Solutions to Chapter Practice Problems
11.10, 11.20, 11.22, 11.49
10)
Apply the first and second conditions for equilibrium to the ladder.
Let n2 be the upward normal force exerted by the ground and let n1 be the horizontal
normal force exerted by the wall. The maximum possible static friction force that can be
exerted by the ground is sn2.
(a) Since the wall is frictionless, the only vertical forces
are the weights of the man and the ladder, and the
normal force n2 . For the vertical forces to balance,
n2  w1  wm  160 N  740 N  900 N, and the maximum
frictional force is sn2  (040)(900N)  360N.
(b) Note that the ladder makes contact with the wall
at a height of 4.0 m above the ground. Balancing
torques about the point of contact with the ground,
(40 m)n1  (15 m)(160 N)  (10 m)(3/5)(740 N)  684 N  m,
so n1  1710 N. This horizontal force must be balanced
by the friction force, which must then be 170 N to two figures.
(c) Setting the friction force, and hence n1, equal to the maximum of 360 N and solving
for the distance x along the ladder, (40 m)(360 N)  (150 m)(160 N)  x(3/5)(740 N),
so x  27 m.
20)
Apply the first and second conditions for equilibrium to the beam.
The free-body diagram for the beam is given
The cable is given as perpendicular to the beam,
so the tension is found by taking torques about the
pivot point;
T (3.00 m)  (1.00 kN)(2.00 m)cos 25.0
and
 (5.00 kN)(4.50 m)cos 25.0,
T  7.40 kN.
The vertical component of the force exerted on the beam by the pivot is the net weight
minus the upward component of T,
6.00 kN  T cos25.0  0.71 kN.
22)
The vertical component is downward. The horizontal force is T sin 25.0  3.13 kN.
The person is in equilibrium, so the torques on him must balance.
The target variable is the force exerted by the deltoid muscle.
The free-body diagram for the arm is
Take the pivot at the shoulder joint and let counterclockwise torques be positive. Use
coordinates as shown. Let F be the force exerted by the deltoid muscle. There are also
the weight of the arm and forces at the shoulder joint, but none of these forces produce
any torque when the arm is in this position. The forces F and T have been replaced by
their x and y components.  z  0.
 z  0 gives:
( F sin12.0)(15.0 cm)  (T cos35)(64.0 cm)  0.
F
49)
(36.0 N)(cos35)(64.0 cm)
 605 N.
(sin12.0)(15.0 cm)
In each case, to achieve balance the center of gravity of the system must be at the
fulcrum. Use the center of mass equation 11.3 to locate xcm , with mi replaced by wi .
Let the origin be at the left-hand end of the rod and take the  x axis to lie along the rod.
Let w1  255 N (the rod) so x1  100 m, let w2  225 N so x2  200 m and let w3  W . In
part (a) x3  0500 m and in part (b) x3  0750 m.
(a) xcm  125 m. xcm 
W
w1x1  w2 x2  w3 x3
(w  w ) x  w x  w x
gives w3  1 2 cm 1 1 2 2 and
w1  w2  w3
x3  xcm
(480 N)(1.25 m)  (255 N)(1.00 m)  (225 N)(2.00 m)
 140 N.
0500 m  1.25 m
(b) Now w3  W  140 N and x3  0750 m.
xcm 
(255 N)(1.00 m)  (225 N)(2.00 m)  (140 N)(0.750 m)
 131 m.
255 N  225 N  140 N
W must be moved 131 m  125 m  6 cm to the right.