12
STATIC EQUILIBRIUM
EXERCISES
Section 12.1 Conditions for Equilibrium
14.
(a)  Fi  (2iˆ  2 ˆj  2iˆ  3 ˆj  ˆj )N  0.
(b) ( i )0  2iˆ  (2iˆ  2 ˆj)  (iˆ)  (2iˆ  3 ˆj)  7iˆ  ˆj)  ˆj] N  m  (4  3  7)kˆ N  m  0.
15.
INTERPRET We have been told that the choice of pivot point does not matter if the sum of forces is zero. Here we
will show that this is true for two different pivot points. Three forces are acting on an object, which is in equilibrium.
DEVELOP
The three forces are F1  2iˆ  2 ˆj N at point ( x, y)  (2 m, 0 m), F2  2iˆ  3 ˆj N at (1 m, 0 m), and
F3  1 ˆj N at (7 m, 1 m). We find the torques due to these three forces around
points (3 m, 2 m) and (7 m, 1 m), using   |r  F | . To find the value of r for a point other than the origin, we
take the vector difference between the point where the force is applied and the point used as the
pivot: r  (rapplied  rpivot ).
EVALUATE For point (3 m, 2 m), the torque due to F1 is
1  (rapplied  rpivot )  F1  ((2  3)iˆ m  (0  2) ˆj m)  (2iˆ  2 ˆj ) N
ˆj kˆ
iˆ
1  1 2 0 Nm  (2  (4))kˆ Nm  2kˆ Nm
2
2
0
Similarly,  2  8kˆ Nm and  3  10kˆ Nm. The sum of these three is total  (2  8 10)kˆ Nm  0.
Around point (–7 m, 1 m), the torques are 1  20kˆ Nm,  2  20kˆ Nm, and  3  0. The sum of these three is
also  total  0.
ASSESS Note that the torque due to force 3 around the second pivot is zero, since the force acts on the pivot.
16.
The conditions for static equilibrium, under the action of three forces, can be written as: F3  ( F1  F2 ) and
r3  F3  (r1  F1  r2  F2 ). (a) In this
case, F1  F ˆj, r1  (2 m) ˆj, F2  Fiˆ, and r2  (1 m) ˆj. Thus, F3  F (iˆ  ˆj ), which is a force of
magnitude 2F , 45 down into the third quadrant (x  225 or 135 CCW from the x axis). The point of
application, r3 , can be found from the second condition, r3  F3  ( x3iˆ  y3 ˆj )(Fiˆ F ˆj ) 
( x3  y3 ) F kˆ  r1  F1  r2  F2  0  (1 m) ˆj  Fiˆ  (1 m)Fkˆ. Thus,  x3  y3  1 m, or the line of action of F3
passes through the point of application of F2 (the point (0, 1m)). Any point on this line is a suitable point of
application for F3 (e.g., the point (0, 1m)). (b) In this case, F1   F2 so F3  0, but r1  F1  r2  F2 
(r2  r1 )  F2  0 so r3  F3  0. Thus there is no single force that can be added to produce static equilibrium.
17.
INTERPRET The problem asks for a set of conditions that must be met for the body to be in static equilibrium.
This means that all the external forces and torques must be zero.
DEVELOP Static equilibrium demands that (see Equations 12.1 and 12.2)
12.1
12.2
Chapter 12
 Fi  0
 i   (ri  Fi )  0
Since all of the forces lie in the same plane, which includes the points O and P, there are two independent
components of the force condition (Equation 12.1) and one component of the torque condition (Equation 12.2).
EVALUATE (a) Taking the x axis to the right, the y axis up and the z axis out of the page in Figure 12.12, we have:
0   Fx   F1  F2 sin   F3
0   Fy   F2 cos   F4
0  ( z )O   L1 F2  L2 F3 sin   L2 F4 cos 
(b) The equation for torque about point P is
0  (  z ) P   L2 F1 sin   ( L2  L1 ) F2
The lever arms of all the forces about either O or P should be evident from Fig. 12.12.
ASSESS In static equilibrium, the torque about any given pivot point must vanish. Otherwise, the body will rotate
about that point.
Section 12.2 Center of Gravity
18.
The center of gravity is at the center of a uniform plate. In calculating the gravitational torque, one may consider
the entire weight as acting at the center of gravity. (a) rA  2L/2 at 135 from the weight of the plate,
so A  ( 2L/2)mg sin135 
1
2
mgL. (b) rB is colinear with the weight, so  B  0. (c) C  12 Lmg sin 90  12 mg L
(but note that  C   A ). (We also assumed that B and C are at the centers of their respective sides. Alternatively,
the torques can be found from the lever arms shown.)
19.
INTERPRET
DEVELOP
In this problem we are asked to find the gravitational torque about various pivot points.
The torque about a point is given by Equation 10.10,   rF sin   r F, where r  r sin  is the lever
arm. In our problem, the center of gravity (CG) is at the center of the triangle, which is at a perpendicular distance
of 2 L3 from any side. We regard CG as the point at which all the mass is concentrated.
EVALUATE (a) The lever arm of the weight about point A is r , A  L /2. Therefore, the gravitational torque
about A is
 A  r, A Fg 
L
1
mg  mgL
2
2
(b) The lever arm about point B is zero (the line of action passes through point B). Therefore, we have  B  0.
(c) The lever arm about point C is r ,C  L /4 (C is halfway up from the base) so
C  r,C Fg 
L
1
mg  mgL
4
4
Static Equilibrium
12.3
ASSESS The torques A and C are in opposite directions. If pivoted at A, the gravitational torque A tends to
rotate the triangle CW. On the other hand, when pivoted at C, C gives rise to a CCW rotation.
20.
The log is in equilibrium under the torques exerted by the cable, gravity, and the wall. Calculating the torques
about the point of contact with the wall (because the other two forces are given), we find (6.2 kN)(23  4) m 
(7.5 kN) xCG , or xCG  15.7 m, from the end on the wall.
Section 12.3 Examples of Static Equilibrium
21.
INTERPRET In this problem we are asked to find where the child should sit on the pivot-supporting board so that
the scale at the right end will read zero.
DEVELOP If we consider torques about the pivot point (so that the force exerted by the pivot does not contribute)
then Equation 12.2,  i  (ri  Fi )  0, is sufficient to determine the position of the child.
EVALUATE As shown on Fig. 12.15, the weight of the board (acting at its center of gravity), the weight of the
child (acting a distance x from the left end), and the scale force, Fs , produce zero torque about the pivot:
(  ) P  0  0  Fs (1.60 m)  (60 kg)(9.8 m/s 2 )(0.40 m)  (40 kg)(9.8 m/s 2 )(0.80 m  x)
Therefore, the relation between the position of the child and the scale force Fs is given by
x
Fs (1.60 m)  78.4 kg  m 2 /s 2
Fs (1.60 m)
 0.20 m
(40 kg)(9.8 m/s 2 )
(40 kg)(9.8 m/s 2 )
If we want the reading of the scale to be zero, i.e., Fs  0, then the child must be at x  0.20 m (distance from the
left end).
ASSESS If the child moves closer to the pivot (increasing x), then Fs will be increased, i.e., the reading of the scale
will go up. One can also show that without the child, the reading would be Fs  147 N.
22.
If we consider torques about the pivot point (so that the force exerted by the pivot does not contribute) then
Equation 12.2 is sufficient to determine the position of the child. As shown on Fig. 12.15, the weight of the
tabletop (acting at its center of gravity), the weight of the child (acting a distance x from the left end), and the scale
force, Fs , produce zero torque about the pivot:
(1/g )(  ) P  0  ( Fs /g )(160 cm)  (60 kg)(40 cm)  (40 kg)(80 cm  x)
Therefore, x  20 cm  ( Fs /9.8 N)4 cm. If (a) Fs  100 N, then x  20 cm  (400/9.8) cm  60.8 cm, and if
(b) Fs  300 N, x  142 cm. (Note that the child is on opposite sides of the pivot in parts (a) and (b), since
without the child, Fs  147 N.)
23.
INTERPRET The problem is about static equilibrium. We want to know where the concrete should be placed so
that the long beam suspended by a cable and with a steelworker standing at one end will be in equilibrium.
DEVELOP At equilibrium, the sum of the torques on the beam (taken about its center, C, so that the cable’s
tension and beam’s weight do not enter the equation), is equal to zero:
( )C  0
 mc gx  ms g Ls
12.4
Chapter 12
where mc and ms are the masses of the concrete and the steelworker, and Ls  2.1 m is the distance from the
steelworker to C.
EVALUATE Using the values given in the problem statement, we have
m 
65 kg
x   s  Ls 
(2.1 m)  0.718 m
190 kg
 mc 
on the opposite side of C from the worker.
ASSESS The lighter the concrete, the longer the lever arm (x) must be for the system to remain in static
equilibrium.
Note also that since sin   sin(   ) will cancel from the torque equation, the beam need not be horizontal to be
in equilibrium; the steelworker’s mental equilibrium is greatest when the beam is horizontal.
24.
Since the board is at rest, the sum of the torques (positive CCW in Fig. 12.16) is zero about the pivot (due to the
weight of the person, the weight of the board acting at its CG, and the scale force, as shown).
Thus, (  z ) P  0  (210 N)(3 m)  (3.4  9.8 N)(1.5 m)  w(1.2 m), or w  483 N.
Section 12.4 Stability
25.
INTERPRET The problem is about the stability of the roller coaster as it moves along the track described by a
height function. We want to identify the equilibrium point and classify its stability.
DEVELOP The potential energy of the roller coaster car, in the equivalent one-dimensional problem, is
U ( x)  mgh( x)  mg (0.94 x  0.01x 2 )
Equilibrium condition is given by Equation 12.3:
according to its second derivative:
dU
dx
d 2U
dx 2
EVALUATE
 0. In addition, the equilibrium condition may be classified
 0, stable

:  0, unstable
 0, neutral

(a) Equation 12.3, the condition for equilibrium, gives
0
dU
 mg (0.94  0.02 x)  x  47 m
dx
(b) Since ddxU2  0.02mg  0, Equation 12.5 implies that this is an unstable equilibrium.
2
ASSESS The point x  47 m with U ( x  47)  22.09mg corresponds to a local maximum where the potentialenergy curve is concaving downward. Therefore, the point is unstable.
Static Equilibrium
26.
INTERPRET
equilibria.
DEVELOP
12.5
We are given a potential function, and are asked to find the positions of any stable and unstable
The condition for equilibrium is that dU
 0, and the equilibrium point is stable if ddxU2  0 and unstable
dx
2
if ddxU2  0. The function we are given is U  2x3  2x2  7 x  10.
EVALUATE We take the derivative and set it equal to zero to find the locations of any equilibrium points.
2
dU
[2 x3  2 x 2  7 x  10]  6 x 2  4 x  7  0
dx
x
(4)  (4) 2  4(6)(7)  2  46 2  46 

,

2(6)
6 
 6
There are equilibrium points at x  2  6 46  0.797 and x  2 6 46  1.46.
Next, we find the second derivative of U, and evaluate it at these two points to ascertain their stability.
d 2U
dU
[2 x3  2 x 2  7 x  10] 
[6 x 2  4 x  7]  12 x  4
2
dx
dx
2  46
d 2U
x
 2  13.6
6
dx
x
2  46
d 2U
 2  13.6
6
dx
So we can see that the first solution is unstable, and the second is stable.
ASSESS The potential function is plotted in the figure, where it can be seen that the points we found are indeed
equilibria. The point at –0.797 is unstable, and the point at 1.46 is metastable.
PROBLEMS
27.
INTERPRET This problem is about static equilibrium. We want to find that tension force the left-hand bolt can
withstand in order to keep the traffic signal system stable.
DEVELOP The forces on the traffic signal structure, and their lever arms about point 0 (on the vertical member’s
centerline between the bolts) are shown on Fig. 12.17. The normal forces exerted by the bolts and the ground on
the vertical member are designated by nL and nr , measured positive upward. (Of course, the ground can only make
a positive contribution, and the bolts only a negative contribution, to these normal forces.)
The two conditions of static equilibrium needed to determine nL and nr are:
0   Fy  nL  nr  (9.8 m/s2 )(320 kg  170 kg  65 kg)
(the vertical component of Equation 12.1, positive up), and
0  ( z )0  (nr  nL )(0.38 m)  [(170 kg)(3.5 m)  (65 kg)(8.0 m)](9.8 m/s 2 )
(the out-of-the-page-component of Equation 12.2, positive CCW).
EVALUATE The above two equations can be rewritten as
nr  nL  5.44 103 N
nr  nL  2.88 104 N
12.6
Chapter 12
Thus, we find nL  1.17  104 N, which is downward and must be exerted by the bolt.
ASSESS The reaction force on the bolt is upward and is a tensile force. Really, n is the difference between the
downward force exerted by the bolt and the upward force exerted by the ground. Tightening the bolt increases the
tensile force it must withstand beyond the minimum value calculated above, under the assumption that the ground
exerts no force.
28.
The magnitude of the (external) torque on the arm is  0  [(4.2 kg)(0.21 m)  (6 kg)(0.56 m)](9.8 m/s 2 )
sin105  40.2 N  m. The direction is clockwise (into the page) about the shoulder joint. (b) The deltoid muscle
exerts a counterclockwise torque of magnitude Fr sin   F (0.18 m)sin170, which, under equilibrium
conditions, equals the magnitude of the torque in part (a).
Thus, F  40.2 N  m/(0.18 m)sin170  1.28 kN, underscoring the comment at the end of Example 12.3. The
skeleto-muscular structure of the human extremities evolved for speed and range of motion, not mechanical
advantage.
29.
INTERPRET The problem is about static equilibrium. The sphere is supported by a rope attached to the wall. The
friction between the sphere and the wall helps keep it in equilibrium. We want to find the smallest possible value
for the coefficient of friction.
DEVELOP In equilibrium, there is no net force acting on the sphere. In addition, the sum of the torques about the
center of the sphere must be zero, so the frictional force is up, as shown. Using Equations 12.1 and 12.2, the
equilibrium conditions can be written as
 Fx  n  T sin   0
 Fy  T cos   f  Mg  0
R
( )c  Rf    T cos   0
2
The above equations can be used to solve for the frictional force f . The coefficient of friction can then be obtained
from f  s n.
EVALUATE The x component of the force equation gives T  n /sin . Substituting this into the torque equation
gives
Static Equilibrium
12.7
1
n cos  n
f  T cos  
 cot 
2
2 sin  2
Since f  s n, the minimum coefficient of friction is
1
2
1
2
s  cot   cot 30 
3
 0.866
2
ASSESS If the angle  is increased, then the minimum coefficient of friction would decrease. Note that s is
independent of the mass and radius of the sphere; it depends only on the angle .
30.
We assume that a horizontal push on the cart results in a horizontal force exerted on the wheels by the axle, as
shown. (We also suppose both wheels share the forces equally, so they can be treated together.) Also shown are the
weight of the cart and the normal force of the ground, both acting through the center of the wheels, and the force of
the step, Fs. If we consider the sum of the torques (positive CCW) about the step, the latter does not contribute, and
the wheels (and cart) will remain stationary as long as (  )step  MgR sin   nR sin   FR cos   0. When n  0,
however, the wheels begin to lose contact with the ground and go over the step. This occurs when F  Mg tan .
From the geometry of the situation, R(1  cos  )  h, the height of the step, so   cos 1 (1  h/R) 
cos 1 (1  8 /30). Then F  (55  9.8 N) tan(cos 1 (11/15))  500 N is the minimum force.
31.
INTERPRET In this problem we want to find the tension in the Achilles tendon and the contact force at the ankle
joint when the foot is in static equilibrium.
DEVELOP If we approximate the bones in the foot as a massless, planar, rigid body, the equilibrium conditions for
the situation depicted in Fig. 12.21 are:
0   Fx  T sin   FC , x
0   Fy  T cos   n  FC , y
0  (  )ankle joint  n(12 cm)  (T cos  )(7 cm)
where   25.
EVALUATE (a) We first note that the normal force is simply equal to the weight of the person, n  mg 
(70 kg)(9.8 m/s 2 )  686 N. Substituting this into the torque equation, the tension in the Achilles tendon is
T
n(12 cm)
(686 N)(12 cm)

 1298 N
(7 cm) cos 
(7 cm) cos 25
(b) Substituting the value of T into the force equations, we find
FC , x  T sin   (1298 N)sin 25  548 N
FC , y  T cos   n  (1298 N) cos 25  686 N  1862 N
Therefore, the contact force at the ankle joint is
12.8
Chapter 12
FC  FC2, x  FC2, y  (548 N)2  (1862 N)2  1941 N
ASSESS The tension in the Achilles tendon is almost twice the weight of the person. This is because the Achilles
tendon is very close to the ankle joint, leading to a small lever arm. The contact force at ankle is roughly three
times the weight of the person. The problem demonstrates that in order to maintain a static equilibrium, many parts
of our body often experience forces that are greater than our own weight.
32.
It is shown in the solution to Problem 45 that the condition for a person of mass m to climb up a fraction of length
of ladder  without the ladder slipping is   s cot   (mL  m)(s cot   12 ), where s is the coefficient of
friction with the floor,  the angle with the vertical frictionless wall, and mL the mass of the ladder. For the
situation in this problem, s cot   0.26cot15  0.970 and   0.970  (5/65)(0.470)  1.01. Therefore
(since   1, by definition) a 65-kg person can climb all the way to the top. However, the right-hand side of the
condition is less than 1 for m  m (s cot   12 )  (1  s cot )  5 kg (0.470) /(0.0297)/79.3 kg, so a person with
mass greater than 79.3 kg causes the ladder to slip before reaching the top.
33.
INTERPRET In this problem we want to find the tension in the cable supporting the boom so that the boom is in
static equilibrium.
DEVELOP For the boom to remain in static equilibrium, the forces must satisfy Equation 12.1. Since we are only
asked about the tension, we can focus only on the torque about point P. The forces on the boom are shown
superposed on the figure. By assumption, T is horizontal and acts at the CM of the boom. To find T, we compute
the torques about P.
EVALUATE The condition for equilibrium implies that that (  ) P  0, or
T (L /2)sin   mb g (L /2) cos   mg L cos   0
which can be solved to give
T  (2m  mb ) g cot   (4400 kg  1700 kg)(9.8 m/s 2 ) cot 50  5.02  104 N
ASSESS To see that our result makes sense, let’s consider the following cases: (i)   90. In this limit, the boom
is vertical and the tension in the cable vanishes (cot 90  0), as expected. (ii) m  0, when no mass is hung from
the end of the boom, the tension in the cable is reduced to T  mb g cot . This comes from the force equations at
P (with m  0 ):
Static Equilibrium
12.9
0   Fx  FP cos   T sin 
0   Fy  FP sin   mb g
34.
The conditions for equilibrium (about the origin drawn on the figure) are:
 Fx  0  T2 sin 60  T1 sin 35,
 Fy  0  T2 cos 60  T1 cos 35  w  W ,
1
( z )0  0  T2 L sin 20.8  W L sin 99.2.
2
Therefore:
T2  0.5W sin 99.2  sin 20.8  1.39W ,
T1  T2 sin 60  sin 35  2.10W ,
and
w  T1 cos35  T2 cos 60  W  1.41W .
35.
INTERPRET In this problem we want to find the force the people apply to pull the car so that it is in static
equilibrium. The applied force is equal to the tension in the rope.
DEVELOP For the car to remain in static equilibrium, the forces must satisfy Equation 12.1. Three forces act on
the car, as shown added to Fig. 12.24. The unknown force, FP, exerted by the edge of the embankment, does not
contribute to Equation 12.2 (positive torques CCW) if evaluated about point P, so the tension necessary to keep the car in
equilibrium can be found directly. Thus, our plan is to compute the torques about P.
EVALUATE The condition for equilibrium implies that that (  ) P  0, or
0  Mg ( L  l ) cos   Tl sin 
where L  2.4 m and l  1.8 m. This gives
 Ll 
T  Mg 
 cot 
 l 
 2.4 m  1.8 m 
 (1250 kg)(9.8 N) 
 cot 34
1.8 m


 6.05  103 N
ASSESS Note that if l  L, (when the CM lies above the edge of the embankment), then the tension becomes zero.
In this situation, the car would remain in equilibrium.
36.
The addition of a frictional force on the ladder where it contacts the wall is shown in the sketch (see Fig. 12.4b).
We assume that f 2  2 n2 is the maximum frictional force and that it acts to oppose the fall of the ladder. (If f 2 is
12.10
Chapter 12
not proportional to n2, there is insufficient data to solve this problem.) The equations of static equilibrium in
Example 12.2 become: f1  n2  0 (horizontal), n1  2 n2  mg  0 (vertical), and n2 L sin   2 n2 L cos  
1
mgL cos   0 (torque about bottom of board). The minimum angle  can be found from the horizontal equation
2
and the requirement that n2  f1  1n1 . n1 can be eliminated from the vertical equation,
n2  1 (mg  2 n2 ) or (1  12 )n2  1mg. n2 can be eliminated from the torque
equation, n2 (tan   2 )  12 mg , so this condition
becomes 12 mg (1  12 ) /(2  tan )  1mg, or tan   (1  1 2 )/21 .
37.
INTERPRET This problem is about static equilibrium. The hanging sign is being supported by a rod and a cable.
Given the maximum tension in the cable, we want to find the minimum height above the pivot for anchoring the
cable to the wall so that the sign remains in static equilibrium.
DEVELOP Suppose that the sign is centered on the rod, so that its CM lies under the center of the rod. Then the
total weight may be considered to act through the center of the rod, as shown. For the sign to remain in static
equilibrium, the forces and the torques must satisfy Equations 12.1 and 12.2. In this situation, since we are only
interested in the minimum height h and not FP, it is sufficient to focus only on the torques about point P.
EVALUATE In equilibrium, we have ( ) P  0. Note that the pivot force FP makes no contribution to the torque
about P. The equation implies
0  ( ) P  TL sin   Mg ( L /2)
Therefore, we find the tension to be
T
where we have used tan  
h
L
Mg
1
L2
 Mg 1  2
2sin  2
h
and the identity 1  cot 2   csc2. Solving for h, we obtain
 2T 2 
h  L 
  1
 Mg 

1/ 2
Given the maximum tension that could be withstood in the cable, Tmax , the condition that must be met by h for the sign to
remain in static equilibrium is
 2T 2 
h  L  max   1

 Mg 
1/ 2
So the minimum height is hmin  1.17 m.
2



2(800 N)
 (2.3 m) 
 1
2 
 (66 kg  8.2 kg)(9.8 m/s ) 

1/ 2
 1.17 m
Static Equilibrium
12.11
ASSESS When the maximum tension which the cable can withstand is such that Tmax  Mg/2, static equilibrium would
not be possible!
38.
INTERPRET In Example 12.1, we saw how to find the tension in the cable. Now, we need to find the magnitude and
direction of the force on the hinge for that same problem. The bridge is in equilibrium, so the sum of forces is zero.
DEVELOP From Figure 12.1, we can see that in order for the net force to be zero, the horizontal component of the
hinge force is equal and opposite to the horizontal component of the tension; and the vertical component of hinge
force is equal and opposite to the sum of the gravitational force and the vertical component of tension. The mass of
the bridge is m  11000 kg, and the tension on the cable, which is angled 15° below horizontal, is T  180 103 N.
EVALUATE The horizontal component of the hinge force is Fx  T cos(15)  173.9  103 N, directed in the
positive x direction. The vertical component is Fy  mg  T sin(15)  154.4 103 N. The magnitude of the hinge
force is F  Fx2  Fy2  233 kN, and the direction is   tan 1
   41.6 above the horizontal.
Fy
F
ASSESS This force has a larger magnitude than either the weight of the bridge or the tension of the cable, as we
might expect from the rough estimate that the hinge balances both these other forces.
39.
INTERPRET This problem is an equilibrium problem, as long as the log does not slip! We use the conditions for
equilibrium to find the maximum mass of the climber at the end of the log.
DEVELOP We start by drawing a free-body diagram showing all of the forces on the log, as shown in the figure.
The maximum frictional force on the end of the log is F f   s FN . This frictional force is balanced by the normal
force from the wall, FN , so if the force from the wall is greater than the maximum frictional force, then the log will
slip. We can find the force from the wall by using   0 with the left end of the log as the pivot point. We also
see from the figure that FN  ( M  m) g, where M  340 kg is the mass of the log, and m is the unknown mass of
the climber. The length of the log is L  6.3 m, and the log forms an angle   27 with the horizontal. The center
of mass of the log is located L3 from the left end, and the coefficient of friction between the left end of the log and
the ground is   0.92.
 m  M3  g cos 
EVALUATE    mg L cos   Mg L3 cos   FN L sin   0  FN 
sin 
This normal force from the wall must be balanced by the static frictional force on the left end of the log:
Ff  FN   FN    m  M  g 
 m  M3  g cos   m  M3 
sin 

tan 
m
M
 m   M 

tan  3 tan 
1 

 1

 m  
 
M
tan  

 3 tan 

  1 
 3 tan   3 
 m  M  1 tan    M 
  3.0 M  1000 kg


 1  3 tan  
 3 tan 

ASSESS This mass limit is large enough that any climber can safely cross.
40.
As in Problem 37, the equilibrium condition for torques about the pivot does not contain the unknown pivot force,
and thus allows the tension to be directly determined without use of the force equations. Thus,
12.12
Chapter 12
TL sin 50  MgL cos50  mg 13 L cos50,
or T  (M  13 m) g cot 50  (2500  13  830)(9.8 N)cot 50  22.8 kN.
41.
INTERPRET In this problem we are given a block that makes an angle  with the horizontal, and we want to find
out the values of  which make it stable.
DEVELOP Let the block be tilted in a plane perpendicular to its thickness, as shown in the figure. The condition
for equilibrium implies that the torque about the pivot point is zero, i.e., (  ) P  0. Since there’s no external force
acting on the block, the only force that contributes to the torque is the weight of the block.
EVALUATE To satisfy the condition that (  ) P  0, we require that the weight force passes through the pivot
point so that the lever arm vanishes. One can readily show that   0 is a possibility and it represents a stable
equilibrium position. On the other hand,   90 is a metastable one. Finally,     90 is an unstable one ( is
the angle a diagonal makes with the longer side, as shown). Since tan   2LL  12 , the unstable equilibrium is at
1
 
  90  tan 1    63.4
2
ASSESS For a rectangular block of length L and width W, one may show that the general expression for the torque
about the pivot point is given by
1
2
  r F  ( L cos   W sin  )mg
Thus, the condition that   0 implies that
tan  
L
W
In our case, we have L /W  2, which gives   63.4.
42.
The equilibrium condition, dU /dx  0 (Equation 12.3), requires 3( x/ x0 ) 2  2a( x/ x0 )  4  0. This quadratic has
two real roots if the discriminant is positive, i.e., a 2  12  0, or |a|  2 3. The roots
are ( x/x0 )  13 (a  a 2  12). The second derivative of the potential energy, evaluated at these roots, is

 U0 
 d 2U  U 0   x 
2
 2   2 6    2a    2 a  12  2 

 dx  x0   x0 
 x0 
Thus, the “plus” root is a position of metastable equilibrium (Equation 12.4), while the “minus” root represents
unstable equilibrium, (Equation 12.5). A plot of the potential energy, which is a cubic, will clarify these remarks.
Static Equilibrium
12.13
For |a|  2 3, U ( x) has no wiggles, as shown. (U passes through the origin, but its position depends on the value
of “a”, and is not shown.)
43.
INTERPRET In this problem a cubical block is placed on an incline. Given the coefficient of friction between the
block and the incline, we’d like to find out whether the block first slides or tips when the angle of the incline is
increased.
DEVELOP We suppose that the block is oriented with two sides parallel to the direction of the incline, and that its
CM is at the center. The condition for sliding is that
mg sin   f smax  s n  s mg cos 
or tan   s. The condition for tipping over is that the CM lie to the left of the lower corner of the block (see
sketch). Thus    , where
  tan 1  wh 
is the diagonal angle of the block.
EVALUATE For a cubical block, w  h, and the cube will tip over when     tan 1 ( w/h)  tan 1 (1)  45 but
will slide when
  tan 1 s  tan 1 0.95  43.5
It thus slides before tipping.
ASSESS We find that sliding happens first if s  w/h. This makes sense because when the coefficient of friction
is small, the block has a greater tendency to slide. On the other hand, when the coefficient of friction is large
( s  w/h), we’d expect tipping to take place first.
44.
The forces on the sign are in a similar configuration to those shown in the figure. (The weight of the sign is shown
acting at its center and the force at the bottom corner could have a vertical component.) To apply the limit on the
tensile force Fx , we need only consider Equation 12.2 about point O : Fx (1.4 m  d ) 
(160  9.8 N)(1.15 m). Then Fx  (1.80 kN  m) /(1.4 m  d )  2.1 kN implies that d  1.4 m  1.80 kN  m/
2.1 kN  54.1 cm.
45.
INTERPRET In this problem a ladder is leaning against the wall and we want to find the mass of the heaviest
person who can climb to the top of the ladder while keeping it in static equilibrium.
DEVELOP The forces on the uniform ladder are shown in the sketch, with the force exerted by the (frictionless)
wall horizontal. The person is up the ladder a fraction  of its length. Equilibrium conditions require:
12.14
Chapter 12
0   Fx  f  Fwall
0   Fy  n  (mL  m) g
0  (  ) A  Fwall L cos   mL g ( L/2)sin   mg L sin 
The ladder will not slip if f  s n. Using the equations above, this condition can be rewritten as
1

f  Fwall   mL   m g tan   sn  s (mL  m) g
2

or

s (mL  m) cot   mL /2
m
 s cot  
mL
m
1

 s cot   
2

Here, we used the horizontal force equation to find f , the torque equation to find Fwall, and the vertical force equation to
find n.
EVALUATE For a person at the top of the ladder,   1 and the condition for no slipping becomes
  cot   1/2 
m  mL  s

 1  s cot  
With the data given for the ladder (note that cot   cot( /2  66)  tan 66), we obtain
m  (9.5 kg)
(0.42) tan 66  1/2
 74.3kg
1  (0.42) tan 66
ASSESS The above equation shows that when the coefficient of friction becomes too small, s cot   1/2,
or s  tan  /2 (see Example 12.2), slipping will occur and it’s no longer possible for the ladder to remain in static
equilibrium. In this situation, nobody can climb up to the top of the ladder without making the ladder slip,
regardless of his or her mass.
46.
For a 95 kg person, the condition that the ladder not slip gives   0.42 tan 66  (9/95)(0.42 tan 66  0.5) 
98.8% as the maximum fraction up along the length of the ladder. (See Problem 45.) The maximum height above
the ground is just  L cos   (0.988)(5 m)sin 66  4.51 m.
47.
INTERPRET We want to find the energy required to bring a cube to an unstable equilibrium. The problem is
equivalent to finding the increase in potential energy of the system.
DEVELOP When resting in a stable equilibrium position, the CM of a uniform cube of side s is at a distance
y0  s /2 above the tabletop. When balancing on a corner, the CM is now a distance
y  ( s/2)2  ( s/ 2)2  ( s/ 2)2 
3
s
2
above the corner resting on the tabletop.
EVALUATE From the above, the potential energy difference is
 3 1
U  mg ycm  mgs 
   0.366 mgs
 2 2
This is the energy required to bring the cube to an unstable equilibrium.
Static Equilibrium
12.15
ASSESS Raising the vertical distance of the CM increases the potential energy of the cube. In general, the stability
of a system decreases as its potential energy is increased.
48.
Example 9.3 shows that the CM of an isosceles triangle is 13 the height from the base, or 32 the height from the apex.
The difference in energy between the unstable and stable equilibrium mentioned is U  mg ycm 
mg ( 32 h  13 h)  13 mgh.
49.
INTERPRET This problem is about a ladder leaning against the wall and we want to verify the condition under
which any person (with any mass) can climb to the top, and also the one in which nobody can.
DEVELOP The forces on the uniform ladder are shown in the sketch, with the force exerted by the (frictionless)
wall horizontal. The person is up the ladder a fraction  of its length. Equilibrium conditions require:
0   Fx  f  Fwall
0   Fy  n  (mL  m) g
0  (  ) A  Fwall L cos   m g ( L /2)sin   mg L sin 
The ladder will not slip if f  s n. Using the equations above, this condition can be rewritten as
1

f  Fwall   mL   m  g tan   sn  s (mL  m) g
2

or

s (mL  m) cot   mL /2
m
 s cot  
mL 
1
 s cot   
m
2
Here, we used the horizontal force equation to find f , the torque equation to find Fwall , and the vertical force equation to
find n.
EVALUATE For a person at the top of the ladder,   1 and the condition for no slipping becomes
  cot   1/2 
 s  tan  /2 
m  mL  s
  mL 

 1  s cot  
 tan   s 
Since m is positive, this condition cannot be fulfilled if s  12 tan , i.e., no one can climb to the top without
causing the ladder to slip, whereas if s  tan , the limit is , so anyone can climb to the top.
ASSESS When the coefficient of friction becomes too small, s cot   1/2, or s  tan /2 (see Example 12.2),
slipping will occur and it’s no longer possible for the ladder to remain in static equilibrium. In this situation,
nobody can climb up to the top of the ladder without making the ladder slip, regardless of his or her mass.
50.
The rod is in static equilibrium under the three vertical forces shown in the sketch, so  Fy  0
implies Fsr  Fsr  Mg , and (  )cm  0 implies Fs xcm  Fsr ( L  xcm ). The solution for the left and right scale
forces is Fs  Mg  Fsr  Mg (1  xcm /L). Equation 9.4 gives
L
L
L
L
1 3 
1 2
 1 2
xcm    xdx    dx   (ax  bx 2 ) dx   (a  bx) dx   a L  b L    aL  b L   L(3a  2bL)  (6a  3bL).
0
0
0
0
3  
2 
 2
For the values given, xcm  L  127 and note that M  aL  12 bL2  4 kg. Thus, Fsr  Mgxcm  L  22.9 N
and FsL  16.3 N.
51.
INTERPRET In this problem a wheel has been placed on a slope. We want to apply a horizontal force at its highest
point to keep it from rolling down.
12.16
Chapter 12
DEVELOP Consider the conditions for static equilibrium of the wheel, under the action of the forces shown. Here
Fapp is the applied horizontal force, Fc is the contact force of the incline, normal plus friction, and we assumed that
the CM is at the center.
For the wheel to remain in static equilibrium, the forces must satisfy Equation 12.1. Our plan is to compute the
torques about P using Equation 12.2. Note that the contact force, Fc does not contribute.
EVALUATE The torques about the point of contact sum to zero, or
0  (  ) P  Fapp R(1  cos  )  MgR sin 
Therefore, the applied force is Fapp  Mg 1sincos  Mg tan  2  .
ASSESS The applied force vanishes when   0 (flat surface), and become maximum when   90. In this
limit, Fapp  Mg , and points vertically upward.
52.
We suppose that the block is oriented with two sides parallel to the direction of the incline, and that its CM is at the
center. The condition for sliding is that mg sin   f smax  s n  s mg cos  , or tan   s . For s  0.63, this
condition is   tan 1 0.63  32.2. The condition for tipping over is that the CM lie to the left of the lower corner
of the block (see sketch). Thus    , where   tan 1 ( w/h) is the diagonal angle of the block.
For h  2w,   tan 1 0.5  26.6. Therefore, this block tips over before sliding.
53.
INTERPRET In this problem a rectangular block is placed on an incline. Given the coefficient of friction between
the block and the incline, we’d like to find out under what condition the block would slide first.
DEVELOP We suppose that the block is oriented with two sides parallel to the direction of the incline, and that its
CM is at the center. The condition for sliding is that
mg sin   f smax  s n  s mg cos 
or tan   s . The condition for tipping over is that the CM lie to the left of the lower corner of the block (see
sketch). Thus    , where
 w
 
  tan 1  
h
is the diagonal angle of the block.
EVALUATE For the rectangular block with w  h /2, it tips over when     tan 1 ( w/h)  tan 1 (1/2) but will
slide when   tan 1 s . Thus, if s  tan   1/2, the block in the Problem 52 will slide before tipping.
ASSESS We find that sliding happens first if s  w/h. This makes sense because when the coefficient of friction
is small, the block has a greater tendency to slide. On the other hand, when the coefficient of friction is large
( s  w/h), we’d expect tipping to take place first.
Static Equilibrium
54.
12.17
The analysis for Problem 52 applies to the cone, where  is the angle between the symmetry axis and a line from
the CM to the edge of the base. The integration to find the CM is fastest when the cone is oriented like the aircraft
wing in Example 9.3, for then, the equation of the sloping side is simple, as shown in the sketch. For mass
elements, take thin disks parallel to the base, so dm   y 2 dx  (3M / h3 ) x 2 dx, where   M / 13  R h is the
2
h
density (assumed constant) and M is the mass of the cone. Then xcm  M 1  x dm  (3/h3 )  0 x3dx  34 h, or the CM
is 14 h above the base. Since tan   ( 16h) /( 14h)  32  0.63  s, this cone will slide before tipping over.
55.
INTERPRET In this problem we want to verify the statement that the choice of pivot point does not matter when
applying the conditions for static equilibrium.
DEVELOP With reference to Fig. 12.29, we follow the hint given in the problem statement and write
 P   rPi  Fi  (rOi  R)  Fi   rOi  Fi  R   Fi  O  R  Fnet
EVALUATE When the system is in static equilibrium, the total force and torque acting on the system vanish:
Fnet  0 and  P  0. Therefore, we have  P  O  0, i.e., the total torque about any two points is the same.
ASSESS If the angle  is increased, then the corresponding coefficient of friction must also be increased in order
to keep the pole from slipping.
56.
In equilibrium, the farthest right the center of mass of the combination of three books can lie is directly above the edge
of the table. (This is unstable equilibrium, since the slightest disturbance to the right would cause the books to fall.)
The center of mass of each book is at its center, so if we take the origin at the edge with positive to the right, this
condition becomes
0  xcm 
1 
1 
1
1 


mx1  m  x1  L   m  x1  L  L  
3m 
4 
4
2 


where x1 is the horizontal position of the center of the bottom book, and the centers of the other books are
displaced as given. Therefore, 3x1  L  0, or x1   13 L. If the center of the bottom book is 13 L to the left of the
edge, then only 12 L  13 L  16 L can overhang on the right. (An argument based on torques is equivalent, since at
the farthest right position, the normal contact force on the books acts essentially just at the table’s edge.)
57.
INTERPRET This problem is about static equilibrium. The forces acting on the pole are the tension in the rope,
gravity acting at the CM at its center, and the contact force of the incline (perpendicular component n and parallel
component f ). We want to find the minimum coefficient of friction that will keep the pole from slipping.
DEVELOP Consideration of Equation 12.2 about the CM shows that a frictional force f must be acting up the plane if
the rod is to remain in static equilibrium. Since the weight of the rod, mg, and the normal force, n, contribute no torques
about the CM, there must be a force to oppose the torque of the tension, T. The equations for static equilibrium
(parallel and perpendicular components of Equation 12.1, and CCW-positive component of Equation 12.2) are:
0   F||  f  T cos   mg sin 
0   F  n  T sin   mg cos 
0  (  )cm  T ( L/2) cos   f ( L/2)
The solutions for the forces are f  mg sin , T  12 mg tan , and
1
2
n

1
sin 2  
mg  2 cos  

2
cos  

12.18
Chapter 12
subject to the condition that f  n. Therefore,

sin 2  
tan 
sin     2cos  
  
cos

2

tan 2 


By use of the identities sin 2  2sin  cos , cos 2  cos2   sin 2 , and sin 2   1  cos 2 , this may
EVALUATE
be rewritten as

sin 2
3  cos 2
ASSESS If the angle  is increased, then the corresponding coefficient of friction must also be increased in order to
keep the pole from slipping.
58.
The minimum coefficient of friction found in the Problem 57, min ( )  tan /(2  tan 2  ), is a positive function
which is zero at   0 and 90 (the limits of its domain). Therefore it has a maximum when d min /d  0, or
2  tan 2   2 tan 2   0, or   tan 1 2  54.7.
59.
INTERPRET We use equilibrium methods to find the horizontal component of force on a bookshelf bracket tab.
The bookshelf is in equilibrium, so the sum of forces and the sum of torques are both zero. Since the sum of forces
is zero, we may use any point as the pivot for calculating the torques. We would expect that the horizontal force is
much larger than the weight of the books, since the books have more leverage than the bracket.
DEVELOP We start by drawing a diagram showing the forces and their approximate locations, as shown in the
figure. The mass of books is m  32 kg, the distance y  4.5 cm, and the distance x  12 cm.
Since we know nothing about the force Fb acting on the bottom corner of the bracket, we will use that point as our
pivot point. The sum of torques around this point must be zero; use this to find Fh .
EVALUATE
   0  Fg x  Fh y  Fh  Fg
x
y
 mg xy  836 N.
ASSESS The force is much larger than the weight of the books, as we expected.
60.
INTERPRET We use equilibrium methods to find the horizontal component of force on a bracket mounting screw.
The bracket is in equilibrium, so the sum of forces and the sum of torques is both zero. Since the sum of forces is
zero, we may use any point as the pivot for calculating the torques. We would expect that the horizontal force is
much larger than the weight of the plant, since the plant has more leverage than the bracket.
DEVELOP We start by drawing a diagram showing the forces and their approximate locations, as shown in the
figure. The mass of the plant is m p  4.2 kg, the mass of the bracket is mb  0.85 kg, the distance y  7.2 cm, the
distance x1  9 cm, and the distance x2  28 cm.
Since we know nothing about the force Fb acting on the bottom corner of the bracket, we will use that point as our
pivot point. The sum of torques around this point must be zero; use this to find Fsh .
EVALUATE
   0  Fg1 x1  Fg 2 x2  Fsh y  Fsh 
( Fg 1 x1  Fg 2 x2 )
y

( m1 x1  m2 x2 ) g
y
 170 N.
Static Equilibrium
12.19
ASSESS The force is much larger than the weight of the plant, as we expected.
61.
INTERPRET We have in this problem a disk with an additional off-center mass, and we need to find the angle at
which it will balance on a slope. We do this by finding the torques on the disk, and setting the angle so that the sum
of torques is zero. We will use the point of contact between the disk and the ramp as the pivot point for calculating
torques.
DEVELOP As we can see from the figure, the magnitude of the clockwise torque around the contact point
is 1  FgM R sin . The magnitude of the counterclockwise torque is  2  Fgm ( R cos   R sin  ). In order to
balance, the two torques must have the same magnitude.
EVALUATE
1   2  FgM R sin   Fgm R(cos   sin  )  M gR sin   m gR (cos   sin  )
(M  m) sin   m cos 
M m

   cos 1 
sin  
 m

ASSESS We can check this equation for some special cases. If m is zero, then the disk will balance anywhere as
long as  is zero also. If  is zero, then the disk will balance at   90.
62.
INTERPRET We return to the problem of the disk with the off-center mass, this time finding the minimum offcenter mass required to balance the disk. Again, we find the torques on the disk, and require that the sum of torques
be zero.
DEVELOP We will use the point of contact between disk and ramp as the pivot for calculating torques. As we see
in the figure. the torque around the contact point due to the smaller mass is a maximum when   0. When this is
the case, the counterclockwise torque is  2  Fgm ( R  R sin  ). The clockwise torque is 1  FgM R sin . The
magnitudes of these two torques must be the same.
sin 
EVALUATE 1   2  M gR sin   m gR (1  sin  )  m  1Msin

ASSESS The torque due to m cannot be larger than this amount, but it could be smaller, if necessary, at other
sin 
angles . So m  1Msin
.
12.20
63.
Chapter 12
INTERPRET This problem involves two masses, separated by a massless rod, hovering far above the Earth. We need
to find the net gravitational force on these masses, the net torque around the center of mass, and the center of gravity.
We will use the equation for gravitational force, as well as the equation for torque. The center of gravity is not the
same as the center of mass, in this case, since there is a change in gravitational field between the two masses.
m m
DEVELOP We start with a diagram, as shown in the figure. We use Fg  G rE2 to find the force on each end of
the spacecraft, then find the vector sum of the forces to obtain the net gravitational force. Use the forces on each
end, and the angle found from the figure, to find the torque on each end and thus the net torque around the center
of mass. The center of gravity is the point on the spacecraft at which the torques will be zero. We have labeled the
distance from the left end of the spacecraft to the center of gravity x.
EVALUATE
m m
(a) The gravitational force on the left mass is F1  G (2 RE )2 ˆj. The gravitational force on the right mass
E
   tan    26.7. We note in passing that sin  
and cos   . We break F into x and y components: F  G
(sin iˆ  cos  ˆj )  G
 iˆ  ˆj .
   ˆj    iˆ . The magnitude of this
Now we add the vector components to find the total force: F  G


is F2  G (2 R
mE m
E)
2
5
2
 ( RE
, directed at an angle   tan 1
)2
1
RE
2 RE
2
2
mE m
mE m
(2 RE )2  ( RE )2
5 RE 2
mE m
RE2
force is | F |  G
1
5
1
2
mE m
[1.229], and the direction of the force is tan 1
2
RE
1
4
2
5
1
5
2
5
1
5
   21.3 left of the negative y axis.
1
5
1 2
4
5
(b) The net torque around the center of mass is
 RE 
m m 2  RE 

G E 2


 2 
5  2 
5RE


m m 1
mE m
1 
  G E  
(0.0356)
G
RE  8 5 5 
RE
   F1
RE
R
m m
 F2 ˆj E  G E 2
2
2
4 RE
(c) To find the center of gravity, repeat the calculation for (b) but use x for the left-hand distance and ( RE  x) for
the right-hand distance, setting   0. Solving this for x will give the distance of the center of gravity from the left
R
end, which we can compare to 2E .
0   F1 x  F2 ˆj ( RE  x)  G
mE m
m m 2
( x)  G E 2
( RE  x)
2
4 RE
5 RE 5
1
2
2 
2
1
0  x
( RE  x)  x  
  RE
4
5 5
5 5
4 5 5
 x  0.417 RE

RE
 x  0.083
2
ASSESS The torque is negative, which in this case means counterclockwise as we would expect. The center of
gravity in this case is not at the center of mass, due to the decrease in gravitational force with altitude.
64.
INTERPRET Use equilibrium techniques to establish whether a child seat failed due to manufacturing flaws or due
to an overweight child. We want to find the forces at points (a) and (b) in Figure 12.36, assuming the child seat is
in equilibrium.
Static Equilibrium
12.21
DEVELOP Since we know that the mass of the child is m  10 kg, we can use the torque around point (b) to find
the force at point (a), using    0. Knowing the force at (a), we can then find the total force at (b) using
 F  0. The design of the seat allows for a maximum force at (a) of Fa  96.2 N and a maximum force at (b)
of Fb  229 N.
Note that the directions of the forces shown in the figure are the directions for the force of the chair on the table.
We need the forces on the chair, so change the direction of both forces shown.
EVALUATE   0  mg (0.16 m)  Fa (0.22 m)  Fa  71.3 N.  F  0   Fa  Fb  mg  Fb  169.3 N.
ASSESS These values are within what the chair should be able to withstand, so we should advise the judge that
the chair was defective.
65.
INTERPRET We analyze the forces on a roof rafter, as shown in Figure 12.37. We use equilibrium techniques,
with the sum of torques equaling zero, to determine whether the tie beam will withstand the force.
DEVELOP We start with a diagram showing the relevant forces, as shown in the figure. We will analyze just half
of the roof, since it is symmetric. We use the point where the tie beam attaches as the pivot point.
Fwall  Fsnow  m2 g , where m is the mass of snow and building materials, m  170 kg. We use half this mass because
the other half is on the other half of the roof. The angle of the roof is   tan 1
  , where the width of the roof
y1  y2
1w
2
is w  9.6 m, so   39.8. This allows us to calculate values of the torque arms for the snow and wall forces:
tan  
y1
y
 x1  1  0.96 m, and similarly x2  3.84 m.
x1
tan 
We see from the figure also that the force on the tie beam is equal to the horizontal force at the peak due to the
other half of the roof. So we can use    0 to find Froof , and if this has a magnitude greater than F  7500 N
then the roof collapses.
EVALUATE
   0  Fwall x2  Fsnow x1  Froof y1  Froof 
Fwall x2  Fsnow x1
y1
 Froof  4170 N
ASSESS The roof holds. (You were rather hoping your professor’s roof would collapse, weren’t you?) Note that
the force increases with decreasing values of y1 , as you would expect.
66.
INTERPRET We find the tension in a cable supporting a boom, and determine if the strength of the cable is
sufficient for the job. This is an equilibrium problem, so we will use   0.
DEVELOP We can use the center of mass of the boom to find the torque due to the boom’s weight. We will use
the left end of the boom as the pivot point. The vertical component of the cable’s tension, acting on the end of the
boom, should balance the torque due to the boom’s weight. We will define the length of the boom as L , and the
mass as m  350 kg. The maximum tension in the cable, at an angle of   24, is FT  4000 N.
EVALUATE
12.22
Chapter 12
L
1
sin(40)  ( FT cos  )( L sin(40) )  mg  FT cos 
2
2
mg
 FT 
 1900 N
2 cos 
  0  mg
ASSESS The cable is strong enough by a safety factor of more than 2. If the angle  increases beyond 24°, though,
there could be problems.