Chapter 6 Solutions
6.
Picture the Problem: The book slides in a straight line across the top of the tabletop.
Strategy: The minimum force required to get the book moving is related to the maximum coefficient of static friction, and the
force required to keep the book sliding at constant speed is equal to the magnitude of the kinetic friction force, from which µk
can be determined.
10.
Solution: 1. When the book begins
sliding, the applied force equals the
maximum static friction force:
F  fs  s mg
F
2.25 N
s  app 
 0.127
mg 1.80 kg   9.81 m/s2 
2. When the book is sliding at
constant speed, the applied force
equals the kinetic friction force:
Fapp  f k  k mg
F
1.50 N
k  app 
 0.0849
mg 1.80 kg   9.81 m/s 2 
app
Picture the Problem: The free-body diagram of this situation is depicted at
right.
Strategy: Write Newton’s Second Law in the vertical direction to determine the
normal force of the floor on the crate. Then write Newton’s Second Law in the
horizontal direction to determine the minimum force necessary to start the crate
moving.
Solution: 1. Write Newton’s
Second Law in the vertical
direction:
F
2. Write Newton’s Second
Law in the horizontal direction:
F
3. Now substitute for N and solve
for the maximum force F that
would produce no acceleration:
y
θ
 N  mg  F sin   ma y  0
N  mg  F sin 
x
 F cos   s N  max  0
F cos   s N
F cos   s  mg  F sin  
F  cos   s sin    s mg
 0.57  32 kg   9.81 m/s 
s mg

 cos   s sin   cos 21   0.57  sin 21
2
F
F  250 N  0.25 kN
20. Picture the Problem: A bucket is lifted straight upward due to the tension in a rope.
Strategy: There are two forces acting on the bucket, the rope tension T acting straight upward and the force of gravity
W  mg acting straight downward. Write Newton’s Second Law for the bucket (with upward being the positive
direction) and solve for T.
Solution: 1. Write Newton’s Second Law for the bucket:
2. Solve the equation for T:
F
y
 T  W  ma
T  ma  W  ma  mg  m  a  g 
  4.35 kg  1.78  9.81 m/s 2   50.4 N
22. Picture the Problem: The weight of the mass exerts a downward force on the rear suspension spring, compressing it.
Strategy: The weight of the 110 kg box is the force responsible for compressing the spring. Use equation 5-5 to find
the weight and equation 6-4 to find the spring constant. Let the x axis point straight upward, with x = 0 at the top of the
uncompressed spring.
Solution: Solve equation 6-4 for the
spring constant and use the weight of
the mass as the compressing force:
26.
k 
110 kg   9.81 m/s
F
mg


x
x
  0.13 m 
2
  8300 N/m  8.3 kN/m
Picture the Problem: The spring is capable of being either stretched or compressed by the external force.
Strategy: Use Hooke’s law to find the force required to stretch the spring to a length that is twice its equilibrium length. In that
case the stretch distance is x  2L  L  L . Then use Hooke’s law again to find the force required to compress the spring an
amount equal to x  12 L  L   12 L .
Solution: 1. (a) Use Hooke’s Law to find the magnitude of the
force required to stretch the spring by the amount x  L :
F  k x  k L   250 N/m  0.18 m   45 N
2. (b) Use Hooke’s Law to find the magnitude of the force
required to compress the spring by the amount x   12 L :
F  k x  k  12 L    250 N/m  12  0.18 m   23 N
3. Because the force depends upon the stretch distance, the magnitude of the force required to compress the spring to half its
equilibrium length is less than the force found in part (a).
32.
Picture the Problem: The free-body diagram for each pulley is depicted at
right.
Lower Pulley
Cupper
Strategy: The tension in the rope is the same everywhere and is equal to F as
long as the pulleys rotate without friction. Use the free-body diagrams of the
pulleys to write Newton’s Second Law in the vertical direction for each pulley,
and use the resulting equations to find the tensions. The acceleration is zero
everywhere in the system.
Solution: 1. (a) Write Newton’s Second
Law for the box:
F
2. Now write Newton’s Second Law for
the lower pulley and solve for F:
F
y
F
F
F
Clower
 Clower  mg  0
Clower  mg
y
 F  F  Clower  0
F  12 Clower  12 mg 
1
2
 52 kg   9.81 m/s 2 
 255 N  0.26 kN downwards
3. (b) Write Newton’s Second Law for
the upper pulley and solve for Cupper
4. (c) Use the result of step 1 to find Clower:
F
y
 Cupper  F  F  0
Cupper  2 F  2  0.255 kN   0.51 kN
Clower  mg   52 kg   9.81 m/s 2   510 N  0.51 kN
F
Upper Pulley
37.
Picture the Problem: The forces exerted on the left and right blocks are depicted
at right.
Strategy: Because the pulley is ideal, the tension in the string is equal to the
weight of the hanging block. This can be verified by Newton’s Second Law in the
vertical direction for the hanging block:  Fy  T  mg  0 . Write Newton’s
Second Law along the direction parallel to the incline for the 6.7 kg block and
substitute T  mg into the resulting equation to find m.
40.
Solution: 1. Write Newton’s Second Law along the
direction parallel to the incline:
F
2. Substitute T  mg into the
resulting equation to find m.
T  mg  Mg sin 
x
 T  Mg sin   0
m  M sin    6.7 kg  sin 42  4.5 kg
Picture the Problem: The free-body diagram for the wood block is depicted at right.
Strategy: Write Newton’s Second Law in the vertical direction for the block to find the
magnitude of the normal force. Then write Newton’s Second Law in the horizontal
direction for the block to find the magnitude of the applied force. In both cases the
acceleration of the block is zero.
F
Solution: 1. (a) Write Newton’s Second
Law in the vertical direction:
y
 fs  mg  s N  mg  0
N
F
2. Write Newton’s Second Law in the
horizontal direction:
x
F
3. Solve for the applied force F:
mg
s
FN F
mg
s
0
1.6 kg   9.81 m/s2 

 20 N 
s
 0.79
mg
0.020 kN  4.5 lb
4. (b) The force of static friction would remain the same if you push with a greater force because it must exactly balance the
weight of the wood board.
44.
Picture the Problem: The physical apparatus is shown at right.
Strategy: Write Newton’s Second Law for each of the three blocks and add the
equations to eliminate the unknowns T1 and T2 . Solve the resulting equation for
the acceleration a. Let x be positive in the direction of each mass’s motion.
Solution: 1. Write Newton’s
Second Law for each of the
three blocks and add the
equations:
F
 T1
 m1a
Fx  T1  T2
 m2 a
x
block 1

block 2

Fx 
 T2  m3 g  m3 a
block 3
m3 g   m1  m2  m3  a
2. Solve the resulting
equation for a:


m3
3.0 kg
a
9.81 m/s 2   4.9 m/s 2

g 
6.0 kg
 m1  m2  m3 
58.
Picture the Problem: The forces acting on the car are depicted at right.
Strategy: Write Newton’s Second Law in the horizontal and vertical directions
and combine them to obtain the radius of the curve. The procedure is similar to
Example 6-9 in the text.
61.
Solution: 1. Write Newton’s
Second Law in the x direction
and solve for r:
F
2. Write Newton’s Second Law
in the y direction and solve for N:
 Fy  N cos   mg  0
3. Substitute for N in the
equation from step 1:
r
x
 N sin   macp 
r
N
mv 2
r
mv 2
N sin 
mg
cos 
 22.7 m/s 
mv 2 cos 
v2


 79.4 m
sin  mg
g tan   9.81 m/s 2  tan 33.5
2
Picture the Problem: Free-body diagrams for the top and bottom
positions on the Ferris wheel are drawn at right.
Top of Ferris Wheel
r
Strategy: Find the speed of the seat by dividing the circumference of
its circular path by the time it takes to complete a cycle. Write
Newton’s Second Law for the passenger at the top of the Ferris wheel
and a second Newton’s Second Law for a passenger at the bottom, and
solve each for the normal force.
Solution: 1. (a) The normal force on you by the Ferris wheel seat is
equal to your apparent weight. At the top of the Ferris wheel your
centripetal acceleration points downward and the normal force is the
smallest, but at the bottom of the Ferris wheel the normal force must
both support your weight and provide the upward centripetal
acceleration, so it is at its maximum there.
r
Bottom of Ferris Wheel
2. (b) Write Newton’s Second Law at the top of the
Ferris wheel and solve for your apparent weight N top :
F
3. Find the speed of the seat:
v
4. Calculate the apparent weight N top :
2

1.6 m/s 
N top  m  g  v 2 r    55 kg   9.81 m/s 2 

7.2 m

 N top  mg   macp   mv 2 r
y
N top  m  g  v 2 r 
2 r 2  7.2 m 

 1.6 m/s
T
28 s




 520 N  0.52 kN
5. Write Newton’s Second Law at the bottom of the
Ferris wheel and solve for your apparent weight Nbottom :
F
y
 N bottom  mg  macp  mv 2 r
2

1.6 m/s 
N bottom  m  g  v 2 r    55 kg   9.81 m/s 2 

7.2 m

 560 N  0.56 kN




64.
Picture the Problem: The bucket swings along a vertical circular path at constant speed.
Strategy: The normal force on the water must be greater than zero for it to remain in the bucket. Write Newton’s Second Law
in the vertical direction for the water in the bucket when it is at the top of the circle. Set the normal force equal to zero and solve
for the minimum speed of the bucket.
Solution: 1. (a) Write Newton’s
Second Law for the passenger, set
N to zero and solve for v:
F
y
  N  mg  m v 2 r
0  mg  m v 2 r
v  rg 
2. (b) The answer to part (a) is independent of the mass of the bucket.
1.3 m   9.81 m/s2  
3.6 m/s