Chapter 11 Solutions
4.
Picture the Problem: The arm extends out either horizontally and the weight of the crab trap is exerted straight downward on
the hand.
Strategy: The torque equals the moment arm times the force according to equation 11-3. In this case the moment arm is the
horizontal distance between the shoulder and the hand, and the force is the downward weight of the crab trap.
  r mg   0.70 m  3.6 kg   9.81 m/s 2   25 N  m
Solution: Multiply the moment arm by the weight:
11.
Picture the Problem: The ceiling fan rotates about its axis, decreasing its angular speed at a constant rate.
Strategy: Determine the angular acceleration using equation 10-6 and then use equation 11-4 to find the moment of inertia of
the fan.
Solution: Solve equation 11-4 for I:
I


 t   0.120 N  m  22.5 s 



 0.982 kg  m2
  t 
 0  2.75 rad/s 
19. Picture the Problem: The fish exerts a torque on the fishing reel and it rotates with constant angular acceleration.
Strategy: Use Table 10-1 to determine the moment of inertia of the fishing reel assuming it is a uniform cylinder
( 12 MR2 ). Find the torque the fish exerts on the reel by using equation 11-1. Then apply Newton’s Second Law for
rotation (equation 11-4) to find the angular acceleration and equations 10-2 and 10-10 to find the amount of line pulled
from the reel.
 0.99 kg  0.055 m 
Solution: 1. (a) Use Table 10-1 to find I:
I  12 MR 2 
2. Apply equation 11-1 directly to find  :
  r F   0.055 m  2.2 N   0.121 N  m
3. Solve equation 11-14 for  :

4. (b) Apply equations 10-2 and 10-10:
s  r  r  12  t 2    0.055 m  12  81 rad/s 2   0.25 s 

I

1
2
2
 0.0015 kg  m 2
0.121 N  m
 81 rad/s 2
0.0015 kg  m2
2
 0.14 m
26. Picture the Problem: The person lies on a lightweight plank
that rests on two scales as shown in the diagram at right.
Strategy: Write Newton’s Second Law in the vertical direction
and Newton’s Second Law for rotation to obtain two equations
with two unknowns, m and xcm . Solve each to find m and xcm .
Using the left side of the plank as the origin, there are two
torques to consider: the positive torque due to the right hand
scale and the negative torque due to the person’s mass.
Solution: 1. (a) Write Newton’s Second
Law in the vertical direction to find m:
F
2. (b) Write Newton’s Second Law for
rotation and solve for xcm:
  r F
 F1  F2  mg  0
F  F2 290  122 N
m 1

 42 kg
g
9.81 m/s 2
y
 xcm mg  0
 2.50 m 122 N 
xF
 2 2 
 0.74 m
mg
 42 kg   9.81 m/s2 
2
xcm
2
36.
Picture the Problem: The horizontal force F is applied to the rod as shown in the figure
at right.
Strategy: Let L = the rod length and write Newton’s Second Law for torques (let the bolt
be the pivot point) in order to determine the wire tension T. Then write Newton’s Second
Law in the horizontal and vertical directions to determine the components of the bolt force
Fb .
Solution: 1. (a) Set
  0 and
solve for T :
2. (b) Set
  L T cos 45   L  F  0
1
2
T
F
x
 0 and solve for Fb,x :
F
x
F
F

2 cos 45
2
 F  Fb,x  T cos 45  0
F


1
Fb,x  T cos 45  F  
 cos 45  F   2 F
 2cos 45 
3. (c) Set
F
y
 0 and solve for Fb, y :
F
y
 Fb,y  T sin 45  0
F


Fb,y  
 sin 45 
 2cos 45 
1
2
F
45. Picture the Problem: The glove hangs on one end of the bat and the balance point is found to be 24.7 cm from the
71.1 cm mark.
Strategy: Write Newton’s Second Law for torque with the glove at 0.00 cm and the pivot point at the
71.1 − 24.7 cm = 46.4 cm mark. Solve the resulting expression for the mass M of the bat. Let m be the mass of the
glove.
Solution: Set
  0 and solve for M:
  d  Mg   r  mg   0
M 
r m  71.1  24.7 cm  0.560 kg 

 1.05 kg
d
 24.7 cm 
46.
Picture the Problem: The mass falls straight down, its speed reduced due to the rotation of the
disk. The physical situation is depicted at right.
Strategy: Write Newton’s Second Law for torque about the axis of the pulley, and Newton’s
Second Law in the vertical direction for the bucket. Combine those two equations together with
the relation, a  r , which comes from the fact that the rope does not slip along the rim of the
pulley, in order to find the linear acceleration, angular acceleration, and distance traveled in
1.50 s. Let m = bucket mass, M = pulley mass,
R = pulley radius, T = rope tension, and note that for the pulley, I  12 MR2 . Let downward be
the positive direction for the bucket.
Solution: 1. (a) Set
  I
for the
  rT  I
pulley and solve for T:
F
 may for the bucket and
substitute the expression for T from step 1:
2. Set
y
3. Now solve for a:
T
F
y
I

r

1
2
MR 2   a R 
R
 12 Ma
 T  mg  ma
  12 Ma   mg  ma
mg   m  12 M  a


 m 
2.85 kg
2
2
a
  9.81 m/s   8.68 m/s
g  
1
1
 2.85  2  0.742  kg 
m 2 M 

5. (c) Use equation 4-3(b) with v0 y  0
y  0  12 ay t 2  12 8.68 m/s2  1.50 s   9.77 m
to find y .
74.
a 8.68 m/s2

 71.7 rad/s2
R 0.121 m
4. (b) Now set a  R and solve for  :
2
Picture the Problem: The student rotates freely on the piano stool with outstretched arms holding the masses away from the
axis of rotation, and then pulls the masses inward toward his body.
Strategy: The moment of inertia of the student-masses system will decrease as he brings the masses inward toward the rotation
axis, but the angular momentum of the system will remain the same because there is no external torque. Use conservation of
angular momentum together with equations 10-18 and 11-11 to find the final distances of the masses from the rotation axis.
Then use equation 10-17 to find the initial and final kinetic energies of the system.
Solution: 1. (a) Find I i for the
system using equation 10-18:
I i  I s-s  2mri 2
2. Set Li  Lf and solve for I f :
I ii  I f f
 5.43 kg  m 2  2 1.25 kg  0.759 m   6.87 kg  m 2
2
 
 2.95 rev/s 
2
I f  I i  i    6.87 kg  m2  
  5.73 kg  m
 3.54 rev/s 
 f 
3. Use equation 10-18 to find rf :
I f  I s-s  2mrf2
rf 
I f  I s-s
5.73  5.43 kg  m 2

 0.35 m
2m
2 1.25 kg 
4. (b) Use equation 10-17 to find Ki :
1
2
Iii2  12  6.87 kg  m2   2.95 rev/s  2 rad/rev   1.18 kJ
5. Use equation 10-17 to find K f :
1
2
If f2  12  5.73 kg  m2  3.54 rev/s  2 rad/rev   1.42 kJ
2
2
80.
Picture the Problem: The torque acting through an angular displacement does work on the ice cream crank.
Strategy: Use equation 11-17 to find the work done by the torque acting through the given angular displacement. One complete
turn corresponds to an angular displacement of 2 radians.
W      3.95 N  m  2 rad   24.8 J
Solution: Apply equation 11-17 directly:
99. Picture the Problem: The person stands on the 60.0-N ladder in the manner depicted by
the figure at right.
Strategy: The problem can be solved by setting the vector sums of the forces and the
torques equal to zero. The only differences between this problem and Active Example 113 are the addition of a vector m g at the center of mass of the ladder, and the modification
of the distance b. The horizontal distance between the base of the ladder and the vector
m g is c 
 12  4.0 m    12  3.8 m 
2
Solution: 1. (a) Set
2
 0.62 m.
  0 and solve
for f 3 . Let b  c because the person is
halfway up the ladder:
2. Determine the numerical value of f 3 :
3. Set
4. Set
F
 0 and solve for f 2 :
F
 0 and solve for f1 :
x
y
  a f
 b mg  c m g  0
bmg  cm g c  mg  m g 
f3 

a
a
3
b
 0.62 m  85 kg   9.81 m/s 2   60.0 N 
f3 
3.8 m
 146 N  0.15 kN
F
x
 f 2  f3  0
f 2  f3  0.15 kN
F
y
 f1  mg  m g  0
f1  mg  m g   85 kg   9.81 m/s 2   60.0 N
 894 N  0.89 kN
5. (b) Set
b
  0 and solve for
 34  4.0 m   34  3.8 m
2
2
f 3 . Let
 0.94 m:
f3 

bmg  cm g
a
 0.94 m 85 kg   9.81 m/s 2    0.62 m  60.0 N 
f3  216 N  0.22 kN
6. Let f 2  f3 as in step 3:
f 2  f3  0.22 kN
7. The force f1 is unchanged:
f1  894 N  0.89 kN
3.8 m
100. Picture the Problem: The sign is supported by a rope as indicated in the figure
at right.
Strategy: Set the net torque about the bolt equal to zero and solve for the tension
in the rope. The torque due to the rope is positive and the torque due to the
weight is negative. Then write Newton’s Second Law in the vertical and
horizontal directions to find the vertical and horizontal components of the force
F exerted by the bolt on the sign.
Solution: 1. (a) Set
 τ  0 and solve for T:
 τ   2L T sin     L  mg   0
T
2. (b) Let horizontal forces sum to zero and solve
for Fx :
16.0 kg   9.81 m/s
mg

2sin 
2sin 20
  229 N
Fx  T cos   0
mg
 mg 
Fx  T cos   
 cos  
2 tan 
 2sin  

3. (c) Let vertical forces sum to zero and
solve for Fy :
2
16.0 kg   9.81 m/s 2 
2 tan 20.0
 216 N
Fy  mg  T sin   0
 mg 
1
Fy  mg  T sin   mg  
 sin   2 mg
 2sin  

1
2
16.0 kg   9.81 m/s2  
78.5 N