PSE9e ISM chapter12 final tủ tài liệu training

12
Static Equilibrium and Elasticity
CHAPTER OUTLINE
12.1

Analysis Model: Rigid Object in Equilibrium

12.2

More on the Center of Gravity

12.3

Examples of Rigid Objects in Static Equilibrium

12.4

Elastic Properties of Solids

* An asterisk indicates an item new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ12.1

Answer (b). The skyscraper is about 300 m tall. The gravitational
field (acceleration) is weaker at the top by about 900 parts in ten
million, by on the order of 10−4 times. The top half of the uniform
building is lighter than the bottom half by about (1/2)(10−4) times.
Relative to the center of mass at the geometric center, this effect
moves the center of gravity down, by about (1/2)(10−4)(150 m) ~ 10
mm.

OQ12.2

Answer (c). Net torque = (50 N)(2 m) − (200 N)(5 m) − (300 N)x = 0;
therefore, x = 3 m.

OQ12.3

Answer (a). Our theory of rotational motion does not contradict our
previous theory of translational motion. The center of mass of the
object moves as if the object were a particle, with all of the forces
applied there. This is true whether the object is starting to rotate or
not.

OQ12.4

Answer (d). In order for an object to be in equilibrium, it must be in
both translational equilibrium and rotational equilibrium. Thus, it

must meet two conditions of equilibrium, namely F net = 0 and

τ net = 0.
632

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Chapter 12

633

OQ12.5

Answer (b). The lower the center of gravity, the more stable the can.
In cases (a) and (c) the center of gravity is above the base by one-half
the height of the can. In case (b), the center of gravity is above the
base by only a bit more than one-quarter of the height of the can.

OQ12.6

Answer (d). Using the left end of the plank as a pivot and requiring
that ∑ τ = 0 gives

−mg ( 2.00 m ) + F2 ( 3.00 m ) = 0
or

(

)

2
2mg 2 ( 20.0 kg ) 9.80 m s
F2 =
=
= 131 N
3
3

OQ12.7

Answer: τD > τC > τE > τB > τA. The force exerts a counterclockwise
torque about pivot D. The line of action of the force passes through
C, so the torque about this axis is zero. In order of increasing negative
(clockwise) values come the torques about F, E and B essentially
together, and A.

OQ12.8

Answer (e). In the problems we study, the forces applied to the object
lie in a plane, and the axis we choose is a line perpendicular to this
plane, so it appears as a point on the force diagram. It can be chosen
anywhere. The algebra of solving for unknown forces is generally
easier if we choose the axis where some unknown forces are acting.

OQ12.9

(i) Answer (b). The extension is directly proportional to the original
dimension, according to F/A = Y∆L/Li.
(ii) Answer (e). Doubling the diameter quadruples the area to make
the extension four times smaller.

OQ12.10

Answer (b). Visualize the ax as like a balanced playground seesaw
with one large-mass person on one side, close to the fulcrum, and a
small-mass person far from the fulcrum on the other side. Different
masses are on the two sides of the center of mass. The mean position
of mass is not the median position.

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634

Static Equilibrium and Elasticity

ANSWERS TO CONCEPTUAL QUESTIONS
CQ12.1

The free-body diagram demonstrates that it is
necessary to have friction on the ground to
counterbalance the normal force of the wall and to
keep the base of the ladder from sliding. If there is
friction on the floor and on the wall, it is not
possible to determine whether the ladder will slip,
from the equilibrium conditions alone.

CQ12.2

A V-shaped boomerang, a barstool, an empty
coffee cup, a satellite dish, and a curving
plastic slide at the edge of a swimming pool
each have a center of mass that is not within
the bulk of the object.

CQ12.3

ANS. FIG. CQ12.1

(a)

Consider pushing up with one hand on one side of a steering
wheel and pulling down equally hard with the other hand on
the other side. A pair of equal-magnitude oppositely-directed
forces applied at different points is called a couple.

(b)

An object in free fall has a nonzero net force acting on it, but a
net torque of zero about its center of mass.

CQ12.4

When one is away from a wall and leans over, one’s back moves
backward so the body’s center of gravity stays over the feet. When
standing against a wall and leaning over, the wall prevents the
backside from moving backward, so the center of gravity shifts
forward. Once your CG is no longer over your feet, gravity
contributes to a nonzero net torque on your body and you begin to
rotate.

CQ12.5

If an object is suspended from some point and allowed to freely
rotate, the object’s weight will cause a torque about that point unless
the line of action of its weight passes through the point of support.
Suspend the plywood from the nail, and hang the plumb bob from
the nail. Trace on the plywood along the string of the plumb bob.
The plywood’s center of gravity is somewhere along that line. Now
suspend the plywood with the nail through a different point on the

plywood, not along the first line you drew. Again hang the plumb
bob from the nail and trace along the string. The center of gravity is
located halfway through the thickness of the plywood under the
intersection of the two lines you drew.

CQ12.6

She can be correct. Consider the case of a bridge supported at both
ends: the sum of the forces on the ends equals the total weight of the
bridge. If the dog stands on a relatively thick scale, the dog’s legs on

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Chapter 12

635

the ground might support more of its weight than its legs on the
scale. She can check for and if necessary correct for this error by
having the dog stand like a bridge with two legs on the scale and two
on a book of equal thickness—a physics textbook is a good choice.
CQ12.7

Yes, it can. Consider an object on a spring oscillating back and forth.
In the center of the motion both the sum of the torques and the sum
of the forces acting on the object are (separately) zero. Again, a
meteoroid flying freely through interstellar space feels essentially no
forces and keeps moving with constant velocity.

CQ12.8

Shear deformation. Its deformations are parallel to its surface.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 12.1
P12.1

Analysis Model: Rigid Object in Equilibrium

Use distances, angles, and forces as shown in ANS. FIG. P12.1. The
conditions of equilibrium are:

∑ Fy = 0 ⇒ Fy + Ry − Fg = 0
∑ Fx = 0 ⇒ Fx − Rx = 0
⎛ ⎞
∑ τ = 0 ⇒ Fy cosθ − Fg ⎜⎝ ⎟⎠ cosθ − Fx sin θ = 0
2

ANS. FIG. P12.1
P12.2

Take torques about P, as shown in ANS. FIG. P12.2.





⎡
⎤

⎡
⎤
∑ τ p = −nO ⎢ + d ⎥ + m1 g ⎢ + d ⎥ + mb gd − m2 gx = 0
⎣2
⎦
⎣2
⎦
We want to find x for which nO = 0:

x=

( m1 g + mb g ) d + m1 g 2 ( m1 + mb ) d + m1 2
m2 g

=

m2

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636

Static Equilibrium and Elasticity
For the values given:

x=

( m1 + mb ) d + m1 2

x=

( 5.00 kg + 3.00 kg )( 0.300 m ) + ( 5.00 kg ) 1.002 m

m2

15.0 m

x = 0.327 m

ANS. FIG. P12.2

The situation is impossible because x is larger than the remaining
portion of the beam, which is 0.200 m long.

Section 12.2
P12.3

More on the Center of Gravity

The coordinates of the center of gravity of
piece 1 are
x1 = 2.00 cm and y1 = 9.00 cm

The coordinates for piece 2 are
x2 = 8.00 cm and y 2 = 2.00 cm

The area of each piece is
A1 = 72.0 cm 2 and A2 = 32.0 cm 2

ANS. FIG. P12.3

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Chapter 12

637

And the mass of each piece is proportional to the area. Thus,

xCG =

2
2
∑ mi xi = ( 72.0 cm )( 2.00 cm ) + ( 32.0 cm )( 8.00 cm )
72.0 cm 2 + 32.0 cm 2
∑ mi

= 3.85 cm
and
2
2
mi y i ( 72.0 cm )( 9.00 cm ) + ( 32.0 cm )( 2.00 cm )
∑
y CG =
=
104 cm 2
∑ mi

= 6.85 cm
P12.4

The definition of the center of gravity as the average position of mass
in the set of objects will result in equations about x and y coordinates
that we can rearrange and solve to find where the last mass must be.




mi ri
∑
, r CG( ∑ mi ) = ∑ mi ri
From rCG =
∑ mi
We require the center of mass to be at the origin; this simplifies the
equation, leaving

∑ mi xi = 0 and ∑ mi yi = 0
To find the x coordinate, we substitute the known values:

( 5.00 kg )( 0 m ) + ( 3.00 kg )( 0 m )
+ ( 4.00 kg )( 3.00 m ) + ( 8.00 kg ) x = 0
Solving for x gives

x = –1.50 m.

Likewise, to find the y coordinate, we solve:

( 5.00 kg )( 0 m ) + ( 3.00 kg )( 4.00 m )

+ ( 4.00 kg )( 0 m ) + ( 8.00 kg ) y = 0
to find

y = –1.50 m

Therefore, a fourth mass of 8.00 kg should be located at

r4 = (–1.50ˆi − 1.50ˆj) m

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638
P12.5

Static Equilibrium and Elasticity
Let σ represent the mass-per-face area. (It would be equal to the
material’s density multiplied by the constant thickness of the wood.) A
( x − 3.00 )2 has
vertical strip at position x, with width dx and height
,
9
mass

σ ( x − 3.00 ) dx
dm =
9
2

The total mass is

M = ∫ dm =

σ ( x − 3 ) dx ⎛ σ ⎞
=⎜ ⎟
∫x=0
⎝ 9⎠
9

3.00

2

∫ (x

2

0

− 6x + 9 ) dx

3.00

⎤
⎛ σ ⎞ ⎡ x 6x
=⎜ ⎟⎢ −
+ 9x ⎥
⎝ 9⎠⎣ 3
2
⎦0

3

3.00

2

=σ

The x coordinate of the center of gravity is

xCG

∫ x dm =
=
M

=

1
9σ

3.00

∫
0

σ
σ x ( x − 3 ) dx =
9σ
2

3.00

1 ⎡ x 6x 9x ⎤
−
+
9 ⎢⎣ 4
3
2 ⎥⎦ 0
4

3

2

=

3.00

∫ (x
0

3

− 6x 2 + 9x ) dx

6.75 m
= 0.750 m
9.00

ANS. FIG. P12.5
P12.6

We can visualize this as a whole pizza with mass m1 and center of
gravity located at x1, plus a hole that has negative mass, –m2, with
center of gravity at x2:
xCG =

m1 x1 − m2 x2
m1 − m2

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Chapter 12

639

Call σ the mass of each unit of pizza area.
2

R
−R ⎞
σπ R 2 0 − σπ ⎛ ⎞ ⎛
⎝ 2⎠ ⎝ 2 ⎠
xCG =
2
R
σπ R 2 − σπ ⎛ ⎞
⎝ 2⎠

xCG =
P12.7

R/8
R
=
3/4
6

In a uniform gravitational field, the center of mass and center of
gravity of an object coincide. Thus, the center of gravity of the triangle
is located at x = 6.67 m, y = 2.33 m (see Example 9.12 on the center of
mass of a triangle in Chapter 9).
The coordinates of the center of gravity of the three-object system are
then:

xCG =
=
=
y CG =
=
=

Section 12.3
P12.8

∑ mi xi
∑ mi

(6.00 kg )( 5.50 m ) + ( 3.00 kg )(6.67 m ) + ( 5.00 kg )( −3.50 m )

( 6.00 + 3.00 + 5.00 ) kg

35.5 kg ⋅ m
= 2.54 m and
14.0 kg

∑ mi yi
∑ mi

(6.00 kg )(7.00 m ) + ( 3.00 kg )( 2.33 m ) + ( 5.00 kg )( +3.50 m )
14.0 kg
66.5 kg ⋅ m
= 4.75 m
14.0 kg

Examples of Rigid Objects in Static Equilibrium

The car’s weight is

Fg = mg = ( 1 500 kg ) ( 9.80 m/s 2 )
= 1 4700 N

Call F the force ofthe ground on each of the
front wheels and R the normal force on each of
the rear wheels. If we take torques around the
front axle, with counterclockwise in the picture

ANS. FIG. P12.8

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640

Static Equilibrium and Elasticity
chosen as positive, the equations are as follows:

∑ Fx = 0: 0 = 0
∑ Fy = 0: 2R – 14 700N + 2F = 0
∑ τ = 0:

+2R(3.00m) – (14 700 N)(1.20m) + 2F(0) = 0

The torque equation gives:

R=

17 640 N ⋅ m
= 2 940 N = 2.94 kN
6.00 m

Then, from the second force equation,

2(2.94 kN) – 14.7 kN + 2F = 0
P12.9

and

F = 4.41 kN

The second condition for equilibrium at the
pulley is
[1]

∑ τ = 0 = mg ( 3r ) − Tr
and from equilibrium at the truck, we
obtain

2T − Mg sin 45.0° = 0
Mg sin 45.0°
T=
2
(1 500 kg ) g sin 45.0°
=
2
= 530g N

ANS. FIG. P12.9

solving for the mass of the counterweight from [1] and substituting
gives

m=
P12.10

(a)

T 530g
=
= 177 kg

3g
3g

For rotational equilibrium of the lowest rod about its point of
support, ∑ τ = 0.
+ ( 12.0 g ) g ( 3.00 cm ) − m1 g ( 4.00 cm ) = 0

which gives

m1 = 9.00 g
(b)

For the middle rod,

+ m2 g ( 2.00 cm ) − ( 12.0 g + 9.0 g ) g ( 5.00 cm ) = 0
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Chapter 12

641

which gives

m2 = 52.5 g
(c)

For the top rod,

( 52.5 g + 12.0 g + 9.0 g ) g ( 4.00 cm ) − m g (6.00 cm ) = 0

3

which gives

m3 = 49.0 g
P12.11

Since the beam is in equilibrium, we choose the center as our pivot
point and require that

∑ τ center = −FSam ( 2.80 m ) + FJoe ( 1.80 m ) = 0
or

FJoe = 1.56FSam

[1]

∑ Fy = 0 ⇒ FSam + FJoe = 450 N

[2]

Also,

Substitute equation [1] into [2] to get the following:
FSam + 1.56FSam = 450 N or FSam =

450 N
= 176 N
2.56

Then, equation [1] yields FJoe = 1.56 ( 176 N ) = 274 N

Sam exerts an upward force of 176 N.
Joe exerts an upward force of 274 N.
P12.12

(a)

To find U, measure distances and forces from point A. Then,
balancing torques,

( 0.750 m )U = ( 29.4 N )( 2.25 )
(b)

U = 88.2 N

To find D, measure distances and forces from point B. Then,
balancing torques,

( 0.750 m ) D = ( 1.50 m )( 29.4 N )

D = 58.8 N

Also, notice that U = D + Fg , so ∑ Fy = 0.
P12.13

(a)

The wall is frictionless, but it does exert a horizontal normal force,
nw.

∑ Fx = f − nw = 0
∑ Fy = ng − 800 N − 500 N = 0
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642

Static Equilibrium and Elasticity
Taking torques about an axis at the foot of the ladder,

( 800 N )( 4.00 m ) sin 30.0° + ( 500 N )( 7.50 m ) sin 30.0°
−nw ( 15.0 cm ) cos 30.0° = 0
Solving the torque equation,

nw =

[( 4.00 m )( 800 N ) + (7.50 m )( 500 N )] tan 30.0° = 268 N
15.0 m

Next substitute this value into the Fx equation to find

f = nw = 268 N in the positive x
direction.
Solving the equation ∑ Fy = 0,

ng = 1 300 N in the positive
y direction
(b)

Refer to ANS. FIG. P12.13(b) on the right. In this case, the torque
equation ∑ τ A = 0 gives:

( 9.00 m ) ( 800 N ) sin 30.0°
+ ( 7.50 m ) ( 500 N ) sin 30.0°
− ( 15.0 m ) ( nw ) sin 60.0° = 0

ANS. FIG. P12.13(b)

or nw = 421 N
Since f = nw = 421 N and f = fmax = µng , we find

µ=
P12.14

(a)

fmax
421 N
=
= 0.324
ng
1 300 N

The wall is frictionless, but it does exert a horizontal normal force,
nw.

∑ Fx = f − nw = 0

[1]

∑ Fy = ng − m1 g − m2 g = 0

[2]

⎛ L⎞

∑ τ A = −m1 g ⎜⎝ ⎟⎠ cosθ − m2 gx cosθ + nw L sin θ = 0
2
From the torque equation,

⎡1
⎤
⎛ x⎞
nw = ⎢ m1 g + ⎜ ⎟ m2 g ⎥ cot θ
⎝ L⎠
⎣2
⎦
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Chapter 12

643

⎡1
⎤
⎛ x⎞
Then, from equation [1]: f = nw = ⎢ m1 g + ⎜ ⎟ m2 g ⎥ cot θ
⎝ L⎠

⎣2
⎦

and from equation [2]: ng =

(b)

(a)

1

+ m2 ) g

Refer to ANS. FIG. P12.13(b) above. If the ladder is on the verge
of slipping when x = d, then

µ=
P12.15

(m

f

x=d

ng

=

( m1 / 2 + m2 d/L ) cot θ

m1 + m2

Vertical forces on one-half of the chain are

Te sin 42.0° = 20.0 N
Te = 29.9 N
(b)

Horizontal forces on one-half of the chain are

Te cos 42.0° = Tm
Tm = 22.2 N
P12.16

(a)

See the force diagram shown in ANS.
FIG. P12.16.

(b)

Select a pivot point where an unknown
force acts so that the force has no torque
about that point. Picking the lower end of
the beam eliminates torque from the
normal force, n, and the friction force, f.

∑ τ lower end = 0:

ANS. FIG. P12.16

⎛L
⎞
0 + 0 − mg ⎜ cosθ ⎟ + T ( L sin θ ) = 0
⎝2
⎠
or

T=
(c)

mg ⎛ cos θ ⎞ mg
cot θ
⎜
⎟=
2 ⎝ sin θ ⎠
2

From the first condition for equilibrium,

∑ Fx = 0 ⇒ − T + µs n = 0 or T = µs n

[1]

∑ Fy = 0 ⇒ n − mg = 0 or n = mg

[2]

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644

Static Equilibrium and Elasticity
Substitute equation [2] into [1] to obtain T = µ s mg .
(d) Equate the results of parts (b) and (c) to obtain µ s =

1
cot θ .
2

This result is valid only at the critical angle θ where the beam is
on the verge of slipping (i.e., where fs = (fs)max is valid).
(e)

The ladder slips. When the base of the ladder is moved to the
left, the angle θ decreases. According to the result in part (b), the
tension T increases. This requires a larger friction force to balance
T, but the static friction force is already at its maximum value in
ANS. FIG. P12.16.

P12.17

(a)

In Figure P12.17, let the “Single point of contact” be point P, the
force the nail exerts on the hammer claws be R, the mass of the
hammer (1.00 kg) be M, and the normal force exerted on the
hammer at point P be n, while the horizontal static friction
exerted by the surface on the hammer at P be f.

Taking moments about P,

( R sin 30.0°) 0 + ( R cos 30.0°) ( 5.00 cm ) + Mg ( 0)
− ( 150 N ) ( 30.0 cm ) = 0
R = 1 039.2 N = 1.04 kN

The force exerted by the hammer on the nail is equal in
magnitude and opposite in direction:

1.04 kN at 60° upward and to the right
(b)

From the first condition for equilibrium,

∑ Fx = f − R sin 30.0° + 150 N = 0 → f = 370 N

∑ Fy = n − Mg − R cos 30.0° = 0

→ n = ( 1.00 kg ) ( 9.80 m/s 2 ) + ( 1 040 N ) cos 30.0° = 910 N

Fsurface = 370ˆi + 910ˆj N

(

P12.18

(a)

)

See the force diagram in
ANS. FIG. P12.18.

(b) The mass M of the beam is
20.0 kg. We consider the
torques acting on the beam,
about an axis perpendicular

ANS. FIG. P12.18

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Chapter 12

645

to the page and through the left end of the horizontal beam.

∑ τ = + (T sin 30.0°) d − Mgd = 0
T=

(c)

Mg
196 N
=
= 392 N
sin 30.0° sin 30.0°

From ∑ Fx = 0, H − T cos 30.0° = 0,
or

H = ( 392 N ) cos 30.0° = 339 N to the right

(d) From ∑ Fy = 0, V + T sin 30.0° − 196 N = 0 ,
or
(e)

V = 196 N − ( 392 N ) sin 30.0° = 0

From the same free-body diagram with the axis chosen at the
right-hand end, we write

∑ τ = H(0) − Vd + T(0) + 196N(0) = 0, so
(f)

From ∑ Fy = 0, V + T sin 30.0° − 196 N = 0 ,
or

(g)

P12.19

T = 0 + 196 N/sin 30.0° = 392 N

From ∑ Fx = 0, H − T cos 30.0° = 0,
or

(h)

V=0

H = ( 392 N ) cos 30.0° = 339 N to the right

The two solutions agree precisely. They are equally accurate.

The bridge has mass M = 2 000 kg and the knight and horse have mass
m = 1 000 kg. Relative to the hinge end of the bridge, the cable is
attached horizontally out a distance x = ( 5.00 m cos 20.0° = 4.70 m and

)

vertically up a distance y = ( 5.00 m ) sin 20.0° = 1.71 m. The cable then
makes the following angle with the vertical wall:

⎡ ( 4.70 ) m ⎤
θ = tan −1 ⎢
⎥ = 24.5°
⎢⎣ 12.0 − 1.71 m ⎥⎦
Call the force components at the hinge Hx (to the right) and Hy
(upward).
(a)

Take torques about the hinge end of the bridge:

H x ( 0 ) + H y ( 0 ) − Mg ( 4.00 m ) cos 20.0°

− (T sin 24.5° ) ( 1.71 m ) + (T cos 24.5° ) ( 4.70 m )
− mg ( 7.00 m ) cos 20.0° = 0

which yields T = 27.7 kN
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646

Static Equilibrium and Elasticity
(b)

∑ Fx = 0 ⇒ H x − T sin 24.5° = 0,
or

(c)

H x = ( 27.7 kN ) sin 24.5° = 11.5 kN ( right )

∑ Fy = 0 ⇒ H y − Mg + T cos 24.5° − mg = 0
Thus,

H y = ( M + m) g − ( 27.7 kN ) cos 24.5° = −4.19 kN
= 4.19 kN down
P12.20

(a)

No time interval. The horse’s feet lose contact with the
drawbridge as soon as it begins to move.
From the result of (b) below, the tangential acceleration of the
point where the horse stands is

at = α r = ( 1.73 rad/s 2 )( 7.00 m ) = 12.1 m/s 2
which has a vertical component at cos 20.0° = 11.4 m/s 2 , greater
than the acceleration of gravity.
(b)

Assuming that the bridge does fall from
under the horse, its angular acceleration will
be caused by torque from the weight of the
bridge—if the bridge does not fall out from
under the horse, there will be additional
torque from the weight of the knight and
horse, and the acceleration will be greater.

∑ τ = Iα

ANS. FIG. P12.20(b)

1
3g cos 20.0ο
⎛ ⎞
Mg ⎜ ⎟ cosθ 0 = M2α → α =
= 1.73 rad/s
⎝ 2⎠
3
2 ( 8.00 m )
As cited in part (a), this results in the bridge falling out from
under the horse, so our assumption was justified.
(c)

Because there is no friction at the hinge, the bridge-Earth system
is isolated, so mechanical energy is conserved. When the bridge
strikes the wall:

Ki + U i = K f + U f
Mgh =

1 2
⎛ ⎞
Iω → Mg ⎜ ⎟ ( 1 + sin 20.0ο ) =
⎝ 2⎠
2

1⎛ 1
2⎞
2
⎜⎝ M ⎟⎠ ω
2 3

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Chapter 12

647

which gives

ω=

3g ( 1 + sin 20.0ο )
8.00 m

= 2.22 rad/s

(d) The tangential acceleration of the center of
mass of the bridge is

1
at = α = ( 8.0 m )( 1.73 rad s 2 )
2
2
= 6.92 m s 2

which is directed 20.0° below the
horizontal. By Newton’s second law:

ANS. FIG. P12.20(d)

∑ Fx = Max

H x = ( 2 000 kg ) ( 6.92 m/s 2 ) sin 20.0°
= 4.72 kN

∑ Fy = May
H y − Mg = May

H y = ( 2 000 kg ) ( 9.80 m/s 2 )

+ ( 2 000 kg ) ( −6.92 m/s 2 ) cos 20.0°

= 6.62 kN

(

)

The force at the hinge is 4.72 ˆi + 6.62 ˆj  kN.
(e)

When the bridge strikes the wall, Hx = 0 and
the hinge supplies a vertical centripetal force:

∑ Fy = May

2

⎞
⎛
H y = Mg + Mω 2 = M ⎜ g + ω 2 ⎟
⎝
2
2⎠
H y − Mg = May = Mω 2

ANS. FIG. P12.20(e)

2 8.00 m ⎞
⎛
H y = ( 2 000 kg ) ⎜ 9.80 m/s 2 + ( 2.22 rad/s )

⎟
⎝
2 ⎠

H y = 59.1 kJ

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648
P12.21

Static Equilibrium and Elasticity
Call the required force F, with components Fx = F cos15.0° and
Fy = −F sin 15.0°, transmitted to the center of the wheel by the handles.

ANS. FIG. P12.21
Just as the wheel leaves the ground, the ground exerts no force on it.

∑ Fx = 0:

F cos15.0° − nx = 0

[1]

∑ Fy = 0:

− F sin 15.0° − 400 N + ny = 0

[2]

Take torques about its contact point with the brick. The needed
distances are seen to be:

b = R − 8.00 cm = ( 20.0 − 8.00 ) cm = 12.0 cm
a = R 2 − b 2 = ( 20.0 cm ) − ( 8.00 cm ) = 16.0 cm
2

(a)

∑ τ = 0:

2

− Fxb + Fy a + ( 400 N ) a = 0, or

F ⎡⎣ − ( 12.0 cm ) cos15.0° + ( 16.0 cm ) sin 15.0° ⎤⎦ + ( 400 N ) ( 16.0 cm ) = 0
so
(b)

F=

6 400 N ⋅ cm
= 859 N
7.45 cm

Then, using equations [1] and [2],

nx = ( 859 N ) cos15.0° = 830 N and
ny = 400 N + ( 859 N ) sin 15.0° = 622 N

n = nx2 + ny2 = 1.04 kN
⎛ ny ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ( 0.749 ) = 36.9° to the left and upward
⎝ nx ⎠

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Chapter 12
P12.22

649

Call the required force F, with components Fx = F cos θ and
Fy = −F sin θ , transmitted to the center of the wheel by the handles.

ANS. FIG. P12.22
Just as the wheel leaves the ground, the ground exerts no force on it.

∑ Fx = 0:
∑ Fy = 0:

F cosθ − nx = 0

[1]

− F sin θ − mg + ny = 0

[2]

Take torques about its contact point with the brick. The needed
distances are seen to be:
b=R−h
a = R 2 − ( R − h ) = 2Rh − h2
2

(a)

∑ τ = 0:

− Fxb + Fy a + mga = 0, or

F ⎡⎣ −b cos θ + a sin θ ⎤⎦ + mga = 0
→F=
(b)

mga
mg 2Rh − h2
=
b cos θ − a sin θ ( R − h ) cos θ − 2Rh − h2 sin θ

Then, using equations [1] and [2],
nx = F cos θ =

mg 2Rh − h2 cos θ

( R − h) cosθ −

2Rh − h2 sin θ

⎡
⎤
2Rh − h2 cos θ
⎥
and ny = F sin θ + mg = mg ⎢1 +
⎢ ( R − h ) cos θ − 2Rh − h2 sin θ ⎥
⎣
⎦

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650
P12.23

Static Equilibrium and Elasticity
When x = xmin , the rod is on the verge
of slipping, so

f = ( f s )max = µ s n = 0.50n
From ∑ Fx = 0, n − T cos 37° = 0
or

ANS. FIG. P12.23

n = 0.799T

Thus,

f = 0.50 ( 0.799T ) = 0.399T

From

∑ Fy = 0, f + T sin 37° − 2Fg = 0,

or

0.399T + 0.602T − 2Fg = 0, giving T = 2.00Fg

Using ∑ τ = 0 for an axis perpendicular to the page and through the
left end of the beam gives

( )

−Fg ⋅ xmin − Fg ( 2.0 m ) + ⎡ 2Fg sin 37° ⎤ ( 4.0 m ) = 0
⎣
⎦
which reduces to xmin = 2.81 m
P12.24

(a)

The force diagram is shown in ANS. FIG. P12.24.

(b)

From ∑ Fy = 0 ⇒ nF − 120 N − mmonkey g = 0

(

)

nF = 120 N + ( 10.0 kg ) 9.80 m s 2 = 218 N

ANS. FIG. P12.24

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Chapter 12
(c)

651

When x = 2L/3, we consider the bottom end of the ladder as our
pivot and obtain
bottom = 0:
∑ τ end

⎛L
⎞
⎛2L
⎞
− ( 120 N ) ⎜ cos60.0°⎟ − ( 98.0 N ) ⎜
cos60.0°⎟
⎝ 2
⎠
⎝ 3
⎠
+ nW ( L sin 60.0° ) = 0

⎡60.0 N + ( 196 3 ) N ⎤⎦ cos60.0°
or nW = ⎣
= 72.4 N
sin 60.0°
Then, ∑ Fx = 0 ⇒ T − nW = 0

or

T = nW = 72.4 N

(d) When the rope is ready to break, T = nW = 80.0 N. Then
∑ τ bottom = 0 yields
end

⎛L
⎞
− ( 120 N ) ⎜ cos60.0°⎟ − ( 98.0 N ) x cos60.0°
⎝2
⎠

+ ( 80.0 N )( L sin 60.0° ) = 0

or

(e)

P12.25

x=

[( 80.0 N ) sin 60.0° − (60.0 N ) cos60.0°] L

( 98.0 N ) cos60.0°
= 0.802 L = 0.802 ( 3.00 m ) = 2.41 m

If the horizontal surface were rough and the rope removed, a
horizontal static friction force directed toward the wall would act
on the bottom end of the ladder. Otherwise, the analysis would be
much as what is done above. The maximum distance the monkey
could climb would correspond to the condition that the friction
force have its maximum value, µ s nF , so you would need to know
the coefficient of static friction between the ladder and the floor to
solve part (d).

Consider the torques about an axis perpendicular to the page and
through the left end of the plank. ∑ τ = 0 gives

− ( 700 N )( 0.500 m ) − ( 294 N )( 1.00 m )

+ (T1 sin 40.0° ) ( 2.00 m ) = 0

or

T1 = 501 N

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652

Static Equilibrium and Elasticity
Then, ∑ Fx = 0 gives
−T3 + T1 cos 40.0° = 0

or

T3 = ( 501 N ) cos 40.0° = 384 N

From ∑ Fy = 0,
T2 − 994 N + T1 sin 40.0° = 0,

or

T2 = 994 N − ( 501 N ) sin 40.0° = 672 N

ANS. FIG. P12.25

Section 12.4
P12.26

Elastic Properties of Solids

Count the wires. If they are wrapped together so that all support
nearly equal stress, the number should be

20.0 kN
= 100
0.200 kN
Since cross-sectional area is proportional to diameter squared, the

diameter of the cable will be

(1 mm )
P12.27

We use B = −

100 ~ 1 cm

ΔPVi
ΔP
=−
.
ΔV / Vi
ΔV

(a)

1.13 × 108 N m 2 ) ( 1 m 3 )
(
ΔPVi
ΔV = −
=−
= −0.053 8 m 3
10
2
B
0.21 × 10 N m

(b)

The quantity of water with mass 1.03 × 103 kg occupies volume at
the bottom: 1 m 3 − 0.053 8 m 3 = 0.946 m 3 .
So its density is

1.03 × 103 kg = 1.09 × 103 kg m 3
0.946 m 3

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Chapter 12
(c)

P12.28

(a)

653

With only a 5% volume change in this extreme case, liquid
water is indeed nearly incompressible.
We find the maximum force from the equation for stress:
stress =

F
F
= 2
A πr

⎛ d⎞
F = ( stress ) π ⎜ ⎟
⎝ 2⎠

2

⎛ 2.50 × 10−2 m ⎞
F = ( 1.50 × 10 N m ) π ⎜
⎟⎠
2
⎝
8

2

2

F = 73.6 kN
(b)

From the definition of Young’s modulus,

stress = Y ( strain ) =

( stress ) Li
ΔL =
Y

P12.29

YΔL
Li

(1.50 × 10
=

8

N m 2 ) ( 0.250 m )

1.50 × 1010 N m 2

= 2.50 mm

From the defining equation for the shear modulus, we find Δx as

(

)

5.00 × 10−3 m ( 20.0 N )
hf
Δx =
=
= 2.38 × 10−5 m
6
2
−4
2
SA

3.0 × 10 N m 14.0 × 10 m

(

or
P12.30

)(

)

Δx = 2.38 × 10−2 mm

The definition of Young’s modulus, Y =

stress
, means that Y is the
strain

slope of the graph:

Y=
P12.31

(a)

300 × 106 N m 2
= 1.0 × 1011 N m 2
0.003

From ANS. FIG. P12.31(a),

F =σA

= ( 4.00 × 108 N/m )
2
× ⎡π ( 0.500 × 10−2 m ) ⎤
⎣
⎦

= 3.14 × 10 4 N

ANS. FIG. P12.31(a)

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654

Static Equilibrium and Elasticity
(b)

Now the area of the molecular layers
sliding over each other is the curved
lateral surface area of the cylinder being
punched out, a cylinder of radius 0.500 cm
and height 0.500 cm. So,

F =σA
ANS. FIG. P12.31(b)

= σ (h)(2π r)

= ( 4.00 × 108 N/m )( 2π )( 0.500 × 10−2 m )
× ( 0.500 × 10−2 m )

= 6.28 × 10 4 N
P12.32

Let V represent the original volume. Then, 0.090 0V is the change in
volume that would occur if the block cracked open. Imagine squeezing
the ice, with unstressed volume 1.09V, back down to its previous
volume, so ΔV = –0.090 0V. According to the definition of the bulk
modulus as given in the chapter text, we have

ΔP = −

B(ΔV)
Vi

( 2.00 × 10   N m )( − 0.090 0V)
=−
9

2

1.09V

= 1.65 × 108 N/m 2
P12.33

Young’s Modulus is given by Y =

F/A
.
ΔL/Li

The load force is
F = (200 kg)(9.80 m/s2) = 1 960 N.
so
P12.34

ΔL =

(1 960 N)(4.00 m)(1 000 mm/m)
FLi
=
= 4.90 mm
AY ( 0.200 × 10 –4 m 2 ) ( 8.00 × 1010 N m 2 )

Part of the load force extends the cable and part compresses the
column by the same distance Δ:

F=

YA AA Δ Ys As Δ
+
A
s

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Chapter 12

655

from which we obtain

Δ =
=

F
YA AA /  A + Ys As /  s

(7 × 10 )π ( 0.162 4
10

8 500 N

2

− 0.161 4 )/4 ( 3.25 ) + 20 × 1010 π ( 0.012 7 ) /4 ( 5.75 )
2

2

= 8.60 × 10−4 m
P12.35

Let the 3.00-kg mass be mass #1, with the 5.00-kg mass, mass # 2.
Applying Newton’s second law to each mass gives

and

m1 a = T − m1 g

[1]

m2 a = m2 g − T

[2]

where T is the tension in the wire.
Solving equation [1] for the acceleration gives
a=

T
−g
m1

and substituting this into equation [2] yields
m2
T − m2 g = m2 g − T
m1

Solving for the tension T gives

)

(

2
2m1m2 g 2 ( 3.00 kg ) ( 5.00 kg ) 9.80 m s
T=
=
= 36.8 N
m2 + m1
8.00 kg

From the definition of Young’s modulus, Y =

FLi

A ( ΔL )

, the elongation of

the wire is:

ΔL =

TLi
( 36.8 N )( 2.00 m )
=
2
YA ( 2.00 × 1011 N m 2 ) π ( 2.00 × 10−3 m )

= 0.029 2 mm
P12.36

A particle under a net force model:
F =

m v f − vi
Δt

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656

Static Equilibrium and Elasticity
Hence,

F =

30.0 kg −10.0 m s − 20.0 m s
0.110 s

= 8.18 × 103 N

By Newton’s third law, this is also the magnitude of the average force
exerted on the spike by the hammer during the blow. Thus, the stress
in the spike is
Stress =

F
8.18 × 103 N
=

= 1.97 × 107 N m 2
A π ( 0.023 0 m )2 / 4

and the strain is
strain =

stress 1.97 × 107 N m 2
=
= 9.85 × 10−5
10
2
Y
20.0 × 10 N m

Additional Problems
P12.37

Let nA and nB be the normal forces at the points of support. Then, from
the translational equilibrium equation in the y direction, we have

∑ Fy = 0:

nA + nB − ( 8.00 × 10 4 kg ) g – ( 3.00 × 10 4 kg ) g = 0

Choosing the axis at point A, we find, from the condition for rotational
equilibrium:

∑ τ = 0:

– ( 3.00 × 10 4 kg ) (15.0 m)g – ( 8.00 × 10 4 kg ) (25.0 m)g

+ nB (50.0 m) = 0

We can solve the torque equation directly to find

⎡⎣( 3.00 × 10 4 kg )( 15.0 m ) + ( 8.00 × 10 4 kg )( 25.0 m )⎤⎦ (9.80 m/s 2 )
nB =
50.0 m
5
= 4.80 × 10 N
Then the force equation gives

nA = (8.00 × 10 4 kg + 3.00 × 10 4 kg)(9.80 m/s 2 ) − 4.80 × 105 N
= 5.98 × 105 N

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