Physics 107 HOMEWORK ASSIGNMENT #8
Cutnell & Johnson, 7th edition
Chapter 9: Problems 16, 22, 24, 66, 68
16 A lunch tray is being held in one hand, as the drawing
illustrates. The mass of the tray itself is 0.200 kg, and its
center of gravity is located at its geometrical center. On the
tray is a 1.00-kg plate of food and a 0.250-kg cup of coffee.
Obtain the force
exerted by the thumb and the force
exerted by the four fingers. Both forces act perpendicular to
the tray, which is being held parallel to the ground.
*22 The drawing shows an outstretched arm (0.61 m in
length) that is parallel to the floor. The arm is pulling
downward against the ring attached to the pulley system, in
order to hold the 98-N weight stationary. To pull the arm
downward, the latissimus dorsi muscle applies the force M in
the drawing, at a point that is 0.069 m from the shoulder joint
and oriented at an angle of 29°. The arm has a weight of 47 N
and a center of gravity (cg) that is located 0.28 m from the
shoulder joint. Find the magnitude of M.
*24 A woman who weighs 500N is leaning against a smooth vertical wall, as the
drawing shows. Find (a) the force FN (directed perpendicular to the wall) exerted
on her shoulders by the wall and the (b) horizontal and (c) vertical components of
the force exerted on her shoes by the ground.
*66 The drawing shows a bicycle wheel resting against a small step whose
height is h = 0.120 m. The weight and radius of the wheel are W = 25.0 N and
r = 0.340 m. A horizontal force
is applied to the axle of the wheel. As the
magnitude of
increases, there comes a time when the wheel just begins to
rise up and loses contact with the ground. What is the magnitude of the force
when this happens?
*68 A wrecking ball (weight = 4800 N) is supported by a boom, which may
be assumed to be uniform and has a weight of 3600 N. As the drawing
shows, a support cable runs from the top of the boom to the tractor. The
angle between the support cable and the horizontal is 32°, and the angle
between the boom and the horizontal is 48°. Find (a) the tension in the
support cable and (b) the magnitude of the force exerted on the lower end of
the boom by the hinge at point P.
16. REASONING AND SOLUTION The net torque about an axis through the contact point
between the tray and the thumb is
 = F(0.0400 m)  (0.250 kg)(9.80 m/s2)(0.320 m)  (1.00 kg)(9.80 m/s2)(0.180 m)
 (0.200 kg)(9.80 m/s2)(0.140 m) = 0
F  70.6 N, up
Similarly, the net torque about an axis through the point of contact between the tray and the
finger is
 = T (0.0400 m)  (0.250 kg)(9.80 m/s2)(0.280 m)  (1.00 kg)(9.80 m/s2)(0.140 m)
 (0.200 kg)(9.80 m/s2)(0.100 m) = 0
T  56.4 N, down
0.61 m
22. REASONING The arm, being stationary, is
in equilibrium, since it has no translational or
angular acceleration. Therefore, the net
0.28 m
external force and the net external torque
98 N
acting on the arm are zero. Using the fact
0.069 m
that the net external torque is zero will allow
us to determine the magnitude of the force
cg
Axis
M. The drawing at the right shows three
29
forces: M, the 47-N weight of the arm acting
47 N
at the arm’s center of gravity (cg), and the
98-N force that acts upward on the right end
M
of the arm. The 98-N force is applied to the
(0.070 m) sin 29
arm by the ring. It is the reaction force that
arises in response to the arm pulling downward on the ring. Its magnitude is 98 N, because
it supports the 98-N weight hanging from the pulley system. Other forces also act on the
arm at the shoulder joint, but we can ignore them. The reason is that their lines of action
pass directly through the axis at the shoulder joint, which is the axis that we will use to
determine torques. Thus, these forces have zero lever arms and contribute no torque.
SOLUTION The magnitude of each individual torque is the magnitude of the force times
the corresponding lever arm. The forces and their lever arms are as follows:
Force
Lever Arm
98 N
0.61 m
47 N
0.28
M
(0.069 m) sin 29
Each torque is positive if it causes a counterclockwise rotation and negative if it causes a
clockwise rotation about the axis. Thus, since the net torque must be zero, we see that
 98 N  0.61 m    47 N  0.28 m   M  0.069 m  sin 29  0
Solving for M gives
M
 98 N  0.61 m    47 N  0.28 m   1400 N
 0.069 m  sin 29
24. REASONING AND SOLUTION
a. The net torque about an axis through the point of contact between the floor and her shoes
is
 =  (5.00  102 N)(1.10 m)sin 30.0° + FN(cos 30.0°)(1.50 m) = 0
FN  212 N
b. Newton's second law applied in the horizontal direction gives
Fh  FN = 0,
so
Fh  212 N .
c. Newton's second law in the vertical direction gives Fv  W = 0, so Fv  5.00 102 N .
66. REASONING When the wheel is resting on the ground it is in equilibrium, so the sum of
the torques about any axis of rotation is zero    0  . This equilibrium condition will
provide us with a relation between the magnitude of F and the normal force that the ground
exerts on the wheel. When F is large enough, the wheel will rise up off the ground, and the
normal force will become zero. From our relation, we can determine the magnitude of F
when this happens.
SOLUTION The free body diagram
shows the forces acting on the wheel: its
weight W, the normal force FN, the
horizontal force F, and the force FE that
the edge of the step exerts on the wheel.
We select the axis of rotation to be at
the edge of the step, so that the torque
produced by FE is zero. Letting
N , W , and F represent the lever
r
FN
F
Axis of Rotation
FE
h
W
arms for the forces FN, W, and F, the sum of the torques is
   FN
r 2   r  h
N
Solving this equation for F gives
2
W
r 2  r  h
W
2
 F r  h  0
F
W  FN 
F
r 2  r  h
2
r h
When the bicycle wheel just begins to lift off the ground the normal force becomes zero
(FN = 0 N). When this happens, the magnitude of F is
W  FN 
F
r 2  r  h
2
 25.0 N  0 N   0.340 m 2   0.340 m  0.120 m 2

 29 N
r h
0.340 m  0.120 m
__________________________________________________________________________
68. REASONING If we assume that the system is in equilibrium, we know that the vector sum
of all the forces, as well as the vector sum of all the torques, that act on the system must be
zero.
The figure below shows a free body diagram for the boom. Since the boom is assumed to be
uniform, its weight WB is located at its center of gravity, which coincides with its
geometrical center. There is a tension T in the cable that acts at an angle  to the horizontal,
as shown. At the hinge pin P, there are two forces acting. The vertical force V that acts on
the end of the boom prevents the boom from falling down. The horizontal force H that also
acts at the hinge pin prevents the boom from sliding to the left. The weight WL of the
wrecking ball (the "load") acts at the end of the boom.
By applying the equilibrium conditions to the boom, we can determine the desired forces.
SOLUTION The directions upward and to the right will be taken as the positive directions.
In the x direction we have
(1)
 Fx  H  T cos   0
while in the y direction we have
 Fy  V  T sin   WL  WB  0
(2)
Equations (1) and (2) give us two equations in three unknown. We must, therefore, find a
third equation that can be used to determine one of the unknowns. We can get the third
equation from the torque equation.
In order to write the torque equation, we must first pick an axis of rotation and determine the
lever arms for the forces involved. Since both V and H are unknown, we can eliminate them
from the torque equation by picking the rotation axis through the point P (then both V and H
have zero lever arms). If we let the boom have a length L, then the lever arm for WL
is L cos  , while the lever arm for WB is ( L / 2) cos  . From the figure, we see that the lever
arm for T is L sin( –  ) . If we take counterclockwise torques as positive, then the torque
equation is
 L cos  
   WB 
  WL L cos   TL sin      0
 2 
Solving for T, we have
1 W W
L
T 2 B
cos 
(3)
sin( – )
a. From Equation (3) the tension in the support cable is
T
1 (3600
2
N)  4800 N
sin(48 – 32)
cos 48  1.6 104 N
b. The force exerted on the lower end of the hinge at the point P is the vector sum of the
forces H and V. According to Equation (1),


H  T cos   1.6 104 N cos 32  1.4 104 N
and, from Equation (2)


V  WL  WB  T sin   4800 N  3600 N  1.6 104 N sin 32  1.7 104 N
Since the forces H and V are at right angles to each other, the magnitude of their vector sum
can be found from the Pythagorean theorem:
FP  H 2  V 2  (1.4 104 N)2  (1.7 104 N) 2  2.2 104 N