```CHAPTER 14
STATIC EQUILIBRIUM
ActivPhysics can help with these Problems:
Activities 7.1–7.6
Section 14-1: Conditions for Equilibrium
Problem
1. Five forces act on a rod, as shown in Fig. 14-25.
Write the torque equations that must be satisfied
for the rod to be in static equilibrium taking the
the center of the rod.
Problem
2. A body is subject to three forces: F1 = 2ı̂ + 2̂ N,
applied at the point x = 2 m, y = 0 m; F2 = −2ı̂ −
3̂ N, applied at x = −1 m, y = 0; and F3 = 1̂ N,
applied at x = −7 m, y = 1 m. (a) Show explicitly
that the net force on the body is zero. (b) Show
explicitly that the net torque about the origin is
zero. (c) To confirm the assertion following
Equation 14-4 that the net torque must be zero
about any other point, evaluate the net torque
about the point (3 m, 2 m), the point (−7 m, 1 m),
Solution
P
P
(a)
Fi = (2ı̂ + 2̂ − 2ı̂ − 3̂ + ̂)N = 0. (b) ( τ i )0 =
[2ı̂ × (2ı̂ + 2̂) + (−ı̂) × (−2ı̂ − 3̂) + (−7ı̂ + ̂) ×
̂]N·m = (4 + 3 − 7)P
k̂ N·m = 0. (c) For any point
r0 = (x0 ı̂ + y0 ̂)m,
(ri − r0 ) × Fı̂ = {[(2 − x0 )ı̂ −
y0 ̂] × (2ı̂ + 2̂) + [(−1 − x0 )ı̂ − y0 ̂] × (−2ı̂ − 3̂) +
[(−7 − x0 )ı̂ + (1 − y0 )̂] × ̂} N·m = [(4 + 3 − 7) +
x0 (−2 + 3 − 1) + y0 (2 − 2 + 0)]k̂ N·m = 0.
Problem
figure 14-25 Problem 1.
3. Suppose the force F3 in the preceding problem is
doubled so the forces no longer balance and the
body is therefore accelerating. Show that (a) the
torque about the point (−7 m, 1 m) is still zero, but
that (b) the torque about the origin is no longer
zero. What is the torque about the origin?
Solution
Solution
All of the forces lie in the plane of Fig. 14-25, so all of
the torques about any point on the rod are into or out
of the page. Suppose the latter direction, out of the
page or counterclockwise, is positive. Moreover, all of
the forces are perpendicular to the rod, so their lever
arms about any point on the rod (recall that the
magnitude of the torque is force times lever arm) can
of the rod, F4 and F5 contribute zero torque, and
Equation 14-2 becomes 0 = 31 ℓF3 − 32 ℓF2 + ℓF1 .
(b) About the center of the rod, the perpendicular
distances to F2 and F3 are 61 ℓ, and to F1 , F4 and F5
are 21 ℓ, so 0 = 12 ℓ(F1 + F4 − F5 ) − 61 ℓ(F2 + F3 ).
(a) Since r3 = (−7ı̂ + ̂) m is the point of application
of F3P
, the total torque about r3 is just due to F1 and
F2 : ( τ i )3 = (r1 − r3 ) × F1 + (r2 − r3 ) × F2 =
[(2ı̂ + 7ı̂ − ̂) × (2ı̂ + 2̂) + (−ı̂ + 7ı̂ − ̂) × (−2ı̂ −
3̂)]N·m = [(9×2)k̂ − (1×2)(−
P k̂) + 6 ×
P(−3) k̂ +
(1×2)(−k̂)] N·m = 0. (b) ( τ i )0 = (ri × Fi ) =
[2ı̂ × (2ı̂ + 2̂) + (−ı̂) × (2ı̂ − 3̂) + (−7ı̂ + ̂) ×
(2̂)] N·m = [4k̂ + 3k̂ − 14k̂] N·m = −7k̂ N·m.
Problem
4. A rod of mass m and length ℓ is falling freely in a
horizontal orientation, with no torque about its
center of mass. Find the magnitude of the torque
violate the point made in Problem 8?
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CHAPTER 14
Solution
The magnitude of the torque on a body of mass M in
a constant gravity field is |τ grav | = |rcm × M g| =
rcm M g sin θ. With origin at either end, rcm = 12 ℓ, and
sin θ = 1 for horizontal orientation, hence τ grav =
1
2 M gℓ. With origin at the CM (rcm = 0) the torque is
zero, but this does
P not contradict the result of
Problem 8 since
F = M g 6= 0 in free fall.
Fı̂ = (1 m)F k̂. Thus, −x3 + y3 = 1 m, or the line of
action of F3 passes through the point of application of
F2 (the point (0, 1 m)). Any point on this line is a
suitable point of application for F3 (e.g. the point
(0, 1 m)). (b) In this case, F1 = −F2 so F3 = 0, but
r1 × F1 + r2 × F2 = (r2 − r1 ) × F2 6= 0 so r3 ×
F3 6= 0. Thus there is no single force that can be
Problem
Problem
5. In Fig. 14-26 the forces shown all have the same
magnitude F. For each of the cases shown, is it
possible to place a third force so the three forces
meet both conditions for static equilibrium? If so,
specify the force and a suitable application point; if
not, why not?
6. Are there any other application points for the force
F3 in Problem 2 that will ensure that both static
equilibrium conditions are met?
Solution
Equation 14-1 does not involve the points of
application of the forces. Equation 14-2 can be
satisfied for any point of application r′3 , for F3 ,
provided r′3 × F3 = r3 × F3 , or (x′3 ı̂ + y3′ ̂) × ̂ =
(−7ı̂ + ̂) × ̂, which implies x′3 = −7. Thus, any point
along the line x = −7 is a possible point of application
for F3 , satisfying the equilibrium conditions.
Problem
7. Four forces act on a body, as shown in Fig. 14-27.
Write the set of scalar equations that must hold for
the body to be in equilibrium, evaluating the
F4
φ
2
P
F3
1
φ
F1
F2
O
figure 14-27 Problem 7.
figure 14-26 Problem 5.
Solution
Solution
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̂, r1 = (2 m)̂, F2 = Fı̂, and r2 = (1 m)̂.
Thus,
F3 = −F (ı̂ + ̂), which is a force of magnitude
√
2F , 45◦ down into the third quadrant (θx = 225◦ or
−135◦ CCW from the x axis). The point of
application, r3 , can be found from the second
condition, r3 × F3 = (x3 ı̂ + y3 ̂) × (−Fı̂ − F̂) =
(−x3 + y3 )F k̂ = −r1 × F1 − r2 × F2 = 0 − (1 m)̂×
All of the forces lie in the same plane, which includes
the points O and P, so there are two independent
components of the force condition (Equation 14-1)
and one component of the torque condition
(Equation 14-2). Taking the x axis to the right, the y
axis up and
P the z axis out of the page in Figure
P 14-27,
we have:
Fx =P
0 = −F1 + F2 sin φ + F3 ,
Fy = 0 =
−F2 cos φ + F4 , (P τz )0 = 0 = −ℓ1 F2 − ℓ2 F3 sin φ +
ℓ2 F4 cos φ, and ( τz )P = −ℓ2 F1 sin φ + (ℓ2 − ℓ1 )×
F2 = 0. (The lever arms of all the forces about either
O or P should be evident from Fig. 14-27).
CHAPTER 14
209
Problem
8. In this problem you prove the statement in Section 14-1 that the choice of pivot point does not
matter when applying the conditions for static
equilibrium. Figure 14-28 shows an object on which
the net force is assumed to be zero. Also, the net
torque about the point O shown is zero. You’re to
show that the net torque about any other point P is
also zero.
P To do so, write the net torque about P as
τP =
rP i × Fi , where the vectors rP are from P
to the force application points, and the index i
labels the different forces. Note in Fig. 14-28 that
rP i = rOi + R, where R is a vector from P to O.
Use this result in your expression for τ P and apply
the distributive law to get two separate sums. Use
the assumptions that Fnet = 0 and τ O = 0 to argue
that both terms are zero. This completes the proof.
figure 14-29(a) Problem 9 Solution.
colinear with the weight, so τ B = 0. (c) τ C =
1
1
◦
2 ℓ mg sin 90 = 2 mg ℓ (but note that τ C = −τA ).
(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.)
Problem
figure 14-28 Problem 8.
10. Figure 14-29b shows a thin, uniform plate of
mass m in the shape of an equilateral triangle of
side ℓ. The plate is in a vertical plane. Find the
magnitude of the gravitational torque on the plate
about each of the three points shown.
Solution
Solution
With referenceP
to Fig. 14-28, we can
P
P write τ p =
rPP
×
F
=
(r
+
R)
×
F
=
rOi × Fi +
i
i
Oi
i
R × Fi = τ O + R × Fnet = τ O . If Fnet = 0, the
total torque about any two points is the same.
The center of gravity is at the√center of the triangle, a
perpendicular distance of ℓ/2 3 from any side.
(a) The lever arm of the weight about point A is ℓ/2
so τ A = 21 mgℓ. (b) The lever arm about point B is
zero, and τ B = 0. (c) The lever arm about point C is
ℓ/4 (C is halfway up from the base) so τ C = 41 mgℓ.
(Of course, τ A and τ C are in opposite directions. The
torque could also be found from Equation 12-12.)
Section 14-2: Center of Gravity
Problem
9. Figure 14-29a shows a thin, uniform square plate of
mass m and side ℓ. The plate is in a vertical plane.
Find the magnitude of the gravitational torque on
the plate about each of the three points shown.
Solution
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 τ A = ( 2ℓ/2)mg sin 135◦ = 12 mg ℓ. (b) rB is
Problem
11. Three identical books of length L are stacked over
the edge of a table as shown in Fig. 14-30. The
top book overhangs the middle one by 21 L, so it
just barely avoids falling. The middle book
overhangs the bottom one by 14 L. How much of
the bottom book can be allowed to overhang the
edge of the table without the books falling?
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CHAPTER 14
Problem
12. A 60-kg uniform tabletop 2.4 m long is supported
by a pivot 80 cm from the left end, and by a scale
at the right end (Fig. 14-31). How far from the
left end should a 40-kg child sit if the scale is to
Solution
See the solution to the next problem, when Fs = 0.
figure 14-29(b) Problem 10 Solution.
figure 14-31 Problem 12 Solution.
Solution
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 = − 31 L. If the center of the bottom book is 13 L to
the left of the edge, then only 12 L − 13 L = 61 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.)
Section 14-3: Examples of Static Equilibrium
Problem
13. Where should the child in Fig. 14-31 sit if the
scale is to read (a) 100 N and (b) 300 N?
Solution
If we consider torques about the pivot point (so that
the force exerted by the pivot does not contribute)
then Equation 14-2 is sufficient to determine the
position of the child. As shown on Fig. 14-31, 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
P
(1/g) ( τ )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.)
Problem
figure 14-30 Problem 11 Solution.
14. A 4.2-m-long beam is supported by a cable at its
center. A 65-kg steelworker stands at one end of
the beam. Where should a 190-kg bucket of
concrete be suspended if the beam is to be in
static equilibrium?
Solution
The sum of the torques on the beam (taken about its
center, C, so that the cable’s tension and beam’s
CHAPTER 14
211
Problem
16. A 23-m-long log of irregular cross section is lying
horizontally, supported by a wall at one end and a
cable attached 4.0 m from the other end, as shown
in Fig. 14-33. The log weighs 7.5×103 N, and the
tension in the cable is 6.2×103 N. Where is the
log’s center of gravity?
Problem 14 Solution.
weight do not enter the equation) is equal to zero.
Therefore, (190 kg)g(x) = (65 kg)g(2.1 m), or
x = 71.8 cm, on the opposite side of C from the
worker. (Note: Since sin θ = sin(π − θ) 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.)
Solution
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.
Problem
15. Two pulleys are mounted on a horizontal axle, as
shown in Fig. 14-32. The inner pulley has a
diameter of 6.0 cm, the outer a diameter of 20 cm.
Cords are wrapped around both pulleys so they
don’t slip. In the configuration shown, with what
force must you pull on the outer rope in order to
support the 40-kg mass?
Solution
Since each cord is tangent to its respective pulley, the
lever arms are just the radii, as shown on the figure.
The two torques are equal in magnitude, R1 F1 =
R2 F2 , so that
figure 14-33 Problem 16 Solution.
Problem
17. Figure 14-34 shows a traffic signal, with masses
and positions of its various members indicated.
The structure is mounted with two bolts, located
centerline, as indicated. What tension force must
the left-hand bolt be capable of withstanding?
F1 = (40 kg)(9.8 m/s2 )(6 cm) = (20 cm) = 118 N.
figure 14-34 Problem 17 Solution.
figure 14-32 Problem 15 Solution.
Solution
The forces on the traffic signal structure, and their
lever arms about point 0 (on the vertical member’s
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CHAPTER 14
centerline between the bolts) are shown on Fig. 14-34.
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
P
determine Nℓ and Nr are:
Fy = 0 = Nℓ + Nr −
(9.8)(320 + 170 + 65) N (the vertical
P component of
Equation 14-1, positive up) and ( τz )0 = 0 =
(Nr − Nℓ )(0.38 m) − (170 × 3.5 + 65 × 8)(9.8 N·m) (the
out-of-the-page-component of Equation 14-2, positive
CCW). These can be written as Nr + Nℓ = 5.44 kN,
and Nr − Nℓ = 28.8 kN. Thus Nℓ = −11.7 kN, which
is downward and must be exerted by the bolt. The
reaction force on the bolt is upward and is a tensile
force. (Really, Nℓ 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.)
Problem
18. Figure 14-35 shows how a scale with a capacity of
only 250 N can be used to weigh a heavier person.
The board is 3.0 m long, has a mass of 3.4 kg, and
is of uniform density. It is free to pivot about the
end farthest from the scale. What is the weight of
a person standing 1.2 m from the pivot end, if the
scale reads 210 N? Assume that the beam remains
nearly horizontal.
Problem
19. Figure 14-36a shows an outstretched arm with a
mass of 4.2 kg The arm is 56 cm long, and its
center of gravity is 21 cm from the shoulder. The
hand at the end of the arm holds a 6.0-kg mass.
(a) What is the torque about the shoulder due to
the weights of the arm and the 6.0-kg mass? (b) If
the arm is held in equilibrium by the deltoid
muscle, whose force on the arm acts 5.0◦ below
the horizontal at a point 18 cm from the shoulder
joint (Fig. 14-36b), what is the force exerted by
the muscle?
Solution
(a) The magnitude of the (external) torque on the
arm is τ0 = [(4.2 kg)(0.21 m) + (6 kg)(0.56 m)] ×
2
(9.8 m/s ) sin 105◦ = 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 F r sin θ = F (0.18 m) sin 170◦,
which, under equilibrium conditions, equals the
magnitude of the torque in part (a). Thus, F =
40.2 N·m/(0.18 m) sin 170◦ = 1.28 kN, underscoring
the comment at the end of Example 14-5. The
skeleto-muscular structure of the human extremities
evolved for speed and range of motion, not mechanical
Solution
Since the board is at rest, the sum of the torques
(positive CCW in Fig. 14-35) 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).
P
Thus, ( τz )P = 0 = (210 N)(3 m) − (3.4×9.8 N)×
(1.5 m) − W (1.2 m), or W = 483 N.
figure 14-36 Problem 19 Solution.
Problem
figure 14-35 Problem 18 Solution.
20. Figure 14-37 shows a portable infant seat that is
supported by the edge of a table. The mass of the
seat is 1.5 kg, and its center of mass is located
16 cm from the table edge. A 12-kg baby is sitting
in the seat with her center of mass over the seat’s
CHAPTER 14
center of mass. Find the forces FA and FB that
the seat exerts on the table.
Solution
The infant seat is in equilibrium under the reaction
forces to FA and FB , and the weight of the seat and
infant. Thus,
P
Fy = 0
213
164FAx (−k̂) − 37.5M g(k̂), we find FAx = −(37.5)×
2
(15 kg)(9.8 m/s )/(164) = −33.6 N. From the
x equation, FBx = −FAx = 33.6 N, and by assumption,
FAy = FBy = 12 M g = 73.5 N. Of course, the forces
exerted by the door on the hinges (by Newton’s third
law) are the reactions to the forces, F A and F B , just
calculated.
= FB − FA − (1.5 kg + 12 kg)(9.8 m/s2 )
(positive up)
(
P
τ )A = 0
= (22 cm)FB − (16 + 22) cm(13.5)(9.8) N
(positive CCW)
Solving for FB from the torque equation and FA
from the force equation, we find FB = 229 N and
FA = 96.2 N, in the directions shown in the figure.
Problem 21 Solution.
Problem
940
figure 14-37 Problem 20 Solution.
Problem
21. A 15.0-kg door measures 2.00 m high by 75.0 cm
wide. It hangs from hinges mounted 18.0 cm from
top and bottom. Assuming that each hinge carries
half the door’s weight, determine the horizontal
and vertical forces that the door exerts on each
hinge.
22. Figure 14-38 shows a popular system for mounting
bookshelves. An aluminum bracket is mounted on
a vertical aluminum support by small tabs inserted
into vertical slots. If each bracket in a shelf system
supports 32 kg of books, with the center of gravity
12 cm out from the vertical support, what is the
horizontal component of the force exerted on the
upper of the two bracket tabs? Assume contact
between the bracket and support occurs only at
the upper tab and at the bottom of the bracket,
4.5 cm below the upper tab.
Solution
Solution
If the door is properly hung, all the forces on the door
are coplanar. We assume that the CM is at the
geometrical center of the door. The conditions for
equilibrium are:
P
Fx = 0 = FAx + FBx ,
P
Fy = 0 = FAy + FBy − M g,
P
( τ )B = 0 = rA × FA + rcm × M g
= (164 cm̂) × (FAx ı̂ + FAy ̂)
+ (37.5 cmı̂ + 82 cm̂) × (−M g̂),
where we chose to calculate torques about the lower
hinge at B. Expanding the cross products, 0 =
The forces on the bracket are shown superposed on
Fig. 14-38, assuming only a normal
force on the
P
bottom. Equilibrium implies
Fx = 0 = FAx + FB ,
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CHAPTER 14
P
and ( τz )A = 0 = (4.5 cm)FB − (12 cm)W. (We
chose to evaluate the z component (out of the page) of
the torques about the upper tab, at A, and note that
P
Fy = 0 is not needed.) The solution for FAx is
immediate: FAx = −FB = −(12/4.5)W = −(2.67)×
(32 kg)(9.8 m/s2 ) = −836 N. The negative
x component means that a tensile force is exerted on
the upper tab, as expected. (Note that the upper tab
must also support the weight of the books.)
5.00 kN. The force on the tie beam (the reaction to
F B is a tension. (The function of the tie beam is
precisely to relieve any horizontal force that the roof
may exert on the walls.)
figure 14-39 Problem 23 Solution.
Problem
figure 14-38 Problem 22 Solution.
24. Repeat Example 14-4, now assuming that the
coefficient of friction at the floor is µ1 and that at
the wall µ2 . Show that the minimum angle at
which the board will not slip is now given by
1 − µ1 µ2
.
φ = tan−1
2µ1
Problem
23. Figure 14-39 shows a house designed to have high
“cathedral” ceilings. Following a heavy snow, the
total mass supported by each diagonal roof rafter
is 170 kg, including building materials as well as
snow. Under these conditions, what is the force in
the horizontal tie beam near the roof peak? Is this
force a compression or a tension? Neglect any
horizontal component of force due to the vertical
walls below the roof. Ignore the widths of the
various structural components, treating contact
forces as though they were concentrated at the
roof peak and at the outside edge of the
rafter/wall junction.
Solution
The forces on one of the diagonal rafters are drawn on
Fig. 14-39. If the rafters are symmetrical (without
internal stress), and we neglect the weight of the tie
beam, FB and FC will be horizontal. W and FA are
vertical, the latter by assumption, and we suppose W
acts at the center of the rafter. The
Pequilibrium
conditionsP
needed to find FB are:
Fx = 0 = FB −
FC , and ( τ z )A = 0, or 0 = (4 m)FC − (3.2 m)FB −
(2.4 m)W. (Point A was chosen to eliminate F A from
the equation.) The solution is FB = 2.4W/(4 − 3.2) =
figure 14-11(b) Problem 24 Solution.
Solution
The addition of a frictional force on the board where it
contacts the wall is shown in the sketch (see
Fig. 14-11b). We assume that f2 = µ2 N2 is the
maximum frictional force and that it acts to oppose
the fall of the board. (If f2 is not proportional to N2 ,
there is insufficient data to solve this Problem.) The
equations of static equilibrium in Example 14-4
CHAPTER 14
become: f1 − N2 = 0 (horizontal), N1 + µ2 N2 − mg = 0
(vertical), and N2 L sin φ + µ2 N2 L cos φ −
1
2 mgL cos φ = 0 (torque about bottom of board). The
minimum angle φ can be found from the horizontal
equation and the requirement that N2 = f1 ≤
µ1 N1 . N1 can be eliminated from the vertical
equation, N2 ≤ µ1 (mg − µ2 N2 ) or (1 + µ1 µ2 )×
N2 ≤ µ1 mg. N2 can be eliminated from the torque
equation, N2 (tan φ + µ2 ) = 12 mg, so this condition
becomes 12 mg(1 + µ1 µ2 )/(µ2 + tan φ) ≤ µ1 mg, or
tan φ ≥ (1 − µ1 µ2 )/2µ1 .
Problem
215
Problem
26. Show that if the wall in the previous problem is
frictionless, then a continuation of the rope line
must pass through the center of the sphere.
Solution
From the preceding solution, it is clear that, in the
absence of friction, the condition that the sum of the
torques about the center of the sphere is zero implies
that the lever arm of the tension is zero, or that the
rope must be in line with the center.
Problem
25. A uniform sphere of radius R is supported by a
rope attached to a vertical wall, as shown in
Fig. 14-40. The point where the rope is attached
to the sphere is located so a continuation of the
rope would intersect a horizontal line through the
sphere’s center a distance R/2 beyond the center,
as shown. What is the smallest possible value for
the coefficient of friction between wall and sphere?
27. A garden cart loaded with firewood is being
pushed horizontally when it encounters a step 8.0
cm high, as shown in Fig. 14-41. The mass of the
cart and its load is 55 kg, and the cart is balanced
so that its center of mass is directly over the axle.
The wheel diameter is 60 cm. What is the
minimum horizontal force that will get the cart up
the step?
Solution
In equilibrium, the sum of the torques about the
center of the sphere must be zero, so the frictional
force is up, as shown. The
√ lever arm of the tension in
the rope is 12 R cos 30◦ = 3R = 4, and the weight and
normal force
exert no torque about the center. Thus,
√
f R = T 3R/4. The sum of the horizontal components
◦
of the forces is zero also, so
√ 0 = N − T sin 30 , or
1
T = 2N. Therefore f = 2 3 N. Since f ≤ µs N, this
√
implies µs ≥ f /N = 12 3 = 0.87.
F
60 cm
8.0 cm
figure 14-41 Problem 27.
figure 14-41 Problem 27 Solution.
Solution
figure 14-40 Problem 25 Solution.
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,
216
CHAPTER 14
the latter does not contribute, and the wheels
(and
P
cart) will remain stationary as long as ( τ )step =
M gR sin θ − N R sin θ − F R cos θ = 0. When N = 0,
however, the wheels begin to lose contact with the
ground and go over the step. This occurs when
F = M g tan θ. From the geometry of the situation,
R(1 − cos θ) = h, the height of the step, so θ =
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.
Problem
28. Figure 14-42 shows the foot and lower leg of a
person standing on the ball of one foot. Three
forces act on the foot to maintain this equilibrium:
the tension force T in the Achilles tendon, contact
force F C at the ankle joint, and the normal force
N of the ground that supports the person’s
weight. The person’s mass is 70 kg, and the force
application points are as indicated in Fig. 14-42.
Find the magnitude of (a) the tension in the
Achilles tendon and (b) the contact force at the
ankle joint.
Treat the tower as a uniform cylinder 7.0 m in
diameter and 55 m high, and assume the ground
supports the tower’s weight but does not provide
any torque.
Solution
The center of mass of the tower must be somewhere
over its footprint on the ground, or it will topple. This
will be so, for a uniform cylindrical model of the
tower, if the angle of tilt φ is less than the angle α
that a diagonal makes with the length, as indicated on
Fig. 14-43. Therefore φ ≤ α = tan−1 (7m/55m) =
7.25◦ .
figure 14-43 Problem 29 Solution.
Problem
figure 14-42 Problem 28 Solution.
Solution
Solution
If we approximate the bones in the foot as a massless,
planar, rigid body, the equilibrium conditions
for the
P
situation depicted in P
Fig. 14-42 are:
Fx =
◦
◦
T sin 25
P − FC,x = 0, Fy = T cos 25 + N − FC,y ◦= 0,
and ( τ )ankle joint = N (12 cm) − T (7 cm) sin 115 =
0. Then with N = 70 × 9.8 N, one finds (a) T =
N (12 cm)/(7 cm) sin 115◦ = 1.30 kN, and (b) FC,x =
T sin 25◦ =q
0.548 kN, FC,y = T cos 25◦ + N = 1.86 kN,
and FC =
30. A uniform 5.0-kg ladder is leaning against a
frictionless vertical wall, with which it makes a
15◦ angle. The coefficient of friction between
ladder and ground is 0.26. Can a 65-kg person
climb to the top of the ladder without it slipping?
If not, how high can the person climb? If so, how
massive a person would make the ladder slip?
2 + F 2 = 1.94 kN.
FC,x
C,y
Problem
29. The leaning Tower of Pisa (Fig. 14-43) currently
leans at a 4.7◦ angle to the vertical. Treating the
tower as a solid cylinder, what is the maximum
angle at which it can lean before falling over?
It is shown in the solution to Problem 45, that the
condition for a person of mass m to climb up a
is α ≤ µs cot θ + (mℓ /m)(µs cot θ − 21 ), where µs is the
coefficient of friction with the floor, θ the angle with
the vertical frictionless wall, and mℓ the mass of the
ladder. For the situation in this problem, µs cot θ =
0.26 cot 15◦ = 0.970 and α ≤ 0.970 + (5/65)(0.470) =
1.01. Therefore (since α ≤ 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 >
mℓ (µs cot θ − 12 )/(1−µs cot θ) = 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.
CHAPTER 14
Problem
217
Therefore:
31. The boom in the crane of Fig. 14-44 is free to
pivot about point P and is supported by the cable
that joins halfway along its 18-m total length. The
cable passes over a pulley and is anchored at the
back of the crane. The boom has mass 1700 kg,
distributed uniformly along its length, and the
mass hanging from the end of the boom is
2200 kg. The boom makes a 50◦ angle with the
horizontal. What is the tension in the cable that
supports the boom?
T2 = 0.5W sin 99.2◦/ sin 20.8◦ = 1.39W,
T1 = T2 sin 60◦ / sin 35◦ = 2.10W,
and w = T1 cos 35◦ + T2 cos 60◦ − W = 1.41W.
figure 14-45 Problem 32 Solution.
Problem
figure 14-44 Problem 31 Solution.
Solution
The forces on the boom are shown superposed on the
figure. By assumption, T is horizontal and acts at the
CM of theP
boom. To find T, we compute the torques
about P, ( τ )P = 0, obtaining:
T 12 ℓ sin 50◦ − mb g 21 ℓ cos 50◦ − mgℓ cos 50◦ = 0,
or
T = (2m + mb )g cot 50◦ = (4400 + 1700)(9.8 N) cot 50◦
= 50.2 kN.
Problem
33. Figure 14-46 shows a 1250-kg car that has slipped
over the edge of an embankment. A group of
people are trying to hold the car in place by
pulling on a horizontal rope, as shown. The car’s
bottom is pivoted on the edge of the embankment,
and its center of mass lies further back, as shown.
If the car makes a 34◦ angle with the horizontal,
what force must the group apply to hold it in
place?
Solution
Three forces act on the car, as shown added to
Fig. 14-46. The unknown force, FP , exerted by the
edge of the embankment, does not contribute to
Equation 14-2 (positive torques CCW) if evaluated
about point P, so the tension necessaryPto keep the car
in equilibrium can be found directly. ( τ )P = 0 =
M g(2.4 m − 1.8 m) cos 34◦ − T (1.8 m) sin 34◦, or
T = (1250×9.8 N)/3 tan 34◦ = 6.05 kN.
32. A uniform board of length ℓ and weight W is
suspended between two vertical walls by ropes of
length ℓ/2 each. When a weight w is placed on the
left end of the board, it assumes the configuration
shown in Fig. 14-45. Find the weight w in terms
of the board weight W.
Solution
The conditions for equilibrium (about the origin
drawn on the figure) are:
P
Fx = 0 = T2 sin 60◦ − T1 sin 35◦ ,
P
Fy = 0 = T2 cos 60◦ + T1 cos 35◦ − w − W,
P
( τz )0 = 0 = T2 ℓ sin 20.8◦ − W 12 ℓ sin 99.2◦ .
figure 14-46 Problem 33 Solution.
Problem
34. A uniform board of length ℓ is dangling over a
frictionless edge, secured by a horizontal rope, as
218
CHAPTER 14
shown in Fig. 14-47. Show that the angle
p it makes
with the horizontal must be θ = sin−1 2d/ℓ,
where d is the distance from the edge to the center
of the board.
Solution
mg(0.94x − 0.01x2 ). (a) Equation 14-3, the condition
for equilibrium, gives dU/dx = 0 = mg(0.94 − 0.02x),
or x = 47 m (b) Since d2 U/dx2 = −0.02mg < 0,
Equation 14-5 implies this is an unstable equilibrium.
(c) h(47) = 0.94(47) − 0.01(47)2 = 22.1 m.
See the solution to the next Problem.
Problem
35. Figure 14-47 shows a uniform board dangling over
a frictionless edge, secured by a horizontal rope. If
the angle θ in Fig. 14-47 were 30◦ , what fraction
would the distance d shown in the figure be of the
board length ℓ?
(a)
(b)
figure 14-48 Problem 37.
Problem
figure 14-47 Problems 34, 35.
Solution
The three forces acting on the board are in the same
configuration as those in Problem 33, so Equation 14-2
about the edge gives M gd cos θ = T ( 21 ℓ − d) sin θ. If
the edge is frictionless, then FP is perpendicular to
the board, so Equation 14-1 requires T = FP sin θ and
M g = FP cos θ. Substituting above, we find
FP d cos2 θ = FP ( 12 ℓ − d) sin2 θ, or d(cos2 θ + sin2 θ) =
p
2
−1
1
2d/ℓ. For θ = 30◦ , d/ℓ =
2 ℓ sin θ and θ = sin
2
1
1
◦
2 sin 30 = 8 .
Section 14-4: Stability of Equilibria
Problem
36. A portion of a roller coaster track is described by
h = 0.94x − 0.010x2 , where h and x are the height
and horizontal position in meters. (a) Find a point
where the roller coaster car could be in static
equilibrium on this track. (b) Is the equilibrium
stable or unstable? (c) What is the height of the
track at the equilibrium point?
Solution
The potential energy of the roller coaster car, in the
equivalent one-dimensional problem, is U (x) = mgh =
37. A roly-poly toy clown is made from part of a
sphere topped by a cone. The sphere is truncated
at just the right point so that there is no
discontinuity in angle as the surface changes from
sphere to cone (Fig. 14-48a). If the clown always
returns to an upright position, what is the
maximum possible height for its center of mass?
Would your answer change if the continuity-ofangle condition were not met, as in Fig. 14-48b?
Solution
Suppose the roly-poly rests on a flat horizontal
surface. Its spherical surface is always tangent to the
horizontal if the continuity-of-angle condition holds, as
in sketch (a) (except when upside-down on the point
of the cone). Gravity will always exert a restoring
torque if the CM lies to the left of the vertical through
the left-most point of contact, B, as shown. Since this
vertical passes through O, the center of the sphere, the
CM should lie “below” O, as measured from the
bottom point, A, on the axis.
If the continuity-of-angle condition is not met, the
cone, in general, intersects the sphere in one of two
circles (through BB ′ or CC ′ , as in sketch (b)). If CC ′
is the actual boundary, the reasoning in the first
paragraph still applies. However, if BB ′ is the
boundary, the maximum distance of the CM from
point A should be < AQ (considerably “lower”
than O).
CHAPTER 14
219
Problem
39. The potential energy as a function of position for
a certain particle is given by
3
x
x2
x
U (x) = U0
,
+
a
+
4
x30
x20
x0
where U0 , x0 , and a are constants. For what values
of a will there be two static equilibria? Comment
on the stability of these equilibria.
Solution
Problem 37 Solution.
Problem
38. A uniform rectangular block is twice as long as it
is wide. Letting θ be the angle that the long
dimension makes with the horizontal, determine
the angular positions of any static equilibria, and
comment on their stability.
Solution
Consideration of the block, tilted in a plane
perpendicular to its thickness (the unspecified
dimension in this question) reveals that θ = 0◦ is a
stable equilibrium position, θ = 90◦ a metastable one,
and θ + α = 90◦ an unstable one (α is the angle a
diagonal makes with the longer side, as shown). Since
tan α = ℓ/2ℓ = 12 , the unstable equilibrium is at
θ = 90◦ − tan−1 ( 12 ) = 63.4◦.
The equilibrium condition, dU/dx = 0 (Equation 14-3), requires 3(x/x0 )2 + 2a(x/x0 ) + 4 = 0. This
quadratic has two real roots if the discriminant
is
√
3.
The
roots
are
positive, i.e., a2 − 12
>
0,
or
|a|
>
2
√
(x/x0 )± = 13 (−a ± a2 − 12). The second derivative of
the potential energy, evaluated at these roots, is
" #
2 d U
U0
x
= 2 6
+ 2a
dx2 ±
x0
x0 ±
p
U0
2
.
= ±2 a − 12
x20
Thus, the “plus” root is a position of metastable
equilibrium (Equation 14-4), while the “minus” root
represents unstable equilibrium, (Equation 14-5). A
plot of the potential energy, which
√is a cubic, will
clarify these remarks. 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.)
750
Problem 39 Solution.
Problem
40. A cubical block rests on an inclined board with
two sides parallel to the direction of the incline.
The coefficient of static friction between block and
board is 0.95. If the inclination angle of the board
is increased, will the block first slide or first tip
over?
Solution
Problem 38 Solution.
Reference to the solution of Problem 57 shows that
the cube will tip over when θ > tan−1 (w/h) = 45◦ , but
220
CHAPTER 14
will slide when θ > tan−1 0.95 = 43.5◦ . It thus slides
before tipping.
horizontal cable that supports the boom.
Solution
Paired Problems
Problem
41. Figure 14-49 shows a 66-kg sign hung centered
from a uniform rod of mass 8.2 kg and length
2.3 m. At one end the rod is attached to the wall
by a pivot; at the other end it’s supported by a
cable that can withstand a maximum tension of
800 N. What is the minimum height h above the
pivot for anchoring the cable to the wall?
As in the previous problem, 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, T ℓ sin 50◦ = M gℓ cos 50◦ + mg 31 ℓ cos 50◦, or
T = (M + 31 m)g cot 50◦ = (2500 + 13 × 830)(9.8 N)×
cot 50◦ = 22.8 kN.
Solution
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. In equilibrium, Equation 14-2
calculated about the pivot (which does not contain the
unknown pivot force) yields 0 = T ℓ sin θ − M g 21 ℓ, or
T = M g/2 sin θ. But, tan θ = h/ℓ, so T = 21 M g×
p
1 + ℓ2 /h2 (use the identity 1 + cot2 θ = csc2 θ).
Therefore, the condition T ≤ p
Tmax implies 1 + ℓ2 ÷
2
(2Tmax /M g)2 − 1 =
h2 ≤ (2T
/M
g)
,
or
h
≥
ℓ/
pmax
2
2.3 m/ (1600/74.2×9.8) − 1 = 1.17 m.
figure 14-50 Problem 42 Solution.
Problem
43. A 4.2-kg plant hangs from the bracket shown in
Fig. 14-51. The bracket has a mass of 0.85 kg, and
its center of mass lies 9.0 cm from the wall. A
single screw holds the bracket to the wall, as
shown. Find the horizontal tension force in the
screw. Hint: Imagine that the bracket is slightly
loose and pivoting about its bottom end. Assume
the wall is frictionless.
Solution
figure 14-49 Problem 41 Solution.
Problem
42. A crane in a marble quarry is mounted on the
rock walls of the quarry and is supporting a
2500-kg slab of marble as shown in Fig. 14-50.
The center of mass of the 830-kg boom is located
one-third of the way from the pivot end of its
15 m length, as shown. Find the tension in the
We assume that the screw provides the total support
for the bracket, exerting a force with horizontal
component Fx (the reaction to which is a tensile force
on the screw) and vertical component Fy (the reaction
to which is a shearing force on the screw equal to the
total weight) as shown. A normal contact force
exerted by the wall could be distributed along the
bracket (e.g., by tightening the screw), but if we only
wish to estimate the minimum Fx , we may consider all
the normal force to act at the lowest point of contact
of the bracket, point O. Then Equation 14-2 about O
gives Fx (7.2 cm) = [(4.2 kg)(28 cm) + (0.85 kg)×
2
(9.0 cm)](9.8 m/s ) or Fx = 170 N.
CHAPTER 14
221
of the heaviest person who can safely ascend to
the top of the ladder? (The center of mass of the
Solution
figure 14-51 Problem 43 Solution.
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 α of
its length. Equilibrium conditions require:
P
Fx = 0 = f − Fwall ,
P
Fy = 0 = N − (mℓ + m)g,
P
( τ )A = 0 = Fwall ℓ cos θ − mℓ g 21 ℓ sin θ − mgαℓ sin θ.
The ladder will not slip if f ≤ µs N, which can be
written as
Problem
44. A 160-kg highway sign of uniform density is 2.3 m
wide and 1.4 m high. At one side it is secured to a
pole with a single bolt, mounted a distance d from
the top of the sign. The only other place where
the sign contacts the pole is at its bottom corner.
If the bolt can sustain a horizontal tension of
2100 N, what is the maximum permissible value
for the distance d?
Solution
The forces on the sign are in a similar configuration to
those in the preceding problem, as shown. (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 14-2 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.
Problem 44 Solution.
Problem
45. A 5.0-m-long ladder has mass 9.5 kg and is leaning
against a frictionless wall, making a 66◦ angle with
the horizontal. If the coefficient of friction between
the ladder and ground is 0.42, what is the mass
f = Fwall = ( 12 mℓ + αm)g tan θ ≤ µs N = µs (mℓ + m)g,
or
α ≤ [µs (mℓ + m) cot θ − 21 mℓ ]/m =
µs cot θ + (mℓ /m)(µs cot θ − 21 ).
(Here, we used the horizontal force equation to find f,
the torque equation to find Fwall , and the vertical
222
CHAPTER 14
force equation to find N.) For a person at the top of
the ladder, α = 1 and the condition for no slipping
becomes m ≤ mℓ (µs cot θ − 12 )/(1 − µs cot θ). With the
data given for the ladder (note that cot θ = tan 66◦ )m
≤ (9.5 kg)(0.42 tan 66◦ − 0.5)/(1 − 0.42 tan 66◦ ) =
74.3 kg.
Problem
48. An isosceles triangular block of mass m and height
h is in stable equilibrium, resting on its base on a
horizontal surface. How much energy does it take
to bring it to unstable equilibrium, resting on its
apex? Hint: Consult Example 10-3.
Solution
Example 10-3 shows that the CM of an isosceles
triangle is 31 the height from the base, or 23 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) = 31 mgh.
Supplementary Problems
Problem
Problem 45 Solution.
49. A uniform pole of mass M is at rest on an incline
of angle θ, secured by a horizontal rope as shown
in Fig. 14-52. What is the minimum coefficient of
friction that will keep the pole from slipping?
Problem
46. To what vertical height on the ladder in the
preceding problem could a 95-kg person reach
before the ladder starts to slip?
Solution
For a 95-kg person, the condition that the ladder not
slip gives α ≤ 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 previous problem.) The
maximum height above the ground is just αℓ cos θ =
(0.988)(5 m) sin 66◦ = 4.51 m.
Problem
47. A uniform, solid cube of mass m and side s is in
stable equilibrium when sitting on a level tabletop.
How much energy is required to bring it to an
unstable equilibrium where it’s resting on its
corner?
Solution
When balancing on a corner, the CM of a
uniform
cube (i.e., its center) √
is a distance
p
(s/2)2 + (s/2)2 + (s/2)2 = 3s/2 above the corner
resting on the tabletop. When in stable equilibrium,
the CM is s/2 above the tabletop. Thus, the√potential
energy difference is ∆U = mg ∆ycm = mgs( 3 − 1)/2.
figure 14-52 Problem 49 Solution.
Solution
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 ) as shown. Consideration
of Equation 14-2 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 14-1, and P
CCW-positive
component of EquationP14-2) are:
F|| = 0 =
f + T cos θ − mg
sin
θ,
F
=
0
=
N
− T sin θ −
⊥
P
mg cos θ, and ( τ )cm = 0 = T 12 ℓ cos θ − f 21 ℓ.
The solutions for the forces are f = 21 mg sin θ, T =
2
1
1
2 mg tan θ, and N = 2 mg(2 cos θ + sin θ/ cos θ),
subject to the condition that f ≤ µN. Therefore
sin θ ≤ µ(2 cos θ + sin2 θ/ cos θ) or µ ≥ tan θ ÷
(2 + tan2 θ). (By use of the identities sin 2θ =
CHAPTER 14
2 sin θ cos θ, cos 2θ = cos2 θ − sin2 θ, and sin2 θ =
1 − cos2 θ, this may be rewritten as µ ≥ sin 2θ ÷
(3 + cos 2θ).)
Problem
50. For what angle does the situation of Problem 49
require the greatest coefficient of friction?
Solution
The minimum coefficient of friction found in the
previous problem, µmin (θ) = tan θ/(2 + tan2 θ), is a
positive function which is zero at θ = 0◦ and 90◦ (the
limits of its domain). Therefore it has a maximum
2
2
when dµmin
√/dθ = 0, ◦or 2 + tan θ − 2 tan θ = 0, or
−1
θ = tan
2 = 54.7 .
Problem
51. One end of a board of negligible mass is attached
to a spring of spring constant k, while its other
end rests on a frictionless surface, as shown in
Fig. 14-53. If a mass m is placed on the middle of
the board, by how much will the spring compress?
Solution
If the frictionless surface is horizontal, the three forces
acting on the board are vertical,P
as shown. For
equilibrium, N + Fs = mg and ( τ )cm = 0. The
latter implies N = Fs , so Fs = 21 mg = k ∆x, and the
compression of the spring is ∆x = mg/2k.
223
to show that anyone can climb to the top if
µ ≥ tan θ, but that no one can if µ < 12 tan θ.
Solution
The condition for a person of mass m to be able to
reach the top of the ladder, which was found in the
solution to Problem 45, can be written as m ≤
mℓ (2µs − tan θ)/2(tan θ − µs ). Since m is positive, this
condition cannot be fulfilled if µs ≤ 12 tan θ, i.e., no
one can climb to the top without causing the ladder to
slip, whereas if µs = tan θ, the limit is ∞, so anyone
can climb to the top. (Note: the original expression,
α ≤ µs cot θ + (mℓ /m)(µs cos θ − 12 ), shows that α = 1
is allowed for any m, provided µs cot θ > 1.)
Problem
53. Figure 14-54 shows a wheel on a slope with
inclination angle θ = 20◦ , where the coefficient of
friction is adequate to prevent the wheel from
slipping; however, it might still roll. The wheel is
a uniform disk of mass 1.5 kg, and it is weighted
at one point on the rim with an additional 0.95-kg
mass m. Find the angle φ shown in the figure such
that the wheel will be in static equilibrium.
Solution
P
ThePwheel doesn’t slide if
F|| = 0 and it doesn’t roll
if ( τ )center = 0. (“||” means parallel to the incline,
and “center” is the center and CM of the wheel. These
are the only equilibrium conditions needed for the
solution of this problem.) With reference to the forces
shown added to Fig. 14-52, these conditions are
f = (M + m)g sin θ and f R = mgR cos φ. Together,
they imply f = mg cos φ = (M + m)g sin θ or φ =
cos−1 [(1 + M/m) sin θ] = cos−1 [(1 + 1.5/0.95)×
sin 20◦ ] = 28.1◦ .
figure 14-53 Problem 51 Solution.
figure 14-54 Problem 53 Solution.
Problem
52. A uniform ladder of mass m is leaning against a
frictionless vertical wall with which it makes an
angle θ. The coefficient of static friction at the
floor is µ. Find an expression for the maximum
mass for a person who is able to climb to the top
Problem
54. The wheel in Fig. 14-54 has mass M and is
weighted with an additional mass m. The slope
angle is θ. Show that static equilibrium is possible
only if m > M sin θ/(1 − sin θ).
224
CHAPTER 14
Solution
Static equilibrium is possible only if the CM of the
weighted wheel lies to the left of the point of contact
with the incline (or the y axis in the sketch). The
smallest additional mass would have to be placed
horizontally to the left of the center, as shown. Then
(M + m)xcm = M R sin θ − m(R − R sin θ) < 0, implies
m > M sin θ/(1 − sin θ).
(1 − xcm /ℓ). Equation 10-5 gives
Z ℓ
Z ℓ
xcm =
λx dx ÷
λ dx
0
=
Z
0
0
ℓ
(ax + bx2 ) dx ÷
Z
ℓ
(a + bx) dx
0
= (a 21 ℓ2 + b 31 ℓ3 )/(aℓ + b 12 ℓ2 )
= ℓ(3a + 2bℓ)/(6a + 3bℓ).
7
and note that
For the values given, xcm /ℓ = 12
1 2
M = aℓ + 2 bℓ = 4 kg. Thus, Fsr = M gxcm /ℓ =
22.9 N and Fsℓ = 16.3 N.
Problem
56. What horizontal force applied at its highest point
is necessary to keep a wheel of mass M from rolling
down a slope inclined at angle θ to the horizontal?
Solution
Problem 54 Solution.
Problem
55. A 2.0-m-long rod has a density described by λ =
a + bx, where λ is the density in kilograms per
2
meter of length, a = 1.0 kg/m, b = 1.0 kg/m , and
x is the distance in meters from the left end of the
rod. The rod rests horizontally with its ends each
supported by a scale. What do the two scales
Consider the conditions for static equilibrium of the
wheel, under the action of the forces shown (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). The torques about the
point of contact sum to zero, or Fapp R(1 + cos θ) =
M gR sin θ. Therefore Fapp = M g sin θ/(1 + cos θ) =
M g tan 21 θ.
Problem 56 Solution.
Problem 55 Solution.
Solution
The rod is in static equilibrium under P
the three
vertical forces shown in the sketch,
so
Fy = 0
P
implies Fsr + Fsr = M g, and ( τ )cm = 0 implies
Fsℓ xcm = Fsr (ℓ − xcm ). The solution for the left and
right scale forces is Fsℓ = M g − Fsr = M g×
Problem
57. A rectangular block twice as high as it is wide is
resting on a board. The coefficient of static
friction between board and block is 0.63. If the
board is tilted as shown in Fig. 14-55, will the
block first tip over or first begin sliding?
Solution
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
CHAPTER 14
mg sin θ > fsmax = µs N = µs mg cos θ, or tan θ > µs .
For µs = 0.63, this condition is θ > 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 θ > α, where α = tan−1 (w/h) is the diagonal
angle of the block. For h = 2w, α = tan−1 0.5 = 26.6◦.
Therefore, this block tips over before sliding.
225
Since tan α = ( 61 h)/( 14 h) = 32 > 0.63 = µs , this cone
will slide before tipping over.
Problem 59 Solution.
Problem
figure 14-55 Problem 57 Solution.
Problem
58. What condition on the coefficient of friction in the
preceding problem will cause the block to slide
before it tips?
Solution
If µs < tan α = 0.5, the block in the previous problem
will slide before tipping.
Problem
59. A uniform solid cone of height h and base
diameter 13 h is placed on the board of Fig. 14-55.
The coefficient of static friction between the cone
and incline is 0.63. As the slope of the board is
increased, will the cone first tip over or begin
sliding? Hint: Begin with an integration to find
the center of mass.
60. In Fig. 14-56 a uniform boom of mass 350 kg is
attached to a vertical wall by a pivot, and its far
end is supported by a cable as shown. If the cable
can withstand a maximum tension of 4.0 kN, what
is the maximum value for the angle φ?
Solution
The situation is like that in Problem 41, but with
different angles. In equilibrium, the net torque about
the pivot point is zero, so M g 2ℓ sin 40◦ =
T ℓ sin(40◦ − φ). (The exterior angle of a triangle
equals the sum of the two alternate interior angles.)
The condition T < Tmax means that sin(40◦ − φ) ≥
(M g/2 Tmax) sin 40◦ = 0.276, for the data given, so
φ ≤ 40◦ − sin−1 (0.276) = 24.0◦ is required.
Solution
The analysis for Problem 57 applies to the cone, where
α 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 10-3, 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 = ρπy 2 dx = (3M/h3 )x2 dx, where
ρ = M/ 13 πR2 h is the density (assumed constant) and
M is the mass of the cone. Then xcm = M −1 ∫ x dm =
(3/h3 ) ∫0h x3 dx = 34 h, or the CM is 14 h above the base.
figure 14-56 Problem 60 Solution.
Problem
61. An interstellar spacecraft from an advanced
civilization is hovering above Earth, as shown in
226
CHAPTER 14
Fig. 14-57. The ship consists of two pods of mass
m separated by a rigid shaft of negligible mass
that is one Earth radius (RE ) long. Find (a) the
magnitude and direction of the net gravitational
force on the ship and (b) the net torque about the
center of mass. (c) Show that the ship’s center of
gravity is displaced approximately 0.083RE from
its center of mass.
Solution
(a) The force on each pod (pod #1 over the North
pole) can be calculated from the law of universal
gravitation, Equation 9-1 (vectors in the x-y
Earth-centered frame shown on Fig. 14-57):
1
GME m
GME m
2
− √ ı̂ − √ ̂ .
F1 =
(−̂), F2 = √
(2RE )2
( 5RE )2
5
5
2
When we replace GME by gRE
, the net force becomes:
ı̂
2
1
F1 + F2 = −mg √ +
̂
+ √
4 5 5
5 5
= −mg(0.894ı̂ + 4.29̂) × 10−1 .
This has magnitude (4.38×10−1 )mg and is directed
11.8◦ clockwise from the negative y axis. (b) The
positions of the pods, relative to their CM midway
between them, are r′1 = − 12 RE ı̂ and r′2 = 12 RE ı̂.
Therefore, the net torque about the CM is:
r′1 × F1 + r′2 × F2
1
mg mg
1
= − RE ı̂ × −
̂ +
RE ı̂ × √ (−ı̂ − 2̂)
2
4
2
5 5
mgRE
mgRE
=
k̂ − √ k̂
8
5 5
= (3.56 × 10−2 )mgRE k̂ (out of page).
(c) The CG is positioned between the pods such that
the net gravitational torque about it is zero. If the
positions of the pods relative to the CG are −α1 RE ı̂
and α2 RE ı̂, respectively, where α1 + α2 = 1, then
mg mg
̂ + (α2 RE ı̂) × √ (−ı̂ − 2̂)
O = (−α1 RE ı̂) × −
4
5 5
α1
2α2
= mgRE
k̂.
− √
4
5 5
Solving
for α1 (or α2 ), we
√
√find α1 = 1 − α2 = 1 −
5 5α1 /8, or α1 = (1 + 5 5/8)−1 = 0.417 (and α2 =
0.583). Thus, the CG is (0.5 − α1 )RE = 0.0829RE
closer to pod #1 than the CM.
lfson0 20261 4A rt : 14 −figure
57.eps 14-57 Problem 61 Solution.
PART 1
CUMULATIVE PROBLEMS
CHAPTER 14
chapters in earlier parts, or they present special
challenges.
227
Fig. 1. Where does the apple hit the ground?
Neglect the effect of air resistance on either object
as well as any friction between apple and post.
Problem
1. A 170-g apple sits atop a 2.8-m-high post. A 45-g
arrow moving horizontally at 130 m/s passes
horizontally through the apple and strikes the
ground 36 m from the base of the post, as shown in
Solution
We can assume that momentum is conserved during
the inelastic collision (in a brief interval at t = 0)
between the arrow (m1 ) and the apple (m2 ). The
228
CHAPTER 14
v = 130 m/s
2.8 m
3°
figure 2 Cumulative Problem 2.
figure 1 Cumulative Problem 1.
velocities of the arrow before and after are specified to
be horizontal (in the x direction), therefore the
velocity of the apple (which was at rest before) is also
horizontal after the collision. Thus, m1 v1i,x =
m1 v1f,x + m2 v2f,x . Since both are moving horizontally
after the collision, the arrow and the apple will each
fall to the ground through the same vertical distance y
(equal
p to the height of the post), in the same time
t = 2y/g. However, they strike the ground at
different horizontal positions, which (in the absence of
air resistance) are x1 = v1f,x t and x2 = v2f,x t, relative
to the base of the post. Since x1 = 36 m, y = 2.8 m,
and v1i,x = 130 m/s are given, v1f,x and v2f,x can be
eliminated and a solution for x2 obtained:
px2 =
(m1 /m2 )(v1i,x − v1f,x )t =q
(m1 /m2 )(v1i,x 2y/g −
2
x1 ) = (45/170)[(130 m/s) 2(2.8 m)/(9.8 m/s ) −
36 m] = 16.5 m. (Alternatively, since external
horizontal forces are neglected, the center of mass of
the arrow/apple system moves horizontally at constant
speed until it reaches ground level, vcm,x = constant =
m1 v1i,x /(m1 + m2 ) (its value before the collision).
Then at ground level, m2 x2 = (m1 + m2 )xcm − m1 x1 =
(m1 + m2 )vcm,x t − m1 x1 = m1 (v1i,x t − x1 ), as before.)
Refer to relevant material in Chapters 4, 10, and 11 if
necessary.
Problem
2. A fire department’s tanker truck has a total mass of
21×103 kg, including 15×103 kg of water. Its
brakes fail at the top of a long 3◦ slope and it
begins to roll downward, starting from rest. In an
attempt to stop the truck, firefighters direct a
stream of water parallel to the slope, as shown in
Fig. 2, beginning as soon as the truck starts to roll.
The water leaves the 6.0-cm-diameter hose nozzle
at 50 m/s. Will the truck stop before it runs out of
water? If so, when? If not, what is the minimum
speed reached?
Solution
The fire truck operates like a rocket, with the stream
of water acting as expelled fuel. The rocket equation
(Equation 10-10a) can be modified, as follows, to
include the additional change in momentum due to the
external force of gravity.
In the text’s derivation, we now set ∆P = Fgrav ∆t
(instead of ∆P = 0), so M (v + ∆v) + ∆m(v − vex ) −
(M + ∆m)v = M ∆v − vex ∆m = Fgrav ∆t, and
Equation 10-10a becomes M (∆v/∆t) = −vex (∆M ÷
∆t) + Fgrav . This is a one-dimensional equation with
positive components in the direction of the thrust,
which is up the slope in this problem. The
gravitational force is opposite to the thrust, and in the
limit ∆m → 0, is just the downslope component of the
fire truck’s weight, or −M g sin θ. Therefore, the
equation of motion of the fire truck (which replaces
Equation 10-10b) is M (dv/dt) = −vex (dM/dt) −
M g sin θ.
This equation can be integrated (after
multiplication by dt/M ) to find the speed vf as a
function of time tf ,
Z vf
Z Mf
Z tf
dM
dv = −vex
vf − vi =
dt
− g sin θ
M
vi
Mi
ti
Mi
= vex ln
− g sin θ(tf − ti )
Mf
(this replaces Equation 10-11). For this fire truck,
Mi = 21×103 kg and vi = 0, at ti = 0, and Mf ≥
(21 − 15)×103 kg = 6×103 kg, which is the truck’s
mass without any water. Since vex is constant, so is
the mass rate of flow of water (see Equation 18-4b),
3
thus dM/dt = −ρw vex πR2 = −(103 kg/m )×
(50 m/s)π(3 cm)2 = −45π kg/s, where R = 3 cm is
the radius of the nozzle, and ρw the density of water.
Then the mass of the fire truck (at any time until the
water runs out) is Mf = Mi − (45π kg/s)tf . At the
start, the thrust, vex |dM/dt| = (50 m/s)×
(45π kg/s) = 7.07 kN, is less than the magnitude of
the downslope component of the weight,
2
Mi g sin θ = (21×103 kg)(9.8 m/s ) sin 3◦ = 10.8 kN,
so the truck begins to roll downslope. When the water
runs out, at tf = (15×103 kg)/(45π kg/s) = 106 s,
vf = vex ln(21/6) − g sin θ(106 s) = 8.22 m/s, which is
positive, upslope. Therefore, the firefighters succeed in
stopping the truck (only instantaneously, unless they
also stop the water and block the wheels) when vf =
0 = vex ln[Mi /(Mi − ρw vex πR2 t)] − (g sin θ)t, or, with
the above numbers, ln[1 − (6.732×10−3 s−1 )t] =
−(1.026×10−2 s−1 )t. A numerical solution for t,
CHAPTER 14
obtained with three iterations of Newton’s method or
widely available PC software, yields t = 88.8 s.
Problem
3. A block of mass m1 is attached to the axle of a
uniform solid cylinder of mass m2 and radius R by
massless strings. The two accelerate down a slope
that makes an angle θ with the horizontal, as shown
in Fig. 3. The cylinder rolls without slipping and
the block slides with coefficient of kinetic friction µ
between block and slope. The strings are attached
to the cylinder’s axle with frictionless loops so that
the cylinder can roll freely without any torque from
the string. Find an expression for the acceleration
of the pair, assuming that the string remains taut.
θ
229
motion of the cylinder is τP = (m2 g sin θ − T )R =
IP α = ( 32 m2 R2 )(a|| /R), or m2 g sin θ − T = 32 m2 a||
(see the parallel-axis theorem in Chapter 12 for IP ). If
T is eliminated by adding the two equations of motion,
one finds a|| = 2g[(m1 + m2 ) sin θ − µm1 cos θ]÷
(2m1 + 3m2 ).
Problem
4. A missile is launched from point A in Fig. 4,
heading for target C. The launch angle is 45.0, and
the launch speed is 1.90 km/s. An antimissile
defense system is located at point B, 310 km
downrange from the launch site. It fires an
interceptor rocket at a 65.0◦ launch angle, with the
intention of hitting the attacking missile when the
latter is 270 km downrange of its launch site.
(a) What should be the interceptor’s launch speed?
(b) How long after the launch of the attacking
missile should the interceptor be launched? (c) At
what altitude does the interception take place?
figure 3 Cumulative Problem 3.
C
270 km
310 km
A
B
figure 4 Cumulative Problem 4.
Solution
Cumulative Problem 3 Solution.
Solution
One must consider the forces on the block and the
cylinder, exerted by gravity, the inclined surface, and
the strings, as sketched (since the strings are assumed
massless and other forces are neglected). If the strings
remain taut, then the downslope acceleration, a|| , of
the block and the cylinder’s center of mass are the
same. If the string tension is parallel to the slope, the
normal force on the block is N1 = m1 g cos θ, and the
parallel component of its equation of motion is
Fnet,|| = m1 g sin θ + T − µm1 g cos θ = m1 a|| . When
rolling without slipping, the point of contact, P, of the
cylinder with the slope, is instantaneously at rest, so
the acceleration of its center of mass is a|| = αR (this
follows from v = ωR and ω = ωc = ωcm ; see
Section 12-5). Since only the string tension and
gravity exert torques about point P, the equation of
It is simplest to answer each part of this question
using a different coordinate system, so that equations
in the text can be used without modification. We are
given enough information to specify the location of the
point of interception; it is 270 km from point A, the
missile launch site, and 310 − 270 = 40 km from point
B, the interceptor rocket launch site.
(c) The trajectory of the missile (Equation 4-9
with origin at point A in Fig. 4) gives the altitude of
the point of interception when x = 270 m and other
data for the missile launch are substituted:
2
y = (270 km) tan 45◦ −
(0.0098 km/s )(270 km)2
2(1.90 km/s)2 cos2 45◦
= 72.1 km
The time of flight of the missile to the point of
interception is t = x/v0x = (270 km)÷
(190 cos 45◦ km/s) = 201 s.
(a) With altitude from part (c) above and the
other initial data for the interceptor rocket, its
trajectory equation (Equation 4-9 with origin at point
B and x axis leftward in Fig. 4) can be solved for the
230
CHAPTER 14
launch speed, as in Example 4-4. (The point of
interception has coordinates x = 40 km and
y = 72.1 km relative to point B.) Thus,
"
#1/2 2
(0.0098 km/s )
40 km
v0 =
2(40 tan 65◦ − 72.1) km
cos 65◦
= 1.79 km/s.
(b) The time of flight for the interceptor rocket is
t = x/v0x = (40 km)/(1.79 cos 65◦ km/s) = 52.8 s.
Therefore, it should be launched 201 − 52.8 = 148 s
after the missile.
Problem
5. A solid ball of radius R is set spinning with angular
speed ω about a horizontal axis. The ball is then
lowered vertically with negligible speed until it just
touches a horizontal surface and is released (Fig. 5).
If the coefficient of kinetic friction between the ball
and the surface is µ, find (a) the linear speed of the
ball once it achieves pure rolling motion and (b) the
distance it travels before its motion is pure rolling.
ω
v
figure 5 Cumulative Problem 5.
Solution
(a) While there is relative motion at the point of
contact between the ball and the horizontal surface,
the force of sliding friction (f = µN = µmg) slows the
ball’s rotation and accelerates its center of mass. The
equation for the latter is f = µmg = macm , or acm =
µg (positive to the right in the sketch and Fig. 5).
The equation for the former is τ = −f R = −µmgR =
Iα = (2mR2 /5)α, where α = −5µg/2R is the angular
acceleration about the horizontal axis through the
center of the ball (positive clockwise in the sketch and
Fig. 5). The accelerations are constant, so the
velocities are given by Equations 2-17 and 12-9 as
vcm = acm t = µgt and ω = ω0 + αt = ω0 − 5µgt/2R,
where the ball is released at t = 0 and the initial
velocities, v0 = 0 and ω0 , are given. The accelerated
motion continues until the point of contact is
instantaneously at rest (no more sliding friction). The
ball rolls without slipping thereafter, at a constant
velocity given by vcm = ωR. This occurs at a time t,
when (ω0 − 5µgt/2R)R = µgt, or t = 2ω0 R/7µg. Thus,
the final velocity is vcm = µgt = 2ω0 R/7. (b) The
distance traveled during this time is ∆x = 12 acm t2 =
1
2
2 2
2
2 (µg)(2ω0 R/7µg) = 2ω0 R /49µg (or ∆x = vcm /2acm
with the same result).
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