# Static Equilibrium and Elasticity

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```					                                               CHAPTER                         12
Static Equilibrium and Elasticity

1* · True or false: (a) Σ F = 0 is sufficient for static equilibrium to exist. (b) Σ F = 0 is necessary for static equilibrium
to exist. (c) In static equilibrium, the net torque about any point is zero. (d) An object is in equilibrium only when there
are no forces acting on it.
(a) False (b) True (c) True (d) False
2   · A seesaw consists of a 4-m board pivoted at the center. A 28-kg child sits on one end of the board. Where should
a 40-kg child sit to balance the seesaw?
Apply Στ = 0 about the pivot                            2 × 28 = 40d; d = 1.4 m from pivot

3   · In Figure 12-23, Misako is about to do a push-up. Her center of gravity lies directly above point P on the floor,
which is 0.9 m from her feet and 0.6 m from her hands. If her mass is 54 kg, what is the force exerted by the floor on
her hands?
Apply Στ = 0 about her feet as a pivot                    0.9 × 54 × 9.81 = F × 1.5 N.m; F = 318 N

4   · Juan and Bettina are carrying a 60-kg block on a 4-m board as shown in Figure 12-24. The mass of the board is 10
kg. Since Juan spends most of his time reading cookbooks, whereas Bettina regularly does push-ups, they place the
block 2.5 m from Juan and 1.5 m from Bettina. Find the force in newtons exerted by each to carry the block.
Apply Στ = 0 about the right end of the board             (60 × 1.5 + 10 × 2)g = 4 × FJ; FJ = 270 N
Apply Σ F = 0                                             70g = 270 N + FB; FB = 417 N

5* · Misako wishes to measure the strength of her biceps muscle by exerting a force on a test strap as shown in Figure
12-25. The strap is 28 cm from the pivot point at the elbow, and her biceps muscle is attached at a point 5 cm from the
pivot point. If the scale reads 18 N when she exerts her maximum force, what force is exerted by the biceps muscle?
Apply Στ = 0 about the pivot                               28 × 18 = 5 × F; F = 101 N

6   · A crutch is pressed against the sidewalk with a force Fc along its own direction as in Figure 12-26. This force is
balanced by the normal force Fn and a frictional force fs. (a) Show that when the force of friction is at its maximum
value, the coefficient of friction is related to the angle θ by µs = tan θ. (b) Explain how this result applies to the forces
Chapter 12     Static Equilibrium and Ela sticity

on your foot when you are not using a crutch. (c) Why is it advantageous to take short steps when walking on ice?
(a) f s,max = µsFn; Fn = Fc cos θ. For equilibrium, f s = Fc sin θ. If f s = f s,max = Fc sin θ = µsFc cos θ, µs = tan θ.
(b) Taking long strides requires a large coefficient of static friction because θ is then large.
(c) If µs is small, i.e., there is ice on the surface, θ must be small to avoid slipping.
7   · True or false: The center of gravity is always at the geometric center of a body.
False; the location depends on the mass distribution.
8   · Must there be any material at the center of gravity of an object?
No
9* · If the acceleration of gravity is not constant over an object, is it the center of mass or the center of gravity that is
the pivot point when the object is balanced?
The center of gravity is then the pivot point for balance.
10 · Two spheres of radius R rest on a horizontal table with their centers a distance 4R apart. One sphere has twice the
weight of the other sphere. Where is the center of gravity of this system?
Take the origin at the center of sphere of mass M           Md cm = 2M(4R − d cm); d cm = 8R/3

11 · An automobile has 58% of its weight on the front wheels. The front and back wheels are separated by 2 m.
Where is the center of gravity located with respect to the front wheels?
Proceed as in Problem 10                                    0.58Md cm = 0.42M(2 − d cm); d cm = 0.84 m

12 · Each of the objects shown in Figure 12-27 is suspended from the ceiling by a thread attached to the point marked
+ on the object. Describe the orientation of each suspended object with a diagram.
The figures are shown on the right.
The center of mass for each is in-
dicated by a small +. At static
equilibrium, the center of gravity
is directly below the point of support.

13* ·· A square plate is produced by welding together four smaller square plates, each of side a as shown in Figure 12-
28. Plate 1 weighs 40 N; plate 2, 60 N; plate 3, 30 N; and plate 4, 50 N. Find the center of gravity (xcg, ycg).
1. Use Equ. 12-3 to find xcg                                (1/2)a × 100 + (3/2)a × 80 = 180 × xcg; xcg = 0.944a
2. Similarly, find ycg                                      (1/2)a × 90 + (3/2)a × 90 = 180 × ycg; ycg = a

14 ·· A uniform rectangular plate has a circular section of radius R cut out as shown in Figure 12-29. Find the center of
gravity of the system. Hint: Do not integrate. Use superposition of a rectangular plate minus a circular plate.
Chapter 12    Static Equilibrium and Elasticity

Take the coordinate origin at the lower left corner of the plate. Let a and b be the length and width of the plate.
By symmetry, ycg = b/2. Next, determine the missing mass. Let σ be the mass per unit area. Then the missing
mass, mh, is given by mh = π R2σ. The center of mass of the plate before the hole was cut was at x = a/2. Now
apply Equ. 12-3.
2               2                   2         2      3           2
1/2a(ab σ) − (a − R)π R σ = (ab σ − π R σ)xcg; xcg = (1/2a b − π aR + π R )/(ab − π R ).
15 · When the tree in front of his house was cut down to widen the road, Jay did not want it to go without ceremony, so
he hauled out his electric guitar and amplifier. All that remained was a uniform 10-m log of mass 100 kg resting on two
supports, waiting to be cut up and taken away the next day. One support was 2 m from the left end, and the other was
4 m from the right end. Find the forces exerted on the log by the supports as Jay played his ear-splitting “Requiem for
a Fallen Tree.”
1. Take moments about the left hand support                   3 × 981 N.m = 4FR; FR = 736 N
2. Use Σ F = 0 to find FL                                     FL = (981 − 736) N = 245 N

16 · Bubba uses a crowbar that is 1 m long to lift a heavy crate off the ground. The crowbar rests on a rigid fulcrum 10
cm from one end as shown in Figure 12-30. (a) If Bubba exerts a downward force of 600 N on one end of the crow-
bar, what is the upward force exerted on the crate by the other end? (b) The ratio of the forces at the ends of the
crowbar is called the mechanical advantage of the crowbar. What is the mechanical advantage here?
(a) Apply Στ = 0 at the fulcrum                            0.90 × 600 N.m = 0.10F N.m; F = 5400 N

17* · Figure 12-31 shows a 25-foot sloop. The mast is a uniform pole of 120 kg and is supported on the deck and held
fore and aft by wires as shown. The tension in the forestay (wire leading to the bow) is 1000 N. Determine the tension
in the backstay and the force that the deck exerts on the mast. Is there a tendency for the mast to slide forward or
aft? If so, where should a block be placed to prevent the mast from moving?
−1                 o
1. Find θF, the angle of the forestay with vertical        θF = tan (2.74/4.88) = 29.3
2. Apply Στ = 0 with pivot at bottom of the mast           2.74 × 1000 × cos 29.3o = 4.88TB cos 45o; TB = 692 N
3. Apply Σ F = 0 to mast; FD = force exerted by the        692 × sin 45o − 1000 × sin 29.3o = FDx; FDx = 0
deck on the mast                                       FDy = 692 × cos 45o + 1000 × cos 29.3o + 120g = 2539 N
Chapter 12    Static Equilibrium and Ela sticity

18 · The sloop in Figure 12-32 is rigged slightly differently from the one in Problem 17. The mass of the mast is 150 kg
and the tension in the forestay is again 1000 N. Find the tension in the backstay and the force that the deck exerts on
the mast. Is there a tendency for the mast to slide forward or aft? If so, where should a block be placed on the deck to
prevent the mast from moving?
−1                 o                       o
1. Find the angles θB and θF of the backstay and             θB = tan (4.57/6.1) = 36.84 ; θF = tan(0.5) = 26.57
the forestay
2. Apply Στ = 0 with pivot at bottom of mast                 2.44 × 1000 × cos 26.57o = 4.57TB cos 36.84o;
TB = 596.7 N
2. Apply Σ F = 0 to the mast; FDx, FDy are horizontal        596.7 sin 36.84o − 1000 sin 26.57o + FDx = 0;
and vertical components of the force of the deck        FDx = 89.5 N
on the mast                                             FDy = 596.7 cos 36.84o + 1000 cos 26.57o + 150g
= 2843 N
4. FDx points toward the stern                               Tendency to slide forward; place block in front of mast

19 ·· A 10-m beam of mass 300 kg extends over a ledge as in Figure 12-33. The beam is not attached, but simply rests
on the surface. A 60-kg student intends to position the beam so that he can walk to the end of it. How far from the
edge of the ledge can the beam extend?
Apply Στ = 0 with the pivot at the ledge                   60x = 300(5 − x); x = 4.17 m

20 ·· A gravity board for locating the center of gravity of a person consists of a horizontal board supported by a fulcrum
at one end and by a scale at the other end. A physics student lies horizontally on the board with the top of his head
above the fulcrum point as shown in Figure 12-34. The scale is 2 m from the fulcrum. The student has a mass of 70
kg, and when he is on the gravity board, the scale advances 250 N. Where is the center of gravity of the student?
Apply Στ = 0 about the fulcrum                              70gxcg = 2 × 250 N.m; xcg = 72.8 cm from the fulcrum

21* ·· A 3-m board of mass 5 kg is hinged at one end. A force F is applied vertically at the other end to lift a 60-kg
block, which rests on the board 80 cm from the hinge, as shown in Figure 12-35. (a) Find the magnitude of the force
needed to hold the board stationary at θ = 30o. (b) Find the force exerted by the hinge at this angle. (c) Find the
magnitude of the force F and the force exerted by the hinge if F is exerted perpendicular to the board when
o
θ = 30 .
(a) Apply Στ = 0 about the hinge; cos 30o factors           3F = (0.8 × 60 + 1.5 × 5)9.81 N.m; F = 181.5 N
cancel
(b) Use Σ F = 0                                             FH + 181.5 − 65 × 9.81 = 0; FH = 456 N
(c) Apply Στ = 0                                            3F = (48 + 7.5) × 9.81 × cos 30o; F = 157.2 N
Use Σ Fx = 0, Σ Fy = 0                                  FHy = 65 × 9.81 − 157.2 cos 30o = 501.5 N;
FHx = 157.2 sin 30o = 78.6 N; FH = (78.6 i + 501.5 j) N

22 ·· A cylinder of weight W is supported by a frictionless trough formed by a plane inclined at 30o to the horizontal on
the left and one inclined at 60o on the right as shown in Figure 12-36. Find the force exerted by each plane on the
Chapter 12     Static Equilibrium and Elasticity

cylinder.
The planes are frictionless; therefore, the force exerted by each plane must be perpendicular to that plane. Let F1
be the force exerted by the 30o plane, and let F2 be the force exerted by the 60o plane.
1. Use Σ Fx = 0                        F1 sin 30o = F2 sin 60o; F1 = F2 ×    3

F1 cos 30o + F2 cos 60o = W = 2F2; F2 = 1/2W, F1=( 3 /2)
2. Use Σ Fy = 0                        W

23 ·· An 80-N weight is supported by a cable attached to a strut hinged at point A as in Figure 12-37. The strut is sup-
ported by a second cable under tension T2. The mass of the strut is negligible. (a) What are the three forces acting on
the strut? (b) Show that the vertical component of the tension T2 must equal 80 N. (c) Find the force exerted on the
strut by the hinge.
(a) The forces acting on the strut are the tensions T1 and T2 and FH, the force exerted on the strut by the hinge.
(b) Take moments about the hinge. Στ = 0 then requires that T2v = T1.
(c) Since T2v = T1 = 80 N, T2h = 80 tan 60o = 138.6 N. Using Σ F = 0, we find FH = T2h = 138.6 N, to the right.
24 ·· A horizontal board 8.0 m long is used by pirates to make their victims walk the plank. A pirate of mass 105 kg
stands on the shipboard end of the plank to prevent it from tipping. Find the maximum distance the plank can overhang
for a 63-kg victim to be able to walk to the end if (a) the mass of the plank is negligible, and (b) the mass of the plank
is 25 kg.
(a) Apply Στ = 0; let x be the overhang                      63x = 105(8.0 − x); x = 5.0 m
(b) Apply Στ = 0                                             63x + 25(x − 4) = 105(8 -x); x = 4.87 m

25* ·· As a farewell prank on their alma mater, Sharika and Chico decide to liberate thousands of marbles in the hallway
during final exams. They place a 2-m × 1-m × 1-m box on a hinged board, as in Figure 12-38, and fill it with marbles.
When the building is perfectly silent, they slowly lift one end of the plank, increasing θ, the angle of the incline. If the
coefficient of static friction is large enough to prevent the box from slipping, at what angle will the box tip? (Assume
that the marbles stay in the box until it tips over.)
The box will tip when its center of mass is no longer above the base of the box. So θ = tan−1(0.5) = 26.6o.
26 ·· A uniform 18-kg door that is 2.0 m high by 0.8 m wide is hung from two hinges that are 20 cm from the top and 20
cm from the bottom. If each hinge supports half the weight of the door, find the magnitude and direction of the
horizontal components of the forces exerted by the two hinges on the door.
The drawing shows the door and its two supports. The center of gravity of the door is 0.8 m above (and below) the
hinge, and 0.4 m from the hinges horizontally. Denote the horizontal and vertical components of the hinge force by FHh
and Fhv.
Chapter 12    Static Equilibrium and Ela sticity

Take moments about the lower hinge, and apply         18 × 9.81 × 0.4 N.m = FHh × 1.6
Στ = 0                                                m;
The upper hinge pulls on the door; the lower pushes   FHh = 44.15 N

27 ·· Find the force exerted by the corner on the wheel in Example 12-4, just as the wheel lifts off the surface.
Applying Σ F = 0 and using the results of Example 12-4, one obtains F = Mg(2Rh − h 2)1/2/(h − R) i + Mg j.
28 ·· Lou is promoting the grand opening of Roswell’s, a new nightclub with an alien theme. One end of a uniform 100-
kg beam, 10 m long, is hinged to a wall, and the other end sticks out horizontally over the dance floor. A cable
connects to the beam 6 m from the wall, as in Figure 12-39. Lou sits at the controls of a mock UFO, which hangs from
the free end of the beam. From there, he sends down abduction beams, hypnotic light effects, and spaceship noises to
the patrons below. If the combined weight of Lou and his UFO is 400 kg, (a) what is the tension in the cable? (b)
What is the horizontal force on the hinge? What is the vertical force of the beam on the hinge?
(a) Στ = 0 about the hinge                                  (400 × 10 + 100 × 5)g = 0.8T × 6; T = 9200 N

(b) Horizontal force on the hinge = 0.6T                   Fh = 5520 N, acting to the left
(c) Vertical force on the hinge = 0.8T − 500g              Fv = 2455 N, acting upward

29* ·· The diving board shown in Figure 12-40 has a mass of 30 kg. Find the force on the supports when a 70-kg diver
stands at the end of the diving board. Give the direction of each support force as a tension or a compression.
1. Use Στ = 0 about the end support as a pivot to            (4.2 × 70 + 2.1 × 30)g = 1.2F; F = 2920 N, compression
find the force on the middle support
2. Στ = 0 about the right support to find F of the end       1.2Fend = (0.9 × 30 + 3 × 70)g; Fend = 1940 N, tension
support

30 ·· Find the force exerted on the strut by the hinge at A for the arrangement in Figure 12-41 if (a) the strut is
weightless, and (b) the strut weighs 20 N.
Let T be the tension in the line attached to the wall and L be the length of the strut.
(a) Take moments about the hinge                             60L sin 45o = TL; T = 42.43 N
Apply Σ F = 0 to the strut; here FH is the force     FHx − T sin 45o = 0; FHy + T cos 45o − 60 = 0;
the hinge exerts on the strut                            FHx = 30 N, FHy = 30 N; FH = (30 i + 30 j) N
(b) Proceed as in part (a)                                   (60L + 20L/2)sin 45o = TL; T = 49.5 N; FHx = T sin 45o
= 35 N; FHy + T cos 45o − 80 = 0; FHy = 45 N;
FH = (35 i + 45 j) N
Chapter 12     Static Equilibrium and Elasticity

31 ·· Julie has been hired to help paint the trim of a building, but she is not convinced of the safety of the apparatus. A
5.0-m plank is suspended horizontally from the top of the building by ropes attached at each end. She knows from
previous experience, however, that the ropes being used will break if the tension exceeds 1 kN. Her 80-kg boss
dismisses Julie’s worries and begins painting while standing 1 m from the end of the plank. If Julie’s mass is 60 kg and
the plank has a mass of 20 kg, then over what range of positions can Julie stand if a colorful plummet is to be avoided?
See the diagram. Note that if the 60 kg mass is at the far left end of the plank,
T1 and T2 are less than 1 kN. Let x be the distance of the 60 kg mass from T1.

1. Take moments about the left end of the plank              (60x + 20 × 2.5 + 80 × 4)g = 5T2
2. Set T2 = 1 kN and solve for x                             x = 2.33 m; safe for 0 < x < 2.33 m

32 ·· The cable in Figure 12-39 must remain attached to the wall 8 m above the hinge, but its length can vary so that it
can be connected to the beam at various distances x from the wall. How far from the wall should it be attached so that
the force on the hinge has no vertical component?
It should be attached to the end of the beam. (see Problem 23)
33* ·· A cylinder of mass M and radius R rolls against a step of height h as shown in Figure 12-42. When a horizontal
force F is applied to the top of the cylinder, the cylinder remains at rest. (a) What is the normal force exerted by the
floor on the cylinder? (b) What is the horizontal force exerted by the corner of the step on the cylinder? (c) What is
the vertical component of the force exerted by the corner on the cylinder?
(a) Take moments about the step’s corner; see                 F(2R − h) = (Mg − Fn)(2Rh − h 2)1/2; solve for Fn
Example 12-4 for the moment arm of Mg and Fn              Fn = Mg − F[(2R − h)/h]1/2
(b) Use Σ Fx = 0                                              Fx,corn = −F
(c) Use Σ Fy = 0                                              Fn − Mg + Fy,corn = 0; Fy,corn = F[(2R − h)/h]1/2

34 ·· For the cylinder in Problem 33, find the minimum horizontal force F that will roll the cylinder over the step if the
cylinder does not slide on the corner.
h
To roll over the step, the cylinder must lift off the floor, i.e., Fn = 0. From 32(a), F = Mg          .
2R − h
35 ·· A strong man holds one end of a 3-m rod of mass 5 kg at rest in a horizontal position. (a) What total force does
the man exert on the rod? (b) What total torque does the man exert on the rod? (c) If you approximate the effort of
the man with two forces that act in opposite directions and are separated by the width of the man’s hand, which is
taken to be 10 cm, what are the magnitudes and directions of the two forces?
(a) Use Σ F = 0                                             Fman = mg = 49.1 N
(b) Στ = 0                                                  τman = mgL/2 = 73.6 N.m
(b) Take the moment about the point of application          Fu = (73.6/0.1) N = 736 N, acting up
of the force acting down. Set τman = Fu × d; Fu is      Fd = 687 N, acting down
the force of the couple acting up; Fd is the force
of the couple acting down = Fu − mg
Chapter 12     Static Equilibrium and Ela sticity

36 ·· A large gate weighing 200 N is supported by hinges at the top and bottom and is further supported by a wire as
shown in Figure 12-44. (a) What must the tension in the wire be for the force on the upper hinge to have no horizontal
component? (b) What is the horizontal force on the lower hinge? (c) What are the vertical forces on the hinges?
(a) Set Στ = 0; pivot is at the lower hinge               mg × 1.5 m = [(T cos 45o) + (T sin 45o)] × 1.5 m
Solve for T                                           T = 141 N
(b) Use Σ Fx = 0; let 1 denote the lower hinge            Fx,1 = T cos 45o = 100 N
(c) Use Σ Fy = 0                                          Fy,1 + Fy,2 − mg + T sin 45o = 0; Fy,1 + Fy,2 = 100 N
Fy,1 and Fy,2 cannot be determined independently

37* ··· A uniform log with a mass of 100 kg, a length of 4 m, and a radius of 12 cm is held in an inclined position, as
shown in Figure 12-45. The coefficient of static friction between the log and the horizontal surface is 0.6. The log is on
the verge of slipping to the right. Find the tension in the support wire and the angle the wire makes with the vertical
wall.
We shall use the following nomenclature: T = the tension
in the wire; Fn = the normal force of the surface; f s =
µsFn = the force of static friction. We shall also use Στ =
0 and use the point where the wire is attached to the log
as the pivot. Taking this as the origin, the center of mass
of the log is at the coordinates
(−2 cos 20o + 0.12 sin 20o, −2 sin 20o − 0.12 cos 20o) =
(−1.838, −0.797).
The point of contact with the floor is at (−3.676, −1.594).
1. Use Σ F = 0; see the free body diagram                      T sin θ − µsFn = 0; T cos θ + Fn − mg = 0
2. Apply Στ = 0 about the origin                               1.838mg − 3.676Fn − 1.142µsFn = 0
3. Solve for Fn; m = 100 kg, µs = 0.6                          Fn = 389 N
4. Insert 225 N = Fn into the force equations                  T sin θ = 233 N; T cos θ = 592 N;
−1             o
θ = tan (0.394) = 21.5
5. Evaluate T                                                  T = (233/sin 21.5o) N = 636 N

38 ··· A tall, uniform, rectangular block sits on an inclined plane as shown in Figure 12-46. A cord is attached to the top
of the block to prevent it from falling down the incline. What is the maximum angle θ for which the block will not slide
on the incline? Let b/a be 4 and µs = 0.8.
Chapter 12      Static Equilibrium and Elasticity

Consider what happens just as θ increases beyond θmax.
Since the top of the block is fixed by the cord, the block
will in fact rotate with only the lower right edge of the
block remaining in contact with the plane. It follows that
just prior to this slipping, Fn and f s = µsFn act at the lower
right edge of the block.

1. Apply Σ F = 0; see the free-body diagram                       T + µsFn − mg sin θ = 0;
Fn − mg cos θ = 0
2. Use Στ = 0 about the lower right edge                          1/2a(mg cos θ) + 1/2b(mg sin θ) − bT = 0
3. Use b = 4a, µs = 0.8, and the force equations                  mg(cos θ + 4 sin θ) = 8T; Fn + 4(T + 0.8Fn) = 8T;
T = 1.05Fn
4. Insert T = 1.05Fn into force equations to find θ               1.85Fn = mg sin θ; Fn = mg cos θ; θ = tan−11.85 = 61.6o

39 ··· A thin rail of length 10 m and mass 20 kg is supported at a 30o incline. One support is 2 m and the other is
6 m from the lower end of the rail. Friction prevents the rail from sliding off the supports. Find the force
(magnitude and direction) exerted on the rail by each support.
The rail and the forces acting on it are shown in the free-body diagram.
Supports are indicated by “1” and “2” as shown. Since the x components of
the forces at the supports are friction forces, they are proportional to the
normal, i.e., y, components of the forces at the supports.

Take x up along the rail, y up and normal to the rail.
1. Apply Στ = 0 about support 1                                   3mg cos 30o = 4F2y; F2y = 127.4 N
2. Apply Στ = 0 about support 2                                   mg cos 30o = 4F1y; F1y = 42.5 N; note F2y = 3F1y
3. Use Σ Fx = 0                                                   F1x + F2x = mg sin 30o = 98.1 N
4. Use F1x/F2x = F1y/F2y = 1/3; F2x = 3F1x                        F1x = 24.5 N; F2x = 73.6 N
−1           o
5. Find θ1 and θ2 (see diagram)                                   θ1 = θ2 = tan (Fy/Fx) = 60 ; forces act vertically up
6. Find the magnitudes of F1 and F2                               F1 = mg/4; F2 = 3mg/4; F1 = 49.1 N; F2 = 147.2 N

40 · Two 80-N forces are applied to opposite corners of a rectangular plate as shown in Figure 12-47. Find the torque
produced by this couple.
Take the torque about the top left corner. τ = b(80 cos 30o) − a(80 sin 30o) = 69.3b − 40a.

41* ·   A uniform cube of side a and mass M rests on a horizontal surface. A horizontal force F is applied to the top of
Chapter 12     Static Equilibrium and Ela sticity

the cube as in Figure 12-48. This force is not sufficient to move or tip the cube. (a) Show that the force of static
friction exerted by the surface and the applied force constitute a couple, and find the torque exerted by the couple.
(b) This couple is balanced by the couple consisting of the normal force exerted by the surface and the weight of the
cube. Use this fact to find the effective point of application of the normal force when F = Mg/3.
(c) What is the greatest magnitude of F for which the cube will not tip?
(a) The cube is stationary. Therefore fs = −F, and the torque of that couple is Fa.
(b) Let x = the distance from the point of application of Fn to the center of the cube. Now, Fn = Mg, so
Mgx = Fa; x = Fa/Mg. If F = Mg/3, then x = a/3.
(c) Note that xmax = a/2. The cube will tip if F > Mg/2.
42 ·· Resolve each force in Problem 40 into its horizontal and vertical components, producing two couples. The
algebraic sum of the two component couples equals the resultant couple. Use this result to find the perpendicular
distance between the lines of action of the two forces.
The vertical components are F cos 30o = F        3 /2; the horizontal components are F sin 30o = F/2. The net torque is,

as in Problem 40, 1/2F( 3 b − a). This is also equal to FD, where D is the moment arm of the couple, i.e., the
distance between the two forces. So D = 1/2( 3 b − a).
43 · Is it possible to climb a ladder placed against a wall where the ground is frictionless but the wall is not? Explain.
No. Since the floor can exert no horizontal force, neither can the wall. Consequently, the friction force between the
wall and the ladder is zero regardless of the coefficient of friction between the wall and the ladder.
44 ·· Romeo takes a uniform 10-m ladder and leans it against the smooth wall of the Capulet residence. The ladder’s
mass is 22.0 kg, and the bottom rests on the ground 2.8 m from the wall. When Romeo, whose mass is 70 kg, gets
90% of the way to the top, the ladder begins to slip. What is the coefficient of static friction between the ground and
The ladder and the forces acting on it at the critical
moment of slipping are shown in the diagram. The force
of static friction is then f s = µsFn, as indicated. The angle
−1               o
θ is given by θ = cos 0.28 = 73.74 .

1. Take moments about the bottom of the ladder                22g × 1.4 + 70g × 0.9 × 2.8
= Fw × 10 sin 73.74o
2. Find Fw; Fw + f s = 0; find f s                            Fn = 211.7 N = f s
3. Σ Fy = 0                                                   Fn = 92g = 902.5 N
4. Find µs                                                    µs = f s/Fn = 211.7/902.5 = 0.235

45* ·· A massless ladder of length L leans against a smooth wall making an angle θ with the horizontal floor. The coef-
ficient of friction between the ladder and the floor is µs. A man of mass M climbs the ladder. What height h can he
Chapter 12    Static Equilibrium and Elasticity

The ladder and the forces acting on it are shown in the
diagram. Since the wall is smooth, the force the wall
exerts on the ladder must be horizontal.

1. Use Σ F = 0                                              Fn = Mg; f s = Fw
2. Apply Στ = 0 about the bottom of the ladder              FwL sin θ = Mgx cos θ
3. Solve for x; use f s = µsFn = µsMg = Fw                  x = µsL tan θ; h = x sin θ = µsL tan θ sin θ

46 ·· A uniform ladder of length L and mass m leans against a frictionless vertical wall with its lower end on the ground.
It makes an angle of 60o with the horizontal ground. The coefficient of static friction between the ladder and ground is
0.45. If your mass is four times that of the ladder, how far up the ladder can you climb before it begins to slip?
The ladder and the forces acting on it are shown in the
drawing. We proceed as in the preceding problems.

1. Use Σ F = 0 and solve for Fw                             µsFn = Fw; Fn = 5mg; Fw = 2.25mg
2. Apply Στ = 0 about the bottom of the ladder              FwL sin 60o = (mgL/2 + 4mgx)cos 60o = 2.25mgL sin 60o
3. Solve for x                                              x = 0.85L; you can go 85% of the way to the top

47 ·· A ladder of mass m and length L leans against a frictionless, vertical wall making an angle θ with the horizontal.
The center of mass is at a height h from the floor. A force F pulls horizontally against the ladder at the midpoint. Find
the minimum coefficient of static friction µs for which the top end of the ladder will separate from the wall while the
lower end does not slip.
Find F required to pull the ladder away from the wall. Apply Στ = 0 about the bottom of the ladder. The moment
due to mg is mgh/tan θ, that due to F is 1/2LF sin θ. This yields F = 2mgh/(L sin θ tan θ). Now Fn = mg and
f s = µsmg = F. One obtains µs = 2h/(L sin θ tan θ).
48 ·· A 900-N boy sits on top of a ladder of negligible weight that rests on a frictionless floor as in Figure 12-49. There
is a cross brace halfway up the ladder. The angle at the apex is θ = 30o. (a) What is the force exerted by the floor on
each leg of the ladder? (b) Find the tension in the cross brace. (c) If the cross brace is moved down toward the
bottom of the ladder (maintaining the same angle θ), will its tension be greater or less?
(a) The force exerted by the frictionless floor must be vertical. By symmetry, each leg carries half the total weight. So
the force on each leg is 450 N.
(b) Consider one of the ladder’s legs and take moments about the apex. Then FnD/2 = Tx, where D is the separation
between the legs at the bottom and x is the distance of the cross brace from the apex. Clearly, if x is increased, i.e.,
the brace moved lower, T will decrease.
Chapter 12     Static Equilibrium and Ela sticity

49* ·· A ladder rests against a frictionless vertical wall. The coefficient of static friction between ladder and the floor is
0.3. What is the smallest angle at which the ladder will remain stationary?
Using the notation of Problem 45, we have Fn = mg and f s = µsmg = Fw. Now take moments about the bottom of
the ladder. This gives 1/2Lmg cos θ = Lf w sin θ = Lmgµs sin θ. Solve for θ; θ = tan−1(1/2µs) = 59o.

50 ··· Having failed in his first attempt, Romeo acquires a new ladder to try once again to get to Juliet’s window. This
one has a length L and a weight of 200 N. He tries placing it on the other side of the window, where the coefficients
of static friction are 0.4 between the ladder and the wall, and 0.7 between the ladder and the ground. Because of
bruises suffered in his last fall, Romeo wears heavy padding, which gives him a total mass of 80 kg. Sure enough,
when he is 4/5 of the way up the ladder, it begins to slip. What was the angle between the ladder and the ground when
Romeo was making his ascent?
The ladder and the forces acting on it are shown in the
drawing.

1. Use Σ F = 0                                                (M + m)g = Fn + f sw
= Fn + µwFw;
Fw = f sg = µgFn
2. Solve for and evaluate Fn                                  Fn = [(M + m)g]/(1 + µwµg); Fn = 769 N

3. Apply Στ = 0 to the top of the ladder                      (MgL/5 + mgL/2)cos θ
= FnL cos θ − FnµgL sin θ
4. Use numerical values to get an equation for θ              512 cos θ = 538.3 sin θ; θ = tan−1(0.951) = 43.6o

51 ··· A ladder leans against a large smooth sphere of radius R that is fixed in place on a horizontal surface. The ladder
makes an angle of 60o with the horizontal surface and has a length 5R/2. (a) What is the force that the sphere exerts
on the ladder? (b) What is the frictional force that prevents the ladder from slipping? (c) What is the normal force that
the horizontal surface exerts on the ladder?
The drawing shows the ladder and sphere. The distance
between the point where the ladder touches the ground
and touches the sphere is D = R/tan 30o = 3 R. The
sphere is smooth; hence the force Fs must be radial.

(a) Apply Στ = 0 to bottom of ladder
Chapter 12    Static Equilibrium and Elasticity

(b) Use Σ Fx = 0 to find f s                               (5R/4)(mg)cos 60o =      3 Rf s;
(c) Use Σ Fy = 0 to find Fn                                Fs = 0.361mg
f s = Fs cos 30o = 0.313mg
Fn = mg − Fs sin 30o = 0.820mg

52 · An aluminum wire and a steel wire of the same length L and diameter d are joined to form a wire of length 2L.
The wire is fastened to the roof and a weight W is attached to the other end. Neglecting the mass of the wires, which
of the following statements is true? (a) The aluminum portion will stretch by the same amount as the steel portion. (b)
The tension in the aluminum portion and the steel portion are the same. (c) The tension in the aluminum portion is
greater than that in the steel portion. (d) None of the above statements is true.
(b)

53* · A 50-kg ball is suspended from a steel wire of length 5 m and radius 2 mm. By how much does the wire stretch?
1. Find the stress                                        F =mg = 490.5 N; A = π r2 = 1.26 × 10−5 m2;
F/A = 3.9 × 107 N/m2
7       11
2. From Equ. 12-7 and Table 12-1 find ∆L                  ∆L = (5 × 3.9 × 10 /2 × 10 ) m = 0.976 mm

54 · Copper has a breaking stress of about 3 × 108 N/m2. (a) What is the maximum load that can be hung from a
copper wire of diameter 0.42 mm? (b) If half this maximum load is hung from the copper wire, by what percentage of
its length will it stretch?
(a) Use Equ. 12-6                                       Fmax = [3 × 108 × π × (2.1 × 0−4)2] N = 41.6 N;
M max = 4.24 kg

8         11          −3
(b) Use Equ. 12-7 to find ∆L/L                             ∆L/L = 1.5 × 10 /1.1 × 10 = 1.36 × 10 = 0.136%

55 · A 4-kg mass is supported by a steel wire of diameter 0.6 mm and length 1.2 m. How much will this wire stretch
1. Find the stress                                       F/A = [4 × 9.81/π × (3 × 10−4)2] N/m2 = 1.39 × 108 N/m2
8       11
2. Use Equ. 12-7 and Table 12-1 to find ∆L               ∆L = (1.2 × 1.39 × 10 /2 × 10 ) m = 0.833 mm

56 · As a runner’s foot touches the ground, the shearing force acting on an 8-mm-thick sole is as shown in Figure 12-
50. If the force of 25 N is distributed over an area of 15 cm2, find the angle of shear θ shown, given that the shear
modulus of the sole is 1.9 × 105 N/m2.
Use Equ. 12-10 to find tan θ                                 tan θ = (25/15 × 10−4)/1.9 × 105 = 0.0877; θ = 5.01o

57* ·· A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a
composite wire of length 3.0 m. What is the length of the composite wire if it is used to support a mass of 5 kg?
1. Find the stress in each wire                            F/A = (5 × 9.81 × 4/π × 10−6) N/m2 = 6.245 × 107 N/m2
7       11
2. Find ∆L = ∆LS + ∆LA                                     ∆LS = 1.5 × 6.245 × 10 /2 × 10 m = 0.468 mm
Chapter 12     Static Equilibrium and Ela sticity

∆LA = 0.468(200/70) mm = 1.338 mm; ∆L = 1.81 mm
L = 3.0m + ∆L = 3.0018 m

58 ·· A force F is applied to a long wire of length L and cross-sectional area A. Show that if the wire is considered to be
a spring, the force constant k is given by k = AY/L and the energy stored in the wire is U = 1/2F∆L, where Y is
Young’s modulus and ∆L is the amount the wire has stretched.
For a spring, ∆L = F/k or k = F/∆L. From Equ. 12-7, F/∆L = AY/L, which is k. The energy stored in a spring is
U = 1/2kx2, or in this case, U = 1/2AY(∆L)2/L = 1/2F∆L.
59 ·· The steel E string of a violin is under a tension of 53 N. The diameter of the string is 0.20 mm, and its length under
tension is 35.0 cm. Find (a) the unstretched length of this string, and (b) the work needed to stretch the string. (see
Problem 58)
(a) L′ = L + ∆L; ∆L = LF/AY; solve for L                      L = L′/(1 + F/AY)
= [0.35/(1 + 53/π × 10−8 × 2 × 1011)] m
L = 0.347 m [∆L = 2.93 mm]
(b) W = 1/2F∆L                                                W = (53 × 2.93 × 10−3/2) J = 0.0776 J

60 ·· When a rubber strip with a cross section of 3 mm × 1.5 mm is suspended vertically and various masses are
attached to it, a student obtains the following data for length versus load:

Load, g          0       100         200          300         400           500
Length, cm       5.0     5.6         6.2          6.9         7.8           10.0

(a) Find Young’s modulus for the rubber strip for small loads. (b) Find the energy stored in the strip when the load is
150 g. (see Problem 58)

(a) Find ∆L/F for small loads                               For loads < 200 g, ∆L/F = 1.2 × 10−2/0.2 × 9.81 m/N ∆L/F
= 6.12 × 10−3 m/N
Y = FL/∆LA                                              Y = (5 × 10−2/4.5 × 10−6 × 6.12 × 10−3) N/m2
= 1.82 × 106 N/m2
(b) U = 1/2F∆L                                              U = 0.15 × 9.81 × 9 × 10−3/2 = 6.62 mJ

61* ·· A building is to be demolished by a 400-kg steel ball swinging on the end of a 30-m steel wire of diameter 5 cm
hanging from a tall crane. As the ball is swung through an arc from side to side, the wire makes an angle of 50o with
the vertical at the top of the swing. Find the amount by which the wire is stretched at the bottom of the swing.
1. Find the tension at bottom of the swing                  F = mg + mv 2/R; v2/R = 2g(1 − cos θ) = 0.714g
F = 1.714mg = 6727 N
−4     2                                              −3        11
2. ∆L = FL/AY; A = π × 25 × 10 /4 m                         ∆L = (6727 × 30/1.96 × 10 × 2 × 10 ) m = 0.515 mm
= 1.96 × 10−3 m2

62 ··   A large mirror is hung from a nail as shown in Figure 12-51. The supporting steel wire has a diameter of
Chapter 12    Static Equilibrium and Elasticity

0.2 mm and an unstretched length of 1.7 m. The distance between the points of support at the top of the mirror’s
frame is 1.5 m. The mass of the mirror is 2.4 kg. What is the distance between the nail and the top of the frame
when the mirror is hung?
1. Find the stress in the wire. The right triangle has     T(0.4/0.85) = mg/2; T = 25 N; A = π × 10−8 m2
sides 0.75 m, 0.4 m, and hypothenuse 0.85 m             T/A = 7.96 × 108 N/m2
8       11            −3
2. Find ∆L/L of the hypothenuse                            ∆L/L = 7.96 × 10 /2 × 10 = 3.98 × 10
3. Using a 2 + h 2 = L2 and differentiating                h ∆h = L∆L; ∆h = 0.852 × 3.98 × 10−3/0.4 = 7.2 mm
4. h = 0.4 m + ∆h                                          h = 40.72 cm

63 ·· Two masses, M 1 and M 2, are supported by wires of equal length when unstretched. The wire supporting M 1 is an
aluminum wire 0.7 mm in diameter, and the one supporting M 2 is a steel wire 0.5 mm in diameter. What is the ratio
M 1/M 2 if the two wires stretch by the same amount?
Note that ∆L1 = M 1gL1/A1Y1 and ∆L2 = M 2gL2/A2Y2. Since L1= L2 and ∆L1 = ∆L2, (M 1/d 12Y1) = (M 2/d 22Y2). With
the data given, M 1/M 2 = (49 × 70)/(25 × 200) = 0.686.
64 ·· A mass of 0.5 kg is attached to an aluminum wire having a diameter of 1.6 mm and an unstretched length of 0.7
m. The other end of the wire is fixed to a post. The mass rotates about the post in a horizontal plane at a rotational
speed such that the angle between the wire and the horizontal is 5.0o. Find the tension in the wire and its length.
1. Find the tension; see Example 5-10                       T = mg/sin 5o = 56.3 N
−8         10
2. Find ∆L and L = L0 +∆L                                   ∆L = (56.3 × 0.7/π × 64 × 10 × 7 × 10 ) m
= 0.28 mm; L = 70.03 cm

65* ··· It is apparent from Table 12-2 that the tensile strength of most materials is two to three orders of magnitude
smaller than Young’s modulus. Consequently, these materials, e.g., aluminum, will break before the strain exceeds
even 1%. For nylon, however, the tensile strength and Young’s modulus are approximately equal. If a nylon line of un-
stretched length L0 and cross section A0 is subjected to a tension T, the cross section may be substantially less than A0
before the line breaks. Under these conditions, the tensile stress T/A may be significantly greater than T/A0. Derive an
expression that relates the area A to the tension T, A0, and Young’s modulus Y.
Assume constant volume of the line. Then LA = constant, and taking differentials L∆A + A∆L = 0 or
∆L/L = −∆A/A. But ∆L/L = T/AY, so ∆A = −T/Y. Thus A = A0 + ∆A = A0 − T/Y.
66 · If the net torque about some point is zero, must it be zero about any other point? Explain.
Yes; if it were otherwise, angular momentum conservation would depend on the choice of coordinates.
67 · The horizontal bar in Figure 12-52 will remain horizontal if (a) L1 = L2 and R1 = R2. (b) L1 = L2 and M 1 = M 2. (c)
R1 = R2 and M 1 = M 2. (d) L1M 1 = L2M 2. (e) R1L1 = R2L2.
(c)
68 · Which of the following could not have units of N/m2? (a) Young’s modulus (b) Shear modulus (c) Stress
(d) Strain
(d) Strain is dimensionless.
69* ·· Sit in a chair with your back straight. Now try to stand up without leaning forward. Explain why you cannot do it.
The body’s center of gravity must be above the feet.
70 ·    A 90-N board 12 m long rests on two supports, each 1 m from the end of the board. A 360-N block is placed on
Chapter 12    Static Equilibrium and Ela sticity

the board 3 m from one end as shown in Figure 12-53. Find the force exerted by each support on the board.
Στ = 0 about the right support; obtain FL, the force on 2 × 360 + 5 × 90 = FL × 10; FL = 117 N
the left support.
Use Σ Fy = 0 to find FR.                                FR = (450 − 117) N = 333 N

71 · The height of the center of gravity of a man standing erect is determined by weighing the man as he lies on a
board of negligible weight supported by two scales as shown in Figure 12-54. If the man’s height is 188 cm and the left
scale reads 445 N while the right scale reads 400 N, where is his center of gravity relative to his feet?
Take moments about feet and use definition of CG          d × 845 = 188 × 445; d = 99 cm

72 · Figure 12-55 shows a mobile consisting of four weights hanging on three rods of negligible mass. Find the value of
each of the unknown weights if the mobile is to balance. Hint: Find the weight w1 first.
Apply the balance condition Στ = 0 successively, starting with the lowest part of the mobile.
1. 4w1 = 6; w1 = 1.5 N. 2. 2w2 = 4 × 3.5; w2 = 7 N. 3. 6w3 = 2 × 10.5; w3 = 3.5 N.
73* · A block and tackle is used to support a mass of 120 kg as shown in Figure 12-56. (a) What is the tension in the
rope? (b) What is the mechanical advantage of this device?
(a) With this arrangement, the mass is supported by three ropes. Thus T = 120g/3 = 392 N.
(b) The mechanical advantage is 3.
74 ·· A plate in the shape of an equilateral triangle of mass M is suspended from one corner and a mass m is suspended
from another corner. What should be the ratio m/M so that the base of the triangle makes an angle of 6.0o with the
horizontal?
The figures below show the equilateral triangle without
the mass m, and then the same triangle with the mass m
and rotated through a small angle θ. We take the side
length of the triangle to be 2a. Then the center of mass of
the triangle is at a distance of 2a/ 3 from each vertex.
As the triangle rotates, its CM shifts by (2a/ 3 )θ, for θ
<< 1. Also the vertex to which m is attached moves
toward the plumb line by the amount     3 a θ (see
drawing).

Take moments about the point of suspension                   M(2a/ 3 )θ = ma(1 −       3 θ); θ = 6o = 0.105 rad
Solve for and evaluate m/M
m/M = (2θ/ 3 )/(1 −      3 θ); m/M = 0.148

75 ··   A standard six-sided pencil is placed on a paper pad (Figure 12-57). Find the minimum coefficient of static friction
µs such that it rolls down rather than slides if the pad is inclined.
Chapter 12    Static Equilibrium and Elasticity

If the hexagon is to roll rather than slide, the incline’s
angle must be sufficient that the center of mass falls just
beyond the support base. From the geometry of the
hexagon, one can see that the critical angle is 30o. It
follows (see Chapter 5) that µs ≥ tan 30o = 0.577.

76 ·· Having lost his job at the post office, Barry decides to explore the possibility that he might be a brilliant sculptor.
Not one to start at the bottom, he borrows the money for a marble slab 3 m × 1 m × 1 m. After loading the marble
onto the back of his truck, he drives off with the slab resting on its square end. But on the way home, a confused
squirrel runs into his path, and Barry slams on the brakes. What deceleration will cause the uniform slab to tip over?
The block and forces on it are shown. At the critical condition for tipping, Fn acts at
the edge of the block, as indicated in the drawing.

Take moments about the block’s center of mass                 0.5Fn = 1.5f s; Fn = mg and f s = ma
Solve for a                                                   a = g/3 = 3.27 m/s2

77* ·· A uniform box of mass 8 kg that is twice as tall as it is wide rests on the floor of a truck. What is the maximum
coefficient of static friction between the box and floor so that the box will slide toward the rear of the truck rather than
tip when the truck accelerates with acceleration a = 0.6g on a level road?
Proceed as in the previous problem. The box will tip if µs > 0.5, so it must have µs < 0.5.
78 ·· Barry’s art exhibit contains many tiny marble sculptures placed around a central piece called “Politics.”
The central piece consists of three identical bars, each of length L and mass m, joined as in Figure 12-58. Two
bars form a fixed ⊥ , and the third bar is suspended on a hinge. Asked to explain the name, Barry said, “It swings
from left to right with hinges flapping, but no matter where you start it from, you end up in the same place.”
When the system is in equilibrium, what is the value of θ?
The figure shows the system. We use the angle α rather than θ for convenience.
The locations of the centers of mass of the three parts of the figure are indicated,
and we can now determine the moment arms and torques about the point of support.
These are also shown in the drawing.
Chapter 12     Static Equilibrium and Ela sticity

Take moments about the pivot point                        (mgL/2)sin α + mgL sin α = (mgL/2)cos α − mgL sin α
This yields the following                            5 sin α = cos α ; α = tan−1 0.2 = 11.3o; θ = 90o − α = 78.7o
79 ·· In the 1996 Olympics, the Russian super-heavyweight weightlifter Andrei Chemerkin broke the world record with
a lift of mass 260 kg. Suppose his grip was slightly asymmetrical as shown in Figure 12-59. Find the maximum mass of
the barbell Chemerkin could have handled with a symmetrical grip, assuming that his arms are equally strong.
1. From the figure it is evident that F > F′; use          0.6F + 0.45 × 130g = 1.15 × 130g
Στ = 0 about the right hand of the weightlifter         F = 152g
2. Find the total mass if each arm supports 152 kg         M tot = 304 kg

80 ·· A balance scale has unequal arms. A 1.5-kg block appears to have a mass of 1.95 kg on the left pan of the scale
(Figure 12-60). Find its apparent mass if the block is placed on the right pan.
1. Find L1/L2                                                1.5L1 = 1.95L2; L1/L2 = 1.3
2. Find M with 1.5 kg at L2                                  1.5L2 = ML1 = 1.3ML2; M = 1.15 kg

81* ·· A cube of mass M leans against a frictionless wall making an angle θ with the floor as shown in Figure
12-61. Find the minimum coefficient of static friction µs between the cube and the floor that allows the cube to
stay at rest.
The figure alongside shows the location of the cube’s
center of mass and the forces acting on the cube. The
moment arm of the couple formed by the normal force,
Fn, and Mg is d = (a/ 2 0)sin(45o − θ). The opposing
couple is formed by the friction force f s and the force
exerted by the wall.

1. Set Στ = 0                                                (Mga/ 2 )sin(45o − θ) = f s a sin θ = µs Mga sin θ
2. Solve for µs; sin(α + β) = sin α cos β + sin β cos α      µs = 1/2(cot θ − 1)

82 ·· Figure 12-62 shows a steel meter stick hinged to a vertical wall and supported by a thin wire. The wire and meter
stick make angles of 45o with the vertical. The mass of the meter stick is 5.0 kg. When a mass M = 10.0 kg is
suspended from the midpoint of the meter stick, the tension T in the supporting wire is 52 N. If the wire will break
should the tension exceed 75 N, what is the maximum distance along the meter stick at which the 10.0-kg mass can be
suspended?
1. Use Στ = 0 about the hinge                              75 × 1 = 5g × 0.5 × cos 45o + 10g × d × cos 45o
2. Solve for d                                             d = 0.83 m

83 ·· Figure 12-63 shows a 20-kg ladder leaning against a frictionless wall and resting on a frictionless horizontal
surface. To keep the ladder from slipping, the bottom of the ladder is tied to the wall with a thin wire; the tension in the
wire is 29.4 N. The wire will break if the tension exceeds 200 N. (a) If an 80-kg person climbs halfway up the ladder,
what force will be exerted by the ladder against the wall? (b) How far up can an 80-kg person climb this ladder?
Chapter 12     Static Equilibrium and Elasticity

(a) Use Στ = 0 about the bottom of the ladder               (20 + 80) × 9.81 × 0.75 = 5 × Fw; Fw = 147 N
(c) Note that the tension in the wire = Fw; let f be        20 × 9.81 × 0.75 + 80 × 9.81 × 1.5f = 5 × 200
the fraction of the total length climbed.               f = 0.724; can climb to height of 3.62 m

84 ··   Suppose that the bar hanging from the end of the ⊥ in Problem 78 is of a different length l ≠ L, although its mass
per unit length is the same as that of the bars of the ⊥ . Find the ratio L/ l such that θ = 75o.
The only change here from Problem 78 is that the mass of the free-hanging bar is not m but M = ( l /L)m.
Set θ = 75o, α = 15o and write the torque equation. m sin 15o + 2m sin 15o = M cos 15o − 2M sin 15o. This gives
M/m = 1.73 or l = 1.73L.
85* ·· A uniform cube can be moved along a horizontal plane either by pushing the cube so that it slips or by turning it
over (“rolling”). What coefficient of kinetic friction µk between the cube and the floor makes both ways equal in terms
of the work needed?
To “roll” the cube one must raise its center of mass from y = a/2 to y =    2 a/2, where a is the cube length. Thus the
work done is W = 1/2mga( 2 − 1) = 0.207mga. Since no work is done as the cube flops down, this is the work done
to move the cube a distance a. Now set 0.207mga = Fa = µkmga = work done against friction in moving a distance a.
Thus µk = 0.207.
86 ·· A tall, uniform, rectangular block sits on an inclined plane as shown in Figure 12-64. If µs = 0.4, does the block
slide or fall over as the angle θ is slowly increased?
1. The condition for sliding (see Chapter 5) is µs ≤ tan θ; so θ ≥ tan−10.4 = 21.8o.
2. The condition for tipping is that the plumb line from the center of mass pass outside of the base. In this case,
that requirement gives tan θ = 1/3 or θ = 18.4o. So as θ is increased, the block will tip before it slides.
87 ·· A 360-kg mass is supported on a wire attached to a 15-m-long steel bar that is pivoted at a vertical wall and
supported by a cable as shown in Figure 12-65. The mass of the bar is 85 kg. (a) With the cable attached to the bar
5.0 m from the lower end as shown, find the tension in the cable and the force exerted by the wall on the steel bar. (b)
Repeat if a somewhat longer cable is attached to the steel bar 5.0 m from its upper end, maintaining the same angle
between the bar and the wall.
(a) The cable is normal to the bar; use Στ = 0 about       85g × 7.5 × cos 30o + 360g × 15 × cos 30o = 5T;
the hinge                                              T = 10260 N
Use Σ F = 0                                            Fy = (445 × 9.81 − 10260 sin 60o) N = −4520 N
Fx = (10260 cos 60o) N = 5130 N
(b) Use Στ = 0; angle between cable and bar = 60o          51300 = (10 sin 60o)T′; T′ = 5924 N
Use Σ F = 0                                            Fy = (4365 − 5924 sin 30o) N = 1403 N
Fx = (5924 cos 30o) N = 5130 N

88 ·· Repeat Problem 77 if the truck accelerates at a = 0.6g up a hill that makes an angle of 9.0o with the horizontal.
Following the same procedure as in Problem 77 gives µs ≤ 0.5 − tan 9o = 0.342.
89* ·· A thin rod 60 cm long is balanced 20 cm from one end when a mass of 2m + 2 g is at the end nearest the pivot and
a mass of m at the opposite end (Figure 12-66a). Balance is again achieved if the mass 2m + 2 g is replaced by the
mass m and no mass is placed at the other end (Figure 12-66b). Determine the mass of the rod.
Chapter 12     Static Equilibrium and Ela sticity

1. Take moments about pivot for initial condition            20(m + 2) = 40m + 10M
2. Take moments about pivot for second condition             20m = 10M; M = 2m
3. Solve for m and M using first equation                    m = 1 g; M = 2 g

90 ·· The planet Mars has two satellites, Phobos and Deimos, in nearly circular orbits. The orbit radii of Phobos and
Deimos are 9.38 × 103 km and 23.46 × 103 km, respectively. The mass of Mars is 6.42 × 1023 kg, that of Phobos is
9.63 × 1015 kg, and that of Deimos is 1.93 × 1015 kg. Find the center of gravity and the center of mass of the two-
satellite system using the center of Mars as the origin when (a) the satellites are in opposition (i.e., on exactly opposite
sides of Mars), and (b) the satellites are in conjunction (i.e., in line on the same side of Mars).
Note: The two satellites (taken as point masses) are in different gravitational fields. Therefore, the center of
gravity of the two-satellite sysem is not the same as the center of mass. We shall consider case (b) first.
(b) 1. The gravitational field on the two satellites is        (mP + mD)g(xCG)xCG = mPg(xP)xP + mDg(xD)xD, where
in the same direction. Write Equ. 12-3.                g(x) = GMM/x
2. Solve for xCG                                           xCG = xPxD(mP + mD)/(mPxD + mDxP)
3. Substitute numerical values                             xCG = 10.43 × 103 km
(a) Now g at Phobos and Deimos are in opposite                 xCG = xPxD(mP − mD)/(mPxD − mDxP) and substituting,
directions                                                 xCG = 8.15 × 103 km, in the direction of Phobos

91 ·· When a picture is hung on a smooth vertical wall using a wire and a nail, as in Figure 12-51, the picture almost
always tips slightly forward, i.e., the plane of the picture makes a small angle with the vertical. (a) Explain why
pictures supported in this manner generally do not hang flush against the wall. (b) A framed picture 1.5 m wide and
1.2 m high and having a mass of 8.0 kg is hung as in Figure 12-51 using a wire of 1.7 m length. The ends of the wire
are fastened to the sides of the frame at the rear and 0.4 m below the top. When the picture is hung, the angle
between the plane of the frame and the wall is 5.0o. Determine the force that the wall exerts on the bottom of the
frame.
(a) Since the center of gravity of the picture is in front of the wall, the torque due to mg about
the nail must be balanced by an opposing torque due to the force of the wall on the picture,
acting horizontally. So that Σ Fx = 0, the tension in the wire must have a horizontal component,
and the picture must therefore tilt forward.
(b) Note that 0.75, 0.4, and 0.85 form a Pythagorean triad. Thus, the nail will be at the same
level as the top of the frame. The drawing now shows the end view of the system. Since the
frame’s width is not specified, we assume it to be negligible.

1. Take moments about the nail                     8g × 0.6 sin 5o = Fw × 1.2 cos 5o
2. Solve for Fw                                    Fw = (8 × 9.81 × 0.6 tan 5o)/1.2 N = 3.43 N
92 ·· Repeat Problem 76 is Barry is driving (a) uphill on a road inclined at 10o with the horizontal or (b) downhill on a
road inclined at 10o with the horizontal.
The figures on the next page show the block and the forces acting on it when traveling (a) uphill and decelerating and
Chapter 12     Static Equilibrium and Elasticity

(b) downhill and decelerating. At the critical acceleration the normal force acts at the edge of the block as indicated.
The force mg acts through the center of mass, CM.

(a) 1. Take moments about the block’s CM                      0.5(mg cos 10o) – 1.5f s = 0; f s = (mg cos 10o)/3
2. Fnet = ma                                           f s + mg sin 10o = ma
3. Solve for a                                         a = g[(cos 10o )/3 + sin 10o ] = 4.92 m/s2
(b) 1. Take moments about the block’s CM                       f s = (mg cos 10o )/3
2. Fnet = ma                                              f s − mg sin 10o = ma
3. Solve for a                                            a = g[(cos 10o )/3 − sin 10o ] = 1.52 m/s2

93* ·· If a train travels around a bend in the railbed too fast, the freight cars will tip over. Assume that the cargo portions
of the freight cars are regular parallelepipeds of uniform density and 1.5 × 104 kg mass, 10 m long, 3.0 m high, and
2.20 m wide, and that their base is 0.65 m above the rails. The axles are 7.6 m apart, each 1.2 m from the ends of the
boxcar. The separation between the rails is 1.55 m. Find the maximum safe speed of the train if the radius of
curvature of the bend is (a) 150 m, and (b) 240 m.
The boxcar and rail are shown in the drawing. At the
critical speed, the normal force is entirely on the outside
rail. As indicated, the center of gravity is 0.775 m from
that rail and 2.15 m above it. To find the speed at which
this situation prevails, we take moments about the center
of gravity.

Set Στ = 0 about the car’s center of gravity                 0.775mg = 2.15mv2/R
(a) Solve for v with R = 150 m                               v = (0.775 × 150 × 9.81/2.15) 1/2 m/s
= 23 m/s = 83 km/h
(b) Find v for R = 240 m                                     v = 29.1 m/s = 105 km/h

94 ·· For balance, a tightrope walker uses a thin rod 8 m long and bowed in a circular arc shape. At each end of the rod
is a lead mass of 20 kg. The tightrope walker, whose mass is 58 kg and whose center of gravity is 0.90 m above the
rope, holds the rod tightly at its center 0.65 m above the rope. What should the radius of curvature of the arc of the rod
Chapter 12    Static Equilibrium and Ela sticity

be so that he will be in stable equilibrium as he slowly makes his way across the rope? Neglect the mass of the rod.

1. For stable equilibrium, the center of mass of the
system must be below the foot of the tightrope walker.
The system is shown in the drawing. We take the
coordinate origin at the rope and will determine the
distance d such that ycm = 0. We then determine the
angle θ (not shown on the diagram) subtended by one
half of the long rod.

2. Find distance d                                           0 = 58 × 0.9 − 40d; d = 1.305 m
3. Length of s = 0.65 m + d                                  s = 1.955 m
1 − cos θ
4. Find expression for θ                                     R(1 − cos θ) = 1.955 m; Rθ = 4 m;              = 0.489
θ
5. Find θ by trial and error                                 θ = 1.08 rad
6. Find the radius of curvature R                            R = (4/1.08) m = 3.7 m

95 ·· A large crate weighing 4500 N rests on four 12-cm-high blocks on a horizontal surface (Figure 12-67). The crate
is 2 m long, 1.2 m high, and 1.2 m deep. You are asked to lift one end of the crate using a long steel bar. The coeffi-
cient of static friction between the blocks and the supporting surface is 0.4. Estimate the length of the steel bar you
will need to lift the end of the crate.
1. Note that when the crowbar lifts the crate, only half the weight of the crate is supported by the bar. Also, that force
has a horizontal component that must be balanced by the frictional force on the other end of the crate’s support.
Assume that the maximum force you can apply is 500 N (about 110 lb). Let l be the distance between the points
of contact of the steel bar with the floor and the crate, and let L be the total length of the bar. We denote the force
that the steel bar exerts on the crate by F.
2. Find Fx and Fy; Fy = 1/2(4500 N)                           Fy = 2250 N = F cos θ; Fx = 2250µs = 900 N = F sin θ
−1          o
3. Find θ and F                                               θ = tan 0.4 = 21.8 ; F = 2423 N
4. Determine the mechanical advantage, ME                     ME = 2423/500 = 4.85
5. Determine the lengths l and L of the crowbar              l = (12/tan θ) cm = 30 cm; L = ME × l ≈ 1.5 m

96 ··· Six identical bricks are stacked one on top of the other lengthwise and slightly offset to produce a stepped tower
with the maximum offset that will still allow the tower to stand. (a) Starting from the top, give the maximum possible
offset for each successive brick. (b) What is the total protrusion or offset of the six bricks?
(a) Let each brick have a length L. The maximum offset of the top brick is L/2 so that its CM is just at the point of
support. The distance of the CM of the top and next lower brick from the center of the lower brick is xcm = (mL/2)/2m
= L/4. It follows that the maximum overhang of the two-brick combination is L/2 − L/4 = L/4. Next, use the same
procedure for the top three bricks, locating the CM of the three-brick combination relative to the center of the third
brick. That distance is L/3 so the overhang of the three-brick combination is L/2 − L/3 = L/6. The next overhang is
Chapter 12     Static Equilibrium and Elasticity

L/8, and the following one is L/10.
(b) The total offset is 1/2L(1 + 1/2 + 1/3 + 1/4 + 1/5) = 1.14L. The total offset is greater than the length of one brick,
so the left edge of the top brick is to the right of the right edge of the bottom brick.
97* ··· A uniform sphere of radius R and mass M is held at rest on an inclined plane of angle θ by a horizontal string, as
shown in Figure 12-68. Let R = 20 cm, M = 3 kg, and θ = 30o. (a) Find the tension in the string.
(b) What is the normal force exerted on the sphere by the inclined plane? (c) What is the frictional force acting
on the sphere?
There are four forces acting on the sphere: its weight, mg; the normal force of the plane, Fn; the frictional force, f,
acting parallel to the plane; and the tension in the string, T.
(a) 1. Take moments about the center of the sphere              TR = fR; T = f
2. Set Σ Fx = 0, where x is along the plane                 T cos θ + f = mg sin θ; T = mg sin θ/(1 + cos θ)
3. Evaluate T                                               T = 3 × 9.81 × 0.5/(1 + 0.866) = 7.89 N
(b) Set Σ Fy = 0                                                Fn = mg cos θ + T sin θ = 29.4 N
(c) f = T                                                       f = 7.89 N

98 ··· The legs of a tripod make equal angles of 90o with each other at the apex, where they join together. A 100-kg
block hangs from the apex. What are the compressional forces in the three legs?
The three legs of the tripod form three sides of a cube. The length of the diagonal is     3 L, where L is the length of
each leg. The downward force of mg is equally distributed over the three legs. Consequently, the compressive force in
each leg is ( 3 /3)mg = 566 N.
99 ··· Figure 12-69 shows a 20-cm-long uniform beam resting on a cylinder of 4 cm radius. The mass of the beam is 5.0
kg, and that of the cylinder is 8.0 kg. The coefficient of friction between beam and cylinder is zero. (a) Find the forces
that act on the beam and on the cylinder. (b) What must the minimum coefficients of static friction be between beam
and floor and between the cylinder and floor to prevent slipping?
The forces that act on the beam are its weight, mg; the force of the cylinder, Fc, acting along the radius of the cylinder;
the normal force of the ground, Fn; and the friction force f s = µsFn. The forces acting on the cylinder are its weight,
Mg; the force of the beam on the cylinder, Fcb = Fc in magnitude, acting inward radially; the normal force of the
ground on the cylinder, Fnc; and the force of friction, f sc = µscFnc.
(a) Take moments about the end of right beam to                 mg × 10 × cos 30o = 15Fc; Fc = 28.3 N
find Fc
Use Σ Fy = 0 to find Fn                                     Fc cos 30o + Fn = mg; Fn = 24.5 N
Use Σ Fx = 0 to find f s                                    f s = Fc sin 30o = 14.15 N, to the left
Fcb = Fc, acting radially and downward                      Fcb = 28.3 N in magnitude
Use Σ Fy = 0 to find Fnc                                    Fnc = Mg + Fcb cos 30o = 103 N
Use Σ Fx = 0 to find f sc                                   f sc = Fcb sin 30o = 14.15 N, to the right
o
(b) µs(beam–floor) = f s /Fn                                    µs(beam–floor) = 0.577 = tan 30
µs(cylinder–floor) = f sc /Fnc                              µs(cylinder–floor) = 0.137

100 ··· Two solid smooth spheres of radius r are placed inside a cylinder of radius R as in Figure 12-70. The mass of
Chapter 12    Static Equilibrium and Ela sticity

each sphere is m. Find the force exerted by the bottom of the cylinder, the force exerted by the wall of the
cylinder, and the force exerted by one sphere on the other.

The geometry of the system is shown in the drawing . We denote the angle between
the vertical center line and the line joining the two centers by θ. The distance x in the
drawing is then x = R − r, and sin θ = (R − r)/r. Also, tan θ = (R − r)/[R(2r − R)]1/2.
The force exerted by the bottom of the cylinder is just 2mg. Let F be the force that the
top sphere exerts on the lower sphere. Since the cylinder wall is smooth, F cos θ =
mg, and F = mg/cos θ. Its x component is Fx = F sin θ = mg tan θ. This is the force
that the      wall of the cylinder exerts. Thus Fw = mg(R − r)/[R(2r − R)]1/2. Note that
as r approaches R/2, Fw → ∞.

101*··· A solid cube of side length a balanced atop a cylinder of diameter d is in unstable equilibrium if d << a and is in
stable equilibrium if d >> a (Figure 12-71). Determine the minimum value of d/a for which the cube is in stable
equilibrium.
Consider a small rotational displacement, δθ, of the cube of the figure below from equilibrium. This shifts the point of
contact between cube and cylinder by Rδθ, where R = d/2. As a result of that motion, the cube itself is rotated through
the same angle δθ, and so its center is shifted in the same direction by the amount (a/2)δθ, neglecting higher order
terms in δθ. If the displacement of the cube’s center of mass is less than that of the point of contact, the torque about
the point of contact is a restoring torque, and the cube will return to its equilibrium position. If, on the other hand,
(a/2)δθ > (d/2)δθ, then the torque about the point of contact due to mg is in the direction of δθ, and will cause the
displacement from equilibrium to increase. We see that the minimum value of d/a for stable equilibrium is d/a = 1.

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