W cos 30° = m
v2
sin 30°
r
or


v2
F = m  − g cos 30° + sin 30° 
r


Now
F = μs N
Then




v2
v2
m  − g cos 30° + sin 30°  = μs × m  g sin 30° + cos30° 
r
r




or
v 2 = gr
1 + μs tan 30°
tan 30° − μ s
= (9.81 m/s 2 )(0.6 m)
or
1 + 0.3tan 30°
tan 30° − 0.3
vmax = 4.99 m/s
For the collar not to slide
2.36 m/s < v < 4.99 m/s 
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390
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 μs = 0.35 and μk = 0.25, determine whether
the block will slide in the slot if it is released in the position corresponding to
(a) θ = 80°, (b) θ = 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 θ ) ft
12
vE = ρφABCD
ρ=
an =
vE2
ρ
= ρ (φABCD ) 2
1

=  (26 − 10 sin θ ) ft  (14 rad/s) 2
12

98
= (13 − 5 sin θ ) ft/s 2
3
Assume that the block is at rest with respect to the plate.
v2
ΣFx = max : N + W cos θ = m E sin θ
ρ
or


v2
N = W  − cos θ + E sin θ 


gρ


v2
ΣFy = ma y : −F + W sin θ = − m E cos θ
ρ
or


v2
F = W  sin θ + E cos θ 


gρ


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391
PROBLEM 12.60 (Continued)
(a)
We have
θ = 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 = μs 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° 
θ = 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 = μs 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.
v2
ΣFx = max : N + W cos 40° = m E sin 40°
ρ
or


v2
N = W  − cos 40° + E sin 40°  (as above)
gρ


= 4.4924 lb
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392
PROBLEM 12.60 (Continued)
Sliding:
F = μk N
= 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 ΣFx 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.
ΣF = ma : W + R = ma n
Then
tan (φ − 10°) =
=
W
W
g
=
=
2

man W vE ρ (φ ABCD ) 2
g ρ
32.2 ft/s 2
98
(13 − 5sin 80°) ft/s 2
3
or
φ − 10° = 6.9588°
and
φ = 16.9588°
Now
so that
tan φs = μs
(from above)
μ s = 0.35
φs = 19.29°
0 < φ < φs  Block does not slide and R is directed as shown.
Now
F = R sin φ
Then
F = (0.8 lb)
and R =
W
sin (φ − 10°)
sin16.9588°
sin 6.9588°
= 1.926 lb
F = 1.926 lb
The block does not slide in the slot and
80° 
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393
PROBLEM 12.60 (Continued)
(b)
Assume that the block is at rest with respect to the plate.
ΣF = ma : W + R = ma n
From Part a (above), it then follows that
tan (φ − 50°) =
g
ρ (φABCD )
or
φ − 50° = 5.752°
and
φ = 55.752°
Now
φs = 19.29°
so that
φ > φs
=
2
32.2 ft/s 2
98
(13 − 5sin 40°) ft/s 2
3
The block will slide in the slot and then
φ = φk , where
φk = 14.0362°
or
tan φk = μ k
μk = 0.25
To determine in which direction the block will slide, consider the free-body diagrams for the two
possible cases.
ΣF = 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
v2
ΣFx = max : W cos 40° + R cos φk = m E sin 40°
ρ
F = R sin φk
Now
Then
or
W cos 40° +
F
W vE2
=
sin 40°
tan φk
g ρ


v2
F = μ kW  − cos 40° + E sin 40° 


gρ


= 1.123 lb
(see the first solution)
F = 1.123 lb
The block slides downward in the slot and
40° 
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394
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 ≤ θ ≤ 150°. Knowing that the block begins to slide
when θ = 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 θ = 150°.
v = at = 0,
g = 9.81 m/s 2
ΣFt = mat : − mg sin 30° + N = 0
N = mg sin 30°
ΣFn = man : mg cos 30° − F = m
F = mg cos30° −
Form the ratio
μs =
v2
ρ
mv 2
ρ
F
, and set it equal to μ s for impending slip.
N
F
g cos 30° − v 2 /ρ
9.81 cos 30° − (1.4)2 /0.3
=
=
N
g sin 30°
9.81 sin 30°
μ s = 0.400 
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395
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 θ = 0 and depositing it at the next station when
θ = 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 θ for which sliding is impending.
SOLUTION
ΣFx = max : F =
W vB2
cos θ
g ρ
+ ΣFy = ma y : N − W = −
or
Now
W vB2
sin θ
g ρ


v2
N = W 1 − B sin θ 


gρ




v2
Fmax = μ s N = μ s W 1 − B sin θ 


gρ


and for the component not to slide
F < Fmax
or
or


v2
W vB2
cos θ < μ s W 1 − B sin θ 


g ρ
gρ


μs > g ρ
vB2
cos θ
− sin θ
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396
PROBLEM 12.62 (Continued)
We must determine the values of θ which maximize the above expression. Thus
d  cos θ  − sin θ
 gρ
=
dθ  vB2 − sin θ 
( − sin θ ) − (cos θ )(− cos θ ) = 0
( − sin θ )
gρ
vB2
gρ
vB2
2
or
sin θ =
vB2
gρ
Now
sin θ =
(2.2 ft/s) 2
= 0.180373
(32.2 ft/s 2 ) ( 10
ft )
12
θ = 10.3915° and θ = 169.609°
or
(a)
μs = (μ s )min
for
From above,
( μs ) min = g ρ
vB2
( μs ) min =
cos θ
where sin θ =
− sin θ
cos θ
1 − sin θ
sin θ
=
vB2
gρ
cos θ sin θ
= tan θ
1 − sin 2 θ
= tan10.3915°
( μ s )min = 0.1834 
or
(b)
We have impending motion
to the left for
θ = 10.39° 
to the right for
θ = 169.6° 
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397
PROBLEM 12.63
Knowing that the coefficients of friction between the component I and member BC of the mechanism of
Problem 12.62 are μs = 0.35 and μk = 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 θ for which sliding is impending.
SOLUTION
ΣFx = max : F =
W vB2
cos θ
g ρ
+ ΣFy = ma y : N − W = −
W vB2
sin θ
g ρ


v2
N = W 1 − B sin θ 


gρ


or
Fmax = μ s N
Now


v2
= μ s W 1 − B sin θ 


gρ


and for the component not to slide
F < Fmax
or
or


v2
W vB2
cos θ < μ s W 1 − B sin θ 


g ρ
gρ


vB2 < μs
gρ
cos θ + μs sin θ
(1)
( )
To ensure that this inequality is satisfied, vB2
must be less than or equal to the minimum value
max
of μs g ρ /(cos θ + μs sin θ ), which occurs when (cos θ + μ s sin θ ) is maximum. Thus
d
(cos θ + μs sin θ ) = − sin θ + μ s cos θ = 0
dθ
or
tan θ = μ s
μ s = 0.35
or
θ = 19.2900°
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398
PROBLEM 12.63 (Continued)
(a)
The maximum allowed value of vB is then
(v )
2
B
max
= μs
gρ
cos θ + μs sin θ
= gρ
tan θ
= g ρ sin θ
cos θ + (tan θ ) sin θ
where tan θ = μs
 10 
= (32.2 ft/s 2 )  ft  sin 19.2900°
 12 
(vB ) max = 2.98 ft/s 
or
(b)
First note that for 90° < θ < 180°, Eq. (1) becomes
vB2 < μs
gρ
cos α + μs sin α
where α = 180° − θ . It then follows that the second value of θ for which motion is impending is
θ = 180° − 19.2900°
= 160.7100°
we have impending motion
to the left for
θ = 19.29° 
to the right for
θ = 160.7° 
Alternative solution.
ΣF = ma : W + R = man
For impending motion, φ = φs . Also, as shown above, the values of θ for which motion is impending
(
v2
)
minimize the value of vB, and thus the value of an is an = ρB . From the above diagram, it can be concluded
that an is minimum when ma n and R are perpendicular.
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399
PROBLEM 12.63 (Continued)
Therefore, from the diagram
θ = φs = tan −1 μs
(as above)
and
man = W sin φs
or
v2
m B = mg sin θ
or
vB2 = g ρ sin θ
(as above)
α = 180° − θ
(as above)
ρ
For 90° ≤ θ ≤ 180°, we have
from the diagram
α = φs
and
man = W sin φs
or
vB2 = g ρ sin θ
(as above)
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400
PROBLEM 12.64
In the cathode-ray tube shown, electrons emitted by the cathode
and attracted by the anode pass through a small hole in the anode
and then travel in a straight line with a speed v0 until they strike
the screen at A. However, if a difference of potential V is
established between the two parallel plates, the electrons will be
subjected to a force F perpendicular to the plates while they travel
between the plates and will strike the screen at Point B, which is at
a distance δ from A. The magnitude of the force F is F = eV /d ,
where −e is the charge of an electron and d is the distance
between the plates. Derive an expression for the deflection d in
terms of V, v0 , the charge −e and the mass m of an electron, and
the dimensions d, , and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate
and x =  at the right edge of the plate. At the screen,
x=

2
+L
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2
when it impacts on the screen. For uniform horizontal motion,
x = v0t ,
so that
t1 =
and
t2 =

v0

2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between plates (0 < t < t1 ), the vertical force on the
electron is Fy = eV /d . After it passes the plates (t1 < t < t2 ), it is zero.
For 0 < t < t1 ,
ΣFy = ma y : a y =
Fy
m
=
eV
md
v y = (v y ) 0 + a y t = 0 +
y = y0 + ( v y ) 0 t +
eVt
md
1
eVt 2
ayt 2 = 0 + 0 +
2
2md
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401
PROBLEM 12.64 (Continued)
At t = t1 ,
(v y )1 =
eVt1
md
and
y1 =
eVt12
2md
For t1 < t < t2 , a y = 0
y = y1 + (v y )1 (t − t1 )
At t = t2 ,
y2 = δ = y1 + (v y )1 (t2 − t1 )
δ=
=
eVt12 eVt1
eVt
1
+
( t2 − t1 ) = 1  t2 − t1 
md 
2md md
2 
eV   
L 1 
+ −


mdv0  2v0 v0 2 v0 
or
δ=
eV L

mdv02
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402
PROBLEM 12.65
In Problem 12.64, determine the smallest allowable value of the
ratio d / in terms of e, m, v0, and V if at x =  the minimum
permissible distance between the path of the electrons and the
positive plate is 0.05d .
Problem 12.64 In the cathode-ray tube shown, electrons emitted by
the cathode and attracted by the anode pass through a small hole in
the anode and then travel in a straight line with a speed v0 until they
strike the screen at A. However, if a difference of potential V is
established between the two parallel plates, the electrons will be
subjected to a force F perpendicular to the plates while they travel
between the plates and will strike the screen at point B, which is at a
distance δ from A. The magnitude of the force F is F = eV /d ,
where −e is the charge of an electron and d is the distance between
the plates. Derive an expression for the deflection d in terms of V,
v0 , the charge −e and the mass m of an electron, and the
dimensions d, , and L.
SOLUTION
Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate
and x =  at the right edge of the plate. At the screen,
x=

2
+L
Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0.
Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2
when it impacts on the screen. For uniform horizontal motion,
x = v0t ,
so that
t1 =
and
t2 =

v0

2v0
+
L
.
v0
Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection
produced by the electric force. While the electron is between the plates (0 < t < t1 ), the vertical force on the
electron is Fy = eV/d . After it passes the plates (t1 < t < t2 ), it is zero.
For 0 < t < t1 ,
ΣFy = ma y : a y =
Fy
m
=
eV
md
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403
PROBLEM 12.65 (Continued)
v y = (v y ) 0 + a y t = 0 +
y = y0 + ( v y ) 0 t +
At t = t1 ,

v0
, y=
eVt
md
eVt 2
1
ayt 2 = 0 + 0 +
2
2md
eV  2
2mdv02
d
− 0.05d = 0.450 d
2
But
y<
so that
eV  2
< 0.450 d
2mdv02
d2
1
eV
eV
>
= 1.111 2
2
2
0.450

mv0
2mv0
d
eV
> 1.054


mv02
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404
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 ω. Draw the
FBDs and KDs to determine the forces on pins P1 and P2.
SOLUTION
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405
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
Where r = 6 m, r = −0.2 m/s, r = 0, θ = −0.08 rad/s, θ = 0
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406
PROBLEM 12.F11
Disk A rotates in a horizontal plane about a vertical axis at the constant
rate θ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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407
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 θ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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408
PROBLEM 12.66
Rod OA rotates about O in a horizontal plane. The motion of the
0.5-lb collar B is defined by the relations r = 10 + 6 cosπ t and
θ = π (4t 2 − 8t ), where r is expressed in inches, t in seconds, and θ
in radians. Determine the radial and transverse components of the
force exerted on the collar when (a) t = 0, (b) t = 0.5 s.
SOLUTION
Use polar coordinates and calculate the derivatives of the coordinates r and θ with respect to time.
Mass of collar:
(a)
r = 10 + 6 cos π t in.
θ = π (4t 2 − 8t ) rad
r = −6π sin π t in./s
θ = π (8t − 8) rad/s

r = −6π 2 cos π t in./s 2
θ = 8π rad/s 2
m=
0.5 lb
= 0.015528 lb s 2 /ft = 1.294 × 10−3 lb ⋅ s 2 /in.
2
32.2 ft/s
t = 0:
r = 16 in.
θ =0
r = 0
θ = −8π = −25.1327 rad/s

r = −6π 2 = −59.218 in./s 2
θ = 8π = −25.1327 rad/s 2
ar = r − rθ 2 = −59.218 − (16)( −25.1327)2 = −10165.6 in./s 2
aθ = rθ + 2rθ = (16)(25.1327) + 0 = 402.12 in./s 2
Fr = mar = (1.294 × 10−3 lb ⋅ s 2 /in.)( − 10165.6 in./s 2 )
Fr = −13.15 lb 
Fθ = maθ = (1.294 × 10−3 lb ⋅ s 2 /in.)(402.12 in./s 2 )
Fθ = 0.520 lb 
θ =0 
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409
PROBLEM 12.66 (Continued)
(b)
t = 0.5 s:
r = 10 + 6 cos(0.5π ) = 10 in.
θ = π [(4)(0.25) − (8)(0.5)] = −9.4248 rad = −540° = 180°
r = −6π sin(0.5π ) = −18.8496 in./s
θ = π [(8)(0.5) − 8] = −12.5664 rad/s

r = −6π 2 cos(0.5π ) = 0
θ = 8π = 25.1327 rad/s 2
ar = r − rθ 2 = 0 − (10)(−12.5664) 2 = −1579.14 in./s 2
aθ = rθ + 2rθ = (10)(25.1327) + (2)(−18.8496)(−12.5664) = 725.07 in./s 2
Fr = mar = (1.294 × 10−3 lb ⋅ s 2 /in.)( −1579.14 in./s 2 )
Fr = −2.04 lb 
Fθ = maθ = (1.294 × 10−3 lb ⋅ s 2 /in.)(725.07 in./s 2 )
Fθ = 0.938 lb 
θ = 180° 
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410
PROBLEM 12.67
Rod OA oscillates about O in a horizontal plane. The motion of the 2-lb
collar B is defined by the relations r = 6(1 − e−2t ) and θ = (3/π )(sin π t ),
where r is expressed in inches, t in seconds, and θ in radians. Determine
the radial and transverse components of the force exerted on the collar
when (a) t = 1 s, (b) t = 1.5 s.
SOLUTION
Use polar coordinates and calculate the derivatives of the coordinates r and θ with respect to time.
r = 6(1 − e −2t ) in.
r = 12e
−2 t
in./s
r = −24e−2t in./s 2
Mass of collar:
(a)
t = 1 s:
m=
θ = (3/π )sin π t radians
θ = 3cos π t rad/s
θ = −3π sin π t rad/s 2
2 lb
= 0.06211 lb ⋅ s 2 /ft = 5.176 × 10−3 lb ⋅ s 2 /in.
2
32.2 ft/s
e−2t = 0.13534,
sin π t = 0,
cos π t = −1
r = 6(1 − 0.13534) = 5.188 in.
θ =0
r = (12)(0.13534) = 1.62402 in./s
θ = −3.0 rad/s

r = ( −24) (0.13534) = −3.2480 in./s 2
θ = 0
ar = 
r − rθ 2 = −3.2480 − (5.188)(−3.0)2 = −49.94 in./s 2
aθ = rθ + 2rθ = 0 + (2)(1.62402) (−3) = −9.744 in./s 2
Fr = mar = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 49.94 in./s 2 )
Fr = −0.258 lb 
Fθ = maθ = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 9.744 in./s 2 )
Fθ = −0.0504 lb 
θ =0 
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411
PROBLEM 12.67 (Continued)
(b)
t = 1.5 s:
e −2t = 0.049787,
sin π t = −1,
cos π t = 0
r = 6 (1 − 0.049787) = 5.7013 in.
θ = (3/π )(−1) = −0.9549 rad = −54.7°
r = (12)(0.049787) = 0.59744 in./s 2
θ = 0

r = −(24)(0.049787) = −1.19489 in./s 2 θ = −(3π )(−1) = 9.4248 rad/s 2
ar = 
r − rθ 2 = −1.19489 − 0 = −1.19489 in./s 2
aθ = rθ + 2rθ = (5.7013)(9.4248) + 0 = 53.733 in./s 2
Fr = mar = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 1.19489 in./s 2 )
Fθ = maθ = (5.176 × 10−3 lb ⋅ s 2 /in.)(53.733 in./s 2 )
Fr = −0.00618 lb 
Fθ = 0.278 lb 
θ = −54.7° 
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412
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 θ = 0.75t , where θ 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
t
0
0
 dr =  0.5 dt
or
r = (0.5t ) m
Also,

r =0
θ = (0.75t ) rad/s
θ = 0.75 rad/s 2
Now
ar = r − rθ 2 = 0 − [(0.5t ) m][(0.75t ) rad/s]2 = −(0.28125t 3 ) m/s 2
and
aθ = rθ + 2rθ
= [(0.5t ) m][0.75 rad/s 2 ] + 2(0.5 m/s)[(0.75t ) rad/s]
= (1.125t ) m/s 2
Kinetics
ΣFr = mar : − T = (3 kg)( −0.28125t 3 ) m/s 2
or
T = (0.84375t 3 ) N
ΣFθ = mB aθ : 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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413
PROBLEM 12.69
The horizontal rod OA rotates about a vertical shaft according to the
relation θ = 10t , where θ and t are expressed in rad/s and seconds,
respectively. A 250-g collar B is held by a cord with a breaking
strength of 18 N. Neglecting friction, determine, immediately after the
cord breaks, (a) the relative acceleration of the collar with respect to
the rod, (b) the magnitude of the horizontal force exerted on the collar
by the rod.
SOLUTION
θ = 10t rad/s, θ = 10 rad/s 2
m = 250 g = 0.250 kg
Before cable breaks: Fr = −T and 
r = 0.
Fr = mar : − T = m(
r − rθ 2 )
mrθ 2 = mr + T
or
θ 2 =
mr + T
0 − 18
=
= 144 rad 2 /s 2
mr
(0.25)(0.5)
θ = 12 rad/s
Immediately after the cable breaks: Fr = 0, r = 0
(a)
Acceleration of B relative to the rod.
m(
r − rθ 2 ) = 0
or

r = rθ 2 = (0.5)(12)2 = 72 m/s 2
a B /rod = 72 m/s 2 radially outward 
(b)
Transverse component of the force.
Fθ = maθ : Fθ = m (rθ + 2rθ)
Fθ = (0.250)[(0.5)(10) + (2)(0)(12)] = 1.25
Fθ = 1.25 N 
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414
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 θ = 15 rad/s
and θ = 250 rad/s 2 for the position θ = 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 sin θ )θ

r = −2b(θsin θ + θ 2 cos θ )
Then
r = 2b cos θ
and
ar = r − rθ 2 = −2b(θsin θ + θ 2 cos θ ) − (2b cos θ )θ 2 = − 2b(θsin θ + 2θ 2 cos θ )
Now
 20 
= − 2
ft  [(250 rad/s 2 )sin 20° + 2(15 rad/s) 2 cos 20°] = −1694.56 ft/s 2
12


aθ = rθ + 2rθ = (2b cos θ )θ + 2(−2bθ sin θ )θ = 2b(θcos θ − 2θ 2 sin θ )
and
 20 
= 2
ft  [(250 rad/s 2 ) cos 20° − 2(15 rad/s) 2 sin 20°] = 270.05 ft/s 2
 12 
Kinetics.
(a)
We have
Fr = mar =
and
Fθ = maθ =
1 lb
4
32.2 ft/s
2
1
lb
4
32.2 ft/s 2
× (−1694.56 ft/s 2 ) = −13.1565 lb
× (270.05 ft/s 2 ) = 2.0967 lb
Fr = −13.16 lb 
Fθ = 2.10 lb 
ΣFr : −Fr = −Q cos 20°
(b)
or
Q=
1
(13.1565 lb) = 14.0009 lb
cos 20°
ΣFθ : Fθ = 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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415
PROBLEM 12.71
The two blocks are released from rest when r = 0.8 m and
θ = 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 θ 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,

r + aB = 0
r = −aB
(1)
ΣFx = m Aa A : T cos θ = m Aa A or T = m Aa A sec θ
(2)
or
+ ΣFy = mB aB : mB g − T = mB aB
For block B,
Adding Eq. (1) to Eq. (2) to eliminate T, mB g = m Aa A secθ + mB aB
(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 − rθ 2 = ar = a A ⋅ er = −a A cos θ
(5)
Noting that initially θ = 0, using Eq. (1) to eliminate r, and changing
signs gives
aB = a A cosθ
(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 secθ + mB cos θ
20sec 30° + 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

a B = 4.75 m/s 2 
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416
PROBLEM 12.72
The velocity of block A is 2 m/s to the right at the instant when r = 0.8 m
and θ = 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 θ 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 = −v A cos30°
= −2cos30° = −1.73205 m/s
rθ = vθ = v A ⋅ eθ = −v A sin 30°
= 2sin 30° = 1.000 m/s 2
θ =
vθ
1.000
=
= 1.25 rad/s
r
0.8
Constraint of cable: r + yB = constant,
r + vB = 0,

r + aB = 0
or
r = −aB
(1)
For block A,
ΣFx = m Aa A : T cosθ = m Aa A or T = m Aa A secθ
(2)
For block B,
+ ΣFy = mB aB : mB g − T = mB aB
(3)
Adding Eq. (1) to Eq. (2) to eliminate T, mB g = m Aa A secθ + mB aB
(4)
Radial and transverse components of a A.
Use a method similar to that used for the components of velocity.
r − rθ 2 = ar = a A ⋅ er = −a A cos θ
(5)
Using Eq. (1) to eliminate r and changing signs gives
aB = a A cosθ − rθ 2
(6)
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417
PROBLEM 12.72 (Continued)
Substituting Eq. (6) into Eq. (4) and solving for a A ,
aA =
(
mB g + rθ 2
)
m A secθ + mB cos θ
=
(25)[9.81 + (0.8)(1.25) 2 ]
= 6.18 m/s 2
20sec30° + 25cos 30°
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

a B = 4.10 m/s 2 
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418
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 θ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 )
Σ Fr = mar : − k (r − l0 ) = m(r − rθ 2 )
kl
k


r =  θ 2 −  r + 0
m
m

(1)
But
d
dr dr
dr
( r) =
= r
dt
dr dt
dr

kl 
k
  = rdr
 =  θ 2 −  r + 0  dr
rdr
m
m


r=
Integrate using the condition r = r0 when r = r0 .
1 2 r  1   2 k  2 kl0  r
r
r
=  θ −  r +
m
m  r0
2 r0  2 
kl
1 2 1 2 1  2 k  2
r − r0 =  θ −  r − r02 + 0 (r − r0 )
2
2
2
m
m
(
)
2kl0
k

r 2 = r02 +  θ 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
θ = 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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419
PROBLEM 12.73* (Continued)
(a)
Components of velocity when r = 12 in.
2.5 

2
2
r 2 = 0 + 102 −
 (1.0 − 1.5 ) + 0
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 
vθ = rθ = (1.0)(10)
vθ = 10.00 ft/s 
Components of acceleration.
Fr = −kr + kl0 = −(2.5)(1.0) + 0 = −2.5 lb
ar =
Fr
2.5
=−
m
0.01553
ar = 161.0 ft/s 2 
aθ = rθ + 2rθ = 0 + (2)( −8.7306)(10)
aθ = −174.6 ft/s 2 
(c)
Transverse component of force.
Fθ = maθ = (0.01553)( −174.6)
Fθ = −2.71 lb 
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420
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 2θ and using Eq. (12.27), express
the radial and transverse components of the velocity v of the particle as
functions of θ .
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2θ = h0 = r0 v0
or
θ = 0 20 = 0 0
rv
r v cos 2θ
r
2
r0
=
v0
cos 2θ
r0
Radial component of velocity.
vr = r =
= r0
r0

sin 2θ
dr  d 
θ=
θ

θ = r0
dθ
dθ  cos 2θ 
(cos 2θ )3/2
sin 2θ v0
cos 2θ
(cos 2θ )3/ 2 r
vr = v0
sin 2θ
cos 2θ

Transverse component of velocity.
vθ =
h r0 v0
=
cos 2θ
r
r0
vθ = v0 cos 2θ 
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421
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 2θ and using
Eq. (12.27), express the radial and transverse components of the velocity
v of the particle as functions of θ.
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2θ = h0 = r0 v0
or
rv
r v cos 2θ
r
2
θ = 0 20 = 0 0
r0
=
v0
cos 2θ
r0
Differentiating the expression for r with respect to time,
r =
r0

dr  d 
sin 2θ
sin 2θ v0
sin 2θ
θ=
θ = r0
cos 2θ = v0

θ = r0
3/2
3/ 2
dθ
dθ  cos 2θ 
r
(cos 2θ )
(cos 2θ )
cos 2θ
0
Differentiating again,
r =
(a)
dr  d 
sin 2θ  
2 cos 2 2θ + sin 2 2θ  v0 2 2 cos 2 2θ + sin 2 2θ
θ=
θ=
 v0
θ = v0
dθ
dθ 
r0
(cos 2θ )3/2
cos 2θ 
cos 2θ
vr = r = v0
sin 2θ
cos 2θ
=
v0 r
sin 2θ
r0
vr
vθ = rθ = 0 cos 2θ
r0
v = (vr ) 2 + (vθ ) 2 =
v0 r
sin 2 2θ + cos 2 2θ
r0
v=
v0 r

r0
v 2 2cos 2 2θ + sin 2 2θ
r0
v0 2
ar = 
r − rθ 2 = 0
−
cos 2 2θ
r0
cos 2θ
cos 2θ r0 2
=
v0 2 cos 2 2θ + sin 2 2θ
v0
v 2r
=
= 02
r0
r0
cos 2θ
r0 cos 2θ
Fr = mar =
mv02 r
r02
Fr =
:
mv02 r
r02

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422
PROBLEM 12.75 (Continued)
Since the particle moves under a central force, aθ = 0.
Magnitude of acceleration.
a = ar 2 + aθ 2 =
v0 2 r
r0 2
Tangential component of acceleration.
at =
v0 2 r
dv d  v0 r  v0

r
= 
=
=
sin 2θ

dt dt  r0  r0
r0 2
Normal component of acceleration.
at = a 2 − at 2 =
(b)
But
r 
cos 2θ =  0 
r 
Hence,
an =
But an =
v2
ρ
or
v0 2 r
r0 2
1 − sin 2 2θ =
v0 2 r cos 2θ
r0 2
2
v0 2
r
ρ=
v 2 v0 2 r 2 r
= 2 ⋅ 2
an
r0
v0
ρ=
r3

r02
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423
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 θ and using
Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ .
SOLUTION
Since the particle moves under a central force, h = constant.
Using Eq. (12.27),
h = r 2θ = h0 = r0 v0
or
θ = 0 20 =
rv
r0 v0
r
r0 cos θ
2
2
=
v0
r0 cos 2 θ
Radial component of velocity.
vr = r =
d
(r0 cos θ ) = −( r0 sin θ )θ
dt
Transverse component of velocity.
vθ = rθ = ( r0 cos θ )θ
Speed.
v = vr 2 + vθ 2 = r0θ =
r0 v0
r0 cos θ
2
v=
v0
cos 2 θ

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424
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) θ = 0, (b) θ = 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 θ and
using Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ .
SOLUTION
Since the particle moves under a central force, h = constant
Using Eq. (12.27),
h = r 2θ = h0 = r0 v0
rv
r0 v0
r
r0 cos θ
θ = 0 20 =
2
2
=
v0
r0 cos 2 θ
Radial component of velocity.
vr = r =
d
(r0 cos θ ) = −( r0 sin θ )θ
dt
Transverse component of velocity.
vθ = rθ = ( r0 cos θ )θ
Speed.
v = vr 2 + vθ 2 = r0θ =
r0 v0
r0 cos θ
2
=
v0
cos 2 θ
Tangential component of acceleration.
at =
v0
dv
(−2)( − sin θ )θ 2v0 sin θ
= v0
=
⋅
3
3
dt
cos θ
cos θ r0 cos 2 θ
=
2v0 2 sin θ
r0 cos5 θ
Tangential component of force.
Ft = mat : Ft =
(a)
θ = 0,
Ft = 0
(b)
θ = 45°,
Ft =
2mv0 2 sin θ
r0 cos5 θ
Ft = 0 
2mv0 sin 45°
Ft =
cos 45°
5
8mv0 2

r0
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425
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
G
v2
r
Mm
v2
=
m
r
r2
r 2
v
G
or
M=
Now
v=
so that
r  2π r 
1  2π  3
M= 
r
= 

G τ 
G  τ 
Noting that
τ = 27.32 days = 2.3604 × 106 s
and
r = 238,910 mi = 1.26144 × 109 ft
we have


2π
9
3
M=

 (1.26144 × 10 ft)
−9 4
4 
6 
34.4 × 10 ft /lb ⋅ s  2.3604 × 10 s 
2π r
τ
2
2
2
1
M = 413 × 1021 lb ⋅ s 2 /ft 
or
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426
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 τ required for the moon to complete one full revolution about
the earth. Compute r knowing that τ = 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
[Eq. (12.28)]
v2
r
Mm
v2
m
=
r
r2
GM
r
or
v2 =
Now
GM = gR 2
so that
v2 =
gR 2
g
or v = R
r
r
For one orbit,
τ=
2π r
2π r
=
v
R g
[Eq. (12.30)]
r
1/ 3
or
 gτ 2 R 2 
r = 
2 

 4π 
Now
τ = 27.3 days = 2.35872 × 106 s

Q.E.D.
R = 3960 mi = 20.9088 × 106 ft
1/3
SI:
 9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m) 2 
r=

4π 2


= 382.81 × 106 m
r = 383 × 103 km 
or
U.S. customary units:
1/3
 32.2 ft/s2 × (2.35872 × 106 s) 2 × (20.9088 × 106 ft) 2 
r=

4π 2


= 1256.52 × 106 ft
r = 238 × 103 mi 
or
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427
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 τ be the period time to complete one orbit.
vτ = 2π r
Then
GM τ 2
r =
4π 2
Data:
τ = 23.934 h = 86.1624 × 103 s
(a)
3
In SI units:
v 2τ 2 =
GM τ 2
= 4π 2 r 2
r
But
or
1/3
or
 GM τ 2 
r = 
2 

 4π 
g = 9.81 m/s 2 , R = 6.37 × 106 m
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
π


altitude h = r − R = 35.775 × 106 m
In U.S. units:
h = 35,800 km 
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
π


altitude h = r − R = 117.425 × 106 ft
h = 22, 200 mi 
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428
PROBLEM 12.80 (Continued)
(b)
In SI units:
v=
GM
398.06 × 1012
=
= 3.07 × 103 m/s
6
r
42.145 × 10
v = 3.07 km/s 
v=
GM
14.077 × 1015
=
= 10.09 × 103 ft/s
r
138.334 × 106
v = 10.09 × 103 ft/s 
In U.S. units:
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429
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 τ 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 τ = 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
G
[Eq. (12.28)]
v2
r
Mm
v2
m
=
r
r2
GM
r
or
v2 =
Now
GM = gR 2
so that
v2 =
gR 2
r
For one orbit,
τ=
2π r
2π r
=
v
R g
[Eq. (12.30)]
or
v=R
g
r
r
1/3
or
 gτ 2 R 2 
r = 
2 

 4π 
Solving for g,
g = 4π 2
Q.E.D.

r3
τ 2 R2
and noting that τ = 3.551 days = 306,806 s, then
g Jupiter = 4π 2
= 4π 2
3
rEur
2
RJup
τ Eur
(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 ≈ 2.53g Earth 
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430
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
=
ma
=
n
r
r2
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
GM
1.3335 × 1020 m3 /s 2
=
G
66.73 × 10−12
(a) Mass of sun.
M =
(b) At the surface of the sun,
R = 695.5 × 103 km = 695.5 × 106 m
M = 1.998 × 1030 kg 
GMm
= mg
R2
g =
GM
1.3335 × 1020
=
R2
(695.5 × 106 ) 2
g = 276 m/s 2 
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431
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
2π rA
Velocity of Atlas.
vA =
where
v A = 85.54 × 103 mi = 451.651 × 106 ft
and
τ A = 0.6017 days = 51,987 s
Gravitational force.
τA
vA =
(2π ) (451.651 × 106 )
= 54.587 × 103 ft/s
51,987
F=
GMm mv 2
=
r
r2
from which
GM = rv 2 = constant
For the satellite,
rs vs2 = rAv 2A

r v2
rs = A 2A 
vs
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.
r v 2 (451.651 × 106 )(54.587 × 103 ) 2
M= A A =
G
34.4 × 10−9
M = 39.1 × 1024 lb ⋅ s 2 /ft 
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432
PROBLEM 12.84
The periodic times (see Problem 12.83) of the planet Uranus’s moons Juliet and Titania have been observed to
be 0.4931 days and 8.706 days, respectively. Knowing that the radius of Juliet’s orbit is 40,000 mi, determine
(a) the mass of Uranus, (b) the radius of Titania’s orbit.
SOLUTION
2π rJ
Velocity of Juliet.
vJ =
where
rJ = 40,000 mi = 2.112 × 108 ft
and
τ J = 0.4931 days = 42, 604 s
Gravitational force.
vJ =
(2π )(2.112 × 108 ft)
= 3.11476 × 10 4 ft/s
42, 604 s
F=
GMm mv 2
=
r
r2
GM = rv 2 = constant
from which
(a)
τJ
r v2
M= J J
G
Mass of Uranus.
(2.112 × 108 )(3.11476 × 104 ) 2
34.4 × 10−9
= 5.95642 × 10 24 lb ⋅ s 2 /ft
M=
M = 5.96 × 1024 lb ⋅ s 2 /ft 
(b)
Radius of Titania’s orbit.
GM = rT vT2 =
4π 2 rT3
τ 
rT3 = rJ3  T 
τJ 
τT 2
2
=
4π 2 rJ3
τJ2
2
 8.706 
27 3
= (2.112 × 108 )3 
 = 2.93663 × 10 ft
 0.4931 
rT = 1.43202 × 109 ft = 2.71216 × 105 mi
rT = 2.71 × 105 mi 
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433
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
First note that
rE = RE + hE = (6.37 × 106 + 4.5 × 106 ) m
Then
(a)
RE = 6.37 × 106 m
= 10.87 × 106 m
We have
F=
GMm
r2
[Eq. (12.28)]
and
GM = gR 2
Then
F = gR 2
For the earth orbit,
 6.37 × 106 m 
F = (500 kg)(9.81 m/s 2 ) 

6
 10.87 × 10 m 
[Eq. (12.29)]
m
R
=W  
2
r
r
2
2
F = 1684 N 
or
(b)
From the solution to Problem 12.78, we have
2
1  2π  3
M= 
r
G  τ 
2π r 3/ 2
GM
Then
τ=
Now
τE =τM 
2π rE3/ 2
GM E
=
2π 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
rM = 2510 km 
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434
PROBLEM 12.85 (Continued)
(c)
GM = gR 2
We have
[Eq.(12.29)]
Substituting into Eq. (1)
2π rE3/2
RE g E
=
2π rM3/ 2
RM g M
2
or
3
2
 R  r 
 R  M 
g M =  E   M  gE =  E   M  gE
 RM   rE 
 RM   M E 
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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435
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,
Σ Fn = man : F = m
F =G
Eq. (12.28):
Then
G
or
v2
r
Mm
r2
Mm
v2
=
m
r
r2
GM
v2 =
r
66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg
2200 × 103 m
Then
2
(v A )circ
=
or
(v A )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 + Δv A
= (1493.0 − 26.3) m/s
= 1466.7 m/s
Conservation of angular momentum requires that
rA m(v A )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 + ΔvB
ΔvB = (1535.5 − 1551.3) m/s
ΔvB = −15.8 m/s 
or
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436
PROBLEM 12.87
Plans for an unmanned landing mission on the planet Mars called
for the earth-return vehicle to first describe a circular orbit at an
altitude dA = 2200 km above the surface of the planet with a
velocity of 2771 m/s. As it passed through Point A, the vehicle
was to be inserted into an elliptic transfer orbit by firing its
engine and increasing its speed by Δv A = 1046 m/s. As it passed
through Point B, at an altitude dB = 100,000 km, the vehicle was to
be inserted into a second transfer orbit located in a slightly
different plane, by changing the direction of its velocity and
reducing its speed by ΔvB = −22.0 m/s. Finally, as the vehicle
passed through Point C, at an altitude dC = 1000 km, its speed
was to be increased by ΔvC = 660 m/s to insert it into its return
trajectory. Knowing that the radius of the planet Mars is
R = 3400 km, determine the velocity of the vehicle after
completion of the last maneuver.
SOLUTION
rA = 3400 + 2200 = 5600 km = 5.60 × 106 m
rB = 3400 + 100, 000 = 103, 400 km = 103.4 × 106 m
rC = 3400 + 1000 = 4400 km = 4.40 × 106 m
First transfer orbit.
v A = 2771 m/s + 1046 m/s = 3817 m/s
Conservation of angular momentum:
rA m v A = rB m vB
(5.60 × 10 )(3817) = (103.4 × 106 )vB
6
vB = 206.7 m/s
Second transfer orbit.
vB′ = vB + ΔvB
= 206.7 − 22.0 = 184.7 m/s
Conservation of angular momentum:
rB mvB′ = rC mvC
(103.4 × 106 )(184.7) = (4.40 × 106 )vC
vC = 4340 m/s
After last maneuver.
v = vC + ΔvC = 4340 + 660
v = 5000 m/s 
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437
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 upper-stage 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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438
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
r (v )
(v A ) tr = B B tr
rA
(138.336 × 106 )(5278)
21.8856 × 106
= 33.362 × 103 ft/s
=
(b)
Change in speed at A.
Δv A = (v A ) tr − (v A )circ
= 33.362 × 103 − 25.362 × 103
= 8.000 × 103
Δv A = 8000 ft/s 
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439
PROBLEM 12.89
A space shuttle S and a satellite A are in the circular orbits shown. In
order for the shuttle to recover the satellite, the shuttle is first placed
in an elliptic path BC by increasing its speed by ΔvB = 280 ft/s as it
passes through B. As the shuttle approaches C, its speed is increased
by ΔvC = 260 ft/s to insert it into a second elliptic transfer orbit CD.
Knowing that the distance from O to C is 4289 mi, determine the
amount by which the speed of the shuttle should be increased as it
approaches D to insert it into the circular orbit of the satellite.
SOLUTION
R = 3960 mi = 20.9088 × 106 ft
First note
rA = (3960 + 380) mi = 4340 mi = 22.9152 × 106 ft
rB = (3960 + 180) mi = 4140 mi = 21.8592 × 106 ft
ΣFn = man :
For a circular orbit,
F =G
Eq. (12.28):
Then
or
G
F =m
v2
r
Mm
r2
Mm
v2
=m
2
r
r
v2 =
GM gR 2
=
r
r
using Eq. (12.29).
32.2 ft/s 2 × (20.9088 × 106 ft) 2
22.9152 × 106 ft
Then
2
(v A )circ
=
or
(v A )circ = 24, 785 ft/s
and
2
(vB )circ
=
or
(vB )circ = 25,377 ft/s
We have
(vB )TRBC = (vB )circ + ΔvB = (25,377 + 280) ft/s
32.2 ft/s 2 × (20.9088 × 106 ft) 2
21.8592 × 106 ft
= 25, 657 ft/s
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440
PROBLEM 12.89 (Continued)
Conservation of angular momentum requires that
From Eq. (1)
BC : rB m (vB )TRBC = rC m (vC )TRBC
(1)
CD : rC m (vC )TRCD = rA m (vD )TRCD
(2)
r
4140 mi
(vC )TRBC = B (vB )TRBC =
× 25, 657 ft/s
4289 mi
rC
= 24, 766 ft/s
Now
(vC )TRCD = (vC )TRBC + ΔvC = (24, 766 + 260) ft/s
= 25, 026 ft/s
From Eq. (2)
r
4289 mi
(vD )TRCD = C (vC )TRCD =
× 25,026 ft/s
4340 mi
rA
= 24, 732 ft/s
Finally,
or
(v A )circ = (vD )TRCD + ΔvD
ΔvD = (24,785 − 24, 732) ft/s
ΔvD = 53 ft/s 
or
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441
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 θ = 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)
Fθ = 0 and at A,
Fr = − Fsp = 0
(a A ) r = 0 
(a A )θ = 0 
ΣFr = mar :
(b)
Noting that
−Fsp = m(
r − rθ 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 (v A )θ = rB m (vB )θ
Then
(vB )θ =
where (v A )θ = rA θ0
150 mm
[(150 mm)(16 rad/s)] = 800 mm/s
450 mm
(vB )θ = 0.800 m/s 
or
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442
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 v A = 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 θ be polar coordinates with the origin lying at the shaft.
Constraint of rod: θ B = θ A + π radians; θB = θA = θ; θB = θA = θ.
(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 :
(a B ) r = 0 
Transverse components of acceleration.
(a A )θ = rAθ + 2rAθ = raθ
(aB )θ = rBθ + 2rBθ
(1)
Since the rod is massless, it must be in equilibrium. Draw its free body
diagram, applying Newton’s 3rd Law.
ΣM 0 = 0: rA ( FA )θ + rB ( FB )θ = rAm A (a A )θ + rB mB (aB )θ = 0
rAmArAθ + rB mB (rBθ + 2rBθ) = 0
θ =
At t = 0,
rB = 0
so that
−2mB rBθ
m ArA2 + mB rB 2
θ = 0.
(aB )θ = 0 
From Eq. (1),
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443
PROBLEM 12.91 (Continued)
(b)
Acceleration of B relative to the rod.
(v )
96
At t = 0, (v A )θ = 8 ft/s = 96 in./s, θ = A θ =
= 9.6 rad/s
rA
10
rB − rBθ 2 = (aB ) r = 0

rB = rBθ 2 = (8) (9.6)2 = 737.28 in./s 2

rB = 61.4 ft/s 2 
(c)
Speed of A.
Substituting
d
(mr 2θ) for rFθ in each term of the moment equation
dt
gives
(
)
(
)
d
d
m ArA2θ +
mB rB 2θ = 0
dt
dt
Integrating with respect to time,
(
2
2
2
m ArA θ + mB rB θ = m ArA θ
) + ( m r θ )
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 θ f =  A rA2 + B (rB )02  θ0

g
g
 g

 g

θ f =
WArA2 + WB (rB )02 
(1)(10) 2 + (2)(8) 2
=
(9.6) = 3.5765 rad/s
θ
0
WArA 2 + WB rC 2
(1)(10) 2 + (2)(16) 2
(v A ) f = rAθ f = (10)(3.5765) = 35.765 in./s
(v A ) f = 2.98 ft/s 
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444
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 θ = 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
m Ar1(vθ )1 = m Ar2 (vθ ) 2
r (v )
r 2θ
(0.6) 2 (12)
(vθ ) 2 = 1 θ 1 = 1 1 =
= 3.6
r2
r2
1.2
(vθ ) 2 = 3.60 ft/s 
θ2 =
(vθ ) 2
3.6
=
= 3.00 rad/s
rA
1.2
(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
and
(a A ) r = r − rθ 2
Collar B:
ΣFy = mB aB : T − WB = mB 
y = mB
r
(1)
Collar A:
ΣFr = mA (a A )r : − T = m A (
r − rθ 2 )
(2)
Adding (1) and (2) to eliminate T,
−WB = (m A + mB )
r + m Arθ 2
a A/rod = 
r=
m Arθ 2 − WB (0.08075)(1.2)(3.00) 2 − (2.6)
=
= −10.70 ft/s 2
m A + mB
0.08075 + 0.08075
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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445
PROBLEM 12.93
A small ball swings in a horizontal circle at the end of a cord of length l1 ,
which forms an angle θ1 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 , θ1 , and θ 2 . (b) If the ball is set in motion so that
initially l1 = 0.8 m and θ1 = 35°, determine the angle θ 2 when l2 = 0.6 m.
SOLUTION
(a)
For state 1 or 2, neglecting the vertical component of acceleration,
ΣFy = 0: T cos θ − W = 0
T = W cos θ
ΣFx = man : T sin θ = W sin θ cos θ =
But ρ =  sin θ
mv 2
ρ
so that
v2 =
ρW
m
sin 2 θ cos θ =  g sin θ tan θ
v1 = 1 g sin θ1 tan θ1
and
v2 =  2 g sin θ 2 tan θ 2
ΣM y = 0: H y = constant
r1mv1 = r2 mv2
v11 sin θ1 = v2  2 sin θ 2
or
3/2
3/2
1 g sin θ1 sin θ1 tan θ1 =  2 sin θ 2 sin θ 2 tan θ 2
31 sin 3 θ1 tan θ1 = 32 sin 3 θ 2 tan θ 2 
(b)
With θ1 = 35°, 1 = 0.8 m, and  2 = 0.6 m
(0.8)3 sin 3 35° tan 35° = (0.6)3 sin 3 θ 2 tan θ 2
sin 3 θ 2 tan θ 2 − 0.31320 = 0
θ 2 = 43.6° 
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446
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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447
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)
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448
PROBLEM 12.94
A particle of mass m is projected from Point A with an initial
velocity v 0 perpendicular to OA and moves under a central force F
along an elliptic path defined by the equation r = r0 /(2 − cos θ ).
Using Eq. (12.37), 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 θ
=
,
r
r0
d 2u
2
F
+u = =
2
r0 mh 2u 2
dθ
Solving for F,
F=
du sin θ
=
,
dθ
r0
d 2 u cos θ
=
r0
dθ 2
by Eq. (12.37).
2mh 2 u 2 2mh 2
=
r0
r0 r 2
Since m, h, and r0 are constants, F is proportional to 12 , or inversely proportional to r 2 . 
r
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449
PROBLEM 12.95
A particle of mass m describes the logarithmic spiral r = r0 ebθ under a central force F directed toward the
center of force O. Using Eq. (12.37) show that F is inversely proportional to the cube of the distance r from
the particle to O.
SOLUTION
u=
1 1 −bθ
= e
r r0
du
b
= − e −bθ
dθ
r0
d 2u b 2 −bθ
e
=
r0
dθ 2
d 2u
b 2 + 1 −bθ
F
+u =
e
=
2
r
dθ
mh 2u 2
0
F =
=
(b 2 + 1)mh 2u 2 −bθ
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 13 , or inversely proportional to r 3.
r
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450
PROBLEM 12.96
For the particle of Problem 12.74, and using Eq. (12.37), show that the
central force F is proportional to the distance r from the particle to the
center of force O.
PROBLEM 12.74 A particle of mass m is projected from Point A with an
initial velocity v 0 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 2θ and using
Eq. (12.27), express the radial and transverse components of the velocity v
of the particle as functions of θ .
SOLUTION
u =
1
=
r
cos 2θ
,
r0
sin 2θ
du
=−
dθ
r0 cos 2θ
d 2u
cos 2θ (2 cos 2θ ) − sin 2θ (− sin 2θ / cos 2θ )
=−
dθ
r0 cos 2θ
=−
Eq. (12.37):
2 cos 2 2θ + sin 2 2θ
(1 + cos 2 2θ )
=
−
r0 (cos 2θ )3/2
r0 (cos 2θ )3/2
d 2u
F
+u =
2
dθ
mh 2u 2
Solving for F,
 d 2u

F = mh 2u 2 
+ u 
 dθ

= mh 2
cos 2θ  1 + cos 2 2θ
+
−
r0 2  r0 (cos 2θ )3/2
cos 2θ 

r0 
= mh 2
cos 2θ 
1
−
−
2 
r0  r0 (cos 2θ )3/ 2
cos 2θ
+
r0
=−
mh 2
mh 2
=
−
3
4
r0 cos 2θ
r0
r0
cos 2θ
cos 2θ 

r0 
F =−
mh 2r

r0 4
The force F is proportional to r. The minus sign indicates that it is repulsive.
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451
PROBLEM 12.97
A particle of mass m describes the path defined by the equation r = r0 sin θ under a central force F directed
toward the center of force O. Using Eq. (12.37), show that F is inversely proportional to the fifth power of the
distance r from the particle to O.
SOLUTION
We have
d 2u
F
+u =
2
dθ
mh 2u 2
where
u=
1
r
Eq. (12.37)
and mh 2 = constant
 d 2u

F × u 2  2 + u 
 dθ

u=
Now
1
1
=
r r0 sin θ
Then
du
1  1 
1 cos θ
=

=−
dθ dθ  r0 sin θ 
r0 sin 2 θ
and
d 2u
1  − sin θ (sin 2 θ ) − cos θ (2 sin θ cos θ ) 
=
−


r0 
dθ 2
sin 4 θ

=
Then
1 1 + cos 2 θ
r0 sin 3 θ
2
1 
 1   1 1 + cos θ
F ×  2  
+

3
r0 sin θ 
 r   r0 sin θ
= mh 2
1 1  1 + cos 2 θ sin 2 θ 
+


r0 r 2  sin 3 θ
sin 3 θ 
2 2
1 1
= mh
r0 r 2 sin 3 θ
= mh 2
r
sin θ =  
 r0 
3
3
2r02
r5
F is proportional to
1
1
F× 5
5
r
r
Q.E.D. 
Note: F > 0 implies that F is attractive.
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452
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 m3 /s 2
1
GM
= 2 (1 + ε )
r0
h
1+ε =
h2
(94.09 × 109 )2
=
= 3.33
r0GM
(6.673 × 106 )(398.06 × 1012 )
ε = 3.33 − 1
ε = 2.33 
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453
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, ε = 1.
Eq. (12.39′) :
At θ = 0,
1 GM
= 2 (1 + cos θ )
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
6
24.077 × 10
v0 = 6.48 mi/s 
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454
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,
 2 × 66.73 × 10−12 m3 /kg ⋅ s 2 × 4.87 × 1024 kg 
(v A )par = 

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 + Δv A
Now
(v A )circ =
Then
GM
r0
=
1
Δv A =
1
2
2
Eq. (12.44)
(v A )par
(v A ) par − (v A )par
 1

=
− 1 (10.1314 km/s)
 2

= −2.97 km/s
| Δv A | = 2.97 km/s 
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455
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,
Eq. (12.44)
v=
GM
r
For the orbit of Tethys,
GM = rT vT2
For Voyager’s trajectory, we have
1 GM
= 2 (1 + ε cos θ )
r
h
where h = r0 v0
At O,
r = r0 , θ = 0
Then
1
GM
(1 + ε )
=
r0 (r0 v0 ) 2
or
ε = 0 0 − 1 = 0 02 − 1
r v2
GM
=
r v2
rT vT
2
185 × 10 km  21.0 km/s 
×
 −1
295 × 103 km  11.35 km/s 
3
ε = 1.147 
or
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456
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 θ A
rA
h
and
1 GM
= 2 + C cos θ B .
rB
h
But
θ B = θ A + 180°,
so that
cos θ A = − cos θ B .
Adding,
1 1 1 1 2GM
+ = + = 2 
rA rB r0 r1
h
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457
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 α 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 β v0 , where β < 1, by firing its engine for a short interval of
time. Determine the smallest permissible value of β if the probe is not to crash on the surface of the planet.
SOLUTION
GM
rA
For the circular orbit,
v0 =
where
rA = R + α R = R(1 + α )
Eq. (12.44),
GM = v02 R(1 + α )
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 + α )]( β v0 )
Then
2v02 R(1 + α )
1
1
+ =
R(1 + α ) rB [ R(1 + α ) β v0 ]2
=
2
β R(1 + α )
2
Now β min corresponds to rB → R.
Then
1
1
2
+ = 2
R(1 + α ) R β min
R(1 + α )
β min =
or
2

2 +α
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458
PROBLEM 12.104
At main engine cutoff of its thirteenth flight, the space
shuttle Discovery was in an elliptic orbit of minimum
altitude 60 km and maximum altitude 500 km above the
surface of the earth. Knowing that at Point A the shuttle had
a velocity v0 parallel to the surface of the earth and that the
shuttle was transferred to a circular orbit as it passed through
Point B, determine (a) the speed v0 of the shuttle at A, (b) the
increase in speed required at B to insert the shuttle into the
circular orbit.
SOLUTION
For earth, R = 6370 km = 6370 × 103 m
GM = gR 2 = (9.81)(6370 × 103 ) 2 = 3.9806 × 1014 m3 /s 2
rA = 6370 + 60 = 6430 km = 6430 × 103 m
rB = 6370 + 500 = 6870 km = 6870 × 103 m
Elliptic trajectory.
Using Eq. (12.39),
1 GM
= 2 + C cos θ A
rA
h
But
θ B = θ A + 180°, so that cos θ A = − cos θ B
Adding,
1 GM
= 2 + C cos θ B .
rB
h
1 1 rA + rB 2GM
+ =
= 2
rA rB
rA rB
h
h=
(a)
and
2GMrA rB
(2)(3.9806 × 1014 )(6430 × 103 )(6870 × 103 )
=
= 51.422 × 109 m 2 /s
rA + rB
6430 × 103 + 6870 × 103
Speed v0 at A.
v0 = v A =
(vB )1 =
h 51.422 × 109
=
rA
6430 × 103
v0 = 8.00 × 103 m/s 
h 51.422 × 109
=
= 7.48497 × 103 m/s
3
rA
6870 × 10
For a circular orbit through Point B,
(vB )circ =
(b)
GM
3.9806 × 1014
=
= 7.6119 × 103 m/s
rB
6870 × 103
Increase in speed at Point B.
ΔvB = (vB )circ − (vB )1 = 126.97 m/s
ΔvB = 127 m/s 
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459
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 Δv A . 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
For Earth,
R = 3690 mi = 20.9088 × 106 ft, g = 32.2 ft/s 2
GM earth = gR 2 = (32.2)(20.9088 × 106 )2 = 14.077 × 1015 ft 3 /s 2
For Venus,
For a parabolic trajectory with
GM = 0.82GM earth = 11.543 × 1015 ft 3 /s 2
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
6
rA
49.104 × 10
rB = 190 × 103 mi = 1003.2 × 106 ft
At Point A, where θ = 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 θ = 0°
Adding,
r + rA 2GM
1
1
+
= B
= 2
rA rB
rA rB
hAB
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460
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
rB + rA
1052.3 × 106
(v A ) 2 =
hAB 1.039575 × 1012
=
= 21.174 × 103 ft/s
6
rA
49.104 × 10
(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 θ = 0
1
GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where θ = 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
Δv A = 509 ft/s 
(b)
At B:
(vB )1 − (vB ) 2 = 1.036 × 103 − 811.69
ΔvB = 224 ft/s 
(c)
At C:
(vC )1 − (vC )2 = 27.539 × 103 − 19.758 × 103 
ΔvC = 7.78 × 103 ft/s 
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461
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 m3 /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 = rAv A = (49.104 × 106 )(20, 000) = 982.08 × 109 ft 2 /s
At Point A, where θ = 180°
1
GM
GM
= 2 + C cos 180° = 2 − C
rA
hAB
hAB
At Point B, where θ = 0°
1
GM
GM
= 2 + C cos 0 = 2 + C
rB
hAB
hAB
Adding,
1
1
2GM
+
= 2
rA rB
hAB
1 2GM 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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462
PROBLEM 12.106 (Continued)
rC = 5600 mi = 29.568 × 106 ft
Second transfer orbit BC.
At Point B, where θ = 0
1
GM
GM
= 2 + C cos 0 = 2 + C
rB
hBC
hBC
At Point C, where θ = 180°
1
GM
GM
= 2 + C cos 180° = 2 − C
rC
hBC
hBC
Adding,
r + rC
1
1
2GM
+
= B
= 2
rB
rC
rB rC
hBC
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
6
rB
280.01 × 10
(vC )1 =
hBC 785.755 × 109
=
= 26.575 × 103 ft/s
6
rC
29.568 × 10
rC = 29.568 × 106 ft
Circular orbit with
(vC )2 =
(b)
GM
11.543 × 1015
=
= 19.758 × 103 ft/s
6
rC
29.568 × 10
Speed reductions at B and C.
At B:
(vB )1 − (vB ) 2 = 3.5073 × 103 − 2.8062 × 103
ΔvB = 701 ft/s 

At C:
(vC )1 − (vC )2 = 26.575 × 103 − 19.758 × 103
ΔvC = 6.82 × 103 ft/s 
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463
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 = rA v A
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
(rAv A )2
1/2
or
 665.6 g × 103 
R

v A = earth 
1

+ r1
rA 
r
A
B


1/ 2
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
v A = 52.4 × 103 ft/s 
or
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464
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
(
1 1 (rA v A ) rA + rB
+ =
r1 r2
h2
2
1
1
)
For the elliptic transfer orbit AB′, we have
r1 = rA , r2 = rB′ , h = htr = rA (v A ) tr
Then
(
2 1
1
1
1 ( rAv A ) rA + rB
+
=
rA rB′
[rA (v A ) tr ]2
1/2
or
 r1 + r1 
(v A ) tr = v A  1A 1B 
r +r 
 A B′ 
)
1/ 2
 1 + rA 
rB

= vA 
 1 + rA 
r
B′ 

1/2
 1 + 202

92
= (52, 431 ft/s) 
 1 + 202 
85.5 

= 51,113 ft/s
htr = ( hA ) tr = (hB′ ) tr : rA (v A ) 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′
2
Then
(
1
1 (rA v A ) rA + rB
+
=
rA′ rB′
(rB′vB′ )2
1
1
)
1/ 2
or
1
1
r  r +r 
vB′ = v A A  1A 1B 
rB′  r ′ + r ′ 
B 
 A
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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465
PROBLEM 12.107 (Continued)
Finally,
(v A ) tr = v A + Δv A
or
Δv A = (51,113 − 52, 431) ft/s
|Δv A | = 1318 ft/s 
or
and
vB′ = (vB′ ) tr + ΔvB
or
ΔvB′ = (116,862 − 120, 758) ft/s
= −3896 ft/s
|ΔvB | = 3900 ft/s 
or
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466
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
τH 
 aH 

 =

 τE 
 aE 
Thus
3
τ 
aH = aE  H 
 τE 
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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467
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 ε = 0.999887.
Knowing that for the 1996 sighting the minimum distance between the comet and the sun was 0.230 RE ,
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 =
2π RE 2π RE 3/ 2
GM
=
, τ0 =
RE
v0
GM
or
GM =
2π RE3/ 2
τ0
(1)
For the comet Hyakutake,
1 GM
= 2 = (1 + ε ),
r0
h
a=
1 GM
1+ ε
= 2 (1 + ε ), r1 =
r0
r1
1− ε
h
r
1
1+ ε
r0
(r0 + r1 ) = 0 , b = r0 r1 =
2
1− ε
1−ε
h = GMr0 (1 + ε )
τ=
=
2π r02 (1 + ε )1/ 2
2π ab
=
h
(1 − ε )3/ 2 GMr0 (1 + ε )
2π r03/ 2
GM (1 − ε )3/2
 r 
= 0 
 RE 
3/2
1
(1 − ε )3/2
= (0.230)3/2
Since
=
2π r03/ 2τ 0
2π RE3 (1 − ε )3/2
τ0
1
τ 0 = 91.8 × 103τ 0
(1 − 0.999887)3/ 2
τ 0 = 1 yr, τ = (91.8 × 103 )(1.000)
τ = 91.8 × 103 yr 
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468
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 cosθ A
rA
hAB
θ B = θ A + 180°,
Adding,
so that
and
1
GM
= 2 + C cosθ B .
rB
hAB
cosθ A = − cosθ 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 =
1
1
(rA + rB ) = (9.0 × 106 + 180 × 106 ) = 94.5 × 106 m
2
2
b=
rArB =
(9.0 × 106 )(180 × 106 ) = 40.249 × 106 m
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469
PROBLEM 12.110 (Continued)
Periodic time for full ellipse:
For half ellipse AB,
τ AB =
τ AB =
τ =
2π ab
h
1
π ab
τ =
2
h
π (94.5 × 106 )(40.249 × 106 )
27.085 × 10
9
= 444.81 × 103 s
τ AB = 122.6 h 
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470
PROBLEM 12.111
A space shuttle is in an elliptic orbit of eccentricity 0.0356 and a minimum altitude of 300 km above the
surface of the earth. Knowing that the radius of the earth is 6370 km, determine the periodic time for the orbit.
SOLUTION
For earth,
g = 9.81 m/s 2 ,
R = 6370 km = 6.370 × 106 m
GM = gR 2 = (9.81)(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2
For the orbit,
r0 = 6370 + 300 = 6670 km = 6.670 × 106 m
1
GM
= 2 (1 − ε )
r1
h
1
GM
= 2 (1 + ε )
r0
h
r1 = r0
1+ε
1.0356
− (6.670 × 106 )
= 7.1624 × 106 m
1−ε
0.9644
a =
1
(r0 + r1) = 6.9162 × 106 m
2
b=
r0r1 = 6.9118 × 106 m
h=
(1 + ε )GMr0 =
(1.0356)(398.06 × 1012 )(6.670 × 106 )
= 52.436400 × 109 m 2 /s
τ =
2π ab
2π (6.9118 × 106 )(6.9162 × 106 )
=
h
52.436400 × 109 m 2 /s
τ = 95.5 min 
= 5.7281 × 103 s
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471
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 θ A
rA
h
But
θ B = θ A + 180°, so that cos θ A = − cos θ B .
Adding,
and
1 GM
= 2 + C cos θ B .
rB
h
1
1 r +r
2GM
+ = A B = 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
a = ( rA + rB ) = 3.402 × 106 m
2
b = rA rB = 3.16145 × 106 m
Periodic time.
τ=
2π ab 2π (3.402 × 106 )(3.16145 × 106 )
=
= 17.831 × 103 s
hAB
3.78983 × 109
τ = 4.95 h 

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472
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 + ε cos θ )
r
hBA
[Eq. (12.39′)]
where ε = 1. Now
at A, θ = 0:
1 GM v
= 2 (1 + 1)
rA
hBA
or
rA =
at B, θ = −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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473
PROBLEM 12.113 (Continued)
Then
A=
2A
1
ht or tBA = BA
2
hBA
tBA =
=
2 × 43 rA2
rAv A
hBA = rAv 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,
π
r
2 A
(v A )circ
=
π 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
t BC = 50 min 55 s 
or
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474
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 β v0, where β < 1, to place the probe on
a crash trajectory. Express in terms of n and β 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 = β v0 =
β 2 GM
nR
h = rAv A = nRv A = β 2 nGMR
GM
1
= 2
h2
β nR
1 GM
1 + ε cos θ
= 2 (1 + ε cos θ ) =
r
β 2 nR
h
At Point A, θ = 180°
1
1
1− ε
=
= 2
rA nR β nR
or β 2 = 1 − ε
or ε = 1 − β 2
At impact Point B, θ = π − φ
1 1
=
rB R
1 1 + ε cos (π − φ ) 1 − ε cos φ
=
=
R
β 2 nR
β 2 nR
ε cos φ = 1 − nβ 2 or cos φ =
1 − nβ 2
ε
=
1 − nβ 2
1− β 2
φ = cos −1[(1 − nβ 2 ) /(1 − β 2 )] 
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475
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 Rφ of the trajectory is 6000 km, determine (a) the velocity
of the projectile at C, (b) the eccentricity ε 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
ϕ =
For an elliptic trajectory,
At A,
θ = 180° −
At C,
θ = 180° ,
ϕ
2
1 GM
= 2 (1 + ε cos θ )
r
h
1
GM
= 2 (1 + ε cos153.016°)
rA
h
= 153.016°,
1
GM
= 2 (1 − ε )
rC
h
(1)
(2)
Dividing Eq. (1) by Eq. (2),
rC
1 + ε cos153.016°
=
= 1.23548
rA
1−ε
ε =
1.23548 − 1
= 0.68384
1.23548 + cos153.016°
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476
PROBLEM 12.115 (Continued)
From Eq. (2), h =
h=
GM (1 − ε )rC
(398.06 × 1012 )(0.31616)(7.87 × 106 ) = 31.471 × 109 m 2 /s
(a) Velocity at C.
(b)
vC =
h
31.471 × 109
=
= 4.00 × 103 m/s
6
rC
7.87 × 10
vC = 4 km/s 
ε = 0.684 
Eccentricity of trajectory.
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477
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 + Δv = 7.5773 × 103 − 152 = 7.4253 × 103 m/s
h = rAv A = (6.933 × 106 )(7.4253 × 103 ) = 51.4795 × 109 m 2 /s
1 GM
= 2 (1 + ε cos θ )
r
h
At Point A,
θ = 180°,
r = rA
1
GM
= 2 (1 − ε )
rA
h
1−ε =
h2
(51.4795 × 109 )2
=
= 0.96028
GM rA
(398.06 × 1012 )(6.933 × 106 )
ε = 0.03972
1
GM
= 2 (1 + ε cos θ B )
rB
h
1 + ε cos θ B =
h2
(51.4795 × 109 ) 2
=
= 1.02567
GM rB
(398.06 × 1012 )(6.491 × 106 )
cosθ B =
θ B = 49.7°
1.02567 − 1
ε
= 0.6463
AOB = 180° − θ B = 130.3°
AOB = 130.3° 
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478
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 ε = 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
First we note
rB = (44.423 × 103 + 280) mi = 44.703 × 103 mi
(a)
We have
1 GM j
= 2 (1 + ε cos θ )
r
h
At A, θ = 0:
1 GM j
= 2 (1 + ε )
rA
h
or
At B, θ = θ B =  AOB :
h2
= rA (1 + ε )
GM j
1 GM j
= 2 (1 + ε cos θ B )
rB
h
or
h2
= rB (1 + ε cos θ B )
GM j
Then
rA (1 + ε ) = rB (1 + ε cos θ B )
or
[Eq. (12.39′)]
cos θ B =
=

1  rA
 (1 + ε ) − 1
ε  rB


1  44.0 × 103 mi
(1 + 1.031) − 1

3
1.031  44.703 × 10 mi

= 0.96902
or
θ B = 14.2988°
 AOB = 14.30° 
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479
PROBLEM 12.117 (Continued)
(b)
From above
where
Then
h 2 = GM j rB (1 + ε cos θ B )
1
| rB × mv B | = rB vB sin φ
m
φ = (θ B + 82.9°) = 97.1988°
h=
(rB vB sin φ )2 = GM j rB (1 + ε cos θ B )
or
1/2
vB =

1  GM j
(1 + ε cos θ B ) 

sin φ  rB

1/2
−9 4
4
26
1
 34.4 × 10 ft /lb ⋅ s × (1.30 × 10 slug)

=
× [1 + (1.031)(0.96902)]

6
sin 97.1988° 
236.03 × 10 ft

vB = 196.2 ft/s 
or
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480
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
ρ
=
1 1 1 
 + 
2  r0 r1 
SOLUTION
Using Eq. (12.39),
1 GM
= 2 + C cos θ A
rA
h
and
1 GM
= 2 + C cos θ B .
rB
h
But
θ B = θ A + 180°,
so that
cos θ A = − cos θ B
Adding,
1 1 2GM
+ = 2
rA rB
h
At Points A and B, the radial direction is normal to the path.
an =
But
Fn =
1
ρ
=
v2
ρ
=
h2
r 2ρ
GMm
mh 2
ma
=
=
n
r2
r 2ρ
GM 1  1 1 
=  + 
2  rA rB 
h2
1
ρ
=
1 1 1 
 +  
2  r0 r1 
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481
PROBLEM 12.119
(a) Express the eccentricity ε 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 + ε cos θ )
r
h
At A, θ = 0:
1 GM
= 2 (1 + ε )
r0
h
or
At B, θ = 180°:
Eq. (12.39′)
h2
= r0 (1 + ε )
GM
1 GM
= 2 (1 − ε )
r1
h
or
h2
= r1 (1 − ε )
GM
Then
r0 (1 + ε ) = r1 (1 − ε )
r − r0

r1 + r0
ε= 1
or
(b)
1+ ε
r0
1−ε
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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482
PROBLEM 12.120
Derive Kepler’s third law of planetary motion from Eqs. (12.39) and (12.45).
SOLUTION
For an ellipse,
2a = rA + rB
and b = rA rB
Using Eq. (12.39),
1 GM
= 2 + C cos θ A
rA
h
and
1 GM
= 2 + C cos θ B .
rB
h
But
θ B = θ A + 180°,
so that
cos θ A = − cos θ B .
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
h=b
GM
a
By Eq. (12.45),
τ=
2π ab 2π ab a 2π a3/ 2
=
=
h
b GM
GM
τ2 =
4π 2 a3
GM
For Orbits 1 and 2 about the same large mass,
and
τ12 =
4π 2 a13
GM
τ 22 =
4π 2 a23
GM
2
3
 τ1   a1 
  =  
 τ 2   a2 
Forming the ratio,
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483
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 ε about a planet of mass M can be expressed as
h = GMa(1 − ε 2 )
SOLUTION
By Eq. (12.39′),
1 GM
= 2 (1 + ε cos θ )
r
h
At A, θ = 0°:
1 GM
= 2 = (1 + ε )
rA
h
or
rA =
h2
GM (1 + ε )
At B, θ = 180°:
1 GM
= 2 = (1 − ε )
rB
h
or
rB =
h2
GM (1 − ε )
h2
1 
2h 2
 1
=
+
=

GM  1 + ε 1 − ε  GM (1 − ε 2 )
Adding,
rA + rB =
But for an ellipse,
rA + rB = 2a
2a =
2h2
GM (1 − ε 2 )
h = GMa(1 − ε 2 ) 
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484
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.
ΣFy = 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/s 2
2(s − s0 )
(2)(170)
For braking without skidding μ = μ s , so that μ s N = m | at |
ΣFt = mat : − μ s N = mat
μs = −
(b)
mat
a
31.001
= − t =
W
g
32.2
μ s = 0.963 
Stopping distance with skidding.
Use μ = μk = (0.80)(0.963) = 0.770
ΣF = mat : μk N = −mat
at = −
μk N
m
= − μ 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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485
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 θ = 30°.
SOLUTION
Initial velocity of drop = velocity of bucket
ΣFy = 0:
T cos 30° = mg
ΣFx = max :
Divide (2) by (1):
(1)
T sin 30° = man
tan 30° =
(2)
an
v2
=
g
pg
Thus
v 2 = ρ g tan 30°
But
ρ = 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 cos 30° = 2(1.2) − 1.2 cos 30° = 1.361 m
Thus
1.361 −
1
(9.81) t 2 = 0
2
t = 0.5275
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486
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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487
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:
ΣFx = max : mB aB /A − mB a A cos 30° = 0
aB/A = a A cos 30°
(1)
ΣFy = ma y : N AB − WB = − mB a A sin 30°
N AB = WB − (WB sin 30°)
Block A:
aA
g
(2)
ΣF = ma : WA sin 30° + N AB sin 30° = WA
WA sin 30° + WB sin 30° − (WB sin 2 30°)
aA =
aA
g
aA
a
= WA A
g
g
(WA + WB ) sin 30°
(30 + 12)sin 30°
g=
(32.2) = 20.49 ft/s 2
2
2
WA + WB sin 30°
30 + 12sin 30°
a A = 20.49 ft/s 2
(a)
30° 
aB/ A = (20.49) cos 30° = 17.75 ft/s 2
a B/A = 17.75 ft/s 2
(b)

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488
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.
ΣFx = mB ax : 0 = −mB ax + mB a A cos 25°
B:
aB/A = (1.2 ft/s 2 ) cos 25°
or
a B/A = 1.088 ft/s 2
or
(b)

For crate B
ΣFy = 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
ΣFx = m A 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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489
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 (μ 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.
ρ = 30 m
an =
v2
ρ
=
(20) 2
= 13.333 m/s 2
30
ΣFy = ma y : N R + N F − mg = man
N R + N F = m( g + an )
F = μk ( N R + N F ) = μ k m( g + an )
ΣFx = max : − F = mat
at = −
F
= − μ k ( g + an )
m
| at | = μk ( g + an ) = 0.25(9.81 + 13.33)
(b)
Between A and B.
| at | = 5.79 m/s 2 
ρ =∞
an = 0
| at | = μk g = (0.25)(9.81)
(c)
Just passed Point B.
| at | = 2.45 m/s 2 
ρ = 45 m
an =
v2
ρ
=
(20)2
= 8.8889 m/s 2
45
ΣFy = ma y : N R + N F − mg = − man
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490
PROBLEM 12.126 (Continued)
or
N R + N F = m ( g − an )
F = μ k ( N R + N F ) = μ k m ( g − an )
ΣFx = max : − F = mat
at = −
F
= − μ k ( g − an )
m
| at | = μk ( g − an ) = (0.25)(9.81 − 8.8889)
| at | = 0.230 m/s 2 
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491
PROBLEM 12.127
The 100-g pin B 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 θ0 = 12 rad/s, determine for any given value of θ
(a) the radial and transverse components of the resultant force F
exerted on pin B, (b) the forces P and Q exerted on pin B by
rod OC and the wall of slot DE, respectively.
SOLUTION
Kinematics
From the drawing of the system, we have
r=
Then
0.2
m
cos θ
sin θ  

r =  0.2
θ  m/s
2
cos
θ 

θ = 12 rad/s
θ = 0
and
cos θ (cos θ ) − sin θ (−2cos θ sin θ )  2
θ
cos 4 θ
 1 + sin 2θ  2 
=  0.2
θ  m/s 2
3
cos
θ


2

r = 0.2
Substituting for θ
Now
r = 0.2

sin θ
sin θ 
(12) =  2.4
 m/s
2
cos θ
cos 2 θ 

r = 0.2

1 + sin 2 θ
1 + sin 2 θ 
2
(12)2 =  28.8
 m/s
3
3
cos θ
cos θ 

ar = 
r − rθ 2

1 + sin 2 θ   0.2 
2
=  28.8
 − 
 (12)
3
θ
cos
cos θ  


2

sin θ 
2
=  57.6
3 
 m/s
θ
cos


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492
PROBLEM 12.127 (Continued)
aθ = rθ + 2rθ
and

sin θ 
= 0 + 2  2.4
 (12)
cos 2 θ 


sin θ 
2
=  57.6
2 
 m/s
θ
cos


Kinetics
(a)
We have


sin 2 θ 
Fr = mB ar = (0.1 kg)  57.6
m/s 2 
3 

cos θ 


Fr = (5.76 N) tan 2 θ sec θ 
or
and


sin θ 
Fθ = mB aθ = (0.1 kg)  57.6
m/s 2 
2 

cos θ 


Fθ = (5.76 N) tan θ sec θ 
or
(b)
Now
ΣFy : Fθ cos θ + Fr sin θ = P cos θ
or
P = 5.76 tan θ sec θ + (5.76 tan 2 θ sec θ ) tan θ
P = (5.76 N) tan θ sec2 θ
or
θ 
ΣFr : Fr = Q cos θ
or
Q = (5.76 tan 2 θ sec θ )
1
cos θ
Q = (5.76 N) tan 2 θ sec2 θ
or

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493
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) θ = 90°, (b) θ = 75°, (c) θ = 45°. Indicate in
each case the direction of the impending motion.
SOLUTION
vC = (r sin θ )φAB
First note
= (0.6 m)(6 rad/s)sin θ
= (3.6 m/s)sin θ
(a)
With θ = 90°,
vC = 3.6 m/s
ΣFy = 0: F − WC = 0
or
F = mC g
Now
F = μs N
or
N=
1
μs
ΣFn = mC an : N = mC
1
or
or
μs
mC g = mC
μs =
mC g
vC2
r
vC2
r
gr (9.81 m/s 2 )(0.6 m)
=
vC2
(3.6 m/s) 2
( μ s ) min = 0.454 
or
The direction of the impending motion is downward. 
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494
PROBLEM 12.128 (Continued)
(b) and (c)
First observe that for an arbitrary value of θ, it is not known whether the impending motion will be upward or
downward. To consider both possibilities for each value of θ, let Fdown correspond to impending motion
downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown,
we have
ΣFy = 0: N cos θ ± F sin θ − WC = 0
F = μs N
Now
N cos θ ± μ s N sin θ − mC g = 0
Then
or
N=
mC g
cos θ ± μs sin θ
and
F=
μ s mC g
cos θ ± μ s sin θ
ΣFn = mC an : N sin θ  F cos θ = mC
vC2
ρ
ρ = r sin θ
Substituting for N and F
mC g
μs mC g
v2
sin θ 
cos θ = mC C
cos θ ± μs sin θ
cos θ ± μ s sin θ
r sin θ
μs
vC2
tan θ

=
1 ± μs tan θ 1 ± μs tan θ gr sin θ
or
v2
μs = ±
or
v2
C
1 + gr sin
θ tan θ
vC2
[(3.6 m/s) sin θ ]2
=
= 2.2018sin θ
gr sin θ (9.81 m/s 2 )(0.6 m)sin θ
Now
Then
(b)
C
tan θ − gr sin
θ
μs = ±
tan θ − 2.2018 sin θ
1 + 2.2018sin θ tan θ
μs = ±
tan 75° − 2.2018sin 75°
= ± 0.1796
1 + 2.2018sin 75° tan 75°
θ = 75°
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495
PROBLEM 12.128 (Continued)
Then
downward:
μs = + 0.1796
upward:
μs < 0
not possible
( μ s )min = 0.1796 
The direction of the impending motion is downward. 
(c)
θ = 45°
μs = ±
tan 45° − 2.2018sin 45°
= ± (− 0.218)
1 + 2.2018sin 45° tan 45°
Then
downward:
μs < 0
upward:
μs = 0.218
not possible
( μ s ) min = 0.218 
The direction of the impending motion is upward. 
Note: When
or
tan θ − 2.2018sin θ = 0
θ = 62.988°,
μs = 0. Thus, for this value of θ , friction is not necessary to prevent the collar from sliding on the rod.
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496
PROBLEM 12.129
Telemetry technology is used to quantify kinematic values of a 200-kg
roller coaster cart as it passes overhead. According to the system,
r = 25 m, r = −10 m/s, r = −2 m/s 2 , θ = 90°, θ = −0.4 rad/s, θ =
−0.32 rad/s 2. 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 = rθ = (25 m)( − 0.4 rad/s) = −10 m/s
θ
So the tangential direction is
45° and v = 10 2 m/s.
ar = 
r − rθ 2 = −2 m/s − (25 m)( − 0.4 rad/s)2
= −6 m/s 2
aθ = rθ + 2rθ
= (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 =
ρ =
v2
ρ
v2
(10 2)2
=
an
6 cos 45°
ρ = 47.1 m 
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497
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 ρ the mean density of the planet, show that the time required by the moon to complete one full revolution
about the planet is (24 π /G ρ )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 τ be the periodic time to complete one orbit.
vτ = 2π r
τ
GM
= 2π r
r
hence,
GM = 2
or
2π r 3/2
GM
Solving for τ,
τ=
But
4
M = π R3 ρ ,
3
Then
τ=
3π  r 
G ρ  R 
π
3
G ρ R3/2
3/2
Using r = 2R as a given leads to
τ = 23/2
3π
=
Gρ
24π
Gρ
τ = (24π /G ρ )1/2 
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498
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
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
(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 θ A
rA
h
and
1 GM
= 2 + C cos θ B
rB
rB
But θ B = θ A + 180°, so that cos θ A = − cos θ B
Adding,
1 1 rA + rB 2a 2GM
+ =
= 2 = 2
rA rB
rA rB
b
h
or
h=
Periodic time.
τ=
τ=
GMb 2
a
2π ab 2π ab a 2π a3/ 2
=
=
h
GM
GMb2
2π (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
τ AB = 0.7334 h
τ AB = 44.0 min 
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499
PROBLEM 12.131 (Continued)
(b)
For the circular orbit,
a = b = rB = 21.8064 × 106 ft
τ circ =
2π a3/2
GM
=
2π (21.8064 × 106 )3/2
14.077 × 1015
τ circ = 1.498 h
= 5393 s
τ circ = 89.9 min 
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500
PROBLEM 12.132
It was observed that as the Galileo spacecraft reached the point on its trajectory closest to Io, a moon of the
planet Jupiter, it was at a distance of 1750 mi from the center of Io and had a velocity of 49.4 × 103 ft/s.
Knowing that the mass of Io is 0.01496 times the mass of the earth, determine the eccentricity of the trajectory
of the spacecraft as it approached Io.
SOLUTION
First note
r0 = 1750 mi = 9.24 × 106 ft
Rearth = 3960 mi = 20.9088 × 106 ft
We have
1 GM
= 2 (1 + ε cos θ )
r
h
At Point O,
r = r0 , θ = 0, h = h0 = r0 v0
Also,
Eq. (12.39)
GM Io = G (0.01496 M earth )
2
= 0.01496gRearth
using Eq. (12.30).
Then
2
1 0.01496 gRearth
=
(1 + ε )
r0
(r0 v0 ) 2
or
ε=
=
r0 v02
2
0.01496 gRearth
−1
(9.24 × 106 ft)(49.4 × 103 ft/s) 2
−1
0.01496(32.2 ft/s 2 )(20.9088 × 106 ft) 2
ε = 106.1 
or
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501
PROBLEM 12.133*
Disk A rotates in a horizontal plane about a vertical axis at the
constant rate θ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
θ = θ0 = 12 rad/s
then
θ = 0
ΣFr = mB ar : − Fsp = mB (
r − rθ02 )
 k


− θ02  r = 0
r +
 mB

ΣFθ = mB aθ : FA = mB (0 + 2rθ0 )
(a)
(1)
(2)
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) dt
0
r = 0
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502
PROBLEM 12.133* (Continued)
and
dr
= r = 0 and at t = 0, r0 = 0.5 m
dt

r
r0
dr =
0.1
 (0) dt
0
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
r = vr
d dr d
d
=
= vr
dt dt dr
dr
dvr
dr
dvr
+ 100r = 0
dr
vr
vr
r
 v d v = −100 r dr
At t = 0, vr = 0, r = r0 :
0
r
r
r0
vr2 = −100(r 2 − r02 )
vr = 10 r02 − r 2
vr =
Now
At t = 0, r = r0 :

r
dr
r0
r02 − r 2
=
dr
= 10 r02 − r 2
dt
t
 10 dt = 10t
0
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503
PROBLEM 12.133* (Continued)
r = r0 sin φ ,
Let
sin −1 ( r/r0 )
r0 cos φ dφ
/2
r02 − r02 sin 2 φ
π
Then
dr = r0 cos φ dφ
sin −1 ( r/r0 )
π
/2
= 10t
dφ = 10t
r π
sin −1   − = 10 t
 r0  2
π

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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504
CHAPTER 13
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 μ
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
Answer: (e)
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507
PROBLEM 13.1
A 400-kg satellite was placed in a circular orbit 1500 km above the surface of the earth. At this elevation the
acceleration of gravity is 6.43 m/s2. Determine the kinetic energy of the satellite, knowing that its orbital
speed is 25.6 × 103 km/h.
SOLUTION
Mass of satellite:
m = 400 kg
Velocity:
v = 25.6 × 103 km/h = 7.111 × 103 m/s
Kinetic energy:
T=
1 2 1
mv = (400 kg)(7.111 × 103 m/s) 2
2
2
T = 10.113 × 109 J
T = 10.11 GJ 
Note: Acceleration of gravity has no effect on the mass of the satellite.
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508
PROBLEM 13.2
A 1-lb stone is dropped down the “bottomless pit” at
Carlsbad Caverns and strikes the ground with a speed of
95ft/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
g
32.2 ft/s 2
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/s 2
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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509
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
2
g
32.2 ft/s
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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510
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=
2π r 2π (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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511
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/s 2 )(1.7157 ft) = 110.49 ft 2 /s 2
v = 10.51 ft/s 
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512
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 
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513
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
μs = 0.75
Front wheel drive:
Maximum friction force without slipping:
F = μ 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/s 2 )(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
μs = 0.75
Rear wheel drive:
Maximum friction force without slipping:
F = μ 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.
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514
PROBLEM 13.8
Skid marks on a drag racetrack indicate that the rear (drive)
wheels of a car slip for the first 20 m of the 400-m track.
(a) Knowing that the coefficient of kinetic friction is 0.60,
determine the speed of the car at the end of the first 20-m
portion of the track if it starts from rest and the front wheels
are just off the ground. (b) What is the maximum
theoretical speed for the car at the finish line if, after
skidding for 20 m, it is driven without the wheels slipping
for the remainder of the race? Assume that while the car is
rolling without slipping, 60 percent of the weight of the car
is on the rear wheels and the coefficient of static friction is
0.75. Ignore air resistance and rolling resistance.
SOLUTION
(a)
For the first 20 m, the normal force at the real wheels is equal to the weight of the car. Since the wheels
are skidding, the friction force is
F = μ k N = μk W = μ k mg
Principle of work and energy:
T1 + U1→ 2 = T2
1 2
mv2
2
1
0 + μk mgd = mv22
2
2
v2 = 2 μk gd = (2)(0.6)(9.81 m/s2 )(20 m) = 235.44 m 2 /s 2
0 + Fd =

(b)
v2 = 15.34 m/s 

Assume that for the remainder of the race, sliding is impending and N = 0.6 W
F = μ s N = μ s (0.6W ) = (0.75)(0.6 mg ) = 0.45 mg
Principle of work and energy:
T2 + U 2→3 = T3
1 2
1
mv2 + (0.45mg )d ′ = mv32
2
2
v32 = v22 (2)(0.45) gd ′
= 235.44 m 2 /s 2 + (2)(0.45)(9.81 m/s 2 )(400 m − 20 m)
= 3590.5 m 2 /s 2
v3 = 59.9 m/s 
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515
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
U A−B = ( −W sin15° − F )d
F = μk N = 0.12 N
TA =
ΣF = 0 N − W cos15° = 0 N = W cos15°
U A−B = −W (sin15° + 0.12 cos15°) d = −Wd (0.3747)
TA + U A−B = TB : 32
W
− Wd (0.3743) = 0
g
d=
(b)
32
(9.81)(0.3747)
d = 8.71 m 
Down the plane from B to A: (F reverses direction)
1W 2
vA
TB = 0
2 g
U B−A = (W sin15° − F )d
TA =
d = 8.71 m/s
= W (sin15° − 0.12cos15°)(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
15° 
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516
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
(b)
T2 =
1 2
mv = U1→2 = 169.0 N ⋅ m
2
v2 =
2U1→2
(2) (169.0)
=
m
1.4
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/s 2 )(27.3 m) = 535.6 m 2 /s 2
v4 = 23.1 m/s 
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517
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 μk = 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 = μk N AB = 0.25 mg cos 30°
U A→ B = mgd sin 30° − FAB d
= mgd (sin 30° − μk cos 30°)
N BC = mg
On level surface BC:
xBC = 7 m
FBC = μk mg
U B →C = − μk mg xBC
At A,
TA =
1 2
mv A
2
and v A = 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
mv A + mgd (sin 30° − μ k cos 30°) − μ k mg xBC = mv02
2
2
Dividing by m and solving for d,
vC2 /2 g + μ k xBC − v A2 /2 g 

d=
(sin 30° − μk cos 30°)
=
(2) 2/(2)(9.81) + (0.25)(7) − (1)2/(2)(9.81)
sin 30° − 0.25cos 30°
d = 6.71 m 
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518
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 μk = 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 = μk N AB = 0.25 mg cos 30°
UA→ B = mgd sin 30° − FAB d
= mgd (sin 30° − μ k cos 30°)
On level surface BC:
N BC = mg
xBC = 7 m
FBC = μ k mg
U B →C = − μk mg xBC
At A,
TA =
1 2
mv A
2
and v A = 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
mv A + mgd (sin 30° − μ k cos 30°) − μ k mg xBC = mv02
2
2
Solving for vC2 ,
vC2 = v 2A + 2 gd (sin 30° − μk cos 30°) − 2 μk g xBC
= (1) 2 + (2)(9.81)(7.5)(sin 30° − 0.25cos 30°) − (2)(0.25)(9.81)(7)
= 8.3811 m 2/s 2
(b)
vC = 2.90 m/s 
Box on belt: Let xbelt be the distance moves by a package as it slides on the belt.
ΣFy = ma y
N − mg = 0 N = mg
Fx = μk N = μk mg
At the end of sliding,
v = vbelt = 2 m/s
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519
PROBLEM 13.12 (Continued)
Principle of work and energy:
1 2
1 2
mvC − μ k mg xbelt = mvbelt
2
2
2
vC2 − vbelt
xbelt =
2 μk g
=
8.3811 − (2) 2
(2)(0.25)(9.81)
xbelt = 0.893 m 
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520
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 μk = 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.
ΣFy = 0: N − W cos15° = 0
N = W cos15°
Box is sliding on AB.
F f = μk N = μkW cos15°
Distance
AB = d = 20 ft
Work of gravity force:
(U A− B ) g = Wd sin15°
− F f d = − μkWd cos15°
Work of friction force:
Total work
U A→ B = Wd (sin15° − μ 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° − μk cos15°) =
vB
2 g
2 g
v02 = vB2 − 2 gd (sin15° − μk cos15°)
= (8) 2 − (2)(32.2)(20)[sin15° − (0.40)(cos15°)]
= 228.29 ft 2 /s 2
v 0 = 15.11 ft/s
15° 
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521
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 μk = 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.
ΣFy = 0: N − W cos15° = 0
N = W cos15°
Box is sliding on AB.
F f = μk N = μkW cos15°
Distance
AB = d = 20 ft
Work of gravity force:
(U A− B ) g = Wd sin15°
− F f d = − μkWd cos15°
Work of friction force:
Total work
U A− B = Wd (sin15° − μk 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° − μk cos15°) =
vB
2 g
2 g
v02 = vB2 − 2 gd (sin15° − μk cos15°)
= 0 − (2)(32.2)(20)[sin15° − (0.40)(cos15°)]
= 164.29 ft 2 /s 2
v 0 = 12.81 ft/s
15° 
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522
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
U1→ 2 = −( FC + FT )d
T1 =
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
U1→ 2 = ( H − FC )d
T1 =
T2 = 0
T1 + U1→ 2 = T2
1
mC v12 + ( H − FC ) d = 0
2
H = FC −
mC v12
(1400)(20) 2
= 5000 −
2d
(2)(57.778)
H = 154 N
Trailer hitch force on car:

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523
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 

300
d  2
2
2


= 7.4126 × 103 N
(b)
Ft = 7.41 kN 
Coupling force Fc . Use the trailer alone as a free body.
m = 5400 kg
W = mg = (5400)(9.81) = 52.974 × 103 N
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 


2
d 2
2

 300  2
= 5.5595 × 103 N
Fc = 5.56 kN (tension) 
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524
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
μk = 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) 
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525
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 = μ 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
2
32.2 ft/s
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) 
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526
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
W
10
MB = B =
= 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
mA v 2 = (0.7764)(9)2 = 31.444 ft ⋅ lb
2
2
1
1
(T2 ) B = mB v 2 = (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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527
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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528
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 − 4 F = 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 /s 2
(b)
Case b:
vB = 7.43 ft/s 
d = ?, vB = 0.
30d − 24 = 0
d = 0.800 ft 
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529
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.
μk = 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 = 0.8 N A
= 2600(9.81)sin 5°
For system: A + B
U1− 2 = [( F1 − 1400 g sin 5° − 1200 g sin 5°) − F ]d
= T2 − T1 = 0 −
Since
1
1
mA + Bv 2 = − (2600)(10) 2
2
2
( F1 − 1400 g 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) cos 5°)11.88 = −
1400
(10) 2
2
(T − 10945) = −5892
= 5.053 kN
T = 5.05 kN 
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530
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
Δx A + 3ΔyB = 0
v A + 3vB = 0
Let F be the tension in the cable.
Block A:
m A = 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 )(Δx 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 )(−ΔyB ) =
2 F − 163.5 = 12.5 vB2
(2)
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531
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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532
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 μ s = 0.25 and μ 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:
3F − mB g = 0
F=
Block A:
mB g (25)(9.81)
=
= 81.75 N
3
3
ΣFy = 0:
N A − mA g = 0
N A = m A g = (30)(9.81) = 294.3 N
ΣFx = 0: 250 − FA − F = 0
FA = 250 − 81.75 = 168.25 N
μs N A = (0.25)(294.3) = 73.57 N
Available static friction force:
Since FA > μs N A , the blocks move.
The friction force, FA, during sliding is
FA = μk N A = (0.20)(294.3) = 58.86 N
Constraint of cable:
x A + 3 yB = constant
Δ x A + 3ΔyB = 0
v A + 3vB = 0
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533
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 2A
2
1
0 + (250 − 58.86 − F )(2) = (30)(3vB ) 2
2
0 + ( P − FA − F )(Δx 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 )(−ΔyB ) =
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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534
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
( mA + mB + mC )v22 = (12 kg)v22 = 6v22
2
2
U1− 2 = (m A + mC − mB ) g (0.9 m)
T2 =
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 .
T3 =
T2′ =
1
9 
(m A + mB )v22 =  kg  (2.943) = 13.244 J
2
2 
1
9
( mA + mB )(v3 ) 2 = v32
2
2
U 2′−3 = ( mA − 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
v A = 1.190 m/s 
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535
PROBLEM 13.25
Four packages, each weighing 6 lb, 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 speed of package 2 as it
leaves the belt at A, (b) the speed of
package 3 as it leaves the belt at A. Neglect
the mass of the belt and rollers.
SOLUTION
Slope angle:
(a)
sin β =
6 ft
15 ft
β = 23.6°
Package falls off the belt and 2, 3, 4 move down
6
= 2 ft.
3
1
 3  6 lb  2
T2 = 3  mv22  = 
v = 0.2795v22
2  2
2
 2  32.2 ft/s 
U1− 2 = (3)(W )( R) = (3)(6 lb)(2 ft) = 36 lb ⋅ ft
T1 + U1− 2 = T2
0 + 36 = 0.2795 v22
v22 = 128.8
v 2 = 11.35 ft/s
(b)



23.6° 
Package 2 falls off the belt and its energy is lost to the system and 3 and 4 move down 2 ft.
1
  6 lb 
(128.8)
T2′ = (2)  m v22  = 
2 
2
  32 ft/s 

T2′ = 24 lb ⋅ ft
1
  6 lb  2
(v3 ) = 0.18634v32
T3 = (2)  m v32  = 
2 

2
  32.2 ft/s 
U 2−3 = (2)(W )(2) = (2)(6 lb)(2 ft) = 24 lb ⋅ ft
T2 + U 2−3 = T3
24 + 24 = 0.18634v32
v32 = 257.6 

v3 = 16.05 ft/s
23.6° 
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536
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)
m A = 2 kg, mB = 3 kg
Position 1: Block B has just been removed.
Spring force:
FS = −(m A + mB ) g = −k x
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 while the spring still contacts block A.
Work of the force exerted by the spring:
(U1→2 )e = −
x2
 k x dx
x1
x
2
1
1
1
= − k x 2 = k x12 − k x22
2
2
2
x1
=
Work of the gravitational force:
1
1
(40)(−1.22625) 2 − (40) x22 = 30.074 − 20 x22
2
2
(U1→2 ) g = − mA 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 Av22 = (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
(1)
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537
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 m 2 /s 2
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
1
1
 kxdx = 2 kx = 2 (40)(1.22625) = 30.074 J
x1
2
1
2
Work of the gravitational force:
(U1→3 ) g = − m A 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
h = 1.533 m 
Maximum height:
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538
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)
m A = 2 kg, mB = 3 kg
Position 1: Block B has just been removed.
Spring force:
FS = −(m A + 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
 kxdx
x1
x
2
1
1
1
= − kx 2 = kx12 − kx22
2
2
2
x1
=
Work of the gravitational force:
1
1
(40)(−1.22625) 2 − (40) x22 = 30.074 − 20 x22
2
2
(U1→2 ) g = − mA 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 Av22 = (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
(1)
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539
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 /s 2
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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540
PROBLEM 13.28
An 8-lb collar C slides on a horizontal rod between springs A and B.
If the collar is pushed to the right until spring B is compressed 2 in.
and released, determine the distance through which the collar will
travel assuming (a) no friction between the collar and the rod,
(b) a coefficient of friction μk = 0.35.
SOLUTION
k B = 144 lb/ft
(a)
k A = 216 lb/ft
Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A.
TA = 0
U A− B =

TB = 0
2/12
0
k B xdx −
y
 k xdx
0
A
2
 144 lb/ft  2   216 lb/ft  2
U A− B = 
 ft  − 
 ( y)
2
2

 12  

TA + U A− B = TB :
0 + 2 − 108 y 2 = 0
y = 0.1361 ft = 1.633 in.
Total distance
d = 2 + 16 − (6 − 1.633)
d = 13.63 in. 
(b)
Assume that C does not reach the spring at B because of friction.
N = W = 6 lb
F f = (0.35)(8 lb) = 2.80 lb
TA = TD = 0
U A− D =

2/12
0
TA + U A− D = TD
144 × dx − F f ( y ) = 2 − 2.80 y
0 + 2 − 2.80 y = 0
y = 0.714 ft = 8.57 in.
The collar must travel 16 − 6 + 2 = 12 in. before it engages the spring at B. Since y = 8.57 in., it stops
before engaging the spring at B.
d = 8.57 in. 
Total distance
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541
PROBLEM 13.29
A 6-lb block is attached to a cable and to a spring as shown. The constant of the spring is
k = 8 lb/in. and the tension in the cable is 3 lb. If the cable is cut, determine (a) the
maximum displacement of the block, (b) the maximum speed of the block.
SOLUTION
k = 8 lb/in. = 96 lb/ft
ΣFy = 0: ( Fs )1 = 6 − 3 = 3 lb C
v1 = 0 T1 = 0: T2 =
For weight:
U1− 2 = (6 lb)x = 6 x
For spring:
U1− 2 = −
1  6 lb  2
2

 v2 = 0.09317v2
2  32.2 
x
 (3 + 96 x)dx = −3x − 48x
2
0
T1 + U1− 2 = T2 : 0 + 6 x − 3x − 48 x 2 = 0.09317v22
3x − 48 x 2 = 0.09317v22
(a)
For xm , v2 = 0:
(1)
3x − 48 x 2 = 0
x = 0,
xm =
3
1
= ft
48 16
x m = 0.75 in. 
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542
PROBLEM 13.29 (Continued)
(b)
For vm we see maximum of
U1− 2 = 3x − 48 x 2
dU1− 2
= 3 − 96 x = 0
dx
x=
3
1
ft =
ft
96
32
2
Eq. (1):
 1 
 1 
3
ft  − 48 
ft  = 0.09317vm2
32
32




vm2 = 0.5031
vm = 0.7093 ft/s
vm = 8.51 in./s  
Note: U1− 2 for the spring may be computed using F6 − x curve
U1− 2 = area
1
= 3x + 96x 2
2
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543
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 ( x A ) 2 − k B ( xB ) 2
2
2
(Gravity) (Spring A) (Spring B)
U1− 2 = W ( x A ) −
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
1
1  x
1
U1− 2 = W ( x) − 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

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544
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 
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545
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
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546
PROBLEM 13.31 (Continued)
Work of the force exerted by the lower spring:
(U1→2 )1 =
xf
 ( F − k x)dx
0
1
= F1 x f −
=
1
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 
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547
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=
pL =
pa
x
pR =
pa
(2a − x)
a
2
T1 = 0
At ,
x = a, T2 =
U1− 2 =
1 2
mv
2
a
a
1
1

 ( p − p ) Adx =  paA  x − 2a − x  dx
a/2
L
R
a/2
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
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548
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 
2
mv1 
 (95.3 ft/s)
2
2  32.2 ft/s 2 
T1 = 317,530 lb ⋅ ft
T1 =
= 317.63 k ⋅ ft
v2 = 0
T2 = 0
U1− 2 = (18 k)(5 ft) + (27 k)( d − 5)
= 90 + 27d − 135
= 27d − 45 k ⋅ ft
T1 + U1− 2 = T2
317.53 = 27d − 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) = 
(a )
2  D
 32.2 ft/s 
aD = 386 ft/s 2 
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549
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
Weight:
W1 = (90 ton)(2000 lb/ton) = 180 × 103 lb
Mass:
m=
Speed:
v1 = 1 mi/h =
Kinetic energy:
T1 =
W 180 × 103
=
= 5590 lb ⋅ s 2 /ft
g
32.2
5280 ft
= 1.4667 ft/s
3600 s
1 2 1
mv1 = (5590)(1.4667) 2
2
2
= 6012 ft ⋅ lb
T2 = 0
Principle of work and energy:
(rest)
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 =
k=
x
x
0
0
Fmax 60
=
= 5 kip/in.
xmax 12
1
 fdx =  kx dx = 2 kx
2
1
(5) x 2 = 72.51
2
x 2 = 28.86 in.2
Fender B: We divide area under curve B into trapezoids
Partial area
x = 5.37 in. 
Total Area
From x = 0
to x = 2 in.:
1
(2 in.)(4 kips) = 4 kip ⋅ in.
2
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.
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550
PROBLEM 13.34 (Continued)
We still need ΔU = 72.15 − 46 = 26.15 kip ⋅ in.
Equation of straight line approximating curve B from x = 6 in. to x = 8 in. is
Δx F − 18
=
F = 18 + 6Δx
2 30 − 18
1
ΔU = 18Δx + (6Δx)Δx = 26.15 kip ⋅ in.
2
2
(Δx) + 6Δx − 8.716 = 0
Δx = 1.209 in.
Thus:
x = 6 in. + 1.209 in. = 7.209 in.
x = 7.21 in. 
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551
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
 Fdx = 0
For F = kx = (300 N/m) x
Eq. (1):
49.05 xm −
xm
 3000 x dx = 0
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)
0
For
Eq. (1)
F = (3000 N/m) x(1 + 160 x 2 )
49.05 xm −
xm
 3000( x + 160 x )dx = 0
3
0
1

49.05 xm − 3000  xm2 + 40 xm4  = 0
2


Solve by trial:
Fm = 98.1 N 
(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 
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552
PROBLEM 13.36
A rocket is fired vertically from the surface of the moon with a speed v0 . Derive a formula for the ratio hn /hu
of heights reached with a speed v, if Newton’s law of gravitation is used to calculate hn and a uniform
gravitational field is used to calculate hu . Express your answer in terms of the acceleration of gravity g m on
the surface of the moon, the radius Rm of the moon, and the speeds v and v0 .
SOLUTION
Newton’s law of gravitation
T1 =
U1− 2 =
1 2
mv0
2

Rm + hn
Rm
T2 =
1 2
mv
2
( − Fn )dr
U1− 2 = − mg m Rm2

Fn =
mg m Rm2
r2
Rm + hn dr
r2
Rm
 1

1
U1− 2 = mg m Rm2 
−

 Rm Rm + hn 
T1 + U1− 2 = T2

Rm  1 2
1 2
mv0 + mg m  Rm −
 = mv
Rm + hn  2
2

hn =
( v − v ) 
2
0
2
Rm
R − (
 m
2 gm
v02 − v 2
2 gm

) 

(1)
Uniform gravitational field
T1 =
U1− 2 =
1 2
mv0 T2 = mv 2
2

Rm + hn
Rm
( − Fu )dr = −mg m ( Rm + hu − Rm ) = −mghu
1 2
1
mv0 − mg m hu = mv 2
2
2
T1 + U1− 2 = T2
(v − v )
h =
2
0
u
2
(2)
2 gm
hn
=
hu
Dividing (1) by (2)
1
( v −v )
1−
2
0
2

(2 g m Rm )
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553
PROBLEM 13.37
Express the acceleration of gravity g h 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 earth is used as its weight at an altitude of
(a) 1 km, (b) 1000 km.
SOLUTION
F=
At earth’s surface, (h = 0)
GM E m / R 2
(h + R )
( Rh + 1)
=
2
2
mg h
GM E m
= mg0
R2
GM E
R
2
GM E
= g0
gh =
Thus,
GM E m
gh =
R2
h 
 +1 
R 
2
g0
h 
 +1
R 
2
R = 6370 km
At altitude h, “true” weight
F = mg h = WT
Assumed weight
W0 = mg0
Error = E =
W0 − WT mg0 − mg h g0 − g h
=
=
W0
mg0
g0
g0 −
gh =
g0
( Rh + 1)
2
E=
g0
(1 + ) = 1 −
h
R
2
g0

1
(
1 + Rh

2
) 
(a)
h = 1 km:
1


P = 100E = 100 1 −
2
1
 (1 + 6370 ) 
P = 0.0314% 
(b)
h = 1000 km:
1


P = 100E = 100 1 −

1000 2
 (1 + 6370 ) 
P = 25.3% 
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554
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
U1− 2 = −mge he
Earth
U1− 2 = −mg m hm
Moon
T1 =
Earth
1 2
1
mv − mge he = mvH2
2
2
Moon
1 2
1
mv − mg m hm = mvH2
2
2
1 2
mvH
2
hm g e
=
he g m
− ge he + g m hm = 0
Subtracting
T2 =
 ge 
hm = (60 m) 

 0.165 ge 
hm = 364 m 
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555
PROBLEM 13.39
The sphere at A is given a downward velocity v 0 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 θ 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
U1− 2 = mg (l ) sin θ
T2 =
T1 + U1− 2 = T2
12.5m + 2mg sin θ =
25 + 4 g sin θ = v 2
1 2
mv
2
(1)
Newton’s law at .
v2
v2
=m

2
2
v = 4 g − 2 g sin θ
2mg − mg sin θ = m
(2)
Substitute for v 2 from Eq. (2) into Eq. (1)
25 + 4 g sin θ = 4 g − 2 g sin θ
(4)(9.81) − 25
sin θ =
= 0.2419
(6)(9.81)
θ = 14.00° 
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556
PROBLEM 13.40
The sphere at A is given a downward velocity v 0 and swings
in a vertical circle of radius l and center O. Determine the
smallest velocity v 0 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
v02 = v 2 + 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 = 3gl 
Force in the rod can support the weight so that v can be zero.
v02 = 0 + 2 gl
Thus,
v0 = 2 gl 
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557
PROBLEM 13.41
A small sphere B of weight W is released from rest in the position shown
and swings freely in a vertical plane, first about O and then about the peg A
after the cord comes in contact with the peg. Determine the tension in the
cord (a) just before the sphere comes in contact with the peg, (b) just after
it comes in contact with the peg.
SOLUTION
Velocity of the sphere as the cord contacts A
vB = 0
TB = 0
1 2
mvC
2
U B −C = ( mg )(1)
TC =
TB + U B −C = TC
0 + 1mg =
1 2
mvC
2
vC2 = (2)( g )
Newton’s law
(a)
Cord rotates about Point O ( R = L)
v2
T − mg (cos 60°) = m C
L
T = mg (cos 60°) +
T=
3
mg
2
m(2) g
2
T = 1.5 W 
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558
PROBLEM 13.41 (Continued)
(b)
L

Cord rotates about A  R = 
2

T − mg (cos 60°) =
T=
mvC2
L
2
mg m(2)( g )
+
2
1
5
1

T =  + 2  mg = mg
2
2

T = 2.5W 
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559
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.
T1 = 0
T2 =
1 2
mvP
2
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
ρP
=
2 gyP
ρP
Rider at Point B.
yB = h = 60 ft
ρ B = r = 20 ft
an =
(2 g )(60)
= 6g
20
ΣF = ma:
N B − mg = m(6 g )
N B = 7 mg = 7W = (7)(160 lb)
N B = 1120 lb 
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560
PROBLEM 13.42 (Continued)
Rider at Point D.
yD = h − 2r = 20 ft
ρ D = 20 ft
an =
(2 g )(20)
= 2g
20
ΣF = 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
ΣF = man :
mg = m ⋅
2 gyE
ρE
ρ E = 2 yE
ρ = 80.0 ft 
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561
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 ρ = 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.
T2 =
T1 = 0
1 2
mvP
2
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
ρP
=
2 g yP
ρP
The condition of loss of contact with the track at P is that the curvature of the path is equal to ρp and the
normal contact force N P = 0.
Car at Point D.
ρ D = r = 20 ft
y D = h − 2r
( aD ) n =
2 g (h − 2r )
r
ΣF = ma
2 g ( h − 2r )
r
2 h − 5r
N D = mg
r
N D + mg = m
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562
PROBLEM 13.43 (Continued)
For N D > 0
2 h − 5r > 0
5
h > r = 50 ft
2
Car at Point E.
ρ E = ρ = 75 ft
yE = h − r = h − 20 ft
( aE ) n =
2 g (h − 20)
75
ΣF = ma
2mg ( h − 20)
75
115 − 2h
N E = mg
75
N E − mg = −
For
N E > 0,
115 − 2h > 0
h < 57.5 ft
50.0 ft ≤ h ≤ 57.5 ft 
Range of values for h:
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563
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 θ = 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 θ = −man = −
mvC2
h
vc2 = gh cos θ
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 θ
2
2
U1→2 = weight × change in vertical distance
= mgh (1 − cos θ )
1 2
1
mvB + mg (1 − cos θ ) = mgh cos θ
2
2
2
vB = gh cos θ − 2 gh(1 − cos θ ) = gh(3cos θ − 2)
T1 + U1→ 2 = T2 :
Data:
g = 9.81 m/s 2 , h = 0.9 m, θ = 30°.
vB2 = (9.81)(0.9)(3cos 30° − 2) = 5.2804 m 2 /s 2
vB = 2.30 m/s 
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564
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 θ 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 θ = man
g cos θ =
vC2
h
vC2 = gh cos θ = gy
(1)
Work-energy principle.
TB =
(a)
1 2 1
mv = m(8) 2 = 32m
2
2
1 2
mvC
2
TB + U B −C = TC
TC =
Use Eq. (1)
32m + mg (h − y ) =
1 2
mvC
2
32 + g (h − yC ) =
1
g yC
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 θ
cos θ =
yC 2.6625
=
= 0.8875
3
h
(3)
θ = 27.4° 
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565
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 θ = (9.259)(cos 27.4°) = 8.220 ft/s
(vC ) y = −vC sin θ = −(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 θ ) + (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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566
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 =
ΔU (1000)(70 kg)(9.81 m/s 2 )(300 m)
N⋅m
=
= 57, 225
Δt
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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567
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 )(v A ) = (mC + mL )( g )(v A )
v A = s/t = (2.8 m)/(15 s) = 0.18667 m/s
( PP ) A = [(1200 kg) + (300 kg)](9.81 m/s 2 )(0.18667 m/s)3
(b)
( PP ) A = 2.747 kJ/s
( PP ) A = 2.75 kW 
( PE ) A = ( PP )/η = (2.75 kW)/(0.82)
( PE ) A = 3.35 kW 
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568
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
ΣF = 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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569
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 θ =
3
100
θ = 1.718°
W = WB + WW = 15 + 120
(a)
W = 135 lb
PW = W ⋅ v = (W sin θ ) (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 θ )(v)
PB = (198)(sin 1.718)(20)
PB = 118.7 ft ⋅ lb/s 




(b)
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570
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 η 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/s 2 ) = [mg (N/h)]
Thus,
ΔU [mg (N/h)][b(m)]  mgb 
=
=
 N ⋅ m/s
Δt
(3600 s/h)
 3600 
1000 N ⋅ m/s = 1 kW
Thus, including motor efficiency, η
P (kw) =
mgb (N ⋅ m/s)
 1000 N ⋅ m/s 
(3600) 
 (η )
kW


P(kW) = 0.278 × 10−6
mgb
η

ΔU [W (tons/h)(2000 lb/ton)][b(ft)]
=
Δt
3600 s/h
(b)
=
With η ,
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  η 
hp =
1.010 × 10−3Wb
η

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571
PROBLEM 13.51
In an automobile drag race, the rear (drive) wheels of
a 1000 kg car skid for the first 20 m and roll with sliding
impending during the remaining 380 m. The front wheels of
the car are just off the ground for the first 20 m, and for the
remainder of the race 80 percent of the weight is on the rear
wheels. Knowing that the coefficients of friction are
μs = 0.90 and μk = 0.68, determine the power developed by
the car at the drive wheels (a) at the end of the 20-m portion
of the race, (b) at the end of the race: Give your answer in kW
and in hp. Ignore the effect of air resistance and rolling
friction.
SOLUTION
(a)
First 20 m.
(Calculate velocity at 20 m.) Force generated by rear wheels = μkW , since car skids.
Thus,
Fs = (0.68)(1000)( g )
Fs = (0.68)(1000 kg)(9.81 m/s 2 ) = 6670.8 N
Work and energy.
T1 = 0, T2 =
1 2
2
mv20 = 500v20
2
T1 + U1− 2 = T2
U1− 2 = (20 m)(Fs ) = (20 m)(6670.8 N)
U1− 2 = 133, 420 J
2
0 + 133,420 = 500v20
2
v20
=
133,420
= 266.83
500
v20 = 16.335 m/s
Power = ( Fs )(v20 ) = (6670.8 N)(16.335 m/s)
Power = 108,970 J/s =108.97 kJ/s
1 kJ/s = 1 kW
1 hp = 0.7457 kW
Power = 109.0 kJ/s = 109.0 kW
Power =

(109.0 kW)
= 146.2 hp 
(0.7457 kW/hp)

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572
PROBLEM 13.51 (Continued)
(b)
End of race. (Calculate velocity at 400 m.) For remaining 380 m, with 80% of weight on rear wheels,
the force generated at impending sliding is ( μ s )(0.80)(mg )
FI = (0.90)(0.80)(1000 kg)(9.81 m/s 2 )
FI = 7063.2 N
Work and energy, from 20 m  to 28 m .
v2 = 16.335 m/s [from part (a)]
T2 =
1
(1000 kg)(16.335 m/s) 2
2
T2 = 133, 420 J
T3 =
1 2
2
mv380 = 500v380
2
U 2 −3 = ( FI )(380 m) = (7063.2 N)(380 m)
U 2 −3 = 2, 684,000 J
T2 + U 2 −3 = T3
2
(133, 420 J) + (2, 684, 000 J) = 500v30
v30 = 75.066 m/s
Power = ( FI )(v30 ) = (7063.2 N)(75.066 m/s)
= 530, 200 J/s
kW Power = 530, 200 J = 530 kW
hp Power =

530 kW
= 711 hp 
(0.7457 kW/hp)
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573
PROBLEM 13.52
The frictional resistance of a ship is known to vary directly as the 1.75 power of the speed v of the ship.
A single tugboat at full power can tow the ship at a constant speed of 4.5 km/h by exerting a constant force of
300 kN. Determine (a) the power developed by the tugboat, (b) the maximum speed at which two tugboats,
capable of delivering the same power, can tow the ship.
SOLUTION
(a)
Power developed by tugboat at 4.5 km/h.
v0 = 4.5 km/h = 1.25 m/s
F0 = 300 kN
P0 = F0 v0 = (300 kN)(1.25 m/s)
(b)
P0 = 375 kW 
Maximum speed.
Power required to tow ship at speed v:
1.75
 v 
F = F0  
 v0 
1.75
 v 
P = Fv = F0 v  
 v0 
 v 
= F0 v0  
 v0 
2.75
(1)
Since we have two tugboats, the available power is twice maximum power F0 v0 developed by one
tugboat.
 v 
2 F0 v0 = F0 v0  
 v0 
 v 
 
 v0 
Recalling that
2.75
2.75
=2
v = v0 (2)1/ 2.75 = v0 (1.2867)
v0 = 4.5 km/h
v = (4.5 km/h)(1.2867) = 5.7902 km/h
v = 5.79 km/h 
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574
PROBLEM 13.53
A train of total mass equal to 500 Mg starts from rest and accelerates uniformly to a speed of 90 km/h in 50 s.
After reaching this speed, the train travels with a constant velocity. The track is horizontal and axle friction
and rolling resistance result in a total force of 15 kN in a direction opposite to the direction of motion.
Determine the power required as a function of time.
SOLUTION
Let FP be the driving force and FR be the resisting force due to axle friction and rolling resistance.
Uniformly accelerated motion. (t < 50 s):
At
v = v0 + at
v0 = 0
t = 50 s,
v = 90 km/h = 25 m/s
25 m/s = 0 + a(50)
a = 0.5 m/s 2
v = (0.5 m/s 2 )t
FP − FR = ma
Newton’s second law:
FR = 15 kN = 15 × 103 N
where
m = 500 Mg = 500 × 103 kg
a = 0.5 m/s 2
FP = FR + ma = 15 × 103 + (500 × 103 )(0.5)
= 265 × 103 N = 265 kN
FP v = (265 × 103 )(0.5t )
Power:
(0 < 50s)
Uniform motion.
(t > 50 s):
Power = (132.5 kW/s)t 
a=0
FP = FR = 15 × 103 N; v = 25 m/s
Power:
FP v = (15 × 103 )(25 m/s) = 375 × 103W
(t > 50 s)
Power = 375 kW 
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575
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/s 2 .
SOLUTION
(a) Acceleration = 0
Elevator
Counterweight
Motor
ΣFy = 0: TW − WW = 0
ΣF = 0: 2TC + TW − 6600 = 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/s 2 , vE = 1 ft/s
(b)
Counterweight
Elevator
Counterweight:
ΣF = Ma : TW − W =
W
(aW )
g
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576
PROBLEM 13.54 (Continued)
TW = (2200 lb) +
(2200 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
TW = 2212 lb
Elevator
ΣF = 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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577
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
Answer: (c)
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578
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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579
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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580
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 k2 
1
1 
+ F
 k1 k2 
δ = x1 + x2 = 
1
1
1 
1 
=  +  v0 m /  + 
 k1 k2 
 k1 k2 
1
1 
= v0 m  + 
 k1 k2 
δ = v0 m(k1 + k2 )/k1k2 
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581
PROBLEM 13.56 (Continued)
(b)
Springs are in parallel.
x1 = x2 = δ
Eq. (1) becomes
(k1 + k2 )δ 2 = mv02
δ = v0
m

k1 + k2
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582
PROBLEM 13.57
A 600-g collar C may slide along a horizontal, semicircular rod
ABD. The spring CE has an undeformed length of 250 mm and a
spring constant of 135 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 D.
SOLUTION
First calculate the lengths of the spring when the collar is at positions A, B, and D.
l A = 4402 + 3002 + 1802 = 562.14 mm
lB = 2402 + 3002 + 202 = 384.71 mm
lD = 402 + 3002 + 1802 = 352.14 mm
The elongations of springs are given by e = l − l0 .
eA = 562.14 − 250 = 312.14 mm = 0.31214 m
eB = 384.71 − 250 = 134.71 mm = 0.13471 m
eD = 352.14 − 250 = 102.14 mm = 0.10214 m
V=
Potential energies:
1 2
ke
2
1
(135 N/m)(0.31214 m) 2 = 6.5767 J
2
1
VB = (135 N/m)(0.13471 m) 2 = 1.2249 J
2
1
VD = (135 N/m)(0.10214 m)2 = 0.7042 J
2
VA =
Since the semicircular rod ABC is horizontal, there is no change in gravitational potential energy.
m = 600 g = 0.600 kg
Mass of collar:
Kinetic energies:
1 2
mv A = 0.300 v A2 = 0
2
1
TB = mvB2 = 0.300 vB2
2
1 2
TD = mvD = 0.300 vD2
2
TA =
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583
PROBLEM 13.57 (Continued)
(a)
Speed of collar at B.
Conservation of energy:
TA + VA = TB + VB
0 + 6.5767 = 0.300vB2 + 1.2249
vB2 = 17.839 m 2 /s 2
(b)
vB = 4.22 m/s 
Speed of collar at D.
Conservation of energy:
TA + VA = TD + VD
0 + 6.5767 = 0.300 v02 + 0.7042
vD2 = 19.575 m 2 /s 2
vD = 4.42 m/s 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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584
PROBLEM 13.58
A 3-lb collar is attached to a spring and slides without friction
along a circular rod in a horizontal plane. The spring has an
undeformed length of 7 in. and a constant k = 1.5 lb/in.
Knowing that the collar is in equilibrium at A and is given a
slight push to get it moving, determine the velocity of the
collar (a) as it passes through B, (b) as it passes through C.
SOLUTION
L0 = 7 in., LDA = 20 in.
LDB =
(8 + 6)2 + 62 = 15.23 in.
LDC = 8 in.
Δ LDA = 20 − 7 = 13 in.
Δ LDB = 15.23 − 7 = 8.23 in. 
Δ LDC = 8 − 7 = 1 in. 
(a)
TA = 0, VA =
1
1
k (ΔLDA )2 = (1.5)(13) 2 = 126.75 lb ⋅ in.
2
2
= 10.5625 lb ⋅ ft
TB =
VB =
1 2 1.5 2
mvB =
vB
2
g
1
(1.5)(8.23) 2 = 50.8 lb ⋅ in. = 4.233 lb ⋅ ft
2
TA + VA = TB + VB : 0 + 10.5625 =
1.5vB2
+ 4.233
32.2
vB = 11.66 ft/s 
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585
PROBLEM 13.58 (Continued)
(b)
TA = 0, VA = 10.5625 lb ⋅ ft, TC =
VC =
1.5 2
vC
32.2
1
(1.5)(1) 2 = 0.75 lb ⋅ in. = 0.0625 lb ⋅ ft
2
TA + VA = TC + VC : 0 + 10.5625 =
1.5 2
vc + 0.0625 
32.2
vC = 15.01 ft/s 
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586
PROBLEM 13.59
A 3-lb collar C may slide without friction along a horizontal rod.
It is attached to three springs, each of constant k = 2 lb/in. and
6 in. undeformed length. Knowing that the collar is released from
rest in the position shown, determine the maximum speed it will
reach in the ensuing motion.
SOLUTION
Maximum velocity occurs at E where collar is passing through position of equilibrium.
Position 
T1 = 0
Note: Undeformed length of springs is 6 in. = 0.5 ft.
Spring AC:
L = (1 ft) 2 + (0.5 ft) 2 = 1.1180 ft
Δ = 1.1180 − 0.50 = 0.6180 ft
Spring CD:
L = (0.5 ft)2 + (0.5 ft)2 = 0.70711 ft
Δ = 0.70711 − 0.50 = 0.20711 ft
Spring BD:
L = 0.50 ft, Δ = 0
Potential energy.
(k = 2 lb/in. = 24 lb/ft for each spring)
1
1
1
V1 = Σ k Δ 2 = k ΣΔ 2 = (24 lb/ft)[(0.6180 ft) 2 + (0.20711 ft) 2 + 0]
2
2
2
V1 = 5.0983 lb ⋅ ft
Position 
3.0 lb
1
1
m=
= 0.093168 slug; T2 = mv22 = (0.093168 slug) v22
2
2
2
32.2 ft/s
Spring AC:
L = (0.5 ft)2 + (0.5 ft)2 = 0.7071067 ft
Δ = 0.70711 − 0.50 = 0.20711 ft
Spring CD:
L = 0.50 ft
Δ=0
Spring BC:
L = (0.5 ft) 2 + (0.5 ft)2 = 0.7071067 ft
Δ = 0.70711 − 0.50 = 0.20711 ft
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587
PROBLEM 13.59 (Continued)
Potential energy.
1
1
V2 = Σ k Δ 2 = k ΣΔ 2
2
2
1
V2 = (24 lb/ft)[(0.20711 ft) 2 + 0 + (0.20711 ft) 2 ] = 1.0294 lb ⋅ ft
2
Conservation of energy. T1 + V1 = T2 + V2 : 0 + 5.0983 lb ⋅ ft =
1
(0.093168 slug)v22 + 1.0294 lb ⋅ ft
2
v22 = 87.345
v 2 = 9.35 ft/s ↔ 
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588
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
ΔLA = 0.150 m − 0.080 m
ΔLA = 0.070 m
1
k ( ΔLA )2
2
1
VA = (400 × 103 N/m)(0.070 m)2
2
VA = 980 J
VA =
vB = 0
TB = 0
ΔLB = 0.200 m − 0.080 m = 0.120 m
1
1
k ( ΔLB )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 2A + 980 = 0 + 2880
(2880 − 980)
(0.25)
v A2 = 7600 m 2 /s 2
vA = 87.2 m/s 
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589
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
ΔLC = 0.100 m − 0.080 m = 0.020 m
TC =
VC =
1
1
k (Δ 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 /s 2
vC = 105.8 m/s 
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590
PROBLEM 13.61
An elastic cord is stretched between two Points A and B, located 800 mm apart in
the same horizontal plane. When stretched directly between A and B, the tension is
40 N. The cord is then stretched as shown until its midpoint C has moved through
300 mm to C ′; a force of 240 N is required to hold the cord at C ′. A 0.1 kg pellet
is placed at C ′, and the cord is released. Determine the speed of the pellet as it
passes through C.
SOLUTION
Let  = undeformed length of cord.
Position 1.
Length AC ′B = 1.0 m; Elongation = x1 = 1.0 − 
3 
ΣFx = 0: 2  F1  − 240 N = 0 F1 = 200 N
5 
Position 2.
Length ACB = 0.8 m; Elongation = x2 = 0.8 − 
Given
F2 = 40 N
F1 = kx, F2 = k x2
F1 − F2 = k ( x1 − x2 )
200 − 40 = k [(1.0 − ) − (0.8 − )] = 0.2k
k=
160
= 800 N/m
0.2
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591
PROBLEM 13.61 (Continued)
Position :
Position:
T1 = 0
x1 =
F1
200 N
=
= 0.25 m
k 800 N/m
x2 =
F2
40 N
=
= 0.05 m
k 800 N/m
V1 =
1 2 1
kx1 = (800 N/m)(0.25 m) 2 = 25.0 N ⋅ m
2
2
m = 0.10 kg
1 2 1
mv2 = (0.1 kg) v22 = 0.05 v22
2
2
1
1
V2 = k x22 = (800 N/m)(0.05 m) 2 = 1 N ⋅ m
2
2
T2 =
Conservation of energy:
T1 + V1 = T2 + V2
0 + 25.0 N ⋅ m = 0.05v22 + 1.0 N ⋅ m
24.0 = 0.05v22
v2 = 21.909 m/s
v2 = 21.9 m/s 
Note: The horizontal force applied at the midpoint of the cord is not proportional to the horizontal distance C ′C.
A solution based on the work of the horizontal force would be rather involved.
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592
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
V1 = (100 ft)(600 lb)
Datum at :
= 6 × 104 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 +
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593
PROBLEM 13.62 (Continued)
d = distance from the ground
T1 + V1 = T2 + V2
0 + (186)(130 − d ) = 0 + (266.67)(45 − d ) 2
266.7 d 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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594
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=
π
4
(diameter)2 =
π
4
(0.4 in.)2 = 0.12566 in 2
AE = (0.12556 in.2 )(29 × 106 lb/in 2 ) = 3.6442 × 106 lb
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
T1 =
1
(124.22 lb ⋅ s 2 /ft)(3 ft/s2 ) = 558.99 ft ⋅ lb
2
Kinetic energy:
k=
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595
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:
δ = x2 − x1 = 0.09594 ft
δ = 1.151 in.
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596
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 m 2 /s 2
v B = 2.48 m/s

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597
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 m 2 /s 2
vC = 1.732 m/s 
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598
PROBLEM 13.65
A 1-kg collar can slide along the rod shown. It is attached to an
elastic cord anchored at F, which has an undeformed length of
250 mm and a spring constant of 75 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
LAF = (0.5) 2 + (0.4) 2 + (0.3) 2
LAF = 0.70711 m
LBF = (0.4) 2 + (0.3) 2
LBF = 0.5 m
LFE = (0.5) 2 + (0.3) 2
LFE = 0.58309 m
V = Ve + Vg
(a)
Speed at B:
v A = 0,
TA = 0
Point A:
(VA )e =
1
k (Δ LAF ) 2
2
Δ LAF = LAF − L0 = 0.70711 − 0.25
Δ LAF = 0.45711 m
1
(75 N/m)(0.45711 m) 2
2
(VA )e = 7.8355 N ⋅ m
(VA )e =
(VA ) g = ( mg )(0.4)
= (1.0 kg)(9.81 m/s 2 )(0.4 m)
= 3.9240 N ⋅ m
VA = (VA )e + (VA ) g
= 7.8355 + 3.9240
= 11.7595 N ⋅ m
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599
PROBLEM 13.65 (Continued)
Point B:
1 2 1
mvB = (1.0 kg)vB2
2
2
2
TB = 0.5 vB
TB =
(VB )e =
1
k (Δ LBF ) 2
2
Δ LBF = LBF − L0 = 0.5 − 0.25
Δ LBF = 0.25 m
1
(75 N/m)(0.25 m) 2 = 2.3438 N ⋅ m
2
(VB ) g = ( mg )(0.4) = (1.0 kg)(9.81 m/s2 )(0.4 m) = 3.9240 N ⋅ m
(VB )e =
VB = (vB )e + (VB ) g = 2.3438 + 3.9240 = 6.2678 N ⋅ m
TA + VA = TB + VB
0 + 11.7595 = 0.5 vB2 + 6.2678
vB2 = (5.49169)/(0.5)
vB2 = 10.983 m 2 /s 2
(b)
vB = 3.31 m/s 
Speed at E:
Point A:
TA = 0
VA = 11.7595 N ⋅ m (from part (a))
Point E:
1 2 1
mvE = (1.0 kg)vE2 = 0.5vE2
2
2
1
(VE )e = k ( Δ LFE ) 2
Δ LFE = LFE − L0 = 0.5831 − 0.25
2
TE =
ΔLFE = 0.3331 m
1
(75 N/m)(0.3331 m)2 = 4.1607 N ⋅ m
2
(VE ) g = 0 VE = 4.1607 N ⋅ m
(VE )e =
TA + VA = TE + VE
0 + 11.7595 = 0.5 vE2 + 4.1607
vE2 = 7.5988/0.5
vE2 = 15.1976 m 2 /s 2
vE = 3.90 m/s  
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600
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 θ with the vertical, determine
(a) the smallest value of θ 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 θ occurs when the velocity at D is close to zero.
vC = 0
vD = 0
TC = 0
TD = 0
V = Ve + Vg
Point C:
Δ LBC = (1 ft)(θ ) = θ ft
1
k (ΔLBC ) 2
2
3
(VC )e = θ 2
2
(VC )e =
R = 12 in. = 1 ft
(VC ) g = WR(1 − cos θ )
 8 oz 
(VC ) g = 
 (1 ft)(1 − cosθ )
 16 oz/lb 
(VC ) g =
1
(1 − cos θ )
2
3
1
VC = (VC )e + (VC ) g = θ 2 + (1 − cos θ )
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 + θ 2 + (1 − cos θ ) = 1
2
2
2
(1.5)θ − (0.5) cos θ = 0.5
θ = 0.7592 rad
θ = 43.5° 
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601
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  
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602
PROBLEM 13.67
The system shown is in equilibrium when φ = 0. Knowing that
initially φ = 90° and that block C is given a slight nudge when
the system is in that position, determine the velocity of the block
as it passes through the equilibrium position φ = 0. Neglect the
weight of the rod.
SOLUTION
Find the unstretched length of the spring.
1.1
0.3
= 1.3045 rad
θ = 74.745°
θ = tan −1
LBD = (1.1) 2 + .32
LBD = 1.140 ft
Equilibrium
ΣM A = (0.3)( Fs sin θ ) − (25)(2.1) = 0
(25 lb)(2.1 ft)
(0.3 ft)(sin 74.745°)
= 181.39 lb
Fs = k ΔLBD
Fs =
181.39 lb = (600 lb/ft)(ΔLBD )
ΔLBD = 0.30232 ft
Unstretched length
L0 = LBD − ΔLBD
L0 = 1.140 − 0.3023
= 0.83768 ft
′ , when φ = 90°.
Spring elongation, ΔLBD
′ = (1.1 ft + 0.3 ft) − L0
ΔLBD
′ = 1.4 ft − 0.8377 ft
ΔLBD
= 0.56232 ft
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603
PROBLEM 13.67 (Continued)
At , (φ = 90°)
v1 = 0, T1 = 0
V1 = (V1 )e + (V1 ) g
1
′ )2
k (ΔLBD
2
1
(V1 )e = (600 lb/ft)(0.5623 ft) 2
2
(V1 )e = 94.86 lb ⋅ ft
(V1 )e =
(V1 ) g = −(25 lb)(2.1 ft) = −52.5 ft ⋅ lb
V1 = 94.86 − 52.5 = 42.36 ft ⋅ lb
At , (φ = 0°)
1
1
k ( ΔLBD ) 2 = (600 lb/ft)(0.3023 ft) 2
2
2
(V2 )e = 27.42 lb ⋅ ft
(V2 )e =
(V2 ) g = 0
T2 =
V2 = 27.42 ft ⋅ lb
1 2 1  25 lb  2
2
mv2 = 
 v2 = 0.3882 v2
2
2  32.2 ft/s 2 
T1 + V1 = T2 + V2
0 + 42.36 = 0.3882 v22 + 27.42
v22 = (14.941)/(0.3882)
v22 = 38.48 ft 2 /s 2
v2 = 6.20 ft/s 
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604
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 =
Position 2:
1 2 1
ke1 = (30 × 103 )(0.050) 2 = 37.5 J
2
2
1 2
mv2 = 0 since v2 = 0.
2
V2 g = mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x
T2 =
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 x 2 + 1332.24 x − 1442.09 = 0
Solving for x,
x = 0.26882 and − 0.357 64
x = 0.269 m  
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605
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
V2 g = mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x
T2 =
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.
ΣFn = 0
N − mg cos 20° = 0
N = mg cos 20°
= (50)(9.81) cos 20°
= 460.92 N
F f = μk N
= (0.2)(460.92)
= 92.184
U1→2 = − F f d
= −92.184(8 + x)
= −737.47 − 92.184 x
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606
PROBLEM 13.69 (Continued)
Principle of work and energy:
T1 + V1 + U1− 2 = T2 + V2
100 + 1342.09 + 37.5 − 737.47 − 92.184 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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607
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
mv A = 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
ρ
=
123.94 m 2 /s 2
= 4.59 m/s 2
27 m
(a B )n = 4.59 m/s 2
50°
Apply Newton’s second law to the car at B.
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608
PROBLEM 13.70 (Continued)
50°ΣFn = man : N − mg cos 40° = − man
N = mg cos 40° − man = m( g cos 40° − an )
= (250 kg)[(9.81 m/s 2 ) cos 40° − 4.59 m/s 2 ]
= 1878.7 − 1147.5
N = 731 N 
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609
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
mv A = 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 θ be the slope angle of the track.
ΣFn = man : N − mg cos θ = man
N = mg cos θ + man
Over portion AB of the track,
and
(2)
h = ρ (1 − cos θ )
an = −
mv 2
ρ
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610
PROBLEM 13.71 (Continued)
where ρ is the radius of curvature. ( ρ = 27 m)
N = mg cos θ −
2mg ρ (1 − cos θ )
ρ
= mg (3cos θ − 2)
At Point A (θ = 0)
N A = mg = (250)(9.81) = 2452.5 N
At Point B (θ = 40°)
N B = (2452.5)(3cos 40° − 2)
N B = 731 N
Over portion BC,
θ = 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 θ )
an =
and
mv 2
r
where r is the radius of curvature. (r = 72 m)
2mgh
r
h

= mg cos θ + 2mg  max − 1 − cos θ 
 r

N = mg cos θ +
2h 

= mg  3cos θ − 2 + max 
r 

which is maximum at Point D, where
2h 

N D = mg 1 + max 
r 

Data:
hmax = 27 + 18 = 45 m,
r = 72 m
 (2) (45) 
N D = (2452.5) 1 +
= 5520 N
72 

Summary:

minimum (just above B):
731 N 
5520 N 
maximum (at D):
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611
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.
At A:
 A = 7 + 5 + 5 = 17 in.
 Δ −  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
mv A = 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
vB = 14.34 ft/s 
vB2 = 205.72 ft 2 /s 2
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612
PROBLEM 13.72 (Continued)
Forces at B.
2
Fs = k ( B −  0 ) = (10)   = 6.6667 lb.
3
5
sinα =
13
5
ρ = 5 in. = ft
12
mvB2
man =
ρ
(0.031056)(205.72)
5/12
= 15.3332 lb
=
ΣFy = ma y : Fs sin α − W + N = man
N = man + W − Fs sin α
 5
= 15.3332 + 1 − (6.6667)  
 13 
N = 13.769 lb
N = 13.77 lb 
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613
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.
(VA )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
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614
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 = 0.15528v 2A + 5.605
Conservation of energy:
T1 + V1 = TB + VB
0 + 49.910 = 0.15528vB2 + 21.0
Normal accelerations at A and B.
v 2A = 285.32 ft 2 /s 2
vB2 = 186.18 ft 2 /s2
an = v 2 /ρ
ρ = 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.)
FA = 23.196 lb
FB = (4 lb/in.)(14 in.)
FB = 56.0 lb
45°
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:
ΣF = 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:
ΣF = m ( a B ) n :
N B = (0.31056)(159.58)
N B = 49.6 lb

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615
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.
ΣF = 0 + mg =
mvB2
r
vB2 = rg = 1.5 ft(32.2 ft/s 2 )
vB2 = 48.30
TA =
At A
1 2
mv0
2
0.5


VA = 0  8 oz = 0.5 lb  =
= 0.01553 
32.2


TB =
At B
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
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616
PROBLEM 13.74 (Continued)
(b)
ΣF = man : N = 0.01553
(144.87)
1.5
N = 1.49989
Loop 2
{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 
TC = 0.007765vC2 , VC = 7.5 mg = 3.75
(b) At C
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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617
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 v 0 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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618
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:
ΣF = ma:
v2
mg = m c
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
mv A + 0 = mvB2 + mgr
2
2
1
1
m(5 gr ) + 0 = mvB2 + mgr 
2
2
vB2 = 3gr

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619
PROBLEM 13.76 (Continued)
Newton’s second law at position B.
v2
3gr
man = m B = m
= 3 mg
r
r
ΣF = ΣFeff :
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.
v2
2 gr
man = m B = m
= 2 mg
r
r
ΣF = Σeff : N = 2 mg
Force exerted by loop:
N = 2W

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620
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
Let position 1 be at A.
v1 = v0
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 θ based on Q = 0. Draw the free body diagram.
ΣF = 0: Q + mg sin θ = man =
With
Q = 0, v22 = gl sin θ
mv22
l
or v2 = gl sin θ
(1)
Relationship among v0 , v2 and θ based on conservation of energy.
T1 + V1 = T2 + V2
1 2
1
mv0 − mgl = mv22 + mgl sin θ
2
2
v02 − v22 = 2 gl (1 + sin θ )
(2)
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621
PROBLEM 13.77 (Continued)
Eliminating v2 from Eqs. (1) and (2),
v02 − gl sin θ = 2 gl (1 + sin θ )
sin θ =
 1  (5) 2

1  v02
− 2  = 0.74912
 − 2 = 
3  gl
 3  (9.81)(0.6)

θ = 48.514°
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 θ = 0.6cos 48.514° = 0.39746 m
y2 = l sin θ = 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 θ = −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 (t B − t2 ) t B − t2 = 0.25267 s
The vertical motion is
y = y2 + v2 cos θ (t − t2 ) −
At Point B,
1
g (t − t2 ) 2
2
yB = y2 + v2 cos θ (t B − t2 ) −
1
g (t B − 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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622
PROBLEM 13.78*
Packages are moved from Point A on the upper floor of a warehouse to
Point B on the lower floor, 12 ft directly below A, by means of a chute,
the centerline of which is in the shape of a helix of vertical axis y and
radius R = 8 ft. The cross section of the chute is to be banked in such a
way that each package, after being released at A with no velocity, will
slide along the centerline of the chute without ever touching its edges.
Neglecting friction, (a) express as a function of the elevation y of a
given Point P of the centerline the angle φ formed by the normal to the
surface of the chute at P and the principal normal of the centerline at
that point, (b) determine the magnitude and direction of the force
exerted by the chute on a 20-lb package as it reaches Point B. Hint: The
principal normal to the helix at any Point P is horizontal and directed
toward the y axis, and the radius of curvature of the helix is
ρ = R[1 – (h/2πR)2].
SOLUTION
(a)
v A = 0 TA = 0
At Point A:
VA = mgh
TP =
At any Point P:
1 2
mv
2
VP = Wy = mgy
TA + VA = TP + VP
1 2
mv + mgy
2
v 2 = 2 g (h − y )
0 + mgh =
en along principal normal, horizontal and directed toward y axis
et tangent to centerline of the chute
eD along binormal
β = tan −1
h
(12 ft)
= tan −1
= 13.427°
2π R
2π (8 ft)
mab = 0
since
ab = 0
Note: Friction is zero,
ΣFt = mat : mg sin β = mat
at = g sin β
ΣFb = mab : N b − W cos β = 0 N B = W cos β
ΣFn = man : N n =
mv 2 m 2 g (h − y )
(h − y)
=
= 2W
e
e
e
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623
PROBLEM 13.78* (Continued)
The total normal force is the resultant of Nb and Nm, it lies in the b–m plane, and forms angle φ with
m axis.
tan φ = Nb /N n
2( w(h − y )
e
tan φ = (e /2(h − y )) cos β
tan φ = W cos β
  h 2 
R
e = R 1 + 
 = R(1 + tan 2 β ) =

cos 2 β
  2π R  
Given:
e
R
cos β =
2( h − y )
2( h − y ) cos β
8 ft
4.112
tan φ =
=
2(12 − y ) cos13.427° 12 − y
tan φ =
Thus,
cot φ = 0.243(12 − y ) 
or
(b)
At Point B:
y=0
for x, y, z axes, we write, with W = 20 lb,
N x = Nb sin β = W cos β sin β = (20 lb) cos 13.427° sin13.427° N x = 4.517 lb
N y = Nb cos β = W cos 2 β = (20 lb) cos 2 13.427° N y = 18.922 lb
N z = − N n = −2w
N z = −2(20 lb)
h− y
h− y
= −2W
e
R/cos 2 β
(12 ft-0)
cos 2 13.427° N z = −56.765 lb
8 ft
N = (4.517) 2 + (18.922) 2 + (−56.765) 2
N = 60.0 lb 
cos θ x =
N x 4.517
=
N
60
θ x = 85.7° 
cos θ y =
Ny
θ y = 71.6° 
cos θ z =
Nz
56.742
=−
N
60
N
=
18.922
60
θ z = 161.1° 
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624
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
∂z
∂ Fz
∂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 = −
∂ Fy
∂x
∂V
∂y
Fz = −
=−
∂ 2V
∂ y∂ x
∂V
∂z
∂ Fx ∂ Fy
=

∂y
∂x
∂ 2V
∂ 2V
=
:
∂ x∂ y ∂ y∂ x
We obtain in a similar way
∂ Fy
∂z
=
∂ Fz
∂y
∂ Fz ∂ Fx
=

∂x
∂z
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625
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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626
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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627
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 in a clockwise sense the path A, B, C, A including the
quarter circle x 2 + y 2 = a 2, if (a) F = kyi, (b) F = k ( yi + xj).
SOLUTION
(a)
Fx = ky
∂ Fy
∂ Fx
=k
∂y
Fy = 0
=0
∂x
∂ Fx ∂ Fy
≠
∂y
∂x
Thus,
F is not conservative.
UABCA =
B
C
A
A
B
B
 F ⋅ dr =  kyi ⋅ dyj +  kyi ⋅ (dxi + dyj) +  kyi ⋅ dxj
ABCA
B
 = 0, F is perpendicular to the path.
A

C
B
kyi ⋅ (dxi + dyj) =
C
 ky dx
B
From B to C, the path is a quarter circle with origin at A.
x2 + y2 = a2
Thus,
y = a2 − x2

Along BC,
C
B
kydx =

a
0
k a 2 − x 2 dx =
π ka 2
4
A
 kyi ⋅ dx j = 0 ( y = 0 on CA)
C
UABCA =
(b)
Fx = k y
Fy = kx
B
C
  
A
+
B
+
A
C
=0+
π ka 2
4
+0
U ABCA =
π ka 2
4

∂ Fy
∂ Fx
=k
=k
∂y
∂x
∂ Fx ∂ Fy
=
,
F is conservative.
∂y
∂x
Since ABCA is a closed loop and F is conservative,
U ABCA = 0 
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628
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
 dz = a
0
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,
x
(x + a 2 )1/ 2
2
xdx
a
 (x + a )
0
2
2 1/2
= 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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629
PROBLEM 13.82* (Continued)
Thus,
U BD =
y
a
 ( y + 2a ) dy = ( y + 2a )
0
2
2
2 1/ 2
2 1/ 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 
ΔVOD = V (a, a, a ) − V (0, 0, 0)
= −(a 2 + a 2 + a 2 )1/ 2 − 0
Thus,
ΔVOD = − a 3 
U OABD = −ΔVOD
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630
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 =
xi + yj + zk
( x + y 2 + z 2 )1/ 2
2
D
 P ⋅ dr
O
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 =
UO−D =
3x
= 3
(3x 2 )1/2

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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631
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
( x + y + z 2 )3/2
y
Fy = 2
2
( x + y + z 2 )1/2
2
2
x ( − 32 ) (2 y )
∂ Fx
= 2
∂ y ( x + y 2 + z 2 )5/2
y ( − 32 ) 2 y
∂ Fy
= 2
∂ x ( x + 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
2
2
∂x
( x + y + 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
( x + y + z 2 )1/2
2
2

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632
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 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/s 2
At the surface of the earth,
r1 = R = 6370 km = 6.37 × 106 m
Centric force at the surface of the earth,
GMm
R2
GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2
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
= mv2 +
mv1 −
r1
r2
2
2
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 m 2 /s 2
m
T1
= 62.5 MJ/kg 
m
(a)
1 2
mvesc = gr
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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633
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 /s 2
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 m 2 /s 2
VA = 9.556 × 103 m/s
v A = 9.56 km/s 
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634
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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635
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 =
GM = gRE2
E =−
GMm
mv 2
GM
=
 v2 =
2
r
r
r
V =−
GMm
r
1 GMm GMm
1 GMm
−
=−
2 r
r
2 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
ΔE300 = EGs − E200 = 5.0141 × 1010
Δ E300 = 50.1 × 109 ft ⋅ lb 
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636
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
Δ EE = EGs − EE = −9.5681 × 109 + 125.453 × 109
Δ E E = 115.9 × 109 ft ⋅ lb 
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637
PROBLEM 13.88
A lunar excursion module (LEM) was used in the Apollo moon-landing missions to save fuel by making it
unnecessary to launch the entire Apollo spacecraft from the moon’s surface on its return trip to earth. Check
the effectiveness of this approach by computing the energy per pound required for a spacecraft (as weighed on
the earth) to escape the moon’s gravitational field if the spacecraft starts from (a) the moon’s surface,
(b) a circular orbit 50 mi above the moon’s surface. Neglect the effect of the earth’s gravitational field. (The
radius of the moon is 1081 mi and its mass is 0.0123 times the mass of the earth.)
SOLUTION
Note:
GM moon = 0.0123 GM earth
By Eq. 12.30,
GM moon = 0.0123 gRE2
At ∞ distance from moon: r2 = ∞, Assume v2 = 0
E2 = T2 + V2
GM M m
∞
=0−0
=0−
=0
(a)
On surface of moon:
RM = 1081 mi = 5.7077 × 106 ft
v1 = 0 T1 = 0
RE = 3960 mi = 20.909 × 106 ft
V1 = −
GM M m
RM
E1 = T1 + V1 = 0 −
E1 = −
0.0123 gRE2 m
RM
(0.0123)(32.2 ft/s2 )(20.909 × 106 ft)2 m
(5.7077 × 106 ft)
WE = Weight of LEM on the earth
E1 = ( −30.336 × 106 ft 2 /s 2 )m
m=
WE
g
 30.336 × 106 2 2 
E1 =  −
ft /s WE
2
 32.2 ft/s

ΔE = E2 − E1
= 0 + (942.1 × 103 ft ⋅ lb/lb)WE
ΔE
= 942 × 103 ft ⋅ lb/lb 
WE
Energy per pound:
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638
PROBLEM 13.88 (Continued)
(b)
r1 = RM + 50 mi
r1 = (1081 mi + 50 mi) = 1131 mi = 5.9717 × 106 ft
Newton’s second law:
F = man :
GM M m
r12
v2
=m 1
r1
v12 =
GMm
r1
V1 = −
T1 =
1 2 1 GM M
mv1 = m
r1
2
2
GM M m
r1
E1 = T1 + V1 =
1 GM M m GM M m
−
r1
r1
2
E1 = −
1 GM M m
1 0.0123 gRE2 m
=−
r1
r1
2
2
E1 = −
1 (0.0123)(32.2 ft/s 2 )(20.909 × 106 ft)2 m
2
5.9717 × 106 ft
E1 =
(14.498 × 106 ft 2 /s 2 )WE
2
(32.2 ft/s )
= 450.2 × 103 ft ⋅ lb/lb WE
ΔE = E2 − E1 = 0 + 450.2 × 103 ft ⋅ lb/lb WE
ΔE
= 450 × 103 ft ⋅ lb/lb 
WE
Energy per pound:
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639
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 2A + 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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640
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 m 2 /s 2
vC = 7112 m/s
Velocity reduced to 60% of vC gives v A = 4267 m/s.
Conservation of energy:
TA + VA = TB + VB
1
GM m
1
GM m
= m vB2 −
m v A2 −
rB
rA
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 
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641
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
GM B m
v2
=
m
r
r2
2
rv
MB =
G
2
GM earth = gRearth
F=
= (32.2 ft/s 2 )(3960 mi × 5280 ft/mi) 2
= 14.077 × 1015 (ft 3 /s 2 )
M sun = 330, 000 M 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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642
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
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643
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, vθ = 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,
(vθ )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 = ?,
(vθ ) 2 = ?,
v2 = ?
x2 = r2 − l0 = (0.6 − 0.5) = 0.1 m
Conservation of angular momentum:
r1m(vθ )1 = v2 m(vθ ) 2
r (v )
(0.3)(2)
(vθ ) 2 = 1 θ 1 =
= 1.000 m/s
0.6
r2
Conservation of energy:
T1 + V1 = T2 + V2
1 2 1 2 1 2 1 2
mv1 + kx1 = mv2 + kx2
2
2
2
2
k 2
2
2
2
v2 = v1 +
x1 − x2
m
1200
(0.2)2 − (0.1)2  = 16 m 2 /s 2
= (2) 2 +

3 
2
2
2
2 2
(vr ) 2 = v2 − (vθ ) = 16 − 1 = 15 m /s
(
)
vr = ±3.87 m/s
vr = ±3.87 m/s 
vθ = 1.000 m/s 
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644
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, vθ = 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,
(vθ )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)(vθ ) 22 = 1.5(vθ ) 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(vθ )1 = v2 m(vθ ) 2
r
(0.3)
0.6
(vθ ) 2 = 1 (vθ ), =
(2) =
r2
rm
rm
Conservation of energy:
T1 + V1 = T2 + V2
6 + 24 = 1.5(vθ ) 22 + 600( rm − 0.5)2
2
 0.6 
2
30 = (1.5) 
 + 600( rm − 0.5)
r
 m 
0.54
f (rm ) = 2 + 600( rm − 0.5) 2 − 30 = 0
rm
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645
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 
(vθ ) 2 =
0.6
= 0.8337 m/s
0.7197
(vr ) 2 = 0
v2 = 0.834 m/s 
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646
PROBLEM 13.95
A 4-lb collar A and a 1.5-lb collar 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. At the instant shown, the velocity vA of collar
A has a magnitude of 6 ft/s and a stop prevents collar B from
moving. If the stop is suddenly removed, determine (a) the
velocity of collar A when it is 8 in. from O, (b) the velocity of
collar A when collar B comes to rest. (Assume that collar B
does not hit O, that collar A does not come off rod OE, and
that the mass of the frame is negligible.)
SOLUTION
4
= 0.12422 lb ⋅ s 2 /ft
32.2
1.5
mB =
= 0.04658 lb ⋅ s 2 /ft
32.2
mA =
Masses:
Constraint of the cord. Let r be the radial distance to the center of collar A and y be the distance that collar B
moves up from its initial level. y = Δr ; y = vr
(a)
Let position 1 be the initial position just after the stop at B is removed and position 2 be when the collar
is 8 in. (0.66667 ft) from O.
r1 = 4 in. = 0.33333 ft
(vr )1 = 0
r2 = 8 in. = 0.66667 ft
Δr = y2 = 8 − 4 = 4 in. = 0.33333 ft
V1 = 0,
Potential energy:
V2 = WB y2 = (1.5)(0.33333) = 0.5 ft ⋅ lb
Conservation of angular momentum of collar A:
m A r1 (vθ )1 = m A r2 (vθ ) 2
r (v )
(0.33333)(6)
(vθ ) 2 = 1 θ 1 =
= 3 ft/s
0.66667
r2
Conservation of energy:
T1 + V1 = T2 + V2
1
1
1
1
m A [(vr )12 + (vθ )12 ] + mB y12 = m A [(vr ) 22 + (vθ ) 22 ] + mB y 2 + V2
2
2
2
2
1
1
1
m A [0 + (vθ )12 ] + 0 + 0 = m A [(vr ) 22 + (vθ ) 22 ] + mB (vr ) 22 + 0.5
2
2
2
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647
PROBLEM 13.95 (Continued)
1
1
1
(0.12422)(6) 2 = (0.12422)[(vr ) 22 + (3) 2 ] + (0.04658)(vr ) 22 + 0.5
2
2
2
2.236 = 0.06211(vr )22 + 0.559 + 0.02329(vr ) 22 + 0.5
0.0854(vr )22 = 1.177
(vr ) 2 = 13.78 ft 2 /s 2
(vr ) 2 = 3.71 ft/s 
(vθ ) 2 = 3.00 ft/s 
v = 4.77 ft/s 
(b)
Let position 3 be when collar B comes to rest.
y3 = r3 − 0.33333,
(vr )3 = 0,
y3 = 0
Conservation of angular momentum of collar A.
m A r1 (vθ )1 = m A r3 (vθ )3
r (v )
(0.33333)(6) 2
(vθ )3 = 1 θ 1 =
=
r3
r3
r3
T1 + V1 = T3 + V3
Conservation of energy:
1
1
1
1
m A [(vr )12 + (vθ )12 ] + mB y12 = m A [(vr )32 + (vθ )32 ] + mB y32 + wB y3
2
2
2
2
  2 2 
1
1
(0.12422)[0 + (6)2 ] + 0 = (0.12422) 0 +    + 0 + (1.5)(r3 − 0.33333)
2
2
  r3  


2.236 =
0.24844
+ 1.5r3 − 0.5
r32
1.5r33 − 2.736r32 + 0.24844 = 0
Solving the cubic equation for r3,
r3 = 1.7712 ft, − 0.2805 ft, 0.33333 ft
Since r3 > r1 = 0.33333 ft, the required root is
r3 = 1.7712 ft
Corresponding velocity of collar A:
(vr )3 = 0 
(vθ )3 =


2
2
=
r3 1.7712
(vθ )3 = 1.129 ft/s 
v3 = 1.129 ft/s 

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648
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 v 0 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
v0 = −
[(3 − 2) 2 + (2 − 2) 2 ]
0.04658
(2) 2
−1.25 v02 = −257.6
(a)
v02 = 206.1 ft 2 /s 2
v0 = 14.36 ft/s 
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649
PROBLEM 13.96 (Continued)
(b)
Let v0 = 12 (14.36 ft/s) = 7.18 ft/s 2 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
(3 − 2) 2 = 309.17 ft 2 /s 2
= (7.18) 2 +
0.04658
v2 = 17.583 ft/s
v22 = v02 +
Conservation of angular momentum:
r1m v1 = r2m v2 sin φ
rv
Lv
r2 sin φ = d = 1 1 = 1 0
v2
v2
d =
(3)(7.18)
17.583
d = 1.225 ft 
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650
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
v02 − v22 = − ( L1 − L0 ) 2
m
2
k
L

v02 −  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
12
3v 
v02 −  0  = −
(3 − 2) 2
0.04658
 0.75 
− 15 v02 = −257.6
v02 = 17.17 ft 2 /s 2
(a)
(b)
Maximum speed.
vm = v2 =
3v0
0.75
v0 = 4.14 ft/s 
vm = 16.58 ft/s 
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651
PROBLEM 13.98
Using the principles of conservation of energy and conservation of angular momentum, solve part a of Sample
Problem 12.9.
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,
r0 = rmin , r1 = rmax
 6.870 × 106 
r 
3
v A′ =  0  v0 = 
 (10.25 × 10 )
r1
 r1 


v A′ =
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)
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652
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 ′ −
r1
2
Substituting for v A′ 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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653
PROBLEM 13.99
Solve sample Problem 13.8, assuming that the elastic cord is
replaced by a central force F of magnitude (80/r2) N directed
toward O.
PROBLEM 13.8 Skid marks on a drag racetrack indicate
that the rear (drive) wheels of a car slip for the first 20 m of
the 400-m track.(a) Knowing that the coefficient of kinetic
friction is 0.60, determine the speed of the car at the end of
the first 20-m portion of the track if it starts from rest and the
front wheels are just off the ground. (b) What is the maximum
theoretical speed for the car at the finish line if, after skidding
for 20 m, it is driven without the wheels slipping for the
remainder of the race? Assume that while the car is rolling
without slipping, 60 percent of the weight of the car is on the
rear wheels and the coefficient of static friction is 0.75.
Ignore air resistance and rolling resistance.
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,
TB =

1 2 1
mvm = (0.6 kg)vm2 = 0.3vm2
2
2
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654
PROBLEM 13.99 (Continued)
VB =
and Point C :
−80
rm
TA + VA = TB + VB
120 − 160 = 0.3vm2 −
80
rm
(2)
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 
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655
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 mv A = rB mvB
Conservation of momentum:
r
8770
vB = A v A =
v A = 0.5492v A
15,970
rB
Conservation of energy:
TA =
1 2
mv A
2
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 m3/s 2
−(398.1 × 1012 ) m
= −45.39 × 106 m
3
8770 × 10
−(398.1 × 1012 ) m
VB =
= −24.93 m
(15,970 × 103 )
VA =
TA + VA = TB + VB :
1
1
m v 2A − 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
v A = 7.65 × 103 m/s
v A = 27.6 × 103 km/h 
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656
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 plae 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: mrA v A sin φ A = mrB vB sin φ A
r v sin 90° 21.886 × 106 v A
vB = A A
=
= 0.15816 v A
rB sin 90°
138.283 × 106
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657
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:
Δv A = 33.325 × 103 − 25.362 × 103 = 7.963 × 103 ft/s
Δv A = 7960 ft/s 
ΔvB = 10.089 × 103 − 5.271 × 103 = 4.818 × 103 ft/s
ΔvB = 4820 ft/s 
Increase in speed at B:
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658
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
1
mv′A2
2
−GM sat m
VA =
rA
TA =
Point A:



M sa = Mass of Saturn, determine GM sa from the speed of Tethys in
its circular orbit.
vcirc =
(Eq. 12.44)
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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659
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
r
607.2 × 106
vB = A 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)
Δv A = v A − v′A = 68,800 − 52, 005
Δv A = 16,795 ft/s 
(b)
r
vB = A v′A = (0.6284)(52, 005)
rB
vB = 32, 700 ft/s 
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660
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 Δv 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 − Δv 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 − Δv A ) 2 − 362.8 × 106 m = mvB2 − 1269.8 × 106 m
2
2
(v A − Δv A ) 2 − vB2 = −1814 × 106
(1)
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661
PROBLEM 13.103 (Continued)
Conservation of angular momentum.
rA = 350 × 106 m
rB = 100 × 106 m
rA m(v A − Δv A ) = rB mvB
r 
vB =  A  (v A − Δv A )
 rB 
 350 
=
 (v A − Δv A )
 100 
(2)
Substitute vB in (2) into (1)
(v A − Δv A ) 2 [1 − (3.5) 2 ] = −1814 × 106
(v A − Δv A ) 2 = 161.24 × 106
(v A − Δv A ) = 12.698 × 103 m/s
(Take positive root; negative root reverses flight direction.)
v A = 26.9 × 103 m/s
(given)
Δv A = (26.9 × 103 m/s − 12.698 × 103 m/s)
Δv A = 14.20 km/s 
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662
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 v A 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/s 2 )(6.37 × 106 m)2 = 1.3247(10)20 m3 /s 2
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
1.3247(10)20
= vB2 −
(1.52273) 2 vB2 −
9
2
2
149.6(10)
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,
v A − vE = 32.695 − 29.758 = 2.94 km/s 
(b)
Increase at B,
vB − vM = 24.115 − 21.471 = 2.64 km/s 
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663
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
mrAv A = mrBvB
vA =
rB
7.170
vB =
vB
6.690
rA
rA = 6370 km + 320 km = 6690 km,
rA = 6.690 × 106 m
v A = 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/s 2 )(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
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664
PROBLEM 13.105 (Continued)
TA + VA = TB + VB
1 2
1
mv A − 59.501 × 106 m = mvB2 − 55.5 × 106 m
2
2
v A2 − vB2 = 8.002 × 106
From (1)
v A = 1.0718vB
vB2 [(1.0718) 2 − 1] = 8.002 × 106
vB2 = 53.79 × 106 m 2 /s 2 ,
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
Δv A = v A − (v A )C = 7861 − 7714 = 147 m/s 
ΔvB = (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 
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665
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 φB .
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 m 2 /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 φB rB = 76.544(10)9
sin φB =
76.544(10)9
7.35(106 )[6.37(10)6 + 8.35(10)6 ]
= 0.707483
φ B = 45.0° 
(b)
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666
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 γ
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
mvcirc
r
2
mv
GMm
= m circ
2
r
r
GM
vcirc =
r
Fn =
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
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667
PROBLEM 13.107 (Continued)
Velocity at C:
Conservation of energy:
TA + VA = TC + VC
1
GM m 1
GM m
m v A2 −
= m vC2 −
2
2
rA
rC
 1 1
vC2 = v 2A + 2GM  − 
 rC rA 
1
1


= (11.165 × 103 ) 2 + 2(398.1 × 1012 ) 
−
6
6 
42.14
10
6.67
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 γ
rv
cos γ = A A
rC vC
(6.67 × 106 )(11.165 × 103 )
(42.14 × 106 )(4.919 × 103 )
cos γ = 0.35926
=
γ = 68.9° 
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668
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 α 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 α with the intended circular path.
For elliptic orbit.
Conservation of angular momentum:
r0 mv0 cos α = rA mv A
r

v A =  0 cos α  v0
 rA

(1)
Conservation of energy:
1 2 GMm 1 2 GMm
mv0 −
= mv A −
2
r0
2
rA
r0 
2GM 
v02 − v 2A =
1 − 
r0  rA 
Substitute for v A from (1)
  r 2
 2GM 
r0 
v02 1 −  0  cos 2 α  =
1 − 
r0  rA 
  rA 



But
v02 =
GM
,
r0
2
thus
r 

r 
1 −  0  cos 2 α = 2 1 − 0 
 rA 
 rA 
2
r 
r 
cos 2 α  0  − 2  0  + 1 = 0
 rA 
 rA 
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669
PROBLEM 13.108 (Continued)
Solving for
r0
rA
r0 + 2 ± 4 − 4cos 2 α
1 ± sin α
=
=
2
rA
2cos α
1 − sin 2 α
rA =
(1 + sin α )(1 − sin α )
r0 = (1  sin α )r0
1 ± sin α
also valid for Point A′
Thus,
rmax = (1 + sin α )r0
rmin = (1 − sin α )r0 
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670
PROBLEM 13.109
Upon the LEM’s return to the command module, the Apollo spacecraft of
Problem 13.88 was turned around so that the LEM faced to the rear. The
LEM was then cast adrift with a velocity of 200 m/s relative to the command
module. Determine the magnitude and direction (angle φ formed with the
vertical OC) of the velocity vC of the LEM just before it crashed at C on the
moon’s surface.
SOLUTION
Command module in circular orbit
rB = 1740 + 140 = 1880 km = 1.88 × 106 m
GM moon = 0.0123 GM earth = 0.0123 gR 2
= 0.0123(9.81)(6.37 × 106 )2 = 4.896 × 1012 m3 /s 2
R = 1740 km
ΣF = man
v0 =
GM m m
=
rB2
GMm
=
rB
mv02
rB
4.896 × 1012
1.88 × 106
v0 = 1614 m/s
vB = 1614 − 200 = 1414 m/s
Conservation of energy between B and C:
1 2 GM m m 1 2 GM m m
mvB −
= mvC −
rB
rC
2
2
vC2 = vB2 +
rC = R
2GMm  rB

− 1

rB  R

vC2 = (1414 m/s)2 + 2

(4.896 × 1012 m3/s 2 )  1.88 × 106
− 1

6
6

(1.88 × 10 m)  1.74 × 10

vC2 = 1.999 × 106 + 0.4191 × 106 = 2.418 × 106 m 2 /s 2
vC = 1555 m/s 
Conservation of angular momentum:
rB mvB = RmvC sin φ
rv
(1.88 × 106 m)(1414 m/s)
sin φ = B B =
= 0.98249
rC vC (1.74 × 106 m)(1555 m/s)
φ = 79.26°
φ = 79.3° 
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671
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 φB = 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/s 2 )(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
(1)
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672
PROBLEM 13.110 (Continued)
Conservation of angular momentum:
rA mv A = rB mvB sin φB
vB =
(rA )v A
4185  1 
=
vA
(rB )(sin φB ) 4000  sin 60° 
vB = 1.208 v A
(2)
Substitute vB from (2) in (1)
v 2A = (1.208v A ) 2 − 58.94 × 106
v A2 [(1.208) 2 − 1] = 58.94 × 106
v 2A = 128.27 × 106 ft 2 /s 2
v A = 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 
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673
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 Δv 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 φB
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 2A
2
2
1
1
ΔE109 = m(25.24 × 103 )2 − m(11.32 × 103 ) 2
2
2
6
ΔE109 = 254.46 × 10 m ft ⋅ lb
ΔE109 =
ΔE110 = (0.50)ΔE109 =
(254.46 × 106 m)
ft ⋅ lb
2
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674
PROBLEM 13.111* (Continued)
Thus, additional kinetic energy at A is
1
(254.46 × 106 m)
m( Δv A ) 2 = ΔE110 =
ft ⋅ lb
2
2
(1)
Conservation of energy between A and B:
TA =
1
2
m[(v A )circ
+ ( Δv A ) 2 ]
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(v A )circ = rB mvB sin φB
 r  (v )
(4185) (25.24 × 103 )
= 0.8565
sin φB =  A  A circ =
(4000) (30.88 × 103 )
 rB  (vB )
φB = 58.9° 
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675
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
r
v A = P vP
rA
(1)
1 2 GMm 1 2 GMm
mvP −
= mv A −
2
rP
2
rA
(2)
Conservation of energy:
Substituting for v A from (1) into (2)
2GM
vP2 −
rP
2
r 
2GM
=  P  vP2 −
rA
 rA 
  r 2 


1 −  P   vP2 = 2GM  1 − 1 
  rA  
 rP rA 


rA2 − rP2
rA2
with
vP2 = 2GM
rA − rP
rA rP
rA2 − rP2 = (rA − rP )( rA + rP )
vP2 =
2GM  rA 
  (3) 
rA + rP  rP 
Exchanging subscripts P and A
v A2 =
2GM  rP 
 
rA + rP  rA 
Q.E.D.

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676
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

−
rP
2  ( rA + r0 ) 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.
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677
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 θ with the vertical, its
velocity must be reduced to α v0, where
α = sin θ
2(n − 1)
n − sin 2 θ
(b) the probe will not hit the planet if α is larger than
2 /(1 + n).
2
SOLUTION
(a)
Conservation of energy:
1
m (α 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
= mv 2 −
m (α v0 ) 2 −
2
nR
2
R
(1)
Conservation of angular momentum:
nR mα v0 = Rmv sin θ
v=
nα v0
sin θ
(2)
Replacing v in (1) by (2)
2
(α v0 )2 −
2GM  nα v0 
2GM
=
−

nR
R
 sin θ 
(3)
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678
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)
α2
GM 2GM
n 2α 2  GM  2GM
−
=

−
nR
nR
R
sin 2 θ  nR 

α 2 1 −

n2 
 = 2(1 − n)
sin 2 θ 
α2 =
2(1 − n)(sin 2 θ ) 2(n − 1)sin 2 θ
= 2
(sin 2 θ − n 2 )
(n − sin 2 θ )
α = sin θ
(b)
2(n − 1)
n − sin 2 θ
2
Q.E.D.

2
n +1

Probe will just miss the planet if θ > 90°,
α = sin 90°
Note:
2( n − 1)
=
n − sin 2 90°
2
n 2 − 1 = ( n − 1)(n + 1) 
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679
PROBLEM 13.115
A missile is fired from the ground with an initial velocity v 0 forming an angle φ 0 with the vertical. If the
missile is to reach a maximum altitude equal to α R, where R is the radius of the earth, (a) show that the
required angle φ 0 is defined by the relation
sin φ0 = (1 + α ) 1 −
α  vesc 


1 + α  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 φ0 = rB mvB
rB = R + α R = (1 + α ) R
vB =
Rv0 sin φ0 v0 sin φ0
=
(1 + α ) R
(1 + α )
(1)
Conservation of energy:
1 2 GMm 1 2
GMm
= mvB −
mv0 −
R
2
2
(1 + α ) R
TA + VA = TB + VB
v02 − vB2 =
2 GMm 
1  2 GMm  α 
=
1−

R  1 + α 
R  1 + α 
Substitute for vB from (1)

sin 2 φ0  2 GMm  α 
v02 1 −
=


 (1 + α ) 2 
R 1+α 


From Equation (12.43):
2
=
vesc
2GM
R

sin 2 φ0  2  α 
v02 1 −
=v
 (1 + α ) 2  esc  1 + α 


2
v  α
= 1 −  esc 
2
(1 + α )
 v0  1 + α
sin 2 φ0
α  vesc 
sin φ0 = (1 + α ) 1 −


1 + α  v0 
(2)
2
Q.E.D.
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680
PROBLEM 13.115 (Continued)
(b)
Allowable values of v0 (for which maximum altitude = α R)
0 < sin 2 φ0 < 1
For sin φ0 = 0, from (2)
2
v  α
0 = 1 −  esc 
 v0  1 + α
v0 = vesc
α
1+ α
For sin φ0 = 1, from (2)
2
v  α
1
= 1 −  esc 
2
(1 + α )
 v0  1 + α
2
 vesc 
1
1  1 + 2α + α 2 − 1 2 + α
=

 = 1 + α −
=
α
α (1 + α )
1+ α 
1+ α
 v0 
v0 = vesc
1+α
2 +α
vesc
α
1+α
< v0 < vesc

1+α
2 +α
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681
PROBLEM 13.116
A spacecraft of mass m describes a circular orbit of radius r1 around
the earth. (a) Show that the additional energy ΔE which must be
imparted to the spacecraft to transfer it to a circular orbit of larger
radius r2 is
ΔE =
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 Δ E A and Δ EB which must be imparted at A and B are,
respectively, proportional to r2 and r1 :
Δ EA =
r2
r1 + r2
ΔE
Δ EB =
r1
r1 + r2
ΔE
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 −
=−
r
2
2 r
(1)
Thus ΔE 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 )
ΔE =
Q.E.D.
2r1r2
Δ E = E1 − E2 = −
(b)
(2)
For an elliptic orbit, we recall Equation (3) derived in
Problem 13.113 (with vP = v1 )
v12 =
2Gm r2
(r1 + r2 ) r1
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682
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
ΔE A = T1 − Tcirc =
Recall Equation (2):
2GMr2
1
1 GM
− m
m
2 r1 (r1 + r2 ) 2
r1
ΔE A =
GMm(2r2 − r1 − r2 ) GMm( r2 − r1 )
=
2r1 (r1 + r2 )
2r1 (r1 + r2 )
ΔE A =
r2  GMm(r2 − r1 ) 


r1 + r2 
2r1r2

ΔE A =
r2
ΔE
(r1 + r2 )
Q.E.D.
r1
ΔE
(r1 + r2 )
Q.E.D.
A similar derivation at Point B yields,
ΔEB =
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683
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 α 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 α + 90° = θ0 , that is if cos θ0 = − sin α . We must therefore prove that
cos θ0 = − sin α
(1)
We recall from Equation (12.39):
1 GM
= 2 + C cos θ
r
h
When θ = 0,
r = rmin
and rmin = r0 (1 − sin α )
1
GM
= 2 +C
r0 (1 − sin α )
h
For θ = 180°,

(2)
(3)
r = rmax = r0 (1 + sin α )
1
GM
= 2 −C
r0 (1 + sin α )
h
(4)
Adding (3) and (4) and dividing by 2:
GM
1 
1
1

=
+

2
2r0  1 − sin α 1 + sin α 
h
1
=
r0 cos 2 α
Subtracting (4) from (3) and dividing by 2:
C=
1 
1
1
  1  2sin α
−


=
2r0  1 − sin α 1 + sin α   2r0  1 − sin 2 α
sin α
C=
r0 cos 2 α

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684
PROBLEM 13.117* (Continued)
Substituting for GM2 and C into Equation (2)
h
1
1
(1 + sin α cos θ )
=
r r0 cos 2 α
(5)
Letting r = r0 and θ = θ0 in Equation (5), we have
cos 2 α = 1 + sin α cos θ0
cos 2 α − 1
sin α
sin 2 α
=−
sin α
= − sin α
cos θ0 =
This proves the validity of Equation (1) and thus P0 is an end of the minor axis of the elliptic orbit.
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685
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
2
GM 
2 E  h  
= 2 1+ 1+


rmin
m  GM  
h 


1
(c) Show that the eccentricity ε of the trajectory of the vehicle can be expressed as
ε = 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
r mv
= min max
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)
(2) 
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 
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686
PROBLEM 13.118* (Continued)
Solving the quadratic:
1
rmin
=
2( )
GM
 GM 
+  2  + 2m
2
h
h
 h 
=
GM 
2E  h 
1+ 1+


2
m  GM 
h 

2
E
Rearranging
1
rmin
(c)
2
(3) 
Eccentricity of the trajectory:
Eq. (12.39′)
1 GM
= 2 (1 + ε cos θ )
r
h
When θ = 0,
cos θ = 1 and r = rmin
Thus,
1
rmin
Comparing (3) and (4),
(d )
=
GM
(1 + ε )
h2
ε = 1+
(4)
2E  h 
m  GM 
2
(5)
Recalling discussion in section 12.12 and in view of Eq. (5)

1. Hyperbola if ε > 1, that is, if E > 0

2. Parabola if ε = 1, that is, if E = 0

3. Ellipse if ε < 1, that is, if E < 0

Note: For circular orbit ε = 0 and
2
2E  h 
1+
=0
m  GM 
GM
r
2
 GM  m
,
E = −

 h  2
or
but for circular orbit
v2 =
thus
1 (GM ) 2
1 GMm
E=− m
=−

2
GMr
2 r
and
h2 = v 2 r 2 = GMr ,
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687
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
Answer: (d) This is Newton’s 3rd Law.
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688
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
Answer: (c) In both cases the car will come to a complete stop, so the applied impulse will be the same.
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689
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 μk = 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 θ = 20°.
SOLUTION
Answer:
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690
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 impulse-momentum 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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691
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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692
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:
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693
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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694
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 × 10 kg)(1.1111 m/s) − (150 × 103 N)t = 0
t = 259.26 s
6
t = 4 min19 s 
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695
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 ( μ k = 0.75), (b) on an
icy road ( μ k = 0.10).
SOLUTION
v1 = 60 mph = 88 ft/s
mv1 − μkWt = 0
t=
(a)
(b)
mv1
mv1
v
=
= 1
μkW μk mg μk g
For μk = 0.75
t=
88 ft/s
(0.75)(32.2 ft/s 2 )
t = 3.64 s 
t=
88 ft/s
(0.10)(32.2 ft/s 2 )
t = 27.3 s 
For μk = 0.10
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696
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 = 178.6 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.
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697
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 + ΣImp1 → 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° − μk Nt = mv2
(T − mg sin 20° − μk mg cos 20°)t = mv2
[T − mg (sin 20° + μk cos 20°)]t = mv2
Data:
T = 2500 N, m = 300 kg,
g = 9.81 m/s 2 ,
μk = 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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698
PROBLEM 13.123
A truck is traveling down a road with a 3-percent grade at a
speed of 55 mi/h when the brakes are applied. Knowing the
coefficients of friction between the load and the flatbed trailer
shown are μs = 0.40 and μk = 0.35, 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:
F f = μs N
mv1 + ΣImp1→2 = mv 2
Components in y-direction:
0 + Nt − mgt cos θ = 0
N = mg cos θ
mv1 + mgt sin θ − μ s Nt = mv2
Components in x-direction:
mv1 + mgt (sin θ − μ s cos θ ) = 0
t=
Data:
v1
g ( μ s cos θ − sin θ )
v1 = 55 mi/h = 80.667 ft/s, v2 = 0,
g = 32.2 ft/s 2 , μ s = 0.40,
tan θ = 3/100
θ = 1.71835°
μs cos θ − sin θ = 0.36983
t=
80.667
(32.2)(0.36983)
t = 6.77 s 
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699
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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700
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 θ
m v1 − μs m gt1− 2 cos θ + m gt1− 2 sin θ = m v2
(176 ft/s) − μs (32.2 ft/s 2 )(12 s)(cos 2.86°) + (32.2 ft/s 2 )(12 s)(sin 2.86°) = 88 ft/s
μs =
176 − 88 + (32.2)(12)(sin 2.86°)
(32.2)(12)(cos 2.86°)
μs = 0.278 
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701
PROBLEM 13.126
A 2-kg particle is acted upon by the force, expressed in newtons, F = (8 − 6t )i + (4 − t 2 ) j + (4 + t )k. Knowing
that the velocity of the particle is v = (150 m/s)i + (100 m/s)j − (250 m/s)k at t = 0, determine (a) the time at
which the velocity of the particle is parallel to the yz plane, (b) the corresponding velocity of the particle.
SOLUTION

mv 0 + Fdt = mv

Where
Fdt =
(1)
t
 [(8 − 6t )i + (4 − t ) j + (4 + t )k ]dt
2
0
1  
1 

= (8t − 3t 2 ) i +  4t − t 3  j +  4t + t 2  k
3  
2 

Substituting m = 2 kg,
v0 = 150i + 100 j − 250k into (1):
1  
1 

(2 kg)(150i + 100 j − 250k ) + (8t − 3t 2 )i +  4t − t 3  j +  4t + t 2  k = (2 kg) v
3  
2 

3  
1  
1 

v = 150 + 4t − t 2  i + 100 + 2t − t 3  j +  −250 + 2t + t 2  k
2  
6  
4 

(a)
v is parallel to yz plane when vx = 0, that is, when
3
150 + 4t − t 2 = 0 t = 11.422 s
2
(b)
t = 11.42 s 
1


v = 100 + 2(11.422) − (11.422)3  j
6


1


+  −250 + 2(11.422) + (11.422) 2  k
4


v = −(125.5 m/s) j − (194.5 m/s)k 
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702
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
θ = tan −1
4
= 2.29°
100
mv1 + Σ imp1− 2 = mv 2
mv1 + Wt sin θ − Ft = mv2
v1 = 60 mi/h = 88 ft/s
N = W cos θ W = mg
v2 = 20 mi/h = 29.33 ft/s
F = μ s N = μ sW cos θ
( m )(88 ft/s) + ( m )(32.2 ft/s 2 )(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 
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703
PROBLEM 13.128
Skid marks on a drag race track indicate that the rear (drive)
wheels of a car slip for the first 20 m of the 400-m track.
(a) Knowing that the coefficient of kinetic friction is 0.60,
determine the shortest possible time for the car to travel the
initial 20-m portion of the track if it starts from rest with its
front wheels just off the ground. (b) Determine the minimum
time for the car to run the whole race if, after skidding for 20 m,
the wheels roll without sliding for the remainder of the race.
Assume for the rolling portion of the race that 65 percent of the
weight is on the rear wheels and that the coefficient of static
friction is 0.85. Ignore air resistance and rolling resistance.
SOLUTION
(a)
First 20 m
Velocity at 20 m. Rear wheels skid to generate the maximum force resulting in maximum velocity and
minimum time since all the weight is on the rear wheel: This force is F = μ k N = 0.60W .
T0 + U 0− 20 = T20
Work and energy.
T0 = 0
U 0 − 20 = ( F )(20)
T20 =
1 2
mv20
2
1
2
m v20
2
2
v20
= (2)(0.60)(20 m)(9.81 m/s 2 )
0 + μk mg (20) =
v20 = 15.344 m/s
Impulse-momentum.
0 + μk mgt0 − 20 = mv20
t0− 20 =
v20 = 15.344 m/s
15.344 m/s
(0.60)(9.81 m/s 2 )
t0− 20 = 2.61 s 
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704
PROBLEM 13.128 (Continued)
(b)
For the whole race:
The maximum force on the wheels for the first 20 m is F = μk mg = 0.60 mg. For remaining 360 m, the
maximum force, if there is no sliding and 65 percent of the weight is on the rear (drive) wheels, is
F = μ s (0.65) mg = (0.85)(0.65) mg = 0.5525 mg
Velocity at 400 m.
T0 + U 0− 20 + U 20 − 400 = T400
Work and energy.
T0 = 0
T400 =
U 0 − 20 = (0.60mg )(20 m),
U 60 − 400 = (0.5525mg )(380 m)
1 2
mv400
2
0 + 12 mg + (0.5525)(380)mg =
1 2
mv400
2
v400 = 65.990 m/s
Impulse–momentum.
From 20 m to 400 m
F = μs N = 0.510mg
v20 = 15.344 m/s
v400 = 65.990 m/s
m (15.344) + 0.5525 mgt20 − 400 = m (65.990); t20 − 400 = 9.3442 s
t0− 400 = t0 − 20 + t20 − 400 = 2.61 + 9.34
t0− 400 = 11.95 s 
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705
PROBLEM 13.129
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 time required to bring the train to a stop, (b) the force in
each coupling.
SOLUTION
Weights of cars:
WA = WC = 80,000 lb, WB = 100, 000 lb
Masses of cars:
m A = mC = 2484 lb ⋅ s 2 /ft, mB = 3106 lb ⋅ s 2 /ft
For each car the normal force (upward) is equal in magnitude to the weight of the car.
N A = NC = 80,000 lb
Friction forces:
FA = 0
N B = 100, 000 lb
(brakes not applied)
FB = (0.35)(100, 000) = 35000 lb
FC = (0.35)(80, 000) = 28, 000 lb
Stopping data:
(a)
v1 = 30 mi/h = 44 ft/s, v2 = 0.
Apply the principle of impulse-momentum to the entire train.
m = m A + mB + mC = 8074 lb ⋅ s 2 /ft
F = FA + FB + FC = 63, 000 lb
− mv1 + Ft = mv2
t =
(b)
m(v1 − v2 ) (8074)(44)
=
= 5.639 s
F
63, 000
t = 5.64 s 
Coupling force FAB:
Apply the principle of impulse-momentum to car A alone.
−m Av1 + FAt + FABt = 0
−(2484)(44) + 0 + FAB (5.639) = 0
FAB = 19,390 lb
FAB = 19,390 lb (tension) 
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706
PROBLEM 13.129 (Continued)
Coupling force FBC :
Apply the principle of impulse-momentum to car C alone.
−mC v1 + FC t − FBC t = 0
−(2484)(44) + (28000)(5.639) − FBC (5.639) = 0
FBC = 8620 lb
FBC = 8620 lb (tension) 
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707
PROBLEM 13.130
Solve Problem 13.129 assuming that the brakes are applied only on the wheels of car A.
PROBLEM 13.129 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 time required to bring the train to a stop, (b) the force in each coupling.
SOLUTION
Weights of cars:
WA = WC = 80,000 lb, WB = 100, 000 lb
Masses of cars:
m A = mC = 2484 lb ⋅ s 2 /ft, mB = 3106 lb ⋅ s 2 /ft
For each car the normal force (upward) is equal in magnitude to the weight of the car.
N A = N C = 80,000 lb
N B = 100, 000 lb
FA = (0.35)(80, 000) = 28, 000 lb
Friction forces:
FB = 0 
 (brakes not applied)
FC = 0 
Stopping data:
(a)
v1 = 30 mi/h = 44 ft/s, v2 = 0.
Apply the principle of impulse-momentum to the entire train.
m = m A + mB + mC = 8074 lb ⋅ s 2 /ft
F = FA + FB + FC = 28, 000 lb
− mv1 + Ft = mv2
t =
(b)
m(v1 − v2 ) (8074)(44)
=
= 12.688 s
F
28, 000
t = 12.69 s 
Coupling force FAB:
Apply the principle of impulse-momentum to car A alone.
−mAv1 + FAt + FABt = 0
−(2484)(44) + (28, 000)(12.688) + FAB (12.688) = 0
FAB = −19,390 lb
FAB = 19,390 lb (compression) 
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708
PROBLEM 13.130 (Continued)
Coupling force FBC :
Apply the principle of impulse-momentum to car C alone.
−mC v1 + FC t − FBC t = 0
−(2484)(44) + (0) − FBC (12.688) = 0
FBC = −8620 lb
FBC = 8620 lb (compression) 
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709
PROBLEM 13.131
A trailer truck with a 2000-kg cab and an 8000-kg trailer is
traveling on a level road at 90 km/h. The brakes on the trailer
fail and the antiskid system of the cab provides the largest
possible force which will not cause the wheels of the cab to
slide. Knowing that the coefficient of static friction is 0.65,
determine (a) the shortest time for the rig to come to a stop,
(b) the force in the coupling during that time.
SOLUTION
v = 90 km/h = 25 m/s
(a)
The shortest time for the rig to come to a stop will be when the friction force on the wheels is maximum.
The downward force exerted by the trailer on the cab is assumed to be zero. Since the trailer brakes fail,
all of the braking force is supplied by the wheels of the cab, which is maximum when the wheels of the
cab are at impending sliding.
Ft1− 2 = μs N C t1− 2
N C = mC g = (2000) g
Ft1− 2 = (0.65)(2000) gt
[( mC + mT )v]2 = − Ft + [(mC + mT )v]1
0 = −(0.65)(2000 kg)(9.81 m/s 2 )(t1− 2 ) = 10, 000 kg(25 m/s)
t1− 2 = 19.60 s 
(b)
For the trailer:
[mT v]2 = −Qt1− 2 + [mT v]1
From (a),
t1− 2 = 19.60 s
0 = −Q(19.60 s) + (8000 kg)(25 m/s)
Q = 10, 204 N
Q = 10.20 kN (compression) 
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710
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 = 2 vB
Masses and weights:
vB =
m A = 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
(1)
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711
PROBLEM 13.132 (Continued)
Principle of impulse and momentum applied to weight A.
: 0 + Tt − WAt = mA (v A ) 2
Tt + WAt = m A (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=
t = 0.549 s 
15
+ 29.43
t
T = 56.8 N 
Tension in the cable.
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712
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 = [(m A + 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/s 2 )(0.8 s)
16 kg
v = 3.92 m/s 
Collar A:
(m Av)1 = 0 0 + ( FC + m A 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 
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713
PROBLEM 13.134
An estimate of the expected load on over-the-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 Δt
ave
Δt = 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 Δt = (3727 lb)(0.110 s)
Impulse = area under F −t diagram =
1
Fm (0.110) = (3727)(0.110)
2
Fm = 7450 lb 
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714
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
Mass:
m = 0.060 kg
Weight:
mg = (0.060)(9.81) = 0.5886 N
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
t1
0
0
 Sdt =  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.
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715
PROBLEM 13.135 (Continued)
(a)
Maximum speed (upward motion):
Apply the principle of impulse-momentum to the rocket over 0 ≤ t ≤ t2 .
0.8
 Pdt = area under the given thrust-time plot.
0
1
1
(0.2)(13) + (0.1)(13 + 5) + (0.8 − 0.3)(5)
2
2
= 4.7 N ⋅ s
=
0.8
 Wdt = (0.5886)(0.8) = 0.47088 N ⋅ S
0
m1v1 +

0.8
0
Pdt +

0.8
0
Sdt −
0.8
 Wdt = mv
2
0
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
t3
0
0
 Pdt +  Sdt − Wt = mv
3
3
0 + 4.7 + 0.00266 − 0.5886t3 = 0
t3 = 7.99 s 
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716
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
t = 0,
p = p0 = c1 − c2t
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
π (10 × 10−3 )2
A=
4
A = 78.54 × 10−6 m 2
0+ A

1.6 ×10−3 s
0
(c1 − c2t ) 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
−3
(1.6 × 10 s)
p0 = 222.8 × 106 N/m 2
p0 = 223 MPa  
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717
PROBLEM 13.137
A 125-lb block initially at rest is acted upon by a force P which
varies as shown. Knowing that the coefficients of friction between
the block and the horizontal surface are μs = 0.50 and μk = 0.40,
determine (a) the time at which the block will start moving, (b) the
maximum speed reached by the block, (c) the time at which the
block will stop moving.
SOLUTION


0 + Pdt − Fdt = mv
v=
At any time:
(a)
1
Pdt − Fdt 

m 


(1)
Block starts moving at t.
P = Fs = μsW = (0.50)(125 lb) = 62.5 lb
t1
t1
8s
8s
;
=
=
Fs 100 lb 62.5 lb 100 lb
(b)
t1 = 5.00 s 
Maximum velocity: At t = tm
P = Fk = μkW = 0.4(125) = 50 lb
where
Block moves at t = 5 s.
Shaded area is maximum net impulse
 Pdt −  F dt
R
when t = tm1 v = vm
Eq. (1):
vm =
1 shaded  1  1
1
 1
(12.5 + 50)(3) + (50)(4)  = (193.75)

=
2
m  area  m  2
 m
1
vm = 125 lb [193.75] = 49.91 ft/s
v m = 49.9 ft/s

32.2
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718
PROBLEM 13.137 (Continued)
(c)
Block stops moving when  Pdt − Fdt  = 0; or




 Qdt =  Fdt
Assume tm > 16 s.
1
 Pdt = 2 (100)(16) = 800 lb ⋅ s
1
 Fdt = 2 (62.5)(5) + (50)(t − 5)
m
 Pdt −  Fdt = 800 − [156.25 + 50(t − 5)] = 0
m
tm > 16 s OK
tm = 17.875 s
tm = 17.88 s 
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719
PROBLEM 13.138
Solve Problem 13.137, assuming that the weight of the block is 175 lb.
PROBLEM 13.137 A 125-lb block initially at rest is acted upon by a
force P which varies as shown. Knowing that the coefficients of
friction between the block and the horizontal surface are μs = 0.50
and μk = 0.40, determine (a) the time at which the block will start
moving, (b) the maximum speed reached by the block, (c) the time at
which the block will stop moving.
SOLUTION
See solution of Problem 13.137.
W = 175 lb
(a)
Block starts moving:
v=
1
Pdt − Fdt 

m 


(1)
P = Fs = μsW = (0.50)(175) = 87.5 lb
See first figure of Problem 13.137.
t1
t1
8s
8s
=
=
;
Fs 100 lb 87.5 lb 100 s
(b)
Maximum velocity:
t1 = 7.00 s 
P = Fk = μkW = 0.4(175) = 70 lb
16 − tm
8s
=
70 lb 100 lb
 8 
16 − tm = 70 
 = 5.6
 100 
tm = 10.40 s
Eq. (1):
vm =
1 shaded 


m  area 
1 1
1

(17.5 + 30)(1.0) + (30)(10.4 − 8) 

m 2
2

1
= (59.75)
m
=
1
vm = 175 lb [59.75]
32.2
= 10.994 ft/s
v m = 10.99 ft/s

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720
PROBLEM 13.138 (Continued)
(c)
Block stops moving when net impulse
 ( P − F )dt  = 0



Assume ts , < 16 s.
1
1
ts
 Pdt = 2 (100)(8) + 2 100 + 100
0
=
ts
(16 − t s ) 
 (t s − 8)
8

1
1  100 
(100)(16) − 
(16 − t s ) 2
2
2  8 
1
 Fdt = 2 (87.5)(7) + (70)(t − 7)
s
0
100
 Pdt −  Fdt = 800 − 16 (16 − t ) − 306.25 − 70(t − 7) = 0
2
s
Solving for ts ,
ts = 13.492 s
s
ts = 13.49 s 
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721
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
m=
5
/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 s
(32.2 ft/s 2 ) ( 60
)
Fav = 76.9 lb 
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722
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 (π 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
t0
0
0
πt
πt
Fm t0
 Fdt =  F sin t dt = − π cos t
=−
t0
m
0
Fm t0
π
(cos π − cos 0) =
0 0
2 Fm t0
π
Principle of impulse and momentum.
mv1 +
t0
 F dt = mv
0
2
with v1 = 0,
0+
Solving for Fm,
Fm =
2 Fm t0
π
π mv2
2t0
= mv2
=
π (3.1444 × 10−3 )(146.67)
(2)(0.5 × 10−3 )
= 1.4488 × 103 lb
Fm = 1.45 kip  
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723
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)Δt = mv 2
Δt = 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 
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724
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
sand-filled 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
Δt = 0.22 s
mv1 + (P − W)Δt = 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 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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725
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
Δt = 2 ms = 0.002 s
(a)
Velocity immediately after impact:
Use principle of impulse and momentum:
v1 = 0
v2 = ?
Imp1→2 = Fave (Δt )
mv1 + Imp1→2 = mv 2
0 + Fave (Δt ) = mv2
v2 =
(b)
Fave (Δt ) (2000)(0.002)
=
m
0.200
v2 = 20.0 m/s 
Average resistance to penetration:
Δx = 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 ( Δx) = 0
2
mv22
(0.200)(20.0) 2
=
= 40 × 103 N
2(Δx)
(2)(0.001)
Rave = 40.0 kN 
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726
PROBLEM 13.144
A 25-g steel-jacketed bullet is fired horizontally with a
velocity of 600 m/s and ricochets off a steel plate along the
path CD with a velocity of 400 m/s. Knowing that the
bullet leaves a 10-mm scratch on the plate and assuming
that its average speed is 500 m/s while it is in contact with
the plate, determine the magnitude and direction of the
average impulsive force exerted by the bullet on the plate.
SOLUTION
Impulse and momentum.
Bullet alone:
mv1 + F Δ t = mv 2
t direction:
mv1 cos15° − Ft Δt = mv2 cos 20°
Ft Δt = (0.025 kg)[600 m/s cos 15° − 400 m/s cos 20°] = 5.092 kg ⋅ m/s
Δt =
S BC 0.010 m
=
= 20 × 10−6 s
v AV 500 m/s
Ft = (5.092 kg ⋅ m/s)/(20 × 10−6 s) = 254.6 × 103 kg ⋅ m/s 2 = 254.6 kN
n direction:
− mv1 sin15° + Fn Δt = mv2 sin 20°
Fn Δt = (0.025 kg)[600 m/s sin 15° + 400 m/s sin 20°] = 7.3025 kg ⋅ m/s
Fn = (43025 kg ⋅ m/s)/(20 × 10−6 ) = 365.1 × 103 kg ⋅ m/s 2 = 365.1 kN
Force on bullet:
F = Fn2 + Ft 2 = 365.12 + 254.62 = 445 kN
tanθ =
Fn 365.1
=
Ft 254.6
θ = 55.1°
θ − 15° = 40.1°
F = 445 kN
Force on plate:
40.1°
F ′ = −F
F ′ = 445 kN
40.1° 
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727
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 ( μk = 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 = m A v′ + mB v′
v′ =
(b)
mB vB
(3106)(3.6667)
=
= 2.444 ft/s
4569
mA + mB
v′ = 1.667 mi/h

After coupling: The friction force acts only on car A.
+ ΣF = 0 A : N A − W A = 0
N A = WA
FA = μk N A = μkWA
(sliding)
FB = 0 (Car B is rolling.)
Apply impuslse-momentum to the coupled cars.
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728
PROBLEM 13.145 (Continued)
: −(m A + mB )v′ + FAt = 0
t=
(m A + mB )v11 mB vB
=
μk W A
FA
t=
(1553)(3.6667)
= 0.1898
(0.30)(100, 000)
t = 0.190 s 
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729
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.
Σmv, x :
m A v A cos 30° = (m A + mB )v cos 10°
(1)
Σmv, y :
m A v A sin 30° − mB vB = (mA + mB )v sin 10°
(2)
Dividing (1) into (2),
mB vB
sin 30°
sin 10°
−
=
cos 30° m Av A cos 30° cos 10°
vB (tan 30° − tan 10°)( mA cos 30°)
=
vA
mB
vB
(1500)
= (0.4010)
cos 30°
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
v A = (2.30)(50)
v A = 115.2 km/h 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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730
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 /s 2
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.
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731
PROBLEM 13.147 (Continued)
Thus,
mH vH = (mH + m p )v′
v′ =
Work and energy:
mH vH
(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 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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732
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
mA = 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 =
: Σmv1 = Σ mv2
mH vH = ( mH + mA )v2
v2 =
mH vH
mH + m A
TA =
1
1 mH2 vH2
(mH + m A )v22 =
2
2 mH + mA
T2 =
mH
T1
mH + m A
(1)
Kinetic energy after impact:
(2)
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733
PROBLEM 13.148 (Continued)
Impulse exerted on the hammer:
: mH vH − F (Δt ) = mH v2
F Δt = 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 (Δt ) = (0.04658)(20 − 0) = 0.932 lb ⋅ s
The impulse exerted on the rivet the same magnitude but opposite to direction.
F Δt = 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 (Δt ) = (0.04658)(20 − 2.857)
F (Δt ) = 0.799 lb ⋅ s 
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734
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 μk = 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:
m A = 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 = m A g = 3.00 lb.
Block C + bullet:
N C = (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 + m Av 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
(2)
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735
PROBLEM 13.149 (Continued)
3.
Block A slides on the plane. Use principle of work and energy.
T1 + U1→2 = T2
1
m Av A2 − μ k N A d A = 0 or v A =
2
4.
2μk N A d A
mA
(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 − μ k N C dC = 0 or vC =
2
2 μk NC 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 
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736
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

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737
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 before impact
Just after 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 + 0.075 g (1.6)
2
1
1
(0.075)(2)2 + 0.075 g (0.6) + (0.4)(1.694) 2
2
2
Energy lost = (1.3272 − 1.1653) J = 0.1619 J 
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738
PROBLEM 13.152
A 2-kg sphere A is connected to a fixed Point O by an inextensible cord of length
1.2 m. The sphere is resting on a frictionless horizontal surface at a distance of 1.5 ft
from O when it is given a velocity v 0 in a direction perpendicular to line OA. It
moves freely until it reaches position A′, when the cord becomes taut. Determine
the maximum allowable velocity v 0 if the impulse of the force exerted on the cord
is not to exceed 3 N ⋅ s.
SOLUTION
For the sphere at A′ immediately before and after the cord becomes taut
mv0 + F Δt = mv A′
mv0 sin θ − F Δt = 0
F Δt = 3 N ⋅ s
m = 2 kg
2(sin 65.38°)v0 = 3
v0 = 1.650 m/s 
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739
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.
Σm v1 + ΣImp1→ 2 = mv 2
mv0 cos 30° + 0 = (m + M )v′
Components :
mv0 cos 30° (0.001941)(1400)cos 30°
=
0.157221
m+M
v′ = 14.968 ft/s
v′ =
v′ = 14.97 ft/s 
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740
PROBLEM 13.153 (Continued)
(b)
Use the principle of impulse and momentum applied to the bullet alone.
x-components:
−mv0 sin 30° + Rx Δt = 0
Rx Δt = mv0 sin 30° = (0.001941)(1400) sin 30°
= 1.3587 lb ⋅ s
y-components:
Rx Δt = 1.359 lb ⋅ s 
−mv0 cos 30° + Ry Δt = −mv′
Ry Δt = m(v0 cos 30° − v′)
= (0.001941)(1400 cos 30° − 14.968)
Ry Δt = 2.32 lb ⋅ s 
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741
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 Δt = 0
 60 
  (17.94) = F Δt
 g 
F Δt = 33.43 lb ⋅ s on the block.
F Δt = 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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742
PROBLEM 13.154 (Continued)
(b)
Elastic columns:
Momentum of system of block and beam is conserved.
mv = ( M + m)v′
m
60
v′ =
v=
(17.94 ft/s)
(m + M )
300
v′ = 3.59 ft/s
Referring to figure in part (a),
− mv + F Δt = − mv′
F Δt = 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 Δt = 26.7 lb ⋅ s 
E=
E = 240 ft ⋅ lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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743
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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744
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:
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745
PROBLEM 13.F7
An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg
flatcar C carrying a 30-Mg load B which can slide along the floor of
the car (μk = 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 impulsemomentum 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.
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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746
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:
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747
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:
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748
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:
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749
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):
Σmv1 + ΣImp1→2 = Σmv 2
Horizontal components
Using data,
: m A v A + mB vB = mA v′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 Av 2A + mB vB2 = (5)(1)2 + (3)(−1.5)2 = 5.875 J
2
2
2
2
1
1
1
1
T2 = m A (v′A )2 + mB (vB′ )2 = (5)( −0.59375) 2 + (3)(1.15625) 2 = 2.8867 J
2
2
2
2
Kinetic energies: T1 =
(b)
T1 − T2 = 2.99 J 
Energy loss:
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750
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):
Σmv1 + ΣImp1→2 = Σmv 2
Horizontal components
: m A v A + mB vB = mA v′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),
−2v A = 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 2A + 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 =
T1 − T2 = (1 − e2 ) mv 2 
Energy loss:
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751
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=
v′
v
e=
h′
h
Height of drop
h = 100 in.
Height of bounce
53 in. ≤ h′ ≤ 58 in.
Thus,
53
58
≤e≤
100
100
0.728 ≤ e ≤ 0.762 
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752
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:
m A v A + mB vB = m A v′A + mB v′
(m A )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,

2.00 lb ≤ WB ≤ 6.00 lb 
Range:
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753
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
m A = 20 kg, mB = ?
First impact: A impacts B.
Σmv + ΣImp1→2 = Σmv 2
Impulse-momentum:
m A v0 = m Av′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
(3)
mB = ?, mC = 1 kg
Σmv 2 + ΣImp 2→3 = Σmv3
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754
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 )
mC vC′′ =
(1)(20) mB 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
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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755
PROBLEM 13.160
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
m A = 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
(3)
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756
PROBLEM 13.160 (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.
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757
PROBLEM 13.161
Three steel spheres of equal weight 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 v 0 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 = 5 and e = 0.9.
SOLUTION
(a)
First collision (between A and B):
The total momentum is conserved:
mv A + mvB = mv′A + mvB′
v0 = v′A + vB′
(1)
Relative velocities:
(v A − vB )e = (vB′ − v′A )
v0 e = vB′ − v′A
(2)
Solving Equations (1) and (2) simultaneously,
(b)
v′A =
v0 (1 − e)

2
vB′ =
v0 (1 + e)

2
Second collision (between B and C):
The total momentum is conserved.
mvB′ + mvC = mvB′′ + mvC′
Using the result from (a) for vB′
v0 (1 + e)
+ 0 = vB′′ + vC′
2
(3)
Relative velocities:
(vB′ − 0)e = vC′ − vB′′
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758
PROBLEM 13.161 (Continued)
Substituting again for vB′ from (a)
v0
(1 + e)
(e) = vC′ − vB′′
2
(4)
Solving equations (3) and (4) simultaneously,
vC′ =
(c)
1  v0 (1 + e)
(e) 
+ v0 (1 + e) 

2
2
2 
vC′ =
v0 (1 + e) 2

4
vB′′ =
v0 (1 − e2 )

4
For n spheres
n balls
(n − 1)th collision,
we note from the answer to part (b) with n = 3
vn′ = v3′ = vC′ =
v3′ =
or
v0 (1 + e)2
4
v0 (1 + e)(3−1)
2(3−1)
Thus, for n balls
vn′ =
(d )
v0 (1 + e)( n −1)
2( n −1)

For n = 5, e = 0.90,
from the answer to part (c) with n = 5
vB′ =
=
v0 (1 + 0.9)(5−1)
2(5−1)
v0 (1.9)4
(2) 4
vB′ = 0.815 v0 
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759
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 Av A + mB vB + mC vC = m Av′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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760
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.
mA 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
(9)
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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761
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.
m A v A + mB vB + mC vC = (240)(2) + 0 + (235)(−1.5)
= 127.5 kg ⋅ m/s
m A v′′′A + mB vB′′′ + mC vC′′ = (240)( −0.95633) + (260)(−0.02955) + (235)(1.55193)
= 127.50 kg ⋅ m/s.
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762
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:
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 = ?
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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763
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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764
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 θ = 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 θ = mv′At  v′At = v0 sin θ
Ball B: t-dir
0 = mB v′Bt  v′Bt = 0
Balls A + B: n-dir
mv0 cosθ + 0 = m v′An + m v′Bn
(1)
Coefficient of restitution
vBn
′ − v′An = e (v An − vBn )
v′Bn − v′An = e (v0 cosθ − 0)
(2)
Solve (1) and (2)
1 − e

1 + e 
v′An = v0 
cosθ  ; v′Bn = v0 
 cosθ
 2

 2 
With numbers
e = 0.8; θ = 45°
v′At = v0 sin 45° = 0.707 v0
 1 − 0.8

cos 45°  = 0.0707 v0
v′An = v0 
 2

v′Bt = 0
 1 + 0.8 
v′Bn = v0 
 cos 45° = 0.6364 v0
 2 
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765
PROBLEM 13.164 (Continued)
(A)
1
v" A = [(0.707 v0 ) 2 + (0.0707v0 )2 ] 2 = 0.711v0

= 0.711v0 
 0.0707 
 = 5.7106°
 0.707 
β = tan −1 
So
θ = 45 − 5.7106 = 39.3°
(B)
v′A = 0.711v0
39.3° 
v′B = 0.636 v0
45° 
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766
PROBLEM 13.165
The coefficient of restitution is 0.9 between the two 2.37-in. diameter
billiard balls A and B. Ball A is moving in the direction shown with
a velocity of 3 ft/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 θ , (b) the velocity of B after impact.
SOLUTION
(a)
Since vB′ 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 θ =
10
6−D
10
6 − 2.37
= 70.04°
θ = tan −1
θ = 70.0° 
(b)
Conservation of momentum in x(n) direction:
mv A cos θ + m(vB ) n = m(v′A ) n + mvB′
(3)(cos 70.04) + 0 = (v′A )n + vB′
1.0241 = (v′A )n + (vB′ )
(1)
Relative velocities in the n direction:
e = 0.9
(v A cos θ − (vB ) n )e = vB′ − (v′A ) n
(1.0241 − 0)(0.9) = vB′ − (v′A ) n
(1) + (2)
2vB′ = 1.0241(1.9)
(2)
v′B = 0.972 ft/s

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767
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:
m A(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)
m A (v A )t = mA (v′A )t −3.857 = (v′A )t
(v′A )t = −3.857 m/s
(2)
Ball A alone:
Momentum conserved:
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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768
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:
m A (v A ) n + mB (vB ) n = m A (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 β =
|(v A )t |
|(v A ) n |
3.857
5.075
β = 37.2°
=
β + 40° = 77.2°
v′A = (3.857) 2 + (5.075) 2
= 6.37 m/s
Velocity of B:
v′A = 6.37 m/s
77.2° 
v′B = 1.802 m/s
40° 
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769
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:
Σmv1 + ΣImp1→2 = Σmv 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:
mv A cos 20° − mvB sin 20° = m(v′A ) n + m(vB′ ) n
(v′A )n + (vB′ ) n = v A cos 20° − vB sin 20°
= 3cos 20° − 3sin 20°
(1)
e = 0.9
Coefficient of restitution:
(vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ]
= 0.9[3cos 20° − (−3)sin 20°]
(2)
Solving Eqs. (1) and (2) simultaneously,
(v′A )n = −0.8338 m/s
(vB′ )n = 2.6268 m/s
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770
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 α =
1.0261
0.8338
α = 50.9°
α + 20° = 70.9°
v′A = 1.322 m/s
70.9° 
vB′ = (2.6268) 2 + (2.8191) 2 = 3.85 m/s
tan β =
2.8191
2.6268
β = 47.0°
β − 20° = 27.0°
v′B = 3.85 m/s
27.0° 
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771
PROBLEM 13.168
Two identical pool balls of 57.15-mm 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 = 50 mm and e = 0.7, determine the velocity of
each ball after impact. (b) Show that if e = 1, the final velocities of the
balls from a right angle for all values of b.
SOLUTION
Geometry at instant of impact:
b
50
=
d 57.15
θ = 61.032°
sin θ =
Directions n and t are shown in the figure.
Principle of impulse and momentum:
Ball B:
Ball A:
Ball A, t-direction:
mv0 sin θ + 0 = m(v A )t
Ball B, t-direction:
0 + 0 = m(v B )t
Balls A and B, n-direction:
Coefficient of restitution:
(a)
(v A )t = v0 sin θ
(1)
(v B )t = 0
(2)
mv0 cos θ + 0 + m(v A ) n + m(vB ) n
(v A )n + (vB ) n = v0 cos θ
(3)
(vB ) n − (v A ) n = e[v0 cos θ ]
(4)
e = 0.7. From Eqs. (1) and (2),
(v A )t = 0.87489v0
(1)′
( vB ) t = 0
(2)′
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772
PROBLEM 13.168 (Continued)
From Eqs. (3) and (4),
(v A ) n + (vB )n = 0.48432v0
(3)′
(vB ) n − (v A ) n = (0.7)(0.48432v0 )
(4)′
Solving Eqs. (5) and (6) simultaneously,
(v A )n = 0.072648v0
(vB ) n = 0.41167v0
v A = (v A )n2 + (v A )t2
= (0.072648v0 ) 2 + (0.87489v02
= 0.87790v0
tan β =
(v A ) n 0.072648v0
=
= 0.083037
(v A )t
0.87489v0
β = 4.7468°
ϕ = 90° − θ − β
= 90° − 61.032° − 4.7468°
= 24.221°
(b)
v A = 0.878v0
24.2° 
v B = 0.412v0
61.0° 
e = 1. Eqs. (3) and (4) become
(v A )n + (vB )n = v0 cos θ
(3)′′
(vB )n − (v A ) n = v0 cosθ
(4)′′
Solving Eqs. (3)′′ and (4)′′ simultaneously,
(v A )n = 0, (vB )t = v0 cos θ
But
(v A )t = v0 sin θ , 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.
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773
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 θ
R
2
=
sin135°
R
θ = 20.705°
α = 45° − 20.705°
= 24.295°
Conservation of momentum for ball in t-direction:
−v sin θ = −v′ sin α
Coefficient of restitution in n:
Dividing,
v(cos θ )e = v′ cos α
tan θ
= tan α
e
e=
tan 20.705°
tan 24.295°
e = 0.837 
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774
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.
Σmv1 + ΣImp1→2 = Σmv 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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775
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
g
3.73 m/s 2
1
y = (10.819)(2.901) − (3.73)(2.901) 2
2
t=
=
y = 15.69 m 
(b)
Length of first bounce:
1
10.819t − (3.73) t 2 = 0
2
y = 0:
t = 5.801 s
x = (12.728)(5.801)
x = 73.8 m 
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776
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 v 0 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 cos 30°)e = v′n
(25)(cos 30°)(0.9) = v′n
v′n = 19.49 ft/s
Write v′ in terms of x and y components
(v′x )0 = v′n (cos 30°) − vt′ (sin 30°) = 19.49(cos 30°) − 12.5(sin 30°)
= 10.63 ft/s
(v′y )0 = vn′ (sin 30°) + vt′ (cos 30°) = 19.49(sin 30°) + 12.5(cos 30°)
= 20.57 ft/s
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777
PROBLEM 13.171 (Continued)
Projectile motion
y = y0 + (v′y )0 t −
At B,
1 2
t2
gt = 3 ft + (20.57 ft/s)t − (32.2 ft/s 2 )
2
2
y = 0 = 3 + 20.57t B − 16.1tB2 ;
t B = 1.4098 s
xB = x0 + (v′x )0 tB = 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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778
PROBLEM 13.172
A sphere rebounds as shown after striking an inclined plane
with a vertical velocity v0 of magnitude v0 = 5 m/s. Knowing
that α = 30° and e = 0.8 between the sphere and the plane,
determine the height h reached by the sphere.
SOLUTION
Rebound at A
Conservation of momentum in the t-direction:
mv0 sin 30° = m(v′A )t
(v′A )t = (5 m/s)(sin 30°) = 2.5 m/s
Relative velocities in the n-direction:
(−v0 cos 30° − 0)e = 0 − (v′A ) n
(v′A ) n = (0.8)(5 m/s)(cos 30°) = 3.4641 m/s
Projectile motion between A and B:
After rebound
(vx )0 = (v′A )t cos 30° + (v′A )n sin 30°
(vx )0 = (2.5)(cos 30°) + (3.4641) sin 30 = 3.8971 m/s
(v y )0 = −(v′A )t sin 30° + (v′A ) n cos30°
(v y )0 = −(2.5)(sin 30°) + (3.4641) cos 30° = 1.750 m/s
x-direction:
x = (v x ) 0 t
v x = (v x ) 0
x = 3.8971t vx = 3.8971 m/s = vB
y-direction:
At A:
1 2
gt
2
v y = (v y )0 − gt
y = (v y ) 0 t −
v y = 0 = (v y )0 − gt AB
t AB = (v y )0 /g =
1.75 m/s
9.81 m/s 2
t A− B = 0.17839 s
At B:
gt A2 − B
2
9.81
(0.17839) 2
h = (1.75)(0.17839) −
2
y = h = (v y ) 0 t A − B −
h = 0.156 m 
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779
PROBLEM 13.173
A sphere rebounds as shown after striking an inclined
plane with a vertical velocity v0 of magnitude v0 = 6 m/s.
Determine the value of α that will maximize the horizontal
distance the ball travels before reaching its maximum
height h assuming the coefficient of restitution between
the ball and the ground is (a) e = 1, (b) e = 0.8.
SOLUTION
Directions x, y, n, and t are shown in the sketch.
Analysis of the impact: Use the principle of impulse and momentum for
components in the t-direction.
mv0 sin α + 0 = m(vt′ )1
(vt )1 = v0 sin α
Coefficient of restitution:
(1)
( v n )1 = −e( v n )0
(vn )1 = ev0 cos α
(2)
x and y components of velocity immediately after impact:
(vx )1 = (vn )1 sin α + (vt )1 cos α = v0 (1 + e)sin α cos α
=
1
v0 (1 + e) sin 2 α
2
(3)
(v y )1 = (vn )1 cos α − (vt )1 sin α = v0 (e cos 2 α − sin 2 α )
1
v0 [e(1 + cos 2α ) − (1 − cos 2α )]
2
1
= v0 [(1 + e) cos 2α − (1 − e)]
2
=
(4)
Projectile motion: Use coordinates x and y with the origin at the point of impact.
x0 = 0
y0 = 0
Vertical motion:
v y = (v y )1 − gt
vy =
1
v0 [(1 + e) cos 2α − (1 − e)] − gt
2
v y = 0 at the position of maximum height where
t2 =
(v y )1
g
=
v0
[(1 + e) cos α − (1 − e)]
2g
(5)
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780
PROBLEM 13.173 (Continued)
vx = (vx )1 =
Horizontal motion:
1
v0 (1 + e)sin 2α
2
x = (vx )1 t
At the point of maximum height,
x2 = (vx )1 t2 =
v02
(1 + e) sin 2α [(1 + e) cos 2α − (1 − e)]
4g
Let θ = 2α and Z = 4 gx2 /v02 (1 + e). To determine the value of θ that maximizes x2 (or Z ), differentiate
Z with respect to θ and set the derivative equal to zero.
Z = sin θ [(1 + e) cos θ − (1 − e)]
dZ
= cos θ [(1 + e) cos θ − (1 − e)] − (1 + e) sin 2 θ
dθ
= (1 + e) cos 2 θ − (1 − e) cos θ − (1 + e)(1 − cos 2 θ ) = 0
2(1 + e) cos 2 θ − (1 − e) cos θ − (1 + e) = 0
This is a quadratic equation for cos θ .
(a)
e =1
4 cos 2 θ − 2 = 0
cos 2 θ =
1
2
cos θ = ±
2
2
θ = ±45°
and
± 135°
α = 22.5°
and
67.5°
Reject the negative values of θ which make x2 negative.
Reject α = 67.5° since it makes a smaller maximum height.
α = 22.5° 
(b)
e = 0.8
3.6 cos 2 θ − 0.2cos θ − 1.8 = 0
cos θ = 0.73543 and −0.67987
θ = ±42.656° and ±132.833°
α = ±21.328° and ±66.417°
α = 21.3° 
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781
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:
mA = mB = m
5 mi/h = 7.333 ft/s
m A v A + mB vB = mA v′A + mB vB′
−7.333 + vB = v′A + vB′
(1)
Work and energy.
Care A (after impact):
1
m A (v′A )2
2
T2 = 0
T1 =
U1− 2 = F f (12)
U1− 2 = μk mA 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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782
PROBLEM 13.174 (Continued)
Car B (after impact):
1
mB (vB′ )2
2
T2 = 0
T1 =
U1− 2 = μk mB g (3)
T1 + U1− 2 = T2
1
mB (vB′ ) 2 − μk 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:
(−v A − vB ) e = vB′ − v′A
(−7.333 − 30.2) e = 7.613 − 15.226
−(7.613)
= 0.2028
e=
−(37.53)
e = 0.203 
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783
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 μk = 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:
m A 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 Av12 V1 = 0
2
1
T1 = m A (2.4 m/s) 2 = 2.88 m A
2
T1 =
T2 = 0
V2 = m A gh
T1 + V1 = T2 + V2
2.88 m A + 0 = 0 + m A (9.81) h
h = 0.294 m 
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784
PROBLEM 13.175 (Continued)
(b)
Work and energy:
1
1
mB v12 = mB (0.8 m/s) 2 = 0.32mB
T2 = 0
2
2
U1− 2 = − F f x = − μk Nx = − μ x mB gx = −(0.6)(mB )(9.81) x
T1 =
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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785
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 − v A )e = v′P + (v′B ) 2
vB = v0 ,
vP = 0
v0e = v′P + (v′B ) 2
(2)
Conservation of momentum
mBvB + mPvP = mBv′B + mP v′P
0.25
0.25
1.5
(−v′B ) 2 +
v0 + 0 =
v′P
g
g
g
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786
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
 v0 
v02 = 204.49
v0 = 14.30 ft/s 
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787
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:
m A v A + mB vB = m A v′A + mB vB′
mA = mB = 40 kg
5 m/s + 0 = v′A + vB′
(1)
Relative velocities:
(v A − 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′′
(4)
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788
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 
ΔTL = (TB′ + TC′ ) − (TB′′ + TC′′)
TB′ =
1
 40 
kg  (4.5 m/s) 2 = 405 J
mB (vB′ )2 = 
2
 2

TC′ = 0
TC′′ =
TB′′ =
1
 40 
kg  (2.90) 2 = 168.72 J
mB (vB′′ ) 2 = 
2
 2

1
 15

kg  (4.2545 m/s) 2 = 135.76 J
mC (vC′′ )2 = 
2
 2

Δ TL = (405 + 0) − (168.72 + 135.76) = 100.5 J
ΔTL = 100.5 J 
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789
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 μk = 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
2 g
U1− 2 = − μk WA (1 ft)
T1 + U1− 2 = T2
( )
1 WA 2
1 WA 2
v0 − μk WA (1) =
vA
2
2 g
2 g
(v A2 )2 = v02 − 2μk 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
U 2−3 = − μk WA (3/12)
T2′ =
T2′ + U 2−3 = T3 ,
1 WA
(v′A ) 22 − ( μk )(WA /4) = 0
2 g
1 
(v′A )22 = 2(0.3)(32.2 ft/s 2 )  ft  = 4.83 ft 2 /s 2
4 
′
=
(v A )2 2.198 ft/s
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790
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
m A (v A ) 2 + mB vB = mB (v′A ) 2 + mB vB′ mA = mB
vB′ = 12.144 ft/s
14.342 + 0 = 2.198 + vB′
Relative velocities A and B:
[(v A )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
(vB′ ) 42 = B (vB′ ) 42
T4 =
2 g
2g
(vB′ ) = 12.144 ft/s
U 2− 4 = − μk WB (1 ft) = (0.3) WB
T2 =
F f = μk 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′
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791
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
(vC )2
T5 = 0
U 4−5 = − μkWC ( x)
2 g
1 WC
(3.774) 2 − (0.3) WC ( x) = 0
T4 + U 4−5 = T5
2 g
T4 =
x=
(3.774) 2
= 0.737 ft
2(32.2)(0.3)
x = 8.84 in. 
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792
PROBLEM 13.179
A 0.5-kg sphere A is dropped from a height of 0.6 m onto a 1.0 kg
plate B, which is supported by a nested set of springs and is initially
at rest. Knowing that the coefficient of restitution between the sphere
and the plate is e = 0.8, determine (a) the height h reached by the
sphere after rebound, (b) the constant k of the single spring
equivalent to the given set if the maximum deflection of the plate is
observed to be equal to 3h.
SOLUTION
Velocity of A and B after impact.
m A = 0.5 kg
mB = 1.0 kg
Sphere A falls. Use conservation of energy to find v A , the speed just before impact. Use the plate surface as
the datum.
1
m Av A2 , V2 = 0
2
1
0 + m A gh0 = m Av A2 + 0
2
T1 = 0, V1 = m A gh0 , T2 =
T1 + V1 = T2 + V2
With
h0 = 0.6 m,
v A = 2 gh0 = (2)(9.81)(0.6)
v A = 3.4310 m/s ↓
Analysis of the impact. Conservation of momentum.
m A v A + mB v B = m A v′A + mB v′B
Dividing by m A and using y-components
with
vB = 0
with (mB /mA = 2)
−3.4310 + 0 = (v′A ) y + 2(vB′ ) y
Coefficient of restitution.
(1)
(vB′ ) y − (v′A ) y = e [(v A ) y − (vB ) y ]
(vB′ ) y − (v′A ) y = e (v A ) y = −3.4310e
(2)
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793
PROBLEM 13.179 (Continued)
Solving Eqs. (1) and (2) simultaneously with e = 0.8 gives
(v′A ) y = 0.68621 m/s
(vB′ ) y = −2.0586 m/s
v′A = 0.68621 m/s
v′B = 2.0586 m/s
(a)
Sphere A rises. Use conservation of energy to find h.
1
m A (v′A )2 , V1 = 0, T2 = 0, V2 = m A gh
2
1
T1 + V1 = T2 + V2:
m A (v′A )2 + 0 = 0 + m A gh
2
T1 =
h=
(b)
(v′A ) 2 (0.68621) 2
=
2g
(2)(9.81)
h = 0.0240 m 
Plate B falls and compresses the spring. Use conservation of energy.
Let δ 0 be the initial compression of the spring and Δ be the additional compression of the spring after
impact. In the initial equilibrium state,
ΣFy = 0: kδ 0 − WB = 0 or kδ 0 = WB
T1 =
Just after impact:
1
1
mB (vB′ )2 , V1 = kδ 02
2
2
T2 = 0
At maximum deflection of the plate,
V2 = (V2 ) g + (V2 )e = −WB Δ +
Conservation of energy:
(3)
1
k (δ 0 + Δ) 2
2
T1 + V1 = T2 + V2
1
1
1
1
mB (vB′ ) 2 + kδ 02 = 0 − WB Δ + kδ 02 + kδ 0 Δ + k Δ 2
2
2
2
2
Invoking the result of Eq. (3) gives
1
1
mB (vB′ )2 = k Δ 2
2
2
Data:
(4)
mB = 1.0 kg, vB′ = 2.0586 m/s
Δ = 3h = (3)(0.024) = 0.072 m
k=
mB (vB′ ) 2
Δ2
=
(1.0)(2.0586) 2
(0.072) 2
k = 817 N/m 
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794
PROBLEM 13.180
A 0.5-kg sphere A is dropped from a height of 0.6 m onto a 1.0-kg
plate B, which is supported by a nested set of springs and is initially
at rest. Knowing that the set of springs is equivalent to a single
spring of constant k = 900 N/m , determine (a) the value of the
coefficient of restitution between the sphere and the plate for which
the height h reached by the sphere after rebound is maximum,
(b) the corresponding value of h, (c) the corresponding value of the
maximum deflection of the plate.
SOLUTION
m A = 0.5 kg
mB = 1.0 kg
k = 900 N/m
Sphere A falls. Use conservation of energy to find v A , the speed just before impact. Use the plate surface as
the datum.
T1 = 0
T2 =
With h0 = 0.6 m,
V1 = m A gh0
1
m Av A2 ,
2
V2 = 0
v A = 2 gh0 = (2)(9.81)(0.6)
v A = 3.4310 m/s
Analysis of impact. Conservation of momentum.
m A v A + mB v B = mA v′A + mB vB′ with v B = 0
Dividing by m A and using y components
with (mB /mA = 2)
−3.4310 = (v′A ) y + 2(vB′ ) y
Coefficient of restitution.
(1)
(vB′ ) y − (v′A ) y = e [(v A ) y − (vB ) y ]
(vB′ ) y − (v′A ) y = e(v A ) y = −3.4310e
(vB′ ) y = −3.4310 + (v′A ) y
(2)
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795
PROBLEM 13.180 (Continued)
Substituting into Eq. (1),
−3.4310 = (v′A ) y + (2)[−3.4310e + (v′A ) y ]
From Eq. (2),
(a)
(v′A ) y = 1.1437 (2e − 1)
(3)
(vB′ ) y = −1.1437(1 + e)
(4)
Sphere A rises. Use conservation of energy to find h.
1
m A (v′A )2 ,
V1 = 0
2
T2 = 0,
V2 = mA gh
T1 =
1
m A (v′A )2 + 0 = 0 + m A gh
2
(v′ ) 2 (1.1437) 2 (2e − 1)2
h= A =
2g
(2)(9.81)
T1 + V1 = T2 + V2 :
Since h is to be maximum, e must be as large as possible.
e = 1.000 
Coefficient of restitution for maximum h:
(b)
Corresponding value of h.
(v′A ) = 1.1437[(2)(1) − 1] = 1.1437 m/s
h=
(c)
(v′A ) 2 (1.1437)2
=
2g
(2)(9.81)
h = 0.0667 m 
Plate B falls and compresses the spring. Use conservation of energy.
Let δ 0 be the initial compression of the spring and Δ be the additional compression of the spring after
impact. In the initial equilibrium state,
ΣFy = 0 kδ 0 − WB = 0 or kδ 0 = WB
Just after impact:
T1 =
1
mB (vB′ )2 ,
2
At maximum deflection of the plate,
V1 =
1 2
kδ 0
2
T2 = 0
V2 = (V2 ) g + (V2 )e = −WB Δ +
Conservation of energy:
(3)
1
k (δ 0 + Δ) 2
2
T1 + V1 = T2 + V2
1
1
1
1
mB (vB′ ) 2 + kδ 02 = 0 − WB Δ + kδ 02 + kδ 0 Δ + k Δ 2
2
2
2
2
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796
PROBLEM 13.180 (Continued)
Invoking the result of Eq. (3) gives
1
1
mB (vB′ )2 = k Δ 2
2
2
Data:
mB = 1.0 kg,
(vB′ ) y = −1.1437(1 + 1) = −2.2874 m/s.
v′B = 2.2874 m/s , k = 900 N/m
Δ2 =
mB (vB′ ) 2 (1.0)(2.2874) 2
=
= 0.0058133 m 2
k
900
Δ = 0.0762 m 
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797
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 μk = 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
mv A + 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
3 = 2vB′
vB′ = 1.5 ft/s
v′A = vB′
v = Final (common) velocity
Block C: Impulse and momentum
WC vC + F f t =
WC
v
g
F f = μkWC
0 + (0.2)t =
v
g
v = (0.2) gt
(1)
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798
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:
v = (0.2)(32.2)(0.0776) = 0.5 ft/s
From Eq. (1)
Block C:
T1 = 0
T2 =
1W
W
(v ) 2 =
(0.5) 2
2 g
2g
U1− 2 = F f xC = μk WxC = 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/s 2 )
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 μkWgx A = −0.8Wgx A
T1 + U1− 2 = T2
2.25W − 4(0.2)W (32.2) x A = 0.25W

x A = 0.07764 ft
x A = 0.0776 ft 
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799
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,
(v A ) y = (vB ) y = (v′A ) y = (vB′ ) y = 0
Conservation of momentum for x-direction:
While block is sliding down:
0 + 0 = m A (v A ) x + mB (vB ) x
(vB ) x = − β (v A ) x
(1)
Impact:
0 + 0 = m A (v′A ) x + mB (vB′ ) x
(vB′ ) x = − β (v′A ) x
(2)
β = mA /mB
where
Conservation of energy during frictionless sliding:
Initial potential energies: m A gh for A,
0 for B.
Potential energy just before impact:
V1 = 0
Initial kinetic energy:
T0 = 0 (rest)
Kinetic energy just before impact:
T1 =
1
1
m Av 2A + mB vB2
2
2
T0 + V0 = T1 + V1
1
1
1
m Av A2 + mB vB2 = (m A + mB β 2 )v A2
2
2
2
1
= m A (1 + β ) v A2
2
m A gh =
v A2 = (v A ) 2x =
2 gh
1+ β
vA =
2 gh
1+ β
(3)
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800
PROBLEM 13.182 (Continued)
Velocities just before impact:
2 gh
1+ β
vA =
vB = β
2 gh
1+ β
Analysis of impact. Use Eq. (2) together with coefficient of restitution.
(vB′ ) x − (v′A ) x = e[(v A ) x − (vB ) x ]
− β (v′A ) x − (v′A ) x = e[(v A ) x + β (v A ) x ]
(v′A ) x = −e(v A ) x
(4)
m A = 10 kg
Data:
mB = 30 kg
h = 0.2 m
g = 9.81 m/s 2
β=
10 kg
= 0.33333
30 kg
(2)(9.81)(0.2)
1.33333
= 1.71552 m/s
From Eq. (3),
vA =
e = 0:
(v′A ) x = 0
(a)
(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

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801
PROBLEM 13.183
A 20-g bullet fired into a 4-kg wooden block suspended from
cords AC and BD penetrates the block at Point E, halfway
between C and D, without hitting cord BD. Determine (a) the
maximum height h to which the block and the embedded bullet
will swing after impact, (b) the total impulse exerted on the block
by the two cords during the impact.
SOLUTION
Total momentum in x is conserved:
′
mbl vbl + mbu vbu cos 20° = mbl vbl′ + mbu vbu
′ )
(vbl′ = vbu
0 + (0.02 kg)(−600 m/s)( cos 20) = (4.02 kg)(vbl′ )
vbl′ = −2.805 m/s
Conservation of energy:
1
(mbl + mbu )(vbl′ ) 2
2
 4.02 kg 
2
T1 = 
 (2.805 m/s)
2


T1 = 15.815 J
T1 =
V1 = 0
T2 = 0
V2 = (mbl + mbu ) gh
V2 = (4.02 kg)(9.81 m/s 2 )(h) = 39.44h
T1 + V1 = T2 + V2
15.815 + 0 = 0 + 39.44h
h = 0.401 m
(b)
h = 401 mm 
Refer to figure in part (a).
Impulse-momentum in y-direction:
mbu vbu sin 20 + F Δt = (mbl + mbu )(vbl′ ) y
(vbl ) y = 0
(0.02 kg) (−600 m/s) ( sin 20°) + FΔt = 0
FΔt = 4.10 N ⋅ s 
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802
PROBLEM 13.184
A 2-lb ball A is suspended from a spring of constant 10 lb/in and is initially at rest when it
is struck by 1-lb ball B as shown. Neglecting friction and knowing the coefficient of
restitution between the balls is 0.6, determine (a) the velocities of A and B after the impact,
(b) the maximum height reached by A.
SOLUTION
2 lb
= 0.062112 lb ⋅ s 2 /ft
2
32.2 ft/s
Masses:
mA =
mB =
Other data:
k = (10 lb/in.)(12 in./ft.) = 120 lb/ft, e = 0.6
1 lb
= 0.031056 lb ⋅ s 2 /ft
2
32.2 ft/s
v A = 0, vB = 2 ft/s
For analysis of the impact use the principle of impulse and momentum.
Σmv1 + ΣImp1 2 = Σmv 2
t-direction for ball A:
0 + 0 = mA (v′A )t
t-direction for ball B:
(v′A )t = 0
mB vB sin 20° + 0 = mB (vB′ )t
(vB′ )t = vB sin 20° = (2)(sin 20°) = 0.6840 ft/s
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803
PROBLEM 13.184 (Continued)
n-direction for balls A and B:
mB vB cos 20° + 0 = mB (vB′ ) n + m A (v′A ) n
(vB′ )n +
mA
(v′A )n = vB cos 20°
mB
(vB′ )n + 2(v′A ) n = (2) cos 20°
Coefficient of restitution:
(1)
(vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ]
= e[0 − (vB cos 20°)]
= −(0.6)(2) cos 20°
(2)
Solving Eqs. (1) and (2) simultaneously,
(v′A )n = 1.00234 ft/s
(a)
(vB′ ) n = −0.12529 ft/s
Velocities after the impact:
v′A = 1.00234 ft/s
v′A = 1.002 ft/s 
v′B = (0.6840 ft/s →) + (0.12529 ft/s )
vB = (0.6840) 2 + (0.12529)2 = 0.695 ft/s
tan β =
0.12529
0.6840
β = 10.4°
v′B = 0.695 ft/s
(b)
10.4° 
Maximum height reached by A:
Use conservation of energy for ball A after the impact.
Position 1: Just after impact.
T1 =
1
1
m A (v′A ) 2 = (0.062112)(1.00234)2 = 0.0312021 ft ⋅ lb
2
2
Force in spring = weight of A
x1 = −
W
2 lb
F
=− A =−
= −0.016667 ft
120 lb/ft
k
k
2
(V1 )e =
1 2 1  WB  WA2
kx1 = k 
=
2
2  k 
2k
(2 lb) 2
= 0.016667 ft ⋅ lb
(2)(120)
(V1 ) g = 0
(datum)
=
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804
PROBLEM 13.184 (Continued)
Position 2: Maximum height h.
V2 = 0
T2 = 0
1
1
k (h + x1 ) 2 = (120)(h − 0.016667) 2
2
2
2
= 60h − 2h + 0.016667
(V2 )e =
(V2 ) g = WA h = (2 lb)h = 2h
Conservation of energy: T1 + V1 = T2 + V2
0.031202 + 0.016667 = 0 + 60h 2 − 2h + 0.016667 + 2h
60h 2 = 0.031202
h = ±0.022804 ft
h = 0.274 in. 
Using the positive root,
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805
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 perfectly
elastic impact (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 2A
2
v A = 2 gh = (2)(32.2)(0.66667) = 6.5524 ft/s
Impact
θ = sin −1
r
= 30°
2r
Impulse-Momentum
Unknowns:
vB′ , v′At , v′An
x-dir
0 + 0 = mB vB′ + m Av′An sin 30° + mA v′At cos 30°
(1)
Noting that m A = mB and dividing by mA
vB′ + v′An sin 30° + v′At cos 30° = 0
(1)
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806
PROBLEM 13.185 (Continued)
Ball A alone:
Momentum in t-direction:
− m Av A sin 30° + 0 = mA v At
v′At = −v A 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(v A 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. 
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807
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
mB
70
− e 2 gh0 = (v3′ ) B − 3vP′
(1)
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808
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 (3e2 − e)
Conservation of energy at ,
Thus,
(v3′ ) B = 2 gh2
4 2 gh2 = 2 gh0 (3e2 − e)
3e 2 − e = 4
h2
0.25
=4
= 1.63299
h0
1.5
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 
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809
PROBLEM 13.187
A 700-g sphere A moving with a velocity v0 parallel to the
ground strikes the inclined face of a 2.1-kg wedge B which can
roll freely on the ground and is initially at rest. After impact the
sphere is observed from the ground to be moving straight up.
Knowing that the coefficient of restitution between the sphere
and the wedge is e = 0.6, determine (a) the angle θ that the
inclined face of the wedge makes with the horizontal, (b) the
energy lost due to the impact.
SOLUTION
(a)
Momentum of sphere A alone is conserved in the t-direction:
m Av0 cos θ = m Av′A sin θ
v0 = v′A tan θ
(1)
Total momentum is conserved in the x-direction:
mB vB + m Av0 = mB vB′ + (v′A ) x
vB = 0, (v′A ) x = 0
0 + 0.700 v0 = 2.1vB′ + 0
vB′ = v0 /3
(2)
Relative velocities in the n-direction:
(−v0 sin θ − 0)e = −vB′ sin θ − v′A cos θ
(v0 )(0.6) = vB′ + v′A cot θ
(3)
Substituting vB′ from Eq. (2) into Eq. (3)
0.6v0 = 0.333 v0 + v′A cot θ
0.267v0 = v′A cot θ
(4)
Divide (4) into (1)
1
tan θ
=
= tan 2 θ
0.267 cot θ
tan θ = 1.935
(b)
From (1)
θ = 62.7° 
v0 = v′A tan θ = v′A (1.935)
v′A = 0.5168v0 , vB′ = v0 /3
Tlost =
(
1
1
mA v 2A − m A (v′A ) 2 + mB vB2
2
2
(2)
)
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810
PROBLEM 13.187 (Continued)
1
1
(0.7)(v0 )2 − [(0.7)(0.5168v0 )2 + (2.1)(v0 /3) 2 ]
2
2
1
Tlost = [0.7 − 0.1870 − 0.2333]v02
2
Tlost = 0.1400v02 J
Tlost =
Tlost = 0.1400v02 
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811
PROBLEM 13.188
When the rope is at an angle of α = 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 θ = 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:
m A = (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:
α = 30°, h0 = l (1 − cos α ) = (2.6)(1 − cos 30°) = 0.34833 ft
V0 = m A gh0 = (1)(0.34833) = 0.34833 lb ⋅ ft
T0 =
Just before impact:
α = 0, h1 = 0, V1 = 0
T1 =
Conservation of energy:
1
1  1.0  2
mA v02 = 
 (4) = 0.24845 lb ⋅ ft
2
2  32.2 
1
1  1.0  2  1.0  2
m A v 2A = 
vA = 
 vA
2
2  32.2 
 64.4 
T0 + V0 = T1 + V1
1 2
vA + 0
64.4
v 2A = 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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812
PROBLEM 13.188 (Continued)
Sphere A: Momentum in t-direction:
m A v A sin θ + 0 = m A (v′A )t
(v′A )t = v A sin θ = 6.1994 sin 20° = 2.1203 m/s
( v A )t = 2.1203 m/s 70°
Both A and B: Momentum in x-direction:
m A v A + 0 = m A (v′A ) n cos θ + mA (v′A )t sin θ + 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 θ − (v′A )n = e[v A cos θ − 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 β =
(v′A )t
−(v′A )n
2.1203
1.0446
β = 63.77°
=
β + 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 (Δx) 2
2
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813
PROBLEM 13.188 (Continued)
The maximum deflection will occur when the block comes to rest (ie, no kinetic energy)
1
1
mB vB2 = k (Δx) 2
2
2
m v2
(Δ x) = B B =
k
2
(
2 lb
32.2 ft/s 2
(Δ x) = 0.1642118 ft
) (3.2279 ft/s)
2
2 lb/in (12 in/ft)
(Δ x) = 1.971 in. 
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814
PROBLEM 13.189
When the rope is at an angle of α = 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 θ = 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:
α = 30°, h0 = l (1 − cos α ) = (0.9)(1 − cos 30°) = 0.12058 m
Initial state:
V0 = m A 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
α = 0, h1 = 0, V1 = 0
Just before impact:
T1 =
Conservation of energy:
1
1
m Av 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.
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815
PROBLEM 13.189 (Continued)
Both A and B:
Momentum in x-direction:
m A (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 θ
Coefficient of restitution:
(vB )n = 0, (v′A )n = (v′A ) x cos θ , (vB′ ) x cos 30°
(vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ]
(vB′ ) x cos θ − (v′A ) x cos θ = e[(v A ) x cos θ ]
Dividing by cos θ 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 (Δx) 2
2
The maximum deflection will occur when the block comes to rest (ie, no kinetic energy)
1
1
mB vB2 = k (Δ x) 2
2
2
(Δ x) 2 =
mB vB2 (2)(0.99059 m/s)2
=
1500 N/m)
k
(Δx) = 0.0362 m
Δ x = 36.2 mm 
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816
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
g 32.2 ft/s 2
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
ΔU = −Q (Δl AC ) − Q (ΔlBC )
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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817
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
U1− 2 = − mg E (300 ft) − EL
T1 =
Earth:
( EL = Loss of energy due to drag)
T1 =
Moon:
1 2
mv0
2
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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818
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 −
r0
r1
2
2
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 
 r1r0 
 r0 r1 
 r2 − r2 
r − r 
h2
h 2  1 2 20  = 2 2 (r1 − r0 )( r1 + r0 ) = 2GM  1 0 
 r1r0 
 r1 r0  r1 r0
1 1
h 2  +  = 2GM
 r0 r1 
 1 1  2GM
 + = 2
h
 r0 r1 
Q.E.D. 
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819
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
=
N ⋅ mm
Vm = −
∞ r2
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
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820
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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821
PROBLEM 13.194
A shuttle is to rendezvous with a space station which is in a circular orbit at
an altitude of 250 mi above the surface of the earth. The shuttle has reached
an altitude of 40 mi when its engine is turned off at Point B. Knowing that
at that time the velocity v 0 of the shuttle forms an angle φ0 = 55° with
the vertical, determine the required magnitude of v 0 if the trajectory of
the shuttle is to be tangent at A to the orbit of the space station.
SOLUTION
Conservation of energy:
TB =
1 2
mv0
2
VB = −
GMm
rB
TA =
1 2
mv A
2
VA = −
GMm
rA
GM = gR 2 (Eq.12.30)
TA + VA = TB + VB
1
gR 2
1
gR 2
m v02 −
m = m v 2A
m
2
rB
2
rA
rA = 3960 + 250 = 4210 mi
v A2 = v02 −
2 gR 2  rB 
1 − 
rB  rA 
rB = 3960 + 40 = 4000 mi
v A2 = v02 −
2(32.2)(3960 × 528)3  4000 
1 − 4210 
(4000 × 5280)


v A2 = v02 − 66.495 × 106
(1)
rA v A = rB vB sin φ0 ;
Conservation of angular momomentum:
v A = (4000/4210)v0 sin 55° = 0.77829 v0
Eqs. (2) and (1)
[1 − (0.77829) 2 ] v02 = 66.495 × 106
(2)
v0 = 12,990 ft/s 
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822
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
mv A = (0.3)v 2A = 0.15v A2
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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823
PROBLEM 13.195 (Continued)
(a)
Newton’s second law at A:
an =
v 2A
ρ
=
35.504 m 2 /s 2
= 44.38 m/s 2
0.800 m
a n = 44.38 m/s 2
ΣF = 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
ρ
=
19.808 m 2 /s 2
= 24.76 m/s 2
0.800 m
a n = 24.76 m/s 2
ΣF = man : N B = mg = man
N B = m( an − g ) = (0.3 kg)(24.76 m/s 2 − 9.81 m/s 2 )
(c)
N B = 4.49 N 
Newton’s second law at C:
an =
vC2
ρ
=
35.504 m 2 /s 2
= 44.38 m/s 2
0.800 m
a n = 44.38 m/s 2
ΣF = man : NC = man
N C = (0.3 kg)(44.38 m/s 2 )
N C = 13.31 N

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824
PROBLEM 13.196
A small sphere B of mass m is attached to an inextensible cord of length 2a,
which passes around the fixed peg A and is attached to a fixed support at O.
The sphere is held close to the support at O and released with no initial
velocity. It drops freely to Point C, where the cord becomes taut, and swings
in a vertical plane, first about A and then about O. Determine the vertical
distance from line OD to the highest Point C ′′ that the sphere will reach.
SOLUTION
Velocity at Point C (before the cord is taut).
Conservation of energy from B to C:
TB = 0
 2
VB = mg (2) 
a = mga 2
 2 


1
TC = mvC2
VC = 0
2
TB + VB = TC + VC
0 + mga 2 =
1 2
mvC + 0
2
vC = 2 2 ga
Velocity at C (after the cord becomes taut).
Linear momentum perpendicular to the cord is conserved:
θ = 45°
− mvC sinθ = mvC′
vC′ =
( 2 2 )  22  ga
vC′ =
2 ga = 2 4 ga
1
Note: The weight of the sphere is a non-impulsive force.
Velocity at C:
C to C ′ (conservation of energy):
1
m(vC′ )2
2
1
TC ′ = m(vC′ ′ ) 2
2
TC =
VC = 0
VC ′ = 0
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825
PROBLEM 13.196 (Continued)
Datum:
TC + VC = TC ′ + VC ′
1
1
m(vC′ ) 2 + 0 = m(vC′ ) 2 + 0
2
2
vC′ = vC′ ′
C ′ to C ′′ (conservation of energy):
1
m(vC′ ′ ) 2
2
2
1
TC ′ = m 21/ 4 ga
2
2
TC ′ =
mga
2
TC ′ + VC ′ = TC ′′ + VC ′′
TC ′ =
(
Datum:
)
VC ′ = 0
TC ′′ = 0
VC ′′ = mgh
2
mga + 0 = 0 + mgh
2
2
h=
a
2
h = 0.707 a 
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826
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
m Av0 = mB vB − m Av A : 0.3v0 = 0.9vB − 0.3v A (1)
Restitution
(v A + vB ) = e(v0 + 0) = 0.9v0 (2)
Substituting for vB from (2) in (1)
0.3v0 = 0.9(0.9v0 − v A ) − 0.3v A
1.2v A = 0.51v0
v A = 1.4582 m/s, vB = 1.6297 m/s
(a)
A moves up the distance d where:
1
m Av A2 = mA gd sin 30°;
2
1
(1.4582 m/s) 2 = (9.81 m/s 2 )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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827
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° + mB vB2 = 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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828
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
m A (v A )2 + mB vB2
2
2
v A = vB (kinematics)
T2 =
1
11
(5 + 6)v A2 = v A2
2
2
V2 = m A 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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829
PROBLEM 13.198 (Continued)
Velocity of A and C after A hits C:
v′A = vC′ (plastic impact)
Impulse-momentum A and C:
m A v A + T Δt = (m A + mC ) v′A
(5)(1.741) + T Δt = 8v′A
(1)
vB = v A ; vB′ = v′A (cord remains taut)
B alone:
mB v A − T Δ t = mB v′A
(6)(1.741) − T Δt = 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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830
PROBLEM 13.198 (Continued)
Position :
T3 = 0
V3 = (m A + 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 − m A gx
V4 = (6 − 5) gx
T2 + V2 = T4 + V4
10.291 + 0 = (1)(9.81) x
x = 1.049 m 
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831
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.
Σmv1 + ΣImp1→2 = Σmv 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 = m Av A + mB (vB ) x + mB (vB )t
(2)(10) + 0 = 2v A + 2[(vB )n cos 40° + 6.4279sin 40°]
2v A + 2(vB ) n cos 40° = 11.7365
(1)
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832
PROBLEM 13.199 (Continued)
(e = 0.8)
Coefficient of restitution.
(vB )n = (v A )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
m Av A2 + (Ve )1 + (Vg )1 = m A v22 + (Ve ) 2 + (Vg ) 2
2
2
Position 1: (Just after impact.)
1
1
m Av 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
m Av22 = (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 m 2 /s 2
v2 = 4.489 m/s
Normal acceleration at lowest point on path:
an =
v22
ρ
=
20.158
= 28.798 m/s 2
0.7
a n = 28.8 m/s 2
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833
PROBLEM 13.199 (Continued)
Apply Newton’s second law to the ball.
ΣF = man : N − mg − Fe = man
N = mg + Fe + man
= (2)(9.81) + 80 + (2)(28.798)
N = 157.2 N 
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834
PROBLEM 13.200
A 2-kg block A is pushed up against a spring compressing it a
distance x = 0.1 m. 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. When α = 40°, determine (a) the speed
of B (b) the tension in the rope.
SOLUTION
Data:
m A = 2 kg, mB = 1 kg, k = 800 N/m, x = 0.1 m, d = 1.5 m
μk = 0.2, e = 0.8, θ = 20°, α = 40°, l = 1.0 m
Block slides down the incline:
ΣFy = 0
N − m A g cos θ = 0
N = m A g cos θ
= (2)(9.81) cos 20°
= 18.4368 N
F f = μk 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 x1 = (800)(0.1) 2 = 4.00 J
2
2
(V1 ) g = m A gh1 = mA g ( x + d ) sin θ = (2)(9.81)(1.6)sin 20° = 10.7367 J
U1→2 = − F f ( x + d ) = −(3.6874)(1.6) = −5.8998 J
T2 =
1
1
m A v 2A = (1)(v A2 ) = 1.000 v A2
2
2
V2 = 0
T1 + V1 + U1→2 = T2 + V2 : 0 + 4.00 + 10.7367 − 5.8998 = 1.000 v 2A + 0
v A2 = 8.8369 m 2 /s 2
v A = 2.9727 m/s
20°
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835
PROBLEM 13.200 (Continued)
Impact: Conservation of momentum.
Both A and B, horizontal components
:
m A v A cos θ + 0 = m Av′A cos θ + mB vB
(2)(2.9727) cos 20° = 2v′A cos 20° + (1.00)vB
(1)
(vB′ )n − (v′A ) n = e[(vB ) n − (v A ) n ]
Relative velocities:
vB′ cos θ − v′A = e[v A − 0]
vB′ cos 20° − v′A = (0.8)(2.9727)
(2)
Solving Eqs. (1) and (2) simultaneously,
v′A = 1.0382 m/s
vB′ = 3.6356 m/s
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 α )
2
1
1
T1 + V1 = T2 + V2 :
mB (vB′ ) 2 + 0 = mB v22 + mB g (1 − cos)
2
2
v22 = (vB′ )2 − 2 gl (1 − cos α )
T1 =
= (3.6356)2 − (2)(9.81)(1 − cos 40°)
= 8.6274 m 2 /s 2
v2 = 2.94 m/s 
(a) Speed of B:
(b) Tension in the rope:
ρ = 1.00 m
an =
v22
ρ
=
8.6274
= 8.6274 m/s 2
1.00
ΣFn = mB an :
T − mB g cos α = mB an
T = mB (an + g cos α )
= (1.0)(8.6274 + 9.81cos 40°)
T = 16.14 N 
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836
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 θ 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 θ based on Q = 0. Draw the free body diagram.
ΣF = 0: Q + mg sin θ = man =
With Q = 0,
v22 = g  sin θ
mv22

or v2 = g  sin θ
(1)
Relationship among v0 , v2 , and θ based on conservation of energy.
T1 + V1 = T2 + V2
1 2
1
mv0 − mg  = mv22 + mg  sin θ
2
2
v02 = v22 + 2 g (1 + sin θ )
(2)
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837
PROBLEM 13.201* (Continued)
x and y coordinates at position 2:
x2 =  cos θ
(3)
y2 =  sin θ
(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 θ
x = x2 − (v2 sin θ )(t − t2 )
At Point B,
x = xB
(t B − t2 ) =
Vertical motion:
and t = t B .
(5)
From Eq. (5),
 cos θ − xB
vθ sin θ
(6)
y = v2 cos θ − g (t − t2 )
y = y2 + (v2 cos θ )(t − t2 ) −
1
g (t − t2 ) 2
2
At Point B,
yB =  sin θ + (v2 cos θ )(t B − t2 ) −
Data:
 = 2 ft, xB = 0.3 ft,
1
g (t B − t2 )2
2
(7)
yB = 0.4 ft, g = 32.2 ft/s 2
With the numerical data,
Eq. (1) becomes
Eq. (6) becomes
Eq. (7) becomes
v2 = 64.4sin θ
(1)′
2 cos θ − 0.3
v2 sin θ
(6)′
t B − t2 =
yB = 2sin θ + (v2 cos θ )(t B − t2 ) − 16.1(t B − t2 ) 2
(7)′
Method of solution. From a trial value of θ, calculate v2 from Eq. (1)′, t B − t2 from Eq. (6)′, and yB
from Eq. (7)′. Repeat until yB = 0.4 ft as required.
Try θ = 30°.
v2 = 64.4sin 30° = 5.6745 ft/s
2 cos30° − 0.3
= 0.50473s
5.6745sin 30°
yB = 2sin 30° + (5.6745cos 30°)(0.50473) − (16.1)(0.50473)2
t B − t2 =
= −0.62116 ft
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838
PROBLEM 13.201* (Continued)
Try θ = 45°.
v2 = 64.4sin 45° = 6.7482
2 cos 45° − 0.3
= 0.23351 s
6.7482sin 45°
yB = 2sin 45° + (6.7482 cos 45°)(0.23351) − (16.1)(0.23351)2
t B − t2 =
= 1.65060 ft
Try θ = 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 = θ − 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
θ = 30° + u = 33.95°.
Try θ = 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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839
PROBLEM 13.201* (Continued)
Solving for u,
θ = 30° + 3.572° = 33.572°
u = 3.572°
Try θ = 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 θ = 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 /s 2
v 0 = 15.35 ft/s

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840
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.
Σ mv1 + Σ Imp1 2 = Σ mv 2
Components
:
m A v0 + 0 = (m A + mB )v′
v′ =
m Av0
(30 × 10−3 kg)(450 m/s)
=
= 4.4554 m/s
m A + 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.
Σ mv 2 + Σ Imp 23 = Σ mv3
Components
:
(m A + mB )v′ + 0 = ( mA + mB + mC )v
v′′ =
=
mA v0
(m A + mB )v′
=
mA + mB + mC m A + 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

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843
PROBLEM 14.2
A 30-g bullet is fired with a horizontal velocity of 450 m/s through 3-kg block B and becomes embedded in
carrier C which has a mass of 30 kg. After the impact, block B slides 0.3 m on C before coming to rest relative
to the carrier. Knowing the coefficient of kinetic friction between B and C is 0.2, determine (a) the velocity of
the bullet immediately after passing through B, (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.
(b)
Final velocity of carrier: Use conservation momentum for all three particles, since the impact forces and
the friction force between B and C are internal forces of the system.
Σ mv1 + Σ Imp1 2 = Σ mv 2
Components
:
m A v0 + 0 = (m A + mB + mC )v′′
v′′ =
m Av0
(0.030 kg)(450 m/s)
=
= 0.40872 m/s
33.03 kg
mA + mB + mC
v′′ = 0.409 m/s
(a)

Velocity v A of the bullet:
The sequence of events described is broken into the following states and processes. The symbols for
velocities of A, B, and C at the various states are given in the following table:
Symbol for velocity
State
A
B
C
Process
(1)
v0
0
0
Initial state
(2)
vA
vB
0
1
2: Bullet passes through block
(3)
v AC
vB
v AC
2
3: Bullet impacts end of carrier
(4)
v′′
v′′
v′′
3
4: Block slides to rest relative to carrier
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844
PROBLEM 14.2 (Continued)
For process 1
2 apply conservation of momentum.
m A v0 = m Av A + mB vB
For process 2
3 apply conservation of momentum to A and C.
m A v A = (m A + mC )v AC
For process 3
(1)
(2)
4 apply conservation of momentum to A, B, and C.
(m A + mC )v AC + mB vB = (m A + mB + mC )v′′
(3)
4 apply the principle of work and energy, since the work U 3→4 of the friction
For process 3
force may be calculated.
Normal force:
N = WB = mB g = (3 kg)(9.81 m/s) = 29.43 N
Friction force:
F f = μk N = (0.2)(29.43) = 5.886 N
U 3→4 = − F f d = −(5.886 N)(0.3 m) = −1.7658 J
Work:
Principle of work and energy:
TAC + TB + U 3→ 4 = T ′′
where
TAC =
1
2
(mA + mC )v AC
2
TB =
1
mB vB2
2
T ′′ =
1
(m A + mB + mC )(v′′) 2
2
(4)
Applying the numerical data gives
(0.030)(450) = 0.030v A + 3vB
(1)′
0.030v A = 30.03v AC
(2)′
30.03v AC + 3vB = (33.03)(0.40872)
(3)′
1
1
1
2
(30.03)v AC
+ (3)vB2 − 1.7658 = (33.03)(0.40872) 2
2
2
2
v AC =
From Eq. (3)′,
(4)′
(33.03)(0.40872) − 3vB
= 0.44955 − 0.0999 vB
30.03
Substituting into Eq. (4)′ gives
(15.015)(0.44955 − 0.0999vB )2 + 1.5vB2 − 1.7658 = 2.7586
which reduces to the quadratic equation
1.64985vB2 − 1.34865vB − 1.48995 = 0
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845
PROBLEM 14.2 (Continued)
Solving,
vB = 1.44319 and − 0.62575
vB = 1.44319 m/s
Using Eq. (1)′ with numerical data,
13.5 = 0.030v A + (3)(1.44319)
v A = 306 m/s

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846
PROBLEM 14.3
Car A weighing 4000 lb and car B weighing 3700 lb
are at rest on a 22-ton flatcar which is also at rest.
Cars A and B then accelerate and quickly reach
constant speeds relative to the flatcar of 7 ft/s and
3.5 ft/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 m A =
4000
3700
(22)(2000)
= 124.2 slugs, mB =
= 114.9 slugs, and mF =
= 1366.5 slugs
32.2
32.2
32.2
Let v A , vB , and vF be the sought after velocities in ft/s, positive to the right.
(v A )0 = (vB )0 = (vF )0 = 0.
Initial values:
m A (v A )0 + mB (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 = m Av A + mB vB + mF vF
124.2v A + 114.9vB + 1366.5vF = 0
(1)
The relative velocities are given as
v A/F = v A − vF = − 7 ft/s
(2)
vB/F = vB − vF = − 3.5 ft/s
(3)
Solving (1), (2), and (3) simultaneously,
v A = − 6.208 ft/s, vB = − 2.708 ft/s, vF = 0.7919 ft/s
v F = 0.792 ft/s

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847
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 m A 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 + m Av A + mB vB
Equating,
mv0 = mvB + mA v A + mB vB
m=
mA v 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 + m Av A
Equating,
mv0 = mv1 + mA v A
v1 =
mv0 − mA v A (0.0500)(1500) − (6)(5)
=
= 900 ft/s
0.0500
m
v1 = 900 ft/s

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848
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.
(Σmv)1 = (Σmv) 2
0 = mA 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)
vB/R = 0
vB = vR
B
0 = m A [2  + v R ] + mB v R + mR v e
A
vR =
(b)
−2 m A
−(2 ft/s)(190 lb)
=
(m A + mB + mR ) (190 lb + 125 lb + 300 lb)
vR = 0.618 ft/s 
From Eq. (1),
0 = m Av A + mB vB + 0
(vR = 0)
vB = −
mA v 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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849
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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850
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)
m A v0 = (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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851
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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852
PROBLEM 14.8
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:
m Av A + mB vB = mA v′A + mB vB′
(8 kg)(2 m/s) + 0 = (8 kg)v′A + (4 kg)vB′
4 = 2v′A + vB′
(1)
(v A − vB )e = (vB′ − v′A )
(2)(0.3) = vB′ − v′A
(2)
Relative velocities.
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′
(3)
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853
PROBLEM 14.8 (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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854
PROBLEM 14.9
A system consists of three particles A, B, and C. We know that
m A = 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 H O 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 vC = −8i + 16 j + 8k
rA = 3j,
Position vectors, (meters):
rB = 1.2i + 2.4 j + 3k ,
rC = 3.6i
Angular momentum about O, (kg ⋅ m 2/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
12 6 6
8 6 0
−8 16 8
= (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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855
PROBLEM 14.10
For the system of particles of Problem 14.9, 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
H G of the system about G. Also verify that the answers to this
problem and to problem 14.9 satisfy the equation given in
Problem 14.27.
PROBLEM 14.9 A system consists of three particles A, B, and C.
We know that m A = 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 H O of the system about O.
SOLUTION
Position vectors, (meters):
(a)
Mass center:
rA = 3j,
rB = 1.2i + 2.4 j + 3k ,
rC = 3.6i
(m A + mB + mC ) r = m ArA + mBrB + mC rC
9r = (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 ⋅ m 2 /s).
m A v A = 12i + 6 j + 6k
mB v B = 8i + 6 j
mC vC = −8i + 16 j + 8k
(b)
Linear momentum of the system, (kg ⋅ m/s.)
mv = m A v A + mB v B + mC vC = 12i + 28 j + 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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856
PROBLEM 14.10 (Continued)
(c)
Angular momentum about G, (kg ⋅ m 2 /s).
H G = rA′ × m Av 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
16
8
−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
H G + r × mv = −(4.8 kg ⋅ m 2 /s) j + (9.6 kg ⋅ m 2/s)k
Angular momentum about O.
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
12 6 6
8 6 0
−8 16 8
= (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
H O = H G + r × mv 
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857
PROBLEM 14.11
A system consists of three particles A, B, and C. We know that WA = 5lb,
WB = 4 lb, and WC = 3 lb, and that the velocities of the particles expressed
in ft/s are, r