Chapter 12 Static Equilibrium; Elasticity and Fracture
65. We consider the right half of the bridge in the diagram in the book. We divide it into two segments
of length d1 and 1 d 2 , and let the mass of those two segments be M. Since the roadway is uniform,
2
the mass of each segment will be in proportion to the length of the θ3 θ2
section, as follows. FT3 FT2
m2 = d1 → d2 = 2 m2
m1 d1 m1
22
d1
The net horizontal force on the right tower is to be 0. From the force FN
diagram for the tower, we write this.
FT3 sinθ3 = FT2 sinθ2
From the force diagram for each segment of the cable, write Newton’s second law for both the
vertical and horizontal directions. θ2
Right segment:
FT2 right segment, d1
∑ Fx = FT1 cosθ1 − FT2 sinθ2 = 0 →
FT1 cosθ1 = FT2 sinθ2
∑ Fy = FT2 cosθ2 − FT1 sinθ1 − m1g = 0 → m1g FT1 θ1
m1g = FT2 cosθ2 − FT1 sinθ1
Left segment: left segment, d2 2 θ3
FT4 FT3
∑ Fx = FT3 sinθ3 − FT4 = 0 → FT3 sinθ3 = FT4
∑ Fy = FT3 cosθ3 − m2g = 0 →
m2g = FT3 cosθ3
We manipulate the relationships to solve for the ratio of the m2g
masses, which will give the ratio of the lengths.
FT1 cosθ1 = FT2 sinθ2 → FT1 = FT2 sin θ 2
cosθ1
m1g = FT2 cosθ2 − FT1 sin θ1 = FT2 cosθ2 − FT2 sin θ 2 sin θ1 = FT2 ⎛ cos θ 2 − sin θ 2 sin θ1 ⎞
cosθ1 ⎜ cosθ1 ⎟
⎝ ⎠
FT3 sinθ3 = FT2 sinθ2 → FT3 = FT2 sin θ 2 → m2 g = FT3 cosθ3 = FT2 sin θ 2 cosθ3
sin θ3 sin θ3
d2 m2 m2 g 2FT2 sin θ 2 cosθ3 2 sinθ2 cosθ3 cosθ1
d1 m1 m1g sin θ3
= 2 = 2 = = (cosθ2 cosθ1 − sinθ2 sinθ1 ) sinθ3
⎛ sin θ 2 ⎞
FT2 ⎜ cos θ2 − cosθ1 sin θ1 ⎟⎠
⎝
= 2 sinθ2 cosθ1 = 2 sin 60° cos19° = 3.821 ≈ 3.8
cos 79° tan 66°
cos (θ1 + θ2 ) tanθ3
66. The radius of the wire can be determined from the θ θ
relationship between stress and strain, expressed by Eq. 12-5. FT mg FT
F = E Δl → A = Fl 0 = πr2 → r= 1 F l0
A l0 EΔl π E Δl
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
Use the free-body diagram for the point of connection of the mass to the wire to determine the
tension force in the wire.
∑ ( )Fy = 2FT sinθ − mg = 0 s2
→ FT = mg = (25 kg) 9.80 m = 589.2 N
2 sin θ
2 sin12o
The fractional change in the length of the wire can be found from the geometry l0/2
θ
of the problem.
l 0 + Δl
cosθ = l0 2 → Δl = 1 −1= 1 −1= 2.234 ×10−2 2
l 0 + Δl l0 cosθ cos12o
2
Thus the radius is
1 FT l 0 1 589.2 N 1 = 3.5 ×10−4 m
π E Δl
( )r = =
π 70 ×109 N m2 2.234 ×10−2
67. The airplane is in equilibrium, and so the net force in d FL
each direction and the net torque are all equal to zero.
First write Newton’s second law for both the horizontal FD h1 h2
and vertical directions, to find the values of the forces.
FT
∑ Fx = FD − FT = 0 → FD = FT = 5.0 ×105 N
∑ Fy = FL − mg = 0 mg
( ) ( )FL = mg = 7.7 ×104 kg 9.80 m s2 = 7.546 ×105 N
Calculate the torques about the CM, calling counterclockwise torques positive.
∑τ = FLd − FDh1 − FT h2 = 0
( ) ( ( ) )h1
= FLd − FT h2 = 7.546 ×105 N (3.2 m) − 5.0 ×105 N (1.6 m) = 3.2 m
FD
5.0 ×105 N
68. Draw a free-body diagram for half of the cable. Write Newton’s FT1
second law for both the vertical and horizontal directions, with the 56°
net force equal to 0 in each direction.
∑ Fy = FT1 sin 56° − 1 mg = 0 → FT1 = 1 mg = 0.603 mg FT2
2 2 sin 56°
∑ Fx = FT2 − FT1 cos 56° = 0 → 1 mg
2
FT2 = 0.603mg (cos 56°) = 0.337 mg
So the results are:
(a) FT2 = 0.34mg
(b) FT1 = 0.60mg
(c) The direction of the tension force is tangent to the cable at all points on the cable. Thus the
direction of the tension force is horizontal at the lowest point , and is
56° above the horizontal at the attachment point .
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400
Chapter 12 Static Equilibrium; Elasticity and Fracture
69. (a) For the extreme case of the beam being ready to tip, CA BD
there would be no normal force at point A from the
support. Use the free-body diagram to write the 3.0 m 7.0 m 5.0 m 5.0 m
equation of rotational equilibrium under that
condition to find the weight of the person, with FA mBg FB W
FA = 0. Take torques about the location of support
B, and call counterclockwise torques positive. W is the weight of the person.
∑τ = mBg (5.0 m) − W (5.0 m) = 0 →
W = mB g = 650 N
(b) With the person standing at point D, we have already assumed that FA = 0 . The net force in
the vertical direction must also be zero.
∑ Fy = FA + FB − mB g − W = 0 → FB = mB g + W = 650 N + 650 N = 1300 N
(c) Now the person moves to a different spot, so the B 2.0 m D
free-body diagram changes as shown. Again use the C A
net torque about support B and then use the net
vertical force. 3.0 m 7.0 m 5.0 m
∑τ = mBg (5.0 m) − W (2.0 m) − FA (12.0 m) = 0 FA mBg FB W
FA = mB g (5.0 m) − W (2.0 m) = (650 N) (3.0 m)
12.0 m 12.0 m
= 162.5 N ≈ 160 N
∑ Fy = FA + FB − mB g − W = 0 → FB = mB g + W − FA = 1300 N − 160 N = 1140 N
(d) Again the person moves to a different spot, so the C A2.0 m B D
free-body diagram changes again as shown. Again
use the net torque about support B and then use the 3.0 m 5.0 m 5.0 m FB
net vertical force.
∑τ = mB g (5.0 m) +W (10.0 m) − FA (12.0 m) = 0 FA W mBg
FA = mB g (5.0 m) +W (10.0 m) = (650 N) (5.0 m) + (650 N) (10.0 m) = 810 N
12.0 m 12.0 m
∑ Fy = FA + FB − mB g − W = 0 → FB = mB g + W − FA = 1300 N − 810 N = 490 N
70. If the block is on the verge of tipping, the normal force will be acting at the L /2 F
lower right corner of the block, as shown in the free-body diagram. The mg
block will begin to rotate when the torque caused by the pulling force is h
larger than the torque caused by gravity. For the block to be able to slide,
the pulling force must be as large as the maximum static frictional force. Ffr
Write the equations of equilibrium for forces in the x and y directions and FN
for torque with the conditions as stated above.
∑ Fy = FN − mg = 0 → FN = mg
∑ Fx = F − Ffr = 0 → F = Ffr = μsFN = μsmg
∑τ = mg l − Fh = 0 → mgl = Fh = μsmgh
2 2
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
Solve for the coefficient of friction in this limiting case, to find μs = l.
2h
(a) If μs < l 2h , then sliding will happen before tipping.
(b) If μs > l 2h , then tipping will happen before sliding.
71. The limiting condition for the safety of the painter is the Fleft = 0 Fright
tension in the ropes. The ropes can only exert an upward
tension on the scaffold. The tension will be least in the 1.0 m
rope that is farther from the painter. The mass of the pail is
1.0 m 1.0 m 1.0 m 2.0 m x
mp, the mass of the scaffold is m, and the mass of the Mg
mpg mg
painter is M .
Find the distance to the right that the painter can walk before the tension in the left rope becomes
zero. Take torques about the point where the right-side rope is attached to the scaffold, so that its
value need not be known. Take counterclockwise torques as positive.
∑τ = mg (2.0 m) + mp g (3.0 m) − Mgx = 0 →
x = m (2.0 m) + mp (3.0 m) = (25kg) (2.0 m) + (4.0 kg) (3.0 m) = 0.9538 m ≈ 0.95 m
M
65.0 kg
The painter can walk to within 5 cm of the right edge of the scaffold.
Now find the distance to the left that the painter can walk Fleft Fright = 0
before the tension in the right rope becomes zero. Take
torques about the point where the left-side tension is 1.0 m
attached to the scaffold, so that its value need not be
known. Take counterclockwise torques as positive. x 1.0 m 1.0 m 2.0 m 1.0 m
∑τ = Mgx − mp g (1.0 m) − mg (2.0 m) = 0 → M g mpg mg
x = m (2.0 m) + mp (1.0 m) = (25kg) (2.0 m) + (4.0 kg) (1.0 m) = 0.8308 m ≈ 0.83 m
M
65.0 kg
The painter can walk to within 17 cm of the left edge of the scaffold. We found that both ends are
dangerous.
72. (a) The man is in equilibrium, so the net force and the net torque on him must 1 mg
be zero. We use half of his weight, and then consider the force just on 2
one hand and one foot, considering him to be symmetric. Take torques
about the point where the foot touches the ground, with counterclockwise
as positive. Fh d1 d2 Ff
∑τ = 1 mgd2 − Fh ( d1 + d2 ) = 0
2
( )Fh
= mgd2 (68 kg) 9.80 m s2 (0.95 m) = 231N ≈ 230 N
2 (1.37 m)
2 (d1 + d2 ) =
(b) Use Newton’s second law for vertical forces to find the force on the feet.
∑ Fy = 2Fh + 2Ff − mg = 0
( )Ff
= 1 mg − Fh = 1 ( 68 kg ) 9.80 m s2 − 231N = 103 N ≈ 100 N
2 2
The value of 100 N has 2 significant figures.
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402
Chapter 12 Static Equilibrium; Elasticity and Fracture
73. The force on the sphere from each plane is a normal force, and so is perpendicular
to the plane at the point of contact. Use Newton’s second law in both the θL θR
FL
horizontal and vertical directions to determine the magnitudes of the forces. FR
mg
∑ Fx = FL sinθL − FR sinθR = 0 → FR = FL sin θ L = FL sin 67°
sin θ R sin 32°
∑ Fy = FL cosθL + FR cosθR − mg = 0 → FL ⎛ cos 67° + sin 67° cos 32° ⎞ = mg
⎜⎝ sin 32° ⎟⎠
( )FL
= mg = (23kg) 9.80 m s2 = 120.9 N ≈ 120 N
⎝⎜⎛ cos 67° + sin 67° cos 32° ⎞⎟⎠ ⎛⎜⎝ cos 67° + sin 67° cos 32° ⎠⎟⎞
sin 32° sin 32°
FR = FL sin 67° = (120.9 N ) sin 67° = 210.0 N ≈ 210 N
sin 32° sin 32°
74. See the free-body diagram. The ball is at rest, and so is in equilibrium. Write FA
Newton’s second law for the horizontal and vertical directions, and solve for the
forces.
∑ Fhoriz = FB sinθB − FA sinθA = 0 → FB = FA sin θ A θA
sin θ B
θB FB
mg
∑ Fvert = FA cosθA − FB cosθB − mg = 0 → FA cosθA = FB cosθB + mg →
FA cosθA = FA sin θ A cosθB + mg → FA ⎛ cosθA − sin θ A cosθB ⎞ = mg →
sin θ B ⎜ sin θ B ⎟
⎝ ⎠
sin θ B sin θ B sin 53°
sinθB − sinθA sin 31°
sin (θB − θA )
( )FA
= mg (cosθA cosθB ) = mg = (15.0 kg) 9.80 m s2
= 228N ≈ 230N
FB = FA sin θ A = ( 228 N) sin 22° = 107 N ≈ 110 N
sin θ B sin 53°
75. Assume a constant acceleration as the person is brought to rest, with up as the positive Fsnow
direction. Use Eq. 2-12c to find the acceleration. From the acceleration, find the average
force of the snow on the person, and compare the force per area to the strength of body
tissue.
v2 = v02 − 2a ( x − x0 ) → a = v2 − v02 = 0 − (55 m s)2 = 1513m s2 mg
2 (−1.0 m)
2 ( x − x0 )
( )F ma (75 kg) 1513m s2 = 3.78 ×105 N m2 < Tissue strength = 5 ×105 N m2
A= A =
0.30m2
Since the average force on the person is less than the strength of body tissue, the person may escape
serious injury. Certain parts of the body, such as the legs if landing feet first, may get more than the
average force, though, and so still sustain injury.
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403
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
76. The mass can be calculated from the equation for the relationship between stress and strain. The
force causing the strain is the weight of the mass suspended from the wire. Use Eq. 12-4.
( )Δl 2
( ) ( )l 0
1 F mg → m = EA Δl = 200 ×109 N m2 π 1.15 ×10−3 m 0.030 = 25 kg
=E A = EA g l0 9.80 m s2 100
77. To find the normal force exerted on the road by the trailer tires, take the mg
torques about point B, with counterclockwise torques as positive.
FB
∑τ = mg (5.5 m) − FA (8.0 m) = 0 →
( )FA ⎛ 5.5 m ⎞ s2 ⎛ 5.5 m ⎞
= mg ⎝⎜ 8.0 m ⎟⎠ = (2500 kg) 9.80 m ⎝⎜ 8.0 m ⎠⎟ = 16, 844 N FA
2.5 m 5.5 m
≈ 1.7 ×104 N
The net force in the vertical direction must be zero.
∑ Fy = FB + FA − mg = 0 →
( )FB = mg − FA = (2500 kg) 9.80 m s2 − 16,844 N = 7656 N ≈ 7.7 ×103 N
78. The number of supports can be found from the compressive strength of the wood. Since the wood
will be oriented longitudinally, the stress will be parallel to the grain.
Compressive Strength Load force on supports Weight of roof
Safety Factor = Area of supports = (# supports) (area per support)
(# supports) = Weight of roof Safety Factor
Compressive Strength
(area per support )
( ) ( )1.36 ×104 kg 9.80 m s2
( )= (0.040 m) (0.090 m)
12 = 12.69 supports
35 ×106 N m2
Since there are to be more than 12 supports, and to have the same number of supports on each side,
there will be 14 supports, or 7 supports on each side . That means there will be 6 support-to-support
spans, each of which would be given by Spacing = 10.0 m = 1.66 m gap .
6 gaps
79. The tension in the string when it breaks is found from the ultimate strength of nylon under tension,
from Table 12-2. θ θ
FT FT
FT = Tensile Strength →
A
FT = A(Tensile Strength) mg
( ) ( )= ⎦⎤2
π ⎡⎣ 1 1.15 ×10−3 m 500 ×106 N m2 = 519.3 N
2
From the force diagram for the box, we calculate the angle of the rope relative to the horizontal from
Newton’s second law in the vertical direction. Note that since the tension is the same throughout the
string, the angles must be the same so that the object does not accelerate horizontally.
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404
Chapter 12 Static Equilibrium; Elasticity and Fracture
∑ Fy = 2FT sinθ − mg = 0 → 3−h θ
h
( )θ s2 θ
= sin−1 mg = sin−1 (25kg) 9.80 m = 13.64° 2.0 m
2FT 2 (519.3 N)
To find the height above the ground, consider the second
diagram.
tan θ = 3.00 m − h → h = 3.00 m − 2.00 m (tanθ ) = 3.00 m − 2.00 m (tan13.64°) = 2.5 m
2.00 m
80. See the free-body diagram. Assume that the ladder is just ready to slip, so
the force of static friction is Ffr = μFN. The ladder is of length l, and so FW θ
d1 = 1 l sinθ , d2 = 3 l sinθ , and d3 =l cos θ . The ladder is in Mg
2 4 mg
equilibrium, so the net vertical and horizontal forces are 0, and the net d3
torque is 0. We express those three equilibrium conditions, along with the FG y
friction condition. Take torques about the point where the ladder rests on
the ground, calling clockwise torques positive.
∑ Fvert = FGy − mg − Mg = 0 → FGy = (m + M ) g
∑ Fhoriz = FGx − FW = 0 → FGx = FW FG x
∑τ = mgd1 + Mgd2 − FWd3 = 0 → FW = mgd1 + Mgd2 d1
d3
d2
Ffr = μFN → FGx = μFGy
These four equations may be solved for the coefficient of friction.
mgd1 + Mgd2
μ = FGx = FW = d3 = md1 + Md2 = m( 1 l sin θ ) + M ( 3 l sinθ )
FGy 2 4
(m + M ) g (m + M ) g d3 (m + M )
(l cosθ ) (m + M)
= ( 1 m + 3 M ) tanθ = [ 1 (16.0 kg) + 3 (76.0 kg )] tan 20.0° = 0.257
2 4 2 4
(m + M ) (92.0 kg)
81. The maximum compressive force in a column will occur at the bottom. The bottom layer supports
the entire weight of the column, and so the compressive force on that layer is mg . For the column to
be on the verge of buckling, the weight divided by the area of the column will be the compressive
strength of the material. The mass of the column is its volume (area x height) times its density.
mg = Compressive Strength = hAρ g → h = Compressive Strength
A A ρg
Note that the area of the column cancels out of the expression, and so the height does not depend on
the cross-sectional area of the column.
Compressive Strength 500 ×106 N m2
ρg 7.8 ×103 kg m3 9.80 m s2
( )( )(a) hsteel = = = 6500 m
Compressive Strength 170 ×106 N m2
ρg 2.7 ×103 kg m3 9.80 m s2
( )( )(b) hgranite = = = 6400 m
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
82. See the free–body diagram. Let M represent the mass of the train, and FA FB
m represent the mass of the bridge. Write the equilibrium conditions x
for torques, taken about the left end, and for vertical forces. These Mg mg
two equations can be solved for the forces. Take counterclockwise l
torques as positive. Note that the position of the train is given by
x = vt.
∑τ = Mgx + mg ( 1 l ) − FBl =0 →
2
FB = ⎛ Mg x + 1 mg ⎞ = ⎛ Mgv t + 1 mg ⎞
⎜⎝ l 2 ⎠⎟ ⎜⎝ l 2 ⎠⎟
⎛ kg) s2 ⎛ h) ⎛ 1m s ⎞⎞ ⎞
⎜ ⎝⎜ 3.6 km ⎟⎠ ⎠⎟ ⎟
( )⎜
(95000 9.80 m (80.0 km h ⎟
⎟
( )= ⎜ ⎝ s2
⎜
280 m t + 1 (23000 kg) 9.80 m
2
⎜⎟
⎝⎠
( ) ( )= 7.388 ×104 N s t + 1.127 ×105 N ≈ 7.4 ×104 N s t + 1.1×105 N
∑ Fvert = FA + FB − Mg − mg = 0 →
( ) ( ) ( )FA = ( M + m) g − FB = 1.18 ×105 kg 9.80 m s2 − ⎡⎣ 7.388 ×104 N s t + 1.127 ×105 N⎦⎤
( ) ( )= − 7.388 ×104 N s t + 1.044 ×106 N ≈ − 7.4 ×104 N s t + 1.0 ×106 N
83. Since the backpack is midway between the two trees, the angles in the θ0 θ0
free-body diagram are equal. Write Newton’s second law for the vertical
direction for the point at which the backpack is attached to the cord, with FT0 FT0
mg
the weight of the backpack being the original downward vertical force.
∑ Fy = 2FT0 sinθ0 − mg = 0 → FT0 = mg
2 sinθ0
Now assume the bear pulls down with an additional force, Fbear . The force equation would be
modified as follows.
∑ Fy = 2FT final sin θfinal − mg − Fbear = 0 →
( )Fbear = 2FT final sin θfinal − mg = 2 ⎛ mg ⎞
2 FT 0 sin θfinal − mg = 4 ⎝⎜ 2 sinθ0 ⎟⎠ sin θfinal − mg
⎛ 2 sin θfinal − 1⎟⎞ ⎛ 2 sin 27° 1⎞⎟⎠
⎜ sin θ0 ⎠ ⎜⎝ sin15°
( )= ⎝
mg = (23.0 kg) 9.80 m s2 − = 565.3 N ≈ 570 N
84. (a) See the free-body diagram. To find the tension in Fhinge
the wire, take torques about the left edge of the vert
beam, with counterclockwise as positive. The net FT
torque must be 0 for the beam to be in equilibrium. Fhinge θ
horiz
∑τ = mgx + Mg ( 1 l ) − FT sinθl =0 → x mg Mg
2
l
g (2mx + Ml ) mg Mg
FT = = l sin θ x + 2 sin θ
2l sinθ
We see that the tension force is linear in x.
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406
Chapter 12 Static Equilibrium; Elasticity and Fracture
(b) Write the equilibrium condition for vertical and horizontal forces.
∑ Fx = Fhinge − FT cosθ = 0 → Fhinge = FT cosθ = g (2mx + Ml ) cosθ = g (2mx + M l )
horiz
horiz 2l sinθ 2l tanθ
∑ Fy = Fhinge + FT sinθ − (m + M ) g = 0 →
vert
Fhinge = (m + M ) g − FT sinθ = (m + M ) g − g (2mx + Ml ) sinθ = mg ⎛⎜⎝1 − x ⎞ + 1 Mg
l ⎟⎠ 2
vert 2l sinθ
85. Draw a free-body diagram for one of the beams. By Newton’s third Fbeam
law, if the right beam pushes down on the left beam, then the left beam l sinθ
pushes up on the right beam. But the geometry is symmetric for the FN
two beams, and so the beam contact force must be horizontal. For the mg
θ Ffr
beam to be in equilibrium, FN = mg and so Ffr = μsFN = μmg is the l cosθ
maximum friction force. Take torques about the top of the beam, so
that Fbeam exerts no torque. Let clockwise torques be positive.
∑τ = FNl cosθ − mg ( 1 l ) cos θ − Ffrl sin θ = 0 →
2
θ = tan −1 1 = tan−1 2 ( 1 = 45°
2μs
0.5)
86. Take torques about the elbow joint. Let clockwise torques be positive. Since the arm is in
equilibrium, the total torque will be 0.
∑τ = (2.0 kg) g (0.15 m) + (35kg) g (0.35 m) − Fmax (0.050 m) sin105° = 0 →
Fmax = (2.0 kg) g (0.15 m) + (35kg) g (0.35 m) = 2547 N ≈ 2500 N
(0.050 m) sin105°
87. (a) Use the free-body diagram in the textbook. To find the magnitude of FM , take torques about an
axis through point S and perpendicular to the paper. The upper body is in equilibrium, so the
net torque must be 0. Take clockwise torques as positive.
∑τ = [wT (0.36 m) + wA (0.48 m) + wH (0.72 m)]cos 30° − FM (0.48m) sin12° = 0 →
FM = [wT (0.36 m) + wA (0.48 m) + wH (0.72 m)]cos 30°
(0.48m) sin12°
= w[(0.46) (0.36 m) + (0.12) (0.48 m) + (0.07) (0.72 m)]cos 30° = 2.374w ≈ 2.4w
(0.48m) sin12°
(b) Write equilibrium conditions for the horizontal and vertical forces.
Use those conditions to solve for the components of FV, and then θ −φ
find the magnitude and direction. Note the free–body diagram for
determining the components of FM. The two dashed lines are FM φ
parallel, and so both make an angle of θ with the heavy line
representing the back. θ
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
∑ ( )Fhoriz = FV horiz − FM cos 30° − 12° = 0 →
FV horiz = FM cos18° = (2.374w) cos18° = 2.258w
∑ ( )Fvert = FV vert − FM sin 30° − 12° − wT − wA − wH = 0 →
FV vert = FM sin18° + wT + wA + wH = (2.374w) sin18° + 0.65w = 1.384w
FV = F + F =2 2 (2.258w)2 + (1.384w)2 = 2.648w ≈ 2.6w
V horiz V vert
θV = tan−1 FV vert = tan−1 1.384w = 31.51° ≈ 32° above the horizontal
FV horiz 2.258w
88. We are given that rod AB is under a compressive force F. Analyze the forces on FAC
A
the pin at point A. See the first free-body diagram. Write equilibrium equations
FAB
for the horizontal and vertical directions.
∑ Fhoriz = FAD cos 45° − FAB = 0 → FAD = FAB = 2F, in tension 45°
cos 45° FAD
∑ Fvert = FAC − FAD sin 45° = 0 →
FAC = FAD sin 45° = 2F 2 = F, in compression
r
By symmetry, the other outer forces must all be the same magnitude as FAB , and the other diagonal
force must be the same magnitude as FAB .
FAC = FAB = FBD = FCD = F, in compression ; FAD = FBC = 2F, in tension
89. (a) The fractional decrease in the rod’s length is the strain Use Eq. 12-5. The force applied is the
weight of the man.
( )Δl
( ) ( )l 0
= F mg (65 kg) 9.80 m s2 m2 = 4.506 ×10−8 = 4.5 ×10−6 %
AE = πr2E = π (0.15)2 200 ×109 N
(b) The fractional change is the same for the atoms as for the macroscopic material. Let d represent
the interatomic spacing.
Δd = Δl = 4.506 ×10−8 →
d0 l0
( ) ( ) ( )Δd = 4.506 ×10−8 d0 = 4.506 ×10−8 2.0 ×10−10 m = 9.0 ×10−18 m 3.25 m θ
3.0 m
90. (a) See the free-body diagram for the system, showing forces on the engine F
and the forces at the point on the rope where the mechanic is pulling (the FT
point of analysis). Let m represent the mass of the engine. The fact that FT
the engine was raised a half-meter means that the part of the rope from
the tree branch to the mechanic is 3.25 m, as well as the part from the FT
mechanic to the bumper. From the free-body diagram for the engine, we
know that the tension in the rope is equal to the weight of the engine. mg
Use this, along with the equations of equilibrium at the point where the
mechanic is pulling, to find the pulling force by the mechanic.
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408
Chapter 12 Static Equilibrium; Elasticity and Fracture
Angle: θ = cos−1 3.0 m = 22.62°
3.25 m
∑Engine: Fy = FT − mg = 0 → FT = mg
∑Point: Fx = F − 2FT sinθ = 0 →
( )F = 2mg sinθ = 2 (280 kg) 9.80 m s2 sin 22.62° = 2111N ≈ 2100 N
( )(b) s2
Mechanical advantage = Load force = mg = (280 kg) 9.80 m = 1.3 N
Applied force F
2111 N
91. Consider the free-body diagram for the box. The box is assumed to F 1.0 m
be in equilibrium, but just on the verge of both sliding and tipping. h 2.0 m
Since it is on the verge of sliding, the static frictional force is at its h
maximum value. Use the equations of equilibrium. Take torques FN
about the lower right corner where the box touches the floor, and W
take clockwise torques as positive. We also assume that the box is
just barely tipped up on its corner, so that the forces are still parallel Ffr
and perpendicular to the edges of the box.
∑ Fy = FN − W = 0 → FN = W
∑ Fx = F − Ffr = 0 → F = Ffr = μW = (0.60) (250 N) = 150 N
∑τ = Fh − W (0.5m) = 0 → h = (0.5 m) W = (0.5 m) 250 N = 0.83 m
F 150 N
92. See the free-body diagram. Take torques about the pivot 0.75 m 0.75 m 1.50 m Mg
point, with clockwise torques as positive. The plank is in
equilibrium. Let m represent the mass of the plank, and M θ Fpivot mg
represent the mass of the person. The minimum nail force Fnails
would occur if there was no normal force pushing up on the
left end of the board.
∑τ = mg (0.75m) cosθ + Mg (2.25m) cosθ
− Fnails (0.75 m) cosθ = 0 →
Fnails = mg (0.75m) + Mg ( 2.25 m ) = mg + 3Mg
( 0.75 m )
( )= (45kg + 3(65 kg)) 9.80 m s2 = 2352 N ≈ 2400 N
93. (a) Note that since the friction is static friction, we may NOT θ l
FT
use Ffr = μFN. It could be that Ffr < μFN. So, we must h
Ffr
determine Ffr by the equilibrium equations. Take an axis of
rotation to be out of the paper, through the point of contact of
the rope with the wall. Then neither FT nor Ffr can cause
any torque. The torque equilibrium equation is as follows. FN
mg
FNh = mgr0 → FN = mgr0
h
Take the sum of the forces in the horizontal direction.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
FN = FT sinθ → FT = FN = mgr0
sin θ h sinθ
Take the sum of the forces in the vertical direction.
FT cosθ + Ffr = mg →
Ffr = mg − FT cosθ = mg − mgr0 cosθ = mg ⎛⎝⎜1 − r0 cot θ ⎞
h sinθ h ⎟⎠
(b) Since the sphere is on the verge of slipping, we know that Ffr = μFN.
Ffr = μFN → mg ⎝⎜⎛1 − r0 cot θ ⎞ = μ mgr0 → ⎛ h − cot θ ⎞ = μ = h − cot θ
h ⎟⎠ h ⎝⎜ r0 ⎟⎠ r0
94. There are upward forces at each support (points A and D) and a B
downward applied force at point C. To find the angles of members
AB and BD, see the free-body diagram for the whole truss.
θA = tan −1 6.0 = 56.3° ; θB = tan−1 6.0 = 45° FA 6.0 m FD
4.0 6.0 θD
Write the conditions for equilibrium for the entire truss by θA 4.0 m C 6.0 m D
A F
considering vertical forces and the torques about point A. Let
clockwise torques be positive.
∑ Fvert = FA + FD − F = 0
∑τ = F (4.0 m) − FD (10.0 m) = 0 → FD = F ⎛ 4.0 ⎞ = (12, 000 N ) ⎛ 4.0 ⎞ = 4800 N
⎝⎜ 10.0 ⎠⎟ ⎜⎝ 10.0 ⎟⎠
FA = F − FD = 12, 000 N − 4800 N = 7200 N FA
Analyze the forces on the pin at point A. See the second free-body diagram. A
Write equilibrium equations for the horizontal and vertical directions. θA FAB
∑ Fvert = FA − FAB sinθA = 0 → FAC
FAB = FA = 7200 N = 8654 N ≈ 8700 N, compression
sinθA sin 56.3°
∑ Fhoriz = FAC − FAB cosθA = 0 →
FAC = FAB cosθA = (8654 N) cos 56.3° = 4802 N ≈ 4800 N, tension
Analyze the forces on the pin at point C. See the third free-body diagram. FBC
Write equilibrium equations for the horizontal and vertical directions.
FAC C FCD
∑ Fvert = FBC − F = 0 → FBC = F = 12, 000 N, tension
∑ Fhoriz = FCD − FAC = 0 → FCD = FAC = 4800 N, tension F
Analyze the forces on the pin at point D. See the fourth free-body diagram. FD
Write the equilibrium equation for the horizontal direction. FCD
∑ Fvert = FBD cosθD − FCD = 0 → D
FBD = FCD = 4800 N = 6788 N ≈ 6800 N, compression θD
cosθD cos 45° FBD
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410
Chapter 12 Static Equilibrium; Elasticity and Fracture
95. (a) See the free-body diagram. We write the equilibrium conditions for
horizontal and vertical forces, and for rotation. We also assume that FWy
both static frictional forces are at their maximum values. Take
clockwise torques as positive. We solve for the smallest angle that FWx
makes the ladder be in equilibrium.
∑ Fhoriz = FGx − FWx = 0 → FGx = FWx l sinθ
∑ Fvert = FGy + FWy − mg = 0 → FGy + FWy = mg FGy
mg
∑τ = (mg 1 l cosθ )− FWxl sinθ − FWyl cosθ = 0
2 θ FGx
l cosθ
FGx = μG FGy ; FWy = μWFWx
Substitute the first equation above into the fourth equation, and simplify
the third equation, to give this set of equations.
( )FGy + FWy = mg ; mg = 2 FWx tanθ + FWy ; FWx = μGFGy ; FWy = μWFWx
Substitute the third equation into the second and fourth equations.
( )FGy + FWy = mg ; mg = 2 μGFGy tanθ + FWy ; FWy = μWμGFGy
Substitute the third equation into the first two equations.
( )FGy + μWμGFGy = mg ; mg = 2 μGFGy tanθ + μWμGFGy
Now equate the two expressions for mg, and simplify.
( )FGy + μWμGFGy = 2 μGFGy tanθ + μWμGFGy → tan θ min = 1 − μWμG
2μG
(b) For a frictional wall: θmin = tan−1 1 − μWμG = tan−1 1 − (0.40)2 = 46.4° ≈ 46°
2μG 2 ( 0.40)
For a frictionless wall: θmin = tan−1 1 − μWμG = tan−1 1− (0)2 = 51.3° ≈ 51°
2μG 2 ( 0.40)
% diff = ⎛ 51.3° − 46.4° ⎞⎠⎟100 = 10.6% ≈ 11%
⎝⎜ 46.4°
96. (a) See the free-body diagram for the Tyrolean 12.5 m x 12.5 m θ
traverse technique. We analyze the point on FT
the rope that is at the bottom of the “sag.” To θ
include the safety factor, the tension must be no FT mg
more than 2900 N.
∑ Fvert = 2FT sinθ − mg = 0 →
( )θmin s2
= sin−1 mg = sin −1 (75kg) 9.80 m = 7.280°
2FT 2 (2900 N)
max
tan θ min = xmin → xmin = (12.5 m) tan (7.280°) = 1.597 m ≈ 1.6 m
12.5 m
(b) Now the sag amount is x= x1 = 1 (1.597 m) = 0.3992 m . Use that distance to find the
4
4 min
tension in the rope.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
θ = tan−1 x = tan−1 0.3992 m = 1.829°
12.5 m 12.5 m
( )FT s2
= mg = (75 kg) 9.80 m = 11,512 N ≈ 12,000 N
2 sinθ
2 sin1.829°
The rope will not break, but the safety factor will only be about 4 instead of 10.
97. (a) The stress is given by F , the applied force divided by the cross-sectional area, and the strain is
A
given by Δl , the elongation over the original length.
l0
3.0
2.5
Stress (108 N/m2) 2.0
1.5
1.0
0.5
0.0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
0
Strain
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH12.XLS,” on tab “Problem 12.97a.”
(b) The elastic 1.6 stress = [2.02 x 1011(strain) - 6.52 x 105] N/m2
region is shown 1.4
in the graph. Stress (108 N/m2) 1.2
1.0
The slope of the 0.8
stress vs. strain 0.6
graph is the 0.4
elastic modulus,
and is
2.02 ×1011 N m2 .
The spreadsheet 0.2
used for this 0.0
problem can be 0 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008
found on the Strain
Media Manager,
with filename “PSE4_ISM_CH12.XLS,” on tab “Problem 12.97b.”
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412
Chapter 12 Static Equilibrium; Elasticity and Fracture
98. See the free-body diagram. We assume that point C is not A θ 2.0 m 2.0 m θ B
accelerating, and so the net force at point C is 0. That net force is
the vector sum of applied force F and two identical spring forces l Felas Felas
Felas. The elastic forces are given by Felas = k (amount of stretch). C F
If the springs are unstretched for θ = 0 , then 2.0 m must be
subtracted from the length of AC and BC to find the amount the
springs have been stretched. Write Newton’s second law for the
vertical direction in order to obtain a relationship between F and θ. Note that cosθ = 2.0 m .
l
∑ Fvert = 2Felas sinθ − F = 0 → F = 2Felas sinθ
Felas = k (l − 2.0 m) = k ⎝⎜⎛ 2.0 m − 2.0 m⎟⎞⎠ →
cosθ
F = 2Felas sinθ = 2k ⎛⎜⎝ 2.0 m − 2.0 m ⎟⎠⎞sinθ = 2 (20.0 N m) ( 2.0 m) ⎜⎝⎛ 1 − 1⎞⎠⎟ sinθ
cosθ cosθ
= 80 N (tanθ − sinθ )
This gives F as a function of θ , but we θ (degrees) 75
60
require a graph of θ as a function of F. 45
To graph this, we calculate F for 30
0 ≤ θ ≤ 75°, and then simply interchange
the axes in the graph.
The spreadsheets used for this problem 15 50 100 150 200 250
can be found on the Media Manager,
with filename “PSE4_ISM_CH12.XLS”, 0 F (N)
on tab “Problem 12.98”. 0
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413
CHAPTER 13: Fluids
Responses to Questions
1. No. If one material has a higher density than another, then the molecules of the first could be heavier
than those of the second, or the molecules of the first could be more closely packed together than the
molecules of the second.
2. The cabin of an airplane is maintained at a pressure lower than sea-level atmospheric pressure, and
the baggage compartment is not pressurized. Atmospheric pressure is lower at higher altitudes, so
when an airplane flies up to a high altitude, the air pressure outside a cosmetics bottle drops,
compared to the pressure inside. The higher pressure inside the bottle forces fluid to leak out around
the cap.
3. In the case of the two non-cylindrical containers, perpendicular forces from the sides of the
containers on the fluid will contribute to the net force on the base. For the middle container, the
forces from the sides (perpendicular to the sides) will have an upward component, which helps
support the water and keeps the force on the base the same as the container on the left. For the
container on the right, the forces from the sides will have a downward component, increasing the
force on the base so that it is the same as the container on the left.
4. The pressure is what determines whether or not your skin will be cut. You can push both the pen and
the pin with the same force, but the pressure exerted by the point of the pin will be much greater than
the pressure exerted by the blunt end of the pen, because the area of the pin point is much smaller.
5. As the water boils, steam displaces some of the air in the can. When the lid is put on, and the water
and the can cool, the steam that is trapped in the can condenses back into liquid water. This reduces
the pressure in the can to less than atmospheric pressure, and the greater force from the outside air
pressure crushes the can.
6. If the cuff is held below the level of the heart, the measured pressure will be the actual blood
pressure from the pumping of the heart plus the pressure due to the height of blood above the cuff.
This reading will be too high. Likewise, if the cuff is held above the level of the heart, the reported
pressure measurement will be too low.
7. Ice floats in water, so ice is less dense than water. When ice floats, it displaces a volume of water
that is equal to the weight of the ice. Since ice is less dense than water, the volume of water
displaced is smaller than the volume of the ice, and some of the ice extends above the top of the
water. When the ice melts and turns back into water, it will fill a volume exactly equal to the original
volume of water displaced. The water will not overflow the glass as the ice melts.
8. No. Alcohol is less dense than ice, so the ice cube would sink. In order to float, the ice cube would
need to displace a weight of alcohol equal to its own weight. Since alcohol is less dense than ice, this
is impossible.
9. All carbonated drinks have gas dissolved in them, which reduces their density to less than that of
water. However, Coke has a significant amount of sugar dissolved in it, making its density greater
than that of water, so the can of Coke sinks. Diet Coke has no sugar, leaving its density, including
the can, less that the density of water. The can of Diet Coke floats.
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414
Chapter 13 Fluids
10. In order to float, a ship must displace an amount of water with a weight equal to its own weight. An
iron block would sink, because it does not have enough volume to displace an amount of water equal
to its weight. However, the iron of a ship is shaped more like a bowl, so it is able to displace more
water. If you were to find the average density of the ship and all its contents, including the air it
holds, you would find that this density would be less than the density of water.
11. The liquid in the vertical part of the tube over the lower container will fall into the container through
the action of gravity. This action reduces the pressure in the top of the tube and draws liquid through
the tube, and into the tube from the upper container. As noted, the tube must be full of liquid initially
for this to work.
12. Sand must be added to the barge. If sand is removed, the barge will not need to displace as much
water since its weight will be less, and it will rise up in the water, making it even less likely to fit
under the bridge. If sand is added, the barge will sink lower into the water, making it more likely to
fit under the bridge.
13. As the weather balloon rises into the upper atmosphere, atmospheric pressure on it decreases,
allowing the balloon to expand as the gas inside it expands. If the balloon were filled to maximum
capacity on the ground, then the balloon fabric would burst shortly after take-off, as the balloon
fabric would be unable to expand any additional amount. Filling the balloon to a minimum value on
take-off allows plenty of room for expansion as the balloon rises.
14. The water level will fall in all three cases.
(a) The boat, when floating in the pool, displaces water, causing an increase in the overall level of
water in the pool. Therefore, when the boat is removed, the water returns to its original (lower)
level.
(b) The boat and anchor together must displace an amount of water equal to their combined weight.
If the anchor is removed, this water is no longer displaced and the water level in the pool will go
down.
(c) If the anchor is removed and dropped in the pool, so that it rests on the bottom of the pool, the
water level will again go down, but not by as much as when the anchor is removed from the
boat and pool altogether. When the anchor is in the boat, the combination must displace an
amount of water equal to their weight because they are floating. When the anchor is dropped
overboard, it can only displace an amount of water equal to its volume, which is less than the
amount of water equal to its weight. Less water is displaced so the water level in the pool goes
down.
15. No. If the balloon is inflated, then the air inside the balloon is slightly compressed by the balloon
fabric, making it more dense than the outside air. The increase in the buoyant force, present because
the balloon is filled with air, is more than offset by the increase in weight due to the denser air filling
the balloon. The apparent weight of the filled balloon will be slightly greater than that of the empty
balloon.
16. In order to float, you must displace an amount of water equal to your own weight. Salt water is more
dense than fresh water, so the volume of salt water you must displace is less than the volume of fresh
water. You will float higher in the salt water because you are displacing a lower volume of water.
17. The papers will move toward each other. When you blow between the sheets of paper, you reduce
the air pressure between them (Bernoulli’s principle). The greater air pressure on the other side of
each sheet will push the sheets toward each other.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
18. As the water falls, it speeds up because of the acceleration due to gravity. Because the volume flow
rate must remain constant, the faster-moving water must have a smaller cross-sectional area
(equation of continuity). Therefore the water farther from the faucet will have a narrower stream
than the water nearer the faucet.
19. As a high-speed train travels, it pulls some of the surrounding air with it, due to the viscosity of the
air. The moving air reduces the air pressure around the train (Bernoulli’s principle), which in turn
creates a force toward the train from the surrounding higher air pressure. This force is large enough
that it could push a light-weight child toward the train.
20. No. Both the cup and the water in it are in free fall and are accelerating downward because of
gravity. There is no “extra” force on the water so it will not accelerate any faster than the cup; both
will fall together and water will not flow out of the holes in the cup.
21. Taking off into the wind increases the velocity of the plane relative to the air, an important factor in
the creation of lift. The plane will be able to take off with a slower ground speed, and a shorter
runway distance.
22. As the ships move, they drag water with them. The moving water has a lower pressure than
stationary water, as shown by Bernoulli’s principle. If the ships are moving in parallel paths fairly
close together, the water between them will have a lower pressure than the water to the outside of
either one, since it is being dragged by both ships. The ships are in danger of colliding because the
higher pressure of the water on the outsides will tend to push them towards each other.
23. Air traveling over the top of the car is moving quite fast when the car is traveling at high speed, and,
due to Bernoulli’s principle, will have a lower pressure than the air inside the car, which is stationary
with respect to the car. The greater air pressure inside the car will cause the canvas top to bulge out.
24. The air pressure inside and outside a house is typically the same. During a hurricane or tornado, the
outside air pressure may drop suddenly because of the high wind speeds, as shown by Bernoulli’s
principle. The greater air pressure inside the house may then push the roof off.
Solutions to Problems
1. The mass is found from the density of granite (found in Table 13-1) and the volume of granite.
( ) ( )m = ρ V = 2.7×103 kg m3 108 m3 = 2.7×1011kg ≈ 3×1011kg
2. The mass is found from the density of air (found in Table 13-1) and the volume of air.
( )m = ρ V = 1.29 kg m3 (5.6 m) (3.8m) (2.8m) = 77 kg
3. The mass is found from the density of gold (found in Table 13-1) and the volume of gold.
( )m = ρ V = 19.3×103 kg m3 (0.56 m) (0.28m) (0.22 m) = 670 kg (≈ 1500lb)
4. Assume that your density is that of water, and that your mass is 75 kg.
V = m = 75 kg m3 = 7.5×10−2 m3 = 75 L
ρ 1.00×103 kg
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416
Chapter 13 Fluids
5. To find the specific gravity of the fluid, take the ratio of the density of the fluid to that of water,
noting that the same volume is used for both liquids.
(( ))ρρSJfluid = mV mfluid 89.22 g − 35.00 g 0.8547
=fluid mV fluid = mwater = 98.44 g − 35.00 g =
water
water
6. The specific gravity of the mixture is the ratio of the density of the mixture to that of water. To find
the density of the mixture, the mass of antifreeze and the mass of water must be known.
m = ρ V = SG ρ Vantifreeze m = ρ Vwater
antifreeze antifreeze antifreeze water antifreeze water water
SG = ρρ = m ρ V = m ρ V+ m = SG ρ ρ V V + ρ Vmixturemixture water
water mixture mixture antifreeze antifreeze water antifreeze water water
water water mixture
water mixture
( ) ( )= SG V + V =antifreeze antifreeze 0.80
water 5.0 L + 4.0 L = 0.89
Vmixture 9.0 L
7. (a) The density from the three-part model is found from the total mass divided by the total volume.
Let subscript 1 represent the inner core, subscript 2 represent the outer core, and subscript 3
represent the mantle. The radii are then the outer boundaries of the labeled region.
( ) ( )ρthree ρ1 π r4 3 + ρ2 4 π r23 − r13 + ρ3 4 π r33 − r23
( ) ( )layers 3 3
= m1 + m2 + m3 = ρ1m1 + ρ2m2 + ρ3m3 = 31
V1 +V2 + V3 V1 + V2 + V3
π r4 3 + 4 π r23 − r13 + 4 π r33 − r23
3 3
31
( ) ( ) ( ) ( )= ρ1r13 + ρ2
r23 − r13 + ρ3 r33 − r23 = r13 ρ1 − ρ2 + r23 ρ2 − ρ3 + r33ρ3
r33 r33
( ) ( ) ( )(1220 km)3 1900 kg m3 + (3480 km)3 6700kg m3 + (6371km)3 4400kg m3
= (6371km)3
= 5505.3kg m3 ≈ 5510 kg m3
M M 5.98 ×1024 kg m3 ≈ m3
ρ =one
Vdensity
( )(b) = 4 π R3 = 3 = 5521kg 5520 kg
3 6371 × 103 m
4 π
3
⎛ ρ ρ−one three ⎞ ⎛ 5521kg m3 − 5505kg m3 ⎞
100 ⎜⎜⎜⎝ ⎟ ⎝⎜ 5505kg m3 ⎟⎠
% diff = density layers ⎟⎠⎟ = 100 = 0.2906 ≈ 0.3%
ρthree
layers
8. The pressure is given by Eq. 13-3.
( )P = ρ gh = (1000) 9.80 m s2 (35m) = 3.4 ×105 N m2 ≈ 3.4atm
9. (a) The pressure exerted on the floor by the chair leg is caused by the chair pushing down on the
floor. That downward push is the reaction to the normal force of the floor on the leg, and the
normal force on one leg is assumed to be one-fourth of the weight of the chair.
( () )Pchair= Wleg = 1 ( 66 kg) 9.80 m s2 = 8.085 ×107 N m2 ≈ 8.1×107 N m2 .
A 4
0.020 cm2 ⎛ 1m ⎞2
⎝⎜ 100cm ⎟⎠
(b) The pressure exerted by the elephant is found in the same way, but with ALL of the weight
being used, since the elephant is standing on one foot.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
( )( )Pelephant
= W Aelephant = (1300 kg) 9.80 m s2 = 1.59 ×105 N m2 ≈ 2 ×105 N m2 .
800 cm2 ⎛ 1 m ⎞2
⎝⎜ 100 cm ⎟⎠
Note that the chair pressure is larger than the elephant pressure by a factor of about 400.
10. Use Eq. 13-3 to find the pressure difference. The density is found in Table 13-1.
( ) ( )P = ρ gh → ΔP = ρ gΔh = 1.05×103 kg m3 9.80 m s2 (1.70 m)
= 1.749 ×104 N m2 ⎛ 1 mm-Hg ⎞ = 132 mm-Hg
⎝⎜ 133 N m2 ⎟⎠
11. The height is found from Eq. 13-3, using normal atmospheric pressure. The density is found in
Table 13-1.
h P 1.013 ×105 N m2 = 13m
ρg 0.79 ×103 kg m3 9.80 m s2
( )( )P = ρgh→= =
That is so tall as to be impractical in many cases.
12. The pressure difference on the lungs is the pressure change from the depth of water.
ΔP (85 mm-Hg) ⎛ 133 N m2 ⎞
ρg ⎝⎜ 1 mm-Hg ⎠⎟
( )( )ΔP = ρgΔh→Δh = = = 1.154 m ≈ 1.2 m
1.00 ×103 kg m3 9.80 m s2
13. The force exerted by the gauge pressure will be equal to the weight of the vehicle.
( )mg = PA = P π r2 →
(17.0 atm) ⎛ 1.013 × 105 N m2 ⎠⎞⎟π ⎡⎣ 1 ( 0.225 m) 2
⎜⎝ 1 atm 2
Pπ r2 ⎤⎦
g
( )m = = = 6990 kg
9.80 m s2
14. The sum of the force exerted by the pressure in each tire is equal to the weight of the car.
( ( )( ) )mg = 4PA ⎛ 1 m2 ⎞
4PA 4 2.40 ×105 N m2 220cm2 ⎝⎜ 104cm2 ⎠⎟ = 2200 kg
g 9.80 m s2
→ m = =
15. (a) The absolute pressure is given by Eq. 13-6b, and the total force is the absolute pressure times
the area of the bottom of the pool.
( ) ( )P = P0 + ρgh = 1.013×105 N m2 + 1.00 ×103 kg m3 9.80 m s2 (1.8m)
= 1.189 ×105 N m2 ≈ 1.2 ×105 N m2
( )F = PA = 1.189 ×105 N m2 (28.0 m) (8.5m) = 2.8 ×107 N
(b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the
bottom. Pressure is not directional. P = 1.2 ×105 N m2
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Chapter 13 Fluids
16. (a) The gauge pressure is given by Eq. 13-3. The height is the height from the bottom of the hill
to the top of the water tank.
( )( )PG = ρgh = 1.00×103 kg m3 9.80 m s2 ⎡⎣5.0 m + (110 m) sin 58o ⎦⎤ = 9.6×105 N m2
(b) The water would be able to shoot up to the top of the tank (ignoring any friction).
h = 5.0 m + (110 m) sin 58o = 98 m
17. The pressure at points a and b are equal since they are the same height in the same fluid. If they
were unequal, the fluid would flow. Calculate the pressure at both a and b, starting with atmospheric
pressure at the top surface of each liquid, and then equate those pressures.
Pa = Pb → P0 + ρoil ghoil = P0 ρ+ ghwater water → ρ h = ρ hoil oil →
water water
( )ρ hwater water m3 (0.272 m − 0.0862 m)
(0.272 m)
ρ = =oil
hoil
1.00 ×103 kg = 683 kg m3
18. (a) The mass of water in the tube is the volume of the tube times the density of water.
( ) ( )m = ρV = ρπ r2h = 1.00 ×103 kg m3 π 0.30 ×10−2 m 2 (12 m) = 0.3393 kg ≈ 0.34 kg
(b) The net force exerted on the lid is the gauge pressure of the water times the area of the lid. The
gauge pressure is found from Eq. 13-3.
( )( )F = Pgauge A = ρghπ R2 = 1.00×103 kg m3 9.80m s2 (12m)π (0.21m)2 = 1.6×104 N
19. We use the relationship developed in Example 13-5.
( )P e ( )− 1.25×10−4 m−1 (8850m) = 3.35 ×104 N
= P e−(ρ0g P0 ) y = 1.013×105 N m2 m2 ≈ 0.331atm
0
Note that if we used the constant density approximation, P = P0 + ρ gh, a negative pressure would
result.
20. Consider the lever (handle) of the press. The net torque Fsample F
on that handle is 0. Use that to find the force exerted by l
the hydraulic fluid upwards on the small cylinder (and l
the lever). Then Pascal’s principle can be used to find
the upwards force on the large cylinder, which is the D1 B
same as the force on the sample.
∑τ = F (2l ) − F1l = 0 → F1 = 2F F2
( ) ( )P1 = P2→ F1 2 = F2 → D1
d1 π d1 2
F1
21 22
π
2 = 2F = F2
( ) ( )F2 = F1d2d1 d2 d1 sample →
2 2 350 N 5 2
= 4.0 ×10−4 m2 =
( ) ( ) ( )P = FAsamplesample=2Fd2 d1 4.4 ×107 N m2 ≈ 430atm
sample Asample
21. The pressure in the tank is atmospheric pressure plus the pressure difference due to the column of
mercury, as given in Eq. 13-6b.
(a) P = P0 + ρgh = 1.04 bar + ρHg gh
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⎛ 1.00 ×105 N m2 ⎞
⎜⎝ 1bar ⎠⎟ +
( )( )=
(1.04 bar ) 13.6 ×103 kg m3 9.80 m s2 (0.210 m) = 1.32 ×105 N m2
⎛ 1.00 ×105 N m2 ⎞
⎜⎝ 1bar ⎠⎟
( )( )(b)
P = (1.04 bar ) + 13.6 ×103 kg m3 9.80 m s2 (−0.052 m) = 9.7 ×104 N m2
22. (a) See the diagram. In the accelerated frame of the beaker, there is a aθ
pseudoforce opposite to the direction of the acceleration, and so there is
a pseudo acceleration as shown on the diagram. The effective θ
acceleration, g′ , is given by g′ = g + a. The surface of the water will be
g′ g
perpendicular to the effective acceleration, and thus makes an angle
hθ
θ = tan−1 a . h′
g
(b) The left edge of the water surface, opposite to the direction of the
acceleration, will be higher.
(c) Constant pressure lines will be parallel to the surface. From the second
diagram, we see that a vertical depth of h corresponds to a depth of h′
perpendicular to the surface, where h′ = h cosθ, and so we have the
following.
P = P0 + ρg′h′ = P0 + ρ g2 + a2 (h cosθ )
= P0 + ρ ⎛ g a2 ⎞ = P0 + ρhg
g2 + a2 ⎝⎜⎜ h g2 + ⎟⎠⎟
And so P = P0 + ρhg , as in the unaccelerated case.
23. (a) Because the pressure varies with depth, the force on the wall b
will also vary with depth. So to find the total force on the
wall, we will have to integrate. Measure vertical distance y
downward from the top level of the water behind the dam.
Then at a depth y, choose an infinitesimal area of width b
and height dy. The pressure due to the water at that depth is
P = ρgy .
P = ρgy ; dF = PdA = (ρgy) (bdy) → y dF
dy
dF = h (ρgy) (bdy) = gbh2
∫ ∫F = 1 ρ
0 2
(b) The lever arm for the force dF about the bottom of the dam h
is h − y, and so the torque caused by that force is
dτ = (h − y) dF . Integrate to find the total torque. t
∫h(h ∫ρgb h
0 0
( )τ
= ∫ dτ = − y) (ρgy) (bdy) = hy − y2 dy
( )= ρgb 1 hy2 − 1 y3 h = 1 ρ gbh3
2 3 0 6
Consider that torque as caused by the total force, applied at a single distance from the bottom d.
τ = 1 ρ gbh3 = Fd = 1 ρ gbh2d → d = 1 h
6 2 3
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420
Chapter 13 Fluids
(c) To prevent overturning, the torque caused by gravity about the lower right front corner in the
diagram must be at least as big as the torque caused by the water. The lever arm for gravity is
half the thickness of the dam.
ρ gbh3 ρwater gbh3
( ) ( ) ( )mg1t1 hbt g 1 t 1
2 ≥ 6 → ρconcrete 2 ≥ 6 →
t ρ1 water 1 1.00 ×103 kg m3 = 0.38
h≥ =3 3 2.3×103 kg m3
ρconcrete
So we must have t ≥ 0.38h to prevent overturning. Atmospheric pressure need not be added in
because it is exerted on BOTH sides of the dam, and so causes no net force or torque. In part
(a), the actual pressure at a depth y is P = P0 + ρgy , and of course air pressure acts on the
exposed side of the dam as well.
24. From section 9-5, the change in volume due to pressure change is ΔV = − ΔP , where B is the bulk
V0 B
modulus of the water, given in Table 12-1. The pressure increase with depth for a fluid of constant
density is given by ΔP = ρ gΔh , where Δh is the depth of descent. If the density change is small,
then we can use the initial value of the density to calculate the pressure change, and so ΔP ≈ ρ0gΔh .
Finally, consider a constant mass of water. That constant mass will relate the volume and density at
the two locations by M = ρV = ρ0V0 . Combine these relationships and solve for the density deep in
the sea, ρ .
ρV = ρ0V0 →
ρ0V0 ρ0V0 ρ0V0 ρ0 1025kg m3
V V0 + ΔV ⎜⎝⎛ −V0 ρ0 gh m3 9.80 m s2 5.4 ×103m
B 2.0 ×109 N m2
( )( )( )ρ= = = ΔP = =
B ⎟⎞⎠ 1− 1025 kg
V0 + 1−
= 1054 kg m3 ≈ 1.05×103 kg m3
ρ ρ0 = 1054 = 1.028
1025
The density at the 6 km depth is about 3% larger than the density at the surface.
25. Consider a layer of liquid of (small) height Δh , and ignore the ω
pressure variation due to height in that layer. Take a cylindrical
ring of water of height Δh , radius r, and thickness dr. See the dFradial
diagram (the height is not shown). The volume of the ring of
r
liquid is (2π rΔh) dr , and so has a mass of dm = (2πρrΔh) dr .
That mass of water has a net centripetal force on it of magnitude
( )dFradial = ω2r dm = ω2rρ (2π rΔh) dr . That force comes from a P
dr
pressure difference across the surface area of the liquid. Let the
pressure at the inside surface be P , which causes an outward P + dP
force, and the pressure at the outside surface be P + dP , which
causes an inward force. The surface area over which these
pressures act is 2π rΔh , the “walls” of the cylindrical ring. Use Newton’s second law.
dFradial = dFouter − dFinner → ω2rρ (2π rΔh) dr = ( P + dP) 2π rΔh − ( P) 2π rΔh →
wall wall
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Pr
∫ ∫dP = ω2rρdr → ω 2 rρ dr ρω 2 r 2 ρω 2 r 2
dP = → P − P0 = 1 → P = P0 + 1
2 2
P0 0
26. If the iron is floating, then the net force on it is zero. The buoyant force on the iron must be equal to
its weight. The buoyant force is equal to the weight of the mercury displaced by the submerged iron.
gV = gV →Hg submerged
ρ ρFbuoyant = mFe g → Fe total
Vsubmerged = ρ Fe = 7.8 ×103 kg m3 = 0.57 ≈ 57%
Vtotal ρ Hg 13.6 ×103 kg m3
27. The difference in the actual mass and the apparent mass is the mass of the water displaced by the
rock. The mass of the water displaced is the volume of the rock times the density of water, and the
volume of the rock is the mass of the rock divided by its density. Combining these relationships
yields an expression for the density of the rock.
m − m = Δm = ρ V = ρ mρactual water rock →
apparent water rock rock
( )ρrock mrock m3 9.28 kg
ρ= water Δm = 1.00 ×103 kg 9.28 kg − 6.18 kg = 2990 kg m3
28. (a) When the hull is submerged, both the buoyant force and the tension force act upward on
the hull, and so their sum is equal to the weight of the hull, if the hill is not accelerated as it is
lifted. The buoyant force is the weight of the water displaced.
T + Fbuoyant = mg →
T = mg − Fbuoyant = mhull g ρ− V gwater sub = mhull g ρ− water mhull g = mhull g ⎛ 1 − ρ water ⎞
ρ hull ⎜ ρ hull ⎠⎟
⎝
( ) ( )= 1.6 ×104 kg s2 ⎝⎛⎜1 − 1.00 ×103 kg m3 ⎞ = 1.367 ×105 N 1.4 ×105 N
9.80 m 7.8 ×103 kg m3 ⎟⎠ ≈
(b) When the hull is completely out of the water, the tension in the crane’s cable must be
equal to the weight of the hull.
( ) ( )T = mg = 1.6 ×104 kg 9.80 m s2 = 1.568 ×105 N ≈ 1.6 ×105 N
29. The buoyant force of the balloon must equal the weight of the balloon plus the weight of the helium
in the balloon plus the weight of the load. For calculating the weight of the helium, we assume it is
at 0oC and 1 atm pressure. The buoyant force is the weight of the air displaced by the volume of the
balloon.
ρF =buoyant V gair balloon = mHe g + m gballoon + mcargo g →
ρ ρ ρ ( ρ ρ )m = V − m − m = V − V − m =cargo
air balloon He balloon air balloon He balloon balloon − V − mair He balloon balloon
( )= m3 m)3
1.29 kg m3 − 0.179 kg 4 π (7.35 − 930 kg = 920 kg = 9.0 ×103 N
3
30. The difference in the actual mass and the apparent mass is the mass of the water displaced by the
legs. The mass of the water displaced is the volume of the legs times the density of water, and the
volume of the legs is the mass of the legs divided by their density. The density of the legs is
assumed to be the same as that of water. Combining these relationships yields an expression for the
mass of the legs.
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422
Chapter 13 Fluids
V =water legs m = 2mwaterlegs
ρ ρ ρmactual − mapparent = Δm = →
leg
legs
mleg = 1 Δm = 1 ( 74 kg − 54 kg ) = 10 kg (2 sig. fig.)
2 2
31. The apparent weight is the actual weight minus the buoyant force. The buoyant force is weight of a
mass of water occupying the volume of the metal sample.
gH2O = mmetal g − mmetal
metal
ρ ρ ρm gapparent = mmetal g − FB = mmetal g − Vmetal gH2O →
m = m − mρ ρapparentmetal metal H2O →
metal
mmetal 63.5 g
ρ =H2O 63.5 g − 55.4 g
( ) ( ) ( )ρ =metal
m − mmetal 1000 kg m3 = 7840 kg m3
apparent
Based on the density value, the metal is probably iron or steel .
32. The difference in the actual mass and the apparent mass of the aluminum is the mass of the air
displaced by the aluminum. The mass of the air displaced is the volume of the aluminum times the
density of air, and the volume of the aluminum is the actual mass of the aluminum divided by the
density of aluminum. Combining these relationships yields an expression for the actual mass.
m − m = ρ V = ρ mρactual actual →
apparent air Al air
Al
mactual ρ= mapparent = 3.0000 kg = 3.0014 kg
1 − air
ρ Al 1 − 1.29 kg m3
2.70 ×103 kg m3
33. The buoyant force on the drum must be equal to the weight of the steel plus the weight of the
gasoline. The weight of each component is its respective volume times density. The buoyant force
is the weight of total volume of displaced water. We assume that the drum just “barely” floats – in
other words, the volume of water displaced is equal to the total volume of gasoline and steel.
( ) ρ ρ ρF = W + W → V + V g = V g + V g →B steel gasoline
gasoline steel water steel steel gasoline gasoline
V ρ + V ρ = V ρ + V ρ →gasoline water
steel water steel steel gasoline gasoline
( )⎜⎝⎛ ρρ ρρ ⎠⎞⎟V = Vsteel gasoline−water gasoline = 230 L ⎛ 1000 kg m3 − 680 kg m3 ⎞ = 10.82 L ≈ 1.1×10−2 m3
⎜⎝ 7800 kg m3 −1000 kg m3 ⎟⎠
−steel water
34. (a) The buoyant force is the weight of the water displaced, using the density of sea water.
F = m g = ρ V gbuoyant
water water displaced
displaced
⎛ 1×10−3 m3 ⎞
⎜⎝ 1L ⎟⎠
( ) ( )= 1.025×103 kg m3 L) s2
(65.0 9.80 m = 653 N
( )(b) The weight of the diver is mdiver g = (68.0 kg) 9.80 m s2 = 666 N . Since the buoyant
force is not as large as her weight, she will sink , although it will be very gradual since the two
forces are almost the same.
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35. The buoyant force on the ice is equal to the weight of the ice, since it floats.
F = Wbuoyantice → mseawater g = mice g → mseawater = mice →
submerged submerged
ρ ρ ( ) ρ ( ) ρV = V →seawater seawater
SG V = SG V →seawater water submerged
ice ice ice water ice
ice
( ) ( )SG V = SG V →seawater submerged
ice ice ice
( )V =submerged SG Vice = 0.917 Vice = 0.895Vice
( )ice ice 1.025
SG
seawater
Thus the fraction above the water is Vabove = Vice − Vsubmerged = 0.105 Vice or 10.5%
36. (a) The difference in the actual mass and the apparent mass of the aluminum ball is the mass of the
liquid displaced by the ball. The mass of the liquid displaced is the volume of the ball times the
density of the liquid, and the volume of the ball is the mass of the ball divided by its density.
Combining these relationships yields an expression for the density of the liquid.
m − m = Δm = ρ V = ρ mρactual ball →
apparent liquid ball liquid Al
Δm (3.80 kg − 2.10 kg)
mball
3.80 kg
( )ρliquid = 1210 kg m3
= ρ Al = 2.70 ×103 kg m3
m − mobject
(b) Generalizing the relation from above, we have ρ ⎝⎛⎜ ⎟⎠⎞ ρ=liquid apparent .object
mobject
37. (a) The buoyant force on the object is equal to the weight of the fluid displaced. The force of
gravity of the fluid can be considered to act at the center of gravity of the fluid (see section 9-8).
If the object were removed from the fluid and that space re-filled with an equal volume of fluid,
that fluid would be in equilibrium. Since there are only two forces on that volume of fluid,
gravity and the buoyant force, they must be equal in magnitude and act at the same point.
Otherwise they would be a couple (see Figure 12-4), exert a non-zero torque, and cause rotation
of the fluid. Since the fluid does not rotate, we may conclude that Fbuoy
the buoyant force acts at the center of gravity.
(b) From the diagram, if the center of buoyancy (the point where the Fbuoy
buoyancy force acts) is above the center of gravity (the point where
gravity acts) of the entire ship, when the ship tilts, the net torque
about the center of mass will tend to reduce the tilt. If the center of mg mg
buoyancy is below the center of gravity of the entire ship, when the
ship tilts, the net torque about the center of mass will tend to increase the tilt. Stability is
achieved when the center of buoyancy is above the center of gravity.
38. The weight of the object must be balanced by the two buoyant forces, one from the water and one
from the oil. The buoyant force is the density of the liquid, times the volume in the liquid, times the
acceleration due to gravity. We represent the edge length of the cube by l.
( ) ( )mg = FB + FB = ρ V goil oil + ρ V gwater water = ρ loil 2 0.28l g + ρ lwater 2 0.72l g →
oil water
( ) ( )m = l 3 (0.28ρoil + 0.72ρwater ) = (0.100 m)3 ⎡⎣0.28 810 kg m2 + 0.72 1000 kg m2 ⎤⎦
= 0.9468kg ≈ 0.95kg
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424
Chapter 13 Fluids
( )The buoyant force is the weight of the object, mg = (0.9468kg) 9.80 m s2 = 9.3N
39. The buoyant force must be equal to the combined weight of the helium balloons and the person. We
ignore the buoyant force due to the volume of the person, and we ignore the mass of the balloon
material.
( )FB = ( )→ g VHe r3 mperson
m + mperson g ρ Vair He g = ρm + Vperson = N 4 π ρair − ρHe →
He He He ( )→ 3 =
3mperson 3 ( 75 kg) = 3587 ≈ 3600 balloons
(4π r3 ρair − 1.29 kg m3 − 0.179 kg
=( )N =
)ρHe 4π ( 0.165 m)3 m3
40. There will be a downward gravity force and an upward buoyant force on the fully submerged tank.
The buoyant force is constant, but the gravity force will decrease as the air is removed. Take
upwards to be positive.
( )Ffull = FB − mtotal g = ρ V gwater tank − mtank + mair g
( ) ( ) ( )= ⎡⎣ 1025kg m3 0.0157 m3 −17.0kg⎦⎤ 9.80m s2 = −8.89 N ≈ 9N downward
( )Fempty = FB − mtotal g = ρ V gwater tank − mtank + mair g
( ) ( ) ( )= ⎣⎡ 1025kg m3 0.0157 m3 −14.0kg⎤⎦ 9.80m s2 = 20.51N ≈ 21N upward
41. The apparent weight is the force required to hold the F = wapparent1 Fbuoy
system in equilibrium. In the first case, the object is
object F = wapparent
held above the water. In the second case, the object 2
is allowed to be pulled under the water. Consider
the free-body diagram for each case. w
∑Case 1: F = w1 − w + Fbuoy − wsinker = 0 Fbuoy w
sinker sinker Fbuoy
∑Case 2: F = w2 + Fbuoy − w + Fbuoy − wsinker = 0 sinker
object sinker wsinker
Since both add to 0, equate them. Also note that the wsinker
specific gravity can be expressed in terms of the
buoyancy force.
g = mwater w
ρ ρ ρF = Vbuoy object object = S.G.
ρ ρobject
g = m gwater water
object object
object
w1 − w + Fbuoy − wsinker = 0 = w2 + Fbuoy − w + Fbuoy − wsinker →
sinker object sinker
w w
S.G. w1 − w2
( )w1 Fbuoy
= w2 + = w2 + → S.G. =
object
42. For the combination to just barely sink, the total weight of the wood and lead must be equal to the
total buoyant force on the wood and the lead.
→ mwood g + mPb g = Vwood water g + VPb gwater
ρ ρF = Fweight →
buoyant
+ mwater
ρ ρmwood + mPb = mwood Pb → mPb ⎛ 1 − ρ water ⎞ = mwood ⎛ ρ water − 1⎟⎞ →
ρ ρwood ⎜ ρ Pb ⎟ ⎜ ρ wood ⎠
water ⎝ ⎠⎝
Pb
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⎛ ρ water − 1⎞⎟⎠ ⎛ 1 − 1⎠⎟⎞ ⎝⎛⎜ 1 − 1⎠⎟⎞
⎜⎝ ρ wood ⎜⎝ SGwood ⎛⎝⎜ 0.50
mPb = mwood = mwood = ( 3.25 kg) = 3.57 kg
⎝⎛⎜1 − ρ water ⎞ ⎛ 1 ⎞ 1 ⎞⎠⎟
ρ Pb ⎟ ⎜ 1 − SGPb ⎟ 1 − 11.3
⎠ ⎝ ⎠
43. We apply the equation of continuity at constant density, Eq. 13-7b.
Flow rate out of duct = Flow rate into room
πA v =duct duct V2 room → vduct π Vroom = (8.2 m) (5.0 m) (3.5 m) = 2.8 m s
r v = tduct to fill =2
π ( 0.15 m ) 2 (12 min ) ⎜⎝⎛ 60 s ⎞⎟⎠
r tto fill 1 min
room
room
44. Use Eq. 13-7b, the equation of continuity for an incompressible fluid, to compare blood flow in the
aorta and in the major arteries.
( ) ( )Av aorta =
Av →
arteries
( ) ( )varteries 2
Aaorta π 1.2 cm
= Aarteries vaorta = 2.0 cm2 40 cm s = 90.5 cm s ≈ 0.9 m s
45. We may apply Torricelli’s theorem, Eq. 13-9.
( )v1 = 2g ( y2 − y1 ) = 2 9.80 m s2 (5.3 m) = 10.2 m s ≈ 10 m s (2 sig. fig.)
46. The flow speed is the speed of the water in the input tube. The entire volume of the water in the tank
is to be processed in 4.0 h. The volume of water passing through the input tube per unit time is the
volume rate of flow, as expressed in the text immediately after Eq. 13-7b.
V = Av → v= V l wh = (0.36 m) (1.0 m) (0.60 m) = 0.02122 m s≈ 2.1cm s
Δt AΔt = π r2Δt
π ( 0.015 m )2 ( 4.0 h ) ⎛ 3600 s ⎞
⎝⎜ 1h ⎠⎟
47. Apply Bernoulli’s equation with point 1 being the water main, and point 2 being the top of the spray.
The velocity of the water will be zero at both points. The pressure at point 2 will be atmospheric
pressure. Measure heights from the level of point 1.
P1 + 1 ρ v12 + ρ gy1 = P2 + 1 ρ v22 + ρ gy2 →
2 2
( ) ( )P1 − Patm = ρ gy2 = 1.00 ×103 kg m3 9.80 m s2 (18 m) = 1.8 ×105 N m2
48. The volume flow rate of water from the hose, multiplied times the time of filling, must equal the
volume of the pool.
( )Av = Vpool → t = Vpool = π (3.05 m)2 (1.2 m) = 4.429 ×105 s
t A vhose hose π
hose ⎡ ⎛ 1m ⎞⎤ 2
⎢⎣ ⎝⎜ 39.37" ⎟⎠⎦⎥
1 ( 5 )" ( 0.40 m s)
2 8
4.429 ×105 s ⎛ 60 1day 24 s ⎞ = 5.1 days
⎝⎜ × 60 × ⎠⎟
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Chapter 13 Fluids
49. We assume that there is no appreciable height difference between the two sides of the roof. Then the
net force on the roof due to the air is the difference in pressure on the two sides of the roof, times the
area of the roof. The difference in pressure can be found from Bernoulli’s equation.
2 2
inside outside
ρ ρ ρ ρP + v + gy = P + v + gy →inside1 1
2 inside outside 2 outside
P − P = ρ v = AFinside
outside 12 air →
2 air outside roof
⎢⎡⎣(180 km ⎛ 1m s ⎞ ⎤ 2
⎜⎝ 3.6 km ⎟⎠ ⎦⎥
( )F = ρ v A =1 2 m3 h) ( 6.2 ) (12.4 )
1 1.29 kg m m
air 2 air outside roof 2 h
= 1.2 ×105 N
50. Use the equation of continuity (Eq. 13-7b) to relate the volume flow of water at the two locations,
and use Bernoulli’s equation (Eq. 13-8) to relate the pressure conditions at the two locations. We
assume that the two locations are at the same height. Express the pressures as atmospheric pressure
plus gauge pressure. Use subscript “1” for the larger diameter, and “2” for the smaller diameter.
A1v1 = A2v2 → v2 = v1 A1 = v1 π r12 = v1 r12
A2 π r22 r22
P0 + P1 + 1 ρ v12 + ρ gy1 = P0 + P2 + 1 ρ v22 + ρ gy2 →
2 2
P1 + 1 ρ v12 = P2 + 1 ρ v22 = P2 + 1 ρ v12 r14 → v1 = 2 ( P1 − P2 ) →
2 2 2 r24
⎛ r14 − 1⎞⎟⎠
ρ ⎜⎝ r24
( )2 ( P1 − P2 ) ( )2 32.0 ×103 Pa − 24.0 ×103 Pa
A1v1 = π r12 =π 3.0 × 10−2 m 2
⎛ r14 − 1⎠⎞⎟ ( ) (( ))1.0 ×103 kg ⎛ 4 ⎞
ρ ⎜⎝ r24 m3 ⎜ 3.0 ×10−2 m 4 1⎟
⎜⎝ 2.25 ×10−2 m − ⎠⎟
= 7.7 × 10−3 m3 s
51. The air pressure inside the hurricane can be estimated using Bernoulli’s equation. Assume the
pressure outside the hurricane is air pressure, the speed of the wind outside the hurricane is 0, and
that the two pressure measurements are made at the same height.
2 2
inside outside
ρ ρ ρ ρP + v + gy = P + v + gy →inside1 1
2 inside outside 2 outside
12
2 air inside
P = P − ρ vinside
outside
⎡⎢( ⎛ 1000 m ⎞ ⎛ 1h ⎞ ⎤ 2
⎜⎝ km ⎟⎠ ⎝⎜ 3600 ⎠⎟ ⎥
( )= ⎣ ⎦
1.013 ×105 Pa − 1 1.29 kg m3 300 km h ) s
2
= 9.7 ×104 Pa ≈ 0.96 atm
52. The lift force would be the difference in pressure between the two wing surfaces, times the area of
the wing surface. The difference in pressure can be found from Bernoulli’s equation. We consider
the two surfaces of the wing to be at the same height above the ground. Call the bottom surface of
the wing point 1, and the top surface point 2.
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( )P1 ρ v22
+ 1 ρ v12 + ρ gy1 = P2 + 1 + ρ gy2 → P1 − P2 = 1 ρ v22 − v12
2 2 2
( )Flift
= ( P1 − P2 ) (Area of wing) = 1 ρ v22 − v12 A
2
( ) ( )=
1 1.29 kg m3 ⎡⎣(280 m s)2 − (150 m s)2 ⎦⎤ 88 m2 = 3.2 ×106 N
2
53. Consider the volume of fluid in the pipe. At each end of the pipe there is a force towards the
contained fluid, given by F = PA . Since the area of the pipe is constant, we have that
Fnet = ( P1 − P2 ) A . Then, since the power required is the force on the fluid times its velocity, and
AV = Q = volume rate of flow, we have P = Fnetv = ( P1 − P2 ) Av = ( P1 − P2 ) Q .
54. Use the equation of continuity (Eq. 13-7b) to relate the volume flow of water at the two locations,
and use Bernoulli’s equation (Eq. 13-8) to relate the conditions at the street to those at the top floor.
Express the pressures as atmospheric pressure plus gauge pressure.
A v = A v →street street
top top
( )( )Astreet ⎤⎦2
⎦⎤2
v = v =top street
( )Atop
0.68 m s π ⎣⎡ 1 5.0 ×10−2 m = 2.168 m s≈ 2.2 m s
2 2.8 ×10−2 m
π ⎡⎣ 1
2
P0 + Pgauge + 1 ρv2 + ρ gystreet = P0 + Pgauge + 1 ρv2 + ρ gytop →
2 street 2 top
street top
( ) ( )ρ ρP = P +gauge 2
gauge 1 v − v +2 top gy y − ystreet top
2 street
top street
⎛ 1.013 ×105 Pa ⎞
⎜⎝ atm ⎠⎟
( )= m3 ⎣⎡(0.68 m s)2 − (2.168 m s)2 ⎤⎦
( 3.8 atm ) + 1 1.00 ×103 kg
2
( ) ( )+ 1.00 ×103 kg m3 9.80 m s2 (−18 m)
= 2.064 × 105 Pa ⎛ 1atm Pa ⎞ ≈ 2.0 atm
⎝⎜ 1.013 ×105 ⎠⎟
55. Apply both Bernoulli’s equation and the equation of continuity between the two openings of the
tank. Note that the pressure at each opening will be atmospheric pressure.
A2v2 = A1v1 → v2 = v1 A1
A2
ρ v12 ρ v22 v12 − v22 = 2g y2 − y1 = 2gh
( )P1 P2
+ 1 + ρ gy1 = + 1 + ρ gy2 →
2 2
⎛ A1 ⎞ v12 ⎜⎛1 − A12 ⎞ 2 gh
⎜ A2 ⎟ ⎝ A22 ⎟ 1 − A12 A22
( )v12− v1 ⎠ = 2 gh → ⎠ = 2 gh → v1 =
⎝
56. (a) Relate the conditions at the top surface and at the opening by Bernoulli’s equation.
ρ v22 ρ v12 ρ v12
( )Ptop
+ 1 + ρ gy2 = Popening + 1 + ρ gy1 → P2 + P0 + ρ g y2 − y1 = P0 + 1 →
2 2 2
v1 = 2 P2 + 2g ( y2 − y1 )
ρ
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428
Chapter 13 Fluids
2 ( 0.85 atm ) ⎛ 1.013 ×105 Pa ⎞
⎜⎝ atm ⎟⎠
( ) ( )(b) 2 P2 y1 ) s2 (2.4 m) = 15 m s
v1 = ρ + 2g ( y2 − = 1.00 ×103 kg m3 + 2 9.80 m
57. We assume that the water is launched from the same level at which it lands. Then the level range
formula, derived in Example 3-10, applies. That formula is R = v02 sin 2θ0 . If the range has
g
increased by a factor of 4, then the initial speed has increased by a factor of 2. The equation of
continuity is then applied to determine the change in the hose opening. The water will have the same
volume rate of flow, whether the opening is large or small.
⎛⎜ ⎟⎞→ A = A v = A 1partly
⎝ ⎠v 2open
( ) ( )Av fully = Av partly fully fully
open open open open
fully
open partly
open
Thus 1 2 of the hose opening was blocked.
58. Use Bernoulli’s equation to find the speed of the liquid as it leaves the opening, assuming that the
speed of the liquid at the top is 0, and that the pressure at each opening is air pressure.
ρ v12 ρ v22
( )P1
+ 1 + ρ gy1 = P2 + 1 + ρ gy2 → v1 = 2g h2 − h1
2 2
(a) Since the liquid is launched horizontally, the initial vertical speed is zero. Use Eq. 2-12b for
constant acceleration to find the time of fall, with upward as the positive direction. Then
multiply the time of fall times v1 , the (constant) horizontal speed.
y = y0 + v0 yt + 1 ayt2 → 0 = h1 + 0 − 1 gt 2 → t= 2h1
2 2 g
Δx = v1t = 2g (h2 − h1 ) 2h1 = 2 ( h2 − h1 ) h1
g
(b) We seek some height h1′ such that 2 (h2 − h1 ) h1 = 2 (h2 − h1′) h1′.
2 (h2 − h1 ) h1 = 2 (h2 − h1′) h1′ → (h2 − h1 ) h1 = (h2 − h1′) h1′ →
( )h1′2 − h2h1′ + h2 − h1 h1 = 0 →
( ) ( )h1′ = h2 ± h22 − 4
2 h2 − h1 h1 = h2 ± h22 − 4h1h2 + 4h12 = h2 ± h2 − 2h1 = 2h2 − 2h1 , 2h1
2 2 2 2
h1′ = h2 − h1
59. (a) Apply Bernoulli’s equation to point 1, the exit hole, and point 2, the top surface of the liquid in
the tank. Note that both points are open to the air and so the pressure is atmospheric pressure.
Also apply the equation of continuity ( A1v1 = A2v2 ) to the same two points.
( ) ( )P1 ρ v12 ρ v22
+ 1 + ρ gy1 = P2 + 1 + ρ gy2 → Patm + 1 ρ v12 − v22 = Patm + ρ g y2 − y1 →
2 2 2
( ) ( )1 ⎛ A22 − 1⎞⎟ v22
⎝⎜ A12 ⎠
2
ρ v12 − v22 = ρ gh → v12 − v22 = 2gh → = 2 gh →
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
2 gh 2 ghA12
A22 − A12
( )v2 = ⎛ A22 =
⎝⎜ A12 − 1⎠⎞⎟
level v2 dh dh 2ghA12 .
dt dt = −
A22 − A12
( )Note the
that since water is decreasing, we have = − , and so
(b) Integrate to find the height as a function of time.
2 ghA12 2gA12 dt h dh 2 gA12 t
A22 − A12 A22 − A12 =− A22 − A12
dt
hh0
( )dh → ( ) ∫ ( ) ∫dh → 0 →
=−
dt = − h
( )⎡ ⎤2
h=⎢ ⎥
⎢⎣ ⎥⎦
2 gA12 gA12
A22 − A12 2 A22 − A12
(2 h − ) ( )h0 = − t → h0 − t
(c) We solve for the time at which h = 0, given the other parameters. In particular,
0.25 ×10−2 m 2 = 1.963 × 10−5 m2 1.3 ×10−3 m3
( )A1 = π ; A2 = 0.106 m = 1.226 × 10−2 m2
( )⎡ h0 − t 2 gA12 ⎤2 →
A22 − A12 ⎥ =0
⎢ ⎦⎥
⎣⎢
( ) ( ( )( ) ( ) )t = ⎡ 2 2⎤
2h0 A22 − A12 2 (0.106 m) ⎣ 1.226 × 10−2 m2 1.963 ×10−5 m2 ⎦ 92 s
gA12 −
= 9.80 m s2 1.963 ×10−5 m2 2 =
60. (a) Apply the equation of continuity and Bernoulli’s equation at the same height to the wide and
narrow portions of the tube.
A2v2 = A1v1 → v2 = v1 A1
A2
( )P1
+ 1 ρ v12 = P2 + 1 ρ v22 → 2 P1 − P2 = v22 − v12 →
2 2 ρ
⎞2 2 ( P1 − P2 ) ( )→ A12 A22
⎛ v1 A1 ⎟ − v12 = v12 ⎛ A22 − A22 ⎞ = 2 P1 − P2 →
⎜ A2 ⎠ ρ ⎜ ⎟ ρ
⎝ ⎝ ⎠
2 A22 P1 − P2
ρ A12 − A22
( ) ( )( ) ( )v12
= → v1 = A2 2 P1 − P2
ρ A12 − A22
( )( )(b)
v1 = A2 2 P1 − P2
ρ A12 − A22
⎛ 133 N m2 ⎞
⎝⎜ mm Hg ⎠⎟
( )( )2(18 mm Hg )
= π [ 1 ( 0.010 m )]2 1000 kg m3 2 [ 1 ( 0.030 m )]4 2 [ 1 ( 0.010 m )]4 = 0.24 m s
2 2 2
π −π
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430
Chapter 13 Fluids
61. (a) Relate the conditions inside the rocket and just outside the exit orifice by means of Bernoulli’s
equation and the equation of continuity. We ignore any height difference between the two
locations.
Pin + 1 ρ v2 + ρ gyin = Pout + 1 ρv2 + ρ gyout → P+ 1 ρ v2 = P0 + 1 ρv2 →
2 in 2 out 2 in 2 out
( )2 P − P0 = v2 − v2 = v2 ⎡ − ⎛ vin ⎞2 ⎤
out in out ⎢1 ⎜ vout ⎟⎥
ρ ⎢⎣ ⎝ ⎠ ⎥⎦
Ainvin = A vout out → Avin = A0vout → vin = A0 1→
vout A
2(P − P0 ) = v2 ⎡ ⎛ vin ⎞2 ⎤ ≈ v2 → vout = v = 2 ( P − P0 )
out ⎢1 − ⎜⎝ vout ⎟⎥ out
ρ ⎢⎣ ⎠ ⎦⎥ ρ
(b) Thrust is defined in section 9-10, by Fthrust = vrel dm , and is interpreted as the force on the rocket
dt
due to the ejection of mass.
( ) ( ) ( )Fthrust
= vrel dm = vout d ρV = vout ρ dV = vout ρ v Aout out = ρv2 A0 = ρ 2 P − P0 A0
dt dt dt ρ
= 2 ( P − P0 ) A0
62. There is a forward force on the exiting water, and so by Newton’s third law there is an equal force
pushing backwards on the hose. To keep the hose stationary, you push forward on the hose, and so
the hose pushes backwards on you. So the force on the exiting water is the same magnitude as the
force on the person holding the hose. Use Newton’s second law and the equation of continuity to find
the force. Note that the 450 L/min flow rate is the volume of water being accelerated per unit time.
Also, the flow rate is the product of the cross-sectional area of the moving fluid, times the speed of
the fluid, and so V = A1v1 = A2v2 .
t
F = m Δv = m v2 − v1 = ρ ⎜⎝⎛ V ⎞⎟⎠ (v2 − v1 ) = ρ ⎜⎝⎛ V ⎟⎞⎠ ⎛ A2v2 − A1v1 ⎞ = ρ ⎛⎝⎜ V ⎞2 ⎛ 1 − 1 ⎞
Δt t t t ⎜⎝ A2 A1 ⎠⎟ t ⎠⎟ ⎝⎜ A2 A1 ⎠⎟
= ρ ⎛V ⎞2 ⎛1 − 1 ⎞
⎝⎜ t ⎟⎠ ⎜⎝ π r22 π r12 ⎟⎠
⎛ 450 L 1 min 1m3 2 ⎛ 1 1 ⎞
⎜⎝ min × 60 s 1000 L ⎜ 0.75 ×10−2 m 7.0 ×10−2 m ⎟
( ) ( ) ( )= ⎞ ⎜⎝ 2 ⎠⎟
1.00 ×103 kg m3 × ⎠⎟ 2 −
π
π 1 1
2 2
= 1259 N ≈ 1300 N
63. Apply Eq. 13-11 for the viscosity force. Use the average radius to calculate the plate area.
⎛τ ⎞
⎝⎜ rinner ⎠⎟
v ( ( )( ))→ Fl r − router inner
l Av
F = ηA η= = 2π ravgh ωrinner
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
⎛ 0.024 miN ⎞
⎝⎜ 0.0510 m ⎠⎟
( )=
0.20 ×10−2 m
⎛⎝⎜ rev 2π rad 1 min ⎞⎟⎠ = 7.9 ×10−2 Pais
min rev 60 s
2π ( 0.0520 m) (0.120 m ) 57 × × (0.0510 m )
64. The relationship between velocity and the force of viscosity is given by Eq. Fvis r
13-11, Fvis =ηA v . The variable A is the area of contact between the lh
l mg
moving surface and the liquid. For a cylinder, A = 2π rh. The variable l is
the thickness of the fluid layer between the two surfaces. See the diagram.
If the object falls with terminal velocity, then the net force must be 0, and so
the viscous force will equal the weight. Note that
l = 1 (1.00 cm − 0.900 cm ) = 0.05 cm.
2
F = Fweight vis → mg = η A v →
l
( ( ) )( ( ) )v
= mgl mgl (0.15 kg) 9.80 m s2 0.050 ×10−2 m
ηA = 2π rhη = 2π 0.450 ×10−2 m (0.300 m) 200 ×10−3 Nis m2
= 0.43 m s
65. Use Poiseuille’s equation (Eq. 13-12) to find the pressure difference.
Q = π R4 ( P2 − P1 ) →
8ηl
⎡ 6.2 × 10−3 L 1 min 1 × 10−3 m 3 ⎤
⎢⎣ min 60s × 1L ⎥⎦
( ) ( )(P2
8Qηl 8 × π ( 0.2 Pais) 8.6 × 10−2 m
π R4
− P1 ) = = 0.9 × 10−3 m 4
= 6900 Pa
66. From Poiseuille’s equation, the volume flow rate Q is proportional to R4 if all other factors are the
same. Thus Q R4 = V 1 is constant. If the volume of water used to water the garden is to be the
t R4
same in both cases, then tR4 is constant.
t1R14 = t2 R24 → t2 = t1 ⎛ R1 ⎞4 = t1 ⎛ 3 8 ⎞4 = 0.13t1
⎜ R2 ⎟ ⎜ 5 8 ⎟⎠
⎝ ⎠ ⎝
Thus the time has been cut by 87% .
67. Use Poiseuille’s equation to find the radius, and then double the radius to the diameter.
Q = π R4 ( P2 − P1 ) →
8ηl
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432
Chapter 13 Fluids
⎡ ⎛ 8.0 × 14.0 × 4.0 m3 ⎞ 1/ 4
⎢ ⎜⎝ 720 s ⎟⎠
( ( ) )( )d ⎤
⎡ 8ηl Q 1/ 4 8 1.8 ×10−5 Pais (15.5 m )
⎢ ⎥
= 2R = 2 ⎣ ⎤ = 2⎢ ⎥ = 0.10 m
π ( P2 − P1 ) ⎥ ⎢ π 0.71 ×10−3atm 1.013 × 105 Pa atm ⎥
⎦
⎣⎢ ⎦⎥
68. Use Poiseuille’s equation to find the pressure difference.
Q = π R4 ( P2 − P1 ) →
8ηl
( )( ) ( )(P2
− P1 ) = 8Qηl 8 650 cm3 s 10−6 m3 cm3 (0.20 Pais) 1.9 ×103 m
π R4 = π (0.145 m)4
= 1423 Pa ≈ 1400 Pa
( )( )69. (a)
Re = 2vrρ = 2 (0.35 m s) 0.80 ×10−2 m 1.05 ×103 kg m3 = 1470
η 4 ×10−3 Pais
The flow is laminar at this speed.
(b) Since the velocity is doubled the Reynolds number will double to 2940. The flow is turbulent
at this speed.
70. From Poiseuille’s equation, Eq. 13-12, the volume flow rate Q is proportional to R4 if all other
factors are the same. Thus Q R4 is constant.
Q Qfinal initial ⎝⎛⎜ ⎞⎠⎟Rfinal = Qfinal 1/ 4 0.15 1/ 4 Rinitial = 0.622Rinitial , a 38% reduction.
=4 4 Qinitial
R =initial
R Rfinal initial ( )→
71. The fluid pressure must be 78 torr higher than air pressure as it exits the needle, so that the blood will
enter the vein. The pressure at the entrance to the needle must be higher than 78 torr, due to the
viscosity of the blood. To produce that excess pressure, the blood reservoir is placed above the level
of the needle. Use Poiseuille’s equation to calculate the excess pressure needed due to the viscosity,
and then use Eq. 13-6b to find the height of the blood reservoir necessary to produce that excess
pressure.
π R4 P2 −
η8 bloodl
( )Q= P1 → P2 = P1 + 8ηbloodl Q = ρ blood gΔh →
π R4
Δh = 1 ⎛⎜⎝ P1 + η8 bloodl Q ⎞⎠⎟
ρ gblood π R4
⎛ (78 mm-Hg) ⎛ 133 N m2 ⎞ + ⎞
⎜ ⎝⎜ 1 mm-Hg ⎟⎠ ⎟
1 ⎜ ⎟
⎜ 2.0 × 10−6 m3 ⎟
⎜ 8 4 ×10−3 Pais 60 s ⎟
⎜
( ) ( )(( ))= ⎛ ⎞ ⎟
⎝⎛⎜1.05 × 103 kg ⎠⎟⎞ 9.80 m s2 2.5 × 10−2 m ⎜⎝ ⎠⎟
m3
⎝⎜ π 0.4 ×10−3 m 4 ⎟⎠
= 1.04 m ≈ 1.0 m
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
72. In Figure 13-35, we have γ = F 2l . Use this to calculate the force.
γ F = 3.4 ×10−3 N = 2.4 × 10−2 N m
= 2l
2 (0.070 m)
73. In Figure 13-35, we have γ = F 2l . Use this relationship to
calculate the force.
γ = F 2l → F = 2γ l = 2 (0.025 N m) (0.245 m) = 1.2 ×10−2 N
74. (a) We assume that the weight of the platinum ring is negligible. Then the surface tension is the
force to lift the ring, divided by the length of surface that is being pulled. Surface tension will
act at both edges of the ring, as in Figure 13-35b. Thus γ FF
= 2 (2π r ) = 4π r
F 5.80 ×10−3 N = 1.6 ×10−2 N m
4π r 4π 2.8 ×10−2 m
γ( )(b)= =
75. As an estimate, we assume that the surface tension force acts vertically. We assume that the free-
body diagram for the cylinder is similar to Figure 13-37(a) in the text. The weight must equal the
total surface tension force. The needle is of length l.
( )mg = 2FT 2 l g = 2γ l
→ ρ πneedle d1 →
2 needle
8γ 8(0.072 N m) = 1.55 ×10−3 m ≈ 1.5 mm
ρ πneedle g
7800 kg m3 π 9.80 m s2
( ) ( )d =needle =
76. Consider half of the soap bubble – a hemisphere. The forces on the FT outer
hemisphere will be the surface tensions on the two circles and the net FT inner
force from the excess pressure between the inside and the outside of
the bubble. This net force is the sum of all the forces perpendicular FP
to the surface of the hemisphere, but must be parallel to the surface
tension. Therefore we can find it by finding the force on the circle FT inner
that is the base of the hemisphere. The total force must be zero. FT outer
Note that the forces FT outer and FT inner act over the entire length of the r
circles to which they are applied. The diagram may look like there
are 4 tension forces, but there are only 2. Likewise, there is only 1
pressure force, FP , but it acts over the area of the hemisphere.
2FT = FP → 2 (2π rγ ) = π r2ΔP → ΔP = 4γ
r
77. The mass of liquid that rises in the tube will have the force of gravity acting down on it, and the
force of surface tension acting upwards. The two forces must be equal for the liquid to be in
equilibrium. The surface tension force is the surface tension times the circumference of the tube,
since the tube circumference is the length of the “cut” in the liquid surface. The mass of the risen
liquid is the density times the volume.
FT = mg → γ 2π r = ρπ r2hg → h = 2γ ρ gr
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434
Chapter 13 Fluids
78. (a) The fluid in the needle is confined, and so Pascal’s principle may be applied.
P = P →plunger F Fplunger needle →
needle A = Aplunger needle
0.10 × 10−3 m 2
0.65 ×10−2 m 2
( ) (( ))F = F AA = F ππrr = F rr =needleplungerneedleplunger2 plunger 2 2.8 N
plunger needle needle
2 2
plunger plunger
= 6.627 ×10−4 N ≈ 6.6 × 10−4 N
⎛ 133 N m2 ⎞ 0.65 ×10−2 m 2 = 1.3 N
⎜⎝ 1 mm-Hg ⎠⎟
( )( )(b)
F = P A =plunger 75 mm-Hg π
plunger plunger
79. The pressures for parts (a) and (b) stated in this problem are gauge pressures, relative to atmospheric
pressure. The pressure change due to depth in a fluid is given by ΔP = ρ gΔh .
ΔP (55 mm-Hg) ⎛ 133 N m2 ⎞
ρg ⎝⎜ 1mm-Hg ⎠⎟
( )(a) Δh = = = 0.75m
⎝⎜⎛1.00 g 1kg 106 cm3 ⎞
cm3 × 1000g × 1m3 ⎠⎟ 9.80 m s2
ΔP ( 650 mm-H2O) ⎛ 9.81N m2 ⎞
ρg ⎜ 1mm-H2O ⎟
( )(b) Δh = = ⎝ ⎠ = 0.65m
⎜⎝⎛1.00 g 1kg 106 cm3 ⎞ 9.80 m s2
cm3 × 1000g × 1m3 ⎟⎠
(c) For the fluid to just barely enter the vein, the fluid pressure must be the same as the blood
pressure.
ΔP (78 mm-Hg) ⎛ 133N m2 ⎞
ρg ⎜⎝ 1mm-Hg ⎟⎠
( )Δh= = = 1.059 m ≈ 1.1m
⎝⎛⎜1.00 g 1kg 106 cm3 ⎞
cm3 × 1000 g × 1m3 ⎠⎟ 9.80 m s2
80. The ball has three vertical forces on it – string tension, buoyant force, and gravity. See FT FB
the free-body diagram for the ball. The net force must be 0.
Fnet = FT + FB − mg = 0 →
3 π ρr g4 3 3
3 water
( )FT mg FB 4 π r g 4 π r g
= − = 3 ρ Cu − = 3 ρ ρ−Cu water mg
( )( )=
4 π ( 0.013 m )3 9.80 m s2 8900 kg m3 − 1000 kg m3 = 0.7125 N ≈ 0.71 N
3
Since the water pushes up on the ball via the buoyant force, there is a downward force on the water
due to the ball, equal in magnitude to the buoyant force. That mass-equivalent of that force
(indicated by mB = FB g ) will show up as an increase in the balance reading.
FB = π ρr g4 3 →
3 water
( )( )mBFB
= g = π ρr4 3 = 4 π 0.013 m 3 1000 kg m3 = 9.203 × 10−3 kg = 9.203g
3
3 water
Balance reading = 998.0 g + 9.2 g = 1007.2 g
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435
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
81. The change in pressure with height is given by ΔP = ρ gΔh .
( )( )ΔP = ρ gΔh →
ΔP = ρ gΔh = 1.29 kg m3 9.80 m s2 (380 m) = 0.047 →
P0 P0 1.013×105 Pa
ΔP = 0.047 atm
82. (a) The input pressure is equal to the output pressure.
P = P →input F Finput output →
output A = Ainput output
( ) ( ) ( )A = A FF = πinput
output input 9.0 ×10−2 m 2 350 N = 9.878 × 10−4 m2
output 920 kg 9.80 m s2
≈ 9.9 ×10−4 m2
(b) The work is the force needed to lift the car (its weight) times the vertical distance lifted.
( )W = mgh = (920 kg) 9.80 m s2 (0.42 m) = 3787 J ≈ 3800 J
(c) The work done by the input piston is equal to the work done in lifting the car.
W = Winput output → F dinput input = F doutput ouptut = mgh →
F dinput input (350 N) (0.13m) = 5.047 ×10−3 m ≈ 5.0 × 10−3 m
mg (920 kg) 9.80 m s2
( )h = =
(d) The number of strokes is the full distance divided by the distance per stroke.
hfull = Nhstroke → N = hfull = 0.42 m = 83strokes
hstroke 5.047 ×10−3 m
(e) The work input is the input force times the total distance moved by the input piston.
W = NF dinput → 83(350 N) (0.13m) = 3777 J ≈ 3800 J
input input
Since the work input is equal to the work output, energy is conserved.
83. The pressure change due to a change in height is given by ΔP = ρ gΔh . That pressure is the excess
force on the eardrum, divided by the area of the eardrum.
ΔP = ρ gΔh = F →
A
( ) ( ) ( )F = ρ gΔhA = 1.29 kg m3 9.80 m s2 (950 m) 0.20 ×10−4 m2 = 0.24 N
84. The change in pressure with height is given by ΔP = ρ gΔh .
( )( )ΔP = ρ gΔh →
ΔP = ρ gΔh = 1.05 ×103 kg m3 9.80 m s2 (6 m) = 0.609 →
P0 P0 1.013×105 Pa
ΔP = 0.6 atm
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436
Chapter 13 Fluids
85. The pressure difference due to the lungs is the pressure change in the column of water.
ΔP (75 mm-Hg) ⎛ 133N m2 ⎞
ρg ⎜⎝ 1 mm-Hg ⎟⎠
( )( )ΔP = ρgΔh → Δh = = = 1.018 m ≈ 1.0 m
1.00 ×103 kg m3 9.80 m s2
86. We use the relationship developed in Example 13-5.
P e−(ρ0g P0 ) y e ( ) =− 1.25×10−4 m−1 (2400 m)
0
( )P 0.74 atm
= = 1.0 atm
87. The buoyant force, equal to the weight of mantle displaced, must be equal to the weight of the
continent. Let h represent the full height of the continent, and y represent the height of the continent
above the surrounding rock.
( )ρ ρW = Wcontinent
displaced → Ah g = Acontinent h − y g →mantle
mantle
y = h ⎜⎛1 − ρ continent ⎞ = (35 km)⎝⎛⎜1− 2800 kg m3 ⎞ = 5.3 km
⎝ ρ mantle ⎟ 3300 kg ⎟
⎠ m3 ⎠
88. The “extra” buoyant force on the ship, due to the loaded fresh water, is the weight of “extra”
displaced seawater, as indicated by the ship floating lower in the sea. This buoyant force is given by
F = V ρbuoyant g. But this “extra” buoyant force is what holds up the fresh water, and so must
displaced sea
water
also be equal to the weight of the fresh water.
( ) ( ) ( )ρF = V g = m gbuoyant
displaced sea fresh → mfresh = 2240 m2 8.50 m 1025kg m3 = 1.95×107 kg
water
This can also be expressed as a volume.
V = mfresh fresh = 1.95 ×107 kg = 1.95 ×104 m3 = 1.95 ×107 L
1.00 ×103 kg m3
ρfresh
89. (a) We assume that the one descending is close enough to the surface of the Earth that constant
density may be assumed. Take Eq. 13-6b, modify it for rising, and differentiate it with respect
to time.
P = P0 − ρ gy →
( )( )dP dy m3
dt
dt
= −ρ g =− 1.29 kg 9.80 m s2 (−7.0 m s) = 88.49 Pa s ≈ 88 Pa s
(b) Δy = vt → t = Δy = 350 m = 50s (2 sig. fig.)
v 7.0 m s
90. The buoyant force must be equal to the weight of the water displaced by the full volume of the logs,
and must also be equal to the full weight of the raft plus the passengers. Let N represent the number
of passengers.
weight of water displaced by logs = weight of people + weight of logs
( ) ( )12 ρV glog water = Nmperson g + 12 ρV glog log →
( ) ( ) ( )ρ ρN r l2
= 12Vlog −water log 12π ρ ρ− SGwater 12π r l ρ 12 − SGlog
mperson = log log log water =
log log water
mperson
mperson
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437
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
( )12π (0.225 m)2 (6.1m) m3 (1 − 0.60) = 68.48
=
1000 kg
68 kg
Thus 68 people can stand on the raft without getting wet. When the 69th person gets on, the raft
will go under the surface.
91. We assume that the air pressure is due to the weight of the atmosphere, with the area equal to the
surface area of the Earth.
P = F → F = PA = mg →
A
6.38 ×106 m 2 1.013×105 N m2
9.80 m s2
( ) ( )m=PA= 4π R2 P = 4π = 5.29 ×1018 kg ≈ 5 ×1018 kg
g Earth
g
92. The work done during each heartbeat is the force on the fluid times the distance that the fluid moves
in the direction of the force.
W = FΔl = PAΔl = PV →
⎛ 133 N m2 ⎞
⎜⎝ 1 mm-Hg ⎟⎠
( )Power
W PV (105 mm-Hg ) 70 ×10−6 m3
t t
= = = ⎝⎜⎛ 1 ⎠⎞⎟ ⎛⎜⎝ 60 s ⎟⎠⎞ = 1.1W ≈ 1W
70 min
min
93. (a) We assume that the water is launched at ground level. Since it also lands at ground level, the
level range formula from Example 3-10 may be used.
( )R
= v02 sin 2θ → v0 = Rg (7.0 m) 9.80 m s2 = 8.544 m s ≈ 8.5 m s
g sin 2θ =
sin 70o
(b) The volume rate of flow is the area of the flow times the speed of the flow. Multiply by 4 for
the 4 heads.
( )Volume flow rate = Av = 4π r2v = 4π 1.5 ×10−3 m 2 (8.544 m s)
= 2.416 ×10−4 m3 s ⎛⎜⎝ 1.0 1L m3 ⎟⎞⎠ ≈ 0.24 L s
× 10−3
(c) Use the equation of continuity to calculate the flow rate in the supply pipe.
( ) ( )→
( ) ( )Av supply = v =supply Av = 2.416 × 10−4 m3 s s
Av heads π 0.95 ×10−2 m = 0.85 m
heads 2
Asupply
94. The buoyant force on the rock is the force that would be on a mass of water with the same volume as
the rock. Since the equivalent mass of water is accelerating upward, that same acceleration must be
taken into account in the calculation of the buoyant force.
Fbuoyant − mwater g = mwatera →
mrock
= ρ water
ρ rock
( ) ( ) ( ) ( )F = mbuoyant
water g+a = V ρwater water g+a = V ρrock water g+a g+a
( )= s2
mrock (g + 1.8g ) = (3.0 kg) 2.8 9.80 m = 30.49 N ≈ 30 N (2 sig. fig.)
SGrock
2.7
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438
Chapter 13 Fluids
For the rock to not sink, the upward buoyant force on the rock minus the weight of the rock must be
equal to the net force on the rock.
( )( ) ( )Fbuoyant − mrock g = mrocka → Fbuoyant = mrock g + a = 3.0 kg 2.8 9.80 m s2 = 82 N
The rock will sink , because the buoyant force is not large enough to “float” the rock.
95. Apply both Bernoulli’s equation and the equation of continuity at the two locations of the stream,
with the faucet being location 0 and the lower position being location 1. The pressure will be air
pressure at both locations. The lower location has y1 = 0 and the faucet is at height y0 = y .
(( ))A0v0 = A1v1 → v1 = v0 A0 = v0 π d0 2 2 = v0 d 2 →
A1 π d1 2 2 0
d12
P0 + 1 ρ v02 + ρ gy0 = P1 + 1 ρ v12 + ρ gy1 → v02 + 2gy = v12 = v02 d 4 →
2 2 0
d14
d1 = d0 ⎛ v02 v02 ⎞1/ 4
⎜ + 2gy ⎟
⎝ ⎠
96. (a) Apply Bernoulli’s equation between the surface of the water in the sink and the lower end of the
siphon tube. Note that both are open to the air, and so the pressure at both is air pressure.
v2 v +2
top bottom
ρ ρ ρ ρPtop
+ 1 + gytop = Pbottom + 1 gybottom →
2 2
( ) ( ) ( )v =bottom 2g y − ytop bottom = 2 9.80 m s2 0.44 m = 2.937 m s ≈ 2.9 m s
(b) The volume flow rate (at the lower end of the tube) times the elapsed time must equal the
volume of water in the sink.
(( ) ( ) )( )Av lower Δt = Vsink
→ ( )Δt = Vsink = 0.38 m2 4.0 ×10−2 m s) = 16.47 s ≈ 16 s
Av π
1.0 × 10−2 m 2 (2.937 m
lower
97. The upward force due to air pressure on the bottom of the wing must be equal to the weight of the
airplane plus the downward force due to air pressure on the top of the wing. Bernoulli’s equation
can be used to relate the forces due to air pressure. We assume that there is no appreciable height
difference between the top and the bottom of the wing.
( )Ptop A + mg = P Abottom → mg
P − Pbottom top =A
P0 + Pbottom + 1 ρv2 + ρ gybottom = P0 + Ptop + 1 ρv2 + ρ gytop
2 bottom 2 top
( )ρv2
top
= 2 P − Pbottom top + v2 →
bottom
( )2 ( )( )2 1.7 ×106 kg
( )( )1.29 kg m3
vtop = P − Pbottom top + v2 = 2mg + v2 = 9.80 m s2 + (95 m s)2
bottom ρA bottom 1200 m2
ρ
= 174.8 m s ≈ 170 m s
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439
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
98. We label three vertical levels. Level 0 is at the pump, and the supply tube h2 r1
has a radius of r0 at that location. Level 1 is at the nozzle, and the nozzle h1
has a radius of r1 . Level 1 is a height h1 above level 0. Level 2 is the
highest point reached by the water. Level 2 is a height h2 above level 1. We
may write Bernoulli’s equation relating any 2 of the levels, and we may
write the equation of continuity relating any 2 of the levels. The desired
result is the gauge pressure of the pump, which would be P0 − Patm. Start by
using Bernoulli’s equation to relate level 0 to level 1.
P0 + ρ gh0 + 1 ρ v02 = P1 + ρ gh`1 + 1 ρ v12 r0
2 2
Since level 1 is open to the air, P1 = Patm. Use that in the above equation.
P0 − Patm = ρ gh`1 + 1 ρ v12 − 1 ρ v02
2 2
Use the equation of continuity to relate level 0 to level 1, and then use that result in the Bernoulli
expression above.
A0v0 = A1v1 → π r02v0 = π r12v1 → v0 = r12 v1 = d12 v1
r02
d 2
0
P0 − Patm = ρ gh`1 + 1 ρ v12 − 1 ρ ⎛ d12 v1 ⎞2 = ρ gh`1 + 1 ρ v12 ⎜⎛1 − d14 ⎞
2 2 ⎜⎝ ⎟ 2 ⎝ ⎠⎟
d 2 ⎠ d 4
0 0
Use Bernoulli’s equation to relate levels 1 and 2. Since both levels are open to the air, the pressures
are the same. Also note that the speed at level 2 is zero. Use that result in the Bernoulli expression
above.
ρ v12 ρ v22 v12 = 2 gh2
( )P1 P2
+ ρ gh1 + 1 = + ρg h1 + h2 + 1 →
2 2
P0 − Patm = ρ gh`1 + 1 ρ v12 ⎛ 1 − d14 ⎞ = ρg ⎡ h`1 + h2 ⎜⎛1 − d14 ⎞⎤
2 ⎜ ⎠⎟ ⎢ ⎝ ⎠⎟⎦⎥
⎝ d 4 ⎣ d 4
0 0
( ) ( ) ( )= 1.00 ×103 kg m3 9.80 m s2 ⎣⎡1.1m + (0.14 m) 1 − 0.54 ⎦⎤
= 12066 N m2 ≈ 1.2 ×104 N m2
99. We assume that there is no appreciable height difference to be considered between the two sides of
the window. Then the net force on the window due to the air is the difference in pressure on the two
sides of the window, times the area of the window. The difference in pressure can be found from
Bernoulli’s equation.
2 2
inside outside
ρ ρ ρ ρP + v + gy = P + v + gyinside1 1 →
2 inside outside 2 outside
P − P = ρ v = AFinside
outside 12 air →
2 air outside roof
⎡⎢(200 km ⎛ 1m s ⎞ ⎤ 2
⎜⎝ 3.6 km ⎠⎟ ⎦⎥
⎣
( ) ( )F = ρ v A1 2
air 2 air outside roof
= 1 1.29 kg m3 h ) h 6.0 m2 = 1.2 ×104 N
2
100. From Poiseuille’s equation, the viscosity can be found from the volume flow rate, the geometry of
the tube, and the pressure difference. The pressure difference over the length of the tube is the same
as the pressure difference due to the height of the reservoir, assuming that the open end of the needle
is at atmospheric pressure.
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440
Chapter 13 Fluids
Q = π R4 ( P2 − P1 ) ; P2 − P1 = ρblood gh →
8ηl
0.20 ×10−3 m 4 1.05 × 103 kg m3
( ) ( )(( ) )η
= π R4 ( P2 − P1 ) = π R4ρblood gh = π 9.80 m s2 (1.30 m)
8Ql
8Ql 8 ⎢⎡⎣4.1 cm3 × 1 min × 10−6 m3 ⎤ 3.8 × 10−2 m
min 60 s cm3 ⎦⎥
= 3.2 ×10−3 Pais
101. The net force is 0 if the balloon is moving at terminal velocity. Therefore the upwards buoyancy
force (equal to the weight of the displaced air) must be equal to the net downwards force of the
weight of the balloon material plus the weight of the helium plus the drag force at terminal velocity.
Find the terminal velocity, and use that to find the time to rise 12 m.
FB = m gballoon + m gHelium + FD → 4 π r3ρair g = m gballoon + 4 π r3ρHe g + 1 CDρ πair r 2vT2 →
3 3 2
⎡⎣( )24π r 3 − m⎦⎤ g CD ρ πair r2
3 ρair − ρ He h
vT = t t=h
CDρ πair r 2 = ( )→ 2 ⎡⎣ 4 r 3 − m⎤⎦ g →
3 π ρair − ρ He
( ( ) ) ( )t = (12 m)
(0.47) 1.29 kg m3 π (0.15 m)2 = 4.9 s
2 ⎣⎡ 4 π (0.15 m)3 1.29 kg m3 − 0.179kg m3 − (0.0028 kg)⎤⎦ 9.80 m s2
3
102. From Poiseuille’s equation, the volume flow rate Q is proportional to R4 if all other factors are the
same. Thus Q R4 is constant. Also, if the diameter is reduced by 15%, so is the radius.
( )Q Qfinal initial 4
final
=4 4 4
R Rfinal initial initial
→ Q Rfinal 0.85 4 = 0.52
Q = R =initial
The flow rate is 52% of the original value.
103. Use the definition of density and specific gravity, and then solve for the fat fraction, f.
( )mfat = mf ρ= Vfat fat ; mfat = m 1 − f ρ= Vfat fat
free free free
ρX water mtotal mfat + mfat m 1
Vtotal = f + 1− f
free
ρ ρfat fat
Vfat + Vfat
free
( ) ( )ρbody= = = = f →
mf m 1−
ρ fat +
ρ fat
free free
ρ ρfat fat
free
( ( ))(( )) (( ))f
= ⎛ ⎞ − ρfat ⎞ = 0.90 g cm3 1.10 g cm3 − 0.90 g cm3
⎜⎝ ⎠⎟ ⎛ ⎟⎠ X 1.0 g cm3 0.20 g cm3 0.20 g cm3
ρX water ρ − ρ fat ⎜⎝ ρ − ρ fat
fat fat
free free
= 4.95 − 4.5 → % Body fat = 100 f = 100 ⎛ 4.95 − 4.5 ⎠⎟⎞ = 495 − 450
X ⎝⎜ X X
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441
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
104. The graph is shown. The best-fit Pressure (kPa) 120
equations as calculated by Excel are 100
also shown. Let P represent the 80 2000 4000 6000 8000 10000
pressure in kPa and y the altitude in m. 60
40 Altitude (m)
The spreadsheet used for this problem 20
can be found on the Media Manager,
with filename 0
“PSE4_ISM_CH13.XLS,” on tab 0
“Problem 13.104.”
(a) Quadratic fit: ( ) ( )Pquad = 3.9409 ×10−7 y2 − 1.1344 ×10−2 y + 100.91 ,
(b) Exponential fit: Pexp = 103.81e (− 1.3390×10−4 ) y
( ) ( )(c) Pquad = 3.9409 ×10−7 (8611)2 − 1.1344 ×10−2 (8611) + 100.91 = 32.45 kPa
Pexp = 103.81e (− 1.3390×10−4 )(8611) = 32.77 kPa
(( )) ( ( ) )% diff
= 100 Pexp − Pquad = 200 Pexp − Pquad = 200 (32.77 kPa − 32.45 kPa) = 0.98 %
Pexp + Pquad (32.77 kPa + 32.45 kPa )
P + P1 quad
2 exp
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442
CHAPTER 14: Oscillations
Responses to Questions
1. Examples are: a child’s swing (SHM, for small oscillations), stereo speakers (complicated motion,
the addition of many SHMs), the blade on a jigsaw (approximately SHM), the string on a guitar
(complicated motion, the addition of many SHMs).
2. The acceleration of a simple harmonic oscillator is momentarily zero as the mass passes through the
equilibrium point. At this point, there is no force on the mass and therefore no acceleration.
3. When the engine is running at constant speed, the piston will have a constant period. The piston has
zero velocity at the top and bottom of its path. Both of these properties are also properties of SHM.
In addition, there is a large force exerted on the piston at one extreme of its motion, from the
combustion of the fuel–air mixture, and in SHM the largest forces occur at the extremes of the
motion.
4. The true period will be larger and the true frequency will be smaller. The spring needs to accelerate
not only the mass attached to its end, but also its own mass. As a mass on a spring oscillates,
potential energy is converted into kinetic energy. The maximum potential energy depends on the
displacement of the mass. This maximum potential energy is converted into the maximum kinetic
energy, but if the mass being accelerated is larger then the velocity will be smaller for the same
amount of energy. A smaller velocity translates into a longer period and a smaller frequency.
5. The maximum speed of a simple harmonic oscillator is given by v = A k . The maximum speed
m
can be doubled by doubling the amplitude, A.
6. Before the trout is released, the scale reading is zero. When the trout is released, it will fall
downward, stretching the spring to beyond its equilibrium point so that the scale reads something
over 5 kg. Then the spring force will pull the trout back up, again to a point beyond the equilibrium
point, so that the scale will read something less than 5 kg. The spring will undergo damped
oscillations about equilibrium and eventually come to rest at equilibrium. The corresponding scale
readings will oscillate about the 5-kg mark, and eventually come to rest at 5 kg.
7. At high altitude, g is slightly smaller than it is at sea level. If g is smaller, then the period T of the
pendulum clock will be longer, and the clock will run slow (or lose time).
8. The tire swing is a good approximation of a simple pendulum. Pull the tire back a short distance and
release it, so that it oscillates as a pendulum in simple harmonic motion with a small amplitude.
Measure the period of the oscillations and calculate the length of the pendulum from the expression
T = 2π l . The length, l, is the distance from the center of the tire to the branch. The height of the
g
branch is l plus the height of the center of the tire above the ground.
9. The displacement and velocity vectors are in the same direction while the oscillator is moving away
from its equilibrium position. The displacement and acceleration vectors are never in the same
direction.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
10. The period will be unchanged, so the time will be (c), two seconds. The period of a simple pendulum
oscillating with a small amplitude does not depend on the mass.
11. The two masses reach the equilibrium point simultaneously. The angular frequency is independent of
amplitude and will be the same for both systems.
12. Empty. The period of the oscillation of a spring increases with increasing mass, so when the car is
empty the period of the harmonic motion of the springs will be shorter, and the car will bounce
faster.
13. When walking at a normal pace, about 1 s (timed). The faster you walk, the shorter the period. The
shorter your legs, the shorter the period.
14. When you rise to a standing position, you raise your center of mass and effectively shorten the
length of the swing. The period of the swing will decrease.
15. The frequency will decrease. For a physical pendulum, the period is proportional to the square root
of the moment of inertia divided by the mass. When the small sphere is added to the end of the rod,
both the moment of inertia and the mass of the pendulum increase. However, the increase in the
moment of inertia will be greater because the added mass is located far from the axis of rotation.
Therefore, the period will increase and the frequency will decrease.
16. When the 264-Hz fork is set into vibration, the sound waves generated are close enough in frequency
to the resonance frequency of the 260-Hz fork to cause it to vibrate. The 420-Hz fork has a
resonance frequency far from 264 Hz and far from the harmonic at 528 Hz, so it will not begin to
vibrate.
17. If you shake the pan at a resonant frequency, standing waves will be set up in the water and it will
slosh back and forth. Shaking the pan at other frequencies will not create large waves. The individual
water molecules will move but not in a coherent way.
18. Examples of resonance are: pushing a child on a swing (if you push at one of the limits of the
oscillation), blowing across the top of a bottle, producing a note from a flute or organ pipe.
19. Yes. Rattles which occur only when driving at certain speeds are most likely resonance phenomena.
20. Building with lighter materials doesn’t necessarily make it easier to set up resonance vibrations, but
it does shift the fundamental frequency and decrease the ability of the building to dampen
oscillations. Resonance vibrations will be more noticeable and more likely to cause damage to the
structure.
Solutions to Problems
1. The particle would travel four times the amplitude: from x = A to x = 0 to x = − A to x = 0 to
x = A. So the total distance = 4A = 4 (0.18 m) = 0.72 m .
2. The spring constant is the ratio of external applied force to displacement.
k = Fext = 180 N − 75 N = 105 N = 525 N m≈ 530 N m
x 0.85 m − 0.65 m 0.20 m
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444
Chapter 14 Oscillations
3. The spring constant is found from the ratio of applied force to displacement.
( )k s2
= Fext = mg = (68 kg) 9.80 m = 1.333 ×105 N m
x x
5.0 ×10−3 m
The frequency of oscillation is found from the total mass and the spring constant.
f = 1 k1 1.333×105 N m = 1.467 Hz ≈ 1.5 Hz
2π m = 2π 1568 kg
4. (a) The motion starts at the maximum extension, and so is a cosine. The amplitude is the
displacement at the start of the motion.
x = Acos (ωt ) = Acos ⎜⎛⎝ 2π t ⎠⎞⎟ = (8.8cm) cos ⎝⎜⎛ 2π t ⎟⎞⎠ = (8.8 cm ) cos ( 9.520t )
T 0.66
≈ (8.8cm) cos(9.5t)
(b) Evaluate the position function at t = 1.8 s.
( )x = (8.8cm) cos 9.520s−1 (1.8s) = −1.252 cm ≈ −1.3cm
5. The period is 2.0 seconds, and the mass is 35 kg. The spring constant can be calculated from Eq. 14-
7b.
T = 2π m → T2 = 4π 2 m → k = 4π 2 m = 4π 2 35 kg = 350 N m
k k T2
( 2.0 s )2
6. (a) The spring constant is found from the ratio of applied force to displacement.
( )k s2
= Fext = mg = (2.4 kg) 9.80 m = 653 N m ≈ 650 N m
x x
0.036 m
(b) The amplitude is the distance pulled down from equilibrium, so A = 2.5cm
The frequency of oscillation is found from the oscillating mass and the spring constant.
f = 1 k1 653 N m = 2.625 Hz ≈ 2.6 Hz
2π m = 2π 2.4 kg
7. The maximum velocity is given by Eq. 14-9a.
vmax = ωA = 2π A = 2π (0.15m) = 0.13 m s
T
7.0 s
The maximum acceleration is given by Eq. 14-9b.
amax = ω2A = 4π 2 A = 4π 2 (0.15m) = 0.1209 m s2 ≈ 0.12 m s2
T2 ( 7.0 s )2
amax = 0.1209 m s2 = 1.2 ×10−2 = 1.2%
g 9.80 m s2
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445
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
8. The table of data is time position position / A 1 0.25 0.5 0.75 1 1.25
shown, along with 0 -A
the smoothed graph. T/4 0 0
Every quarter of a T/2 A 0
period, the mass
moves from an 3T/4 0
extreme point to the T -A
equilibrium. The
5T/4 0
graph resembles a cosine wave (actually, -1
the opposite of a cosine wave).
time / T
9. The relationship between frequency, mass, and spring constant is Eq. 14-7a, f = 1 k.
2π m
( )(a) 1 k
f = 2π m → k = 4π 2 f 2m = 4π 2 ( 4.0 Hz)2 2.5×10−4 kg = 0.1579 N m ≈ 0.16 N m
(b) f = 1 k1 0.1579 N m = 2.8 Hz
2π m = 2π 5.0 ×10−4 kg
10. The spring constant is the same regardless of what mass is attached to the spring.
f = 1 k → k = mf 2 = constant → m1 f12 = m2 f12 →
2π m 4π 2
(m kg) (0.83Hz)2 = (m kg + 0.68 kg) (0.60 Hz)2 → m = (0.68 kg) (0.60 Hz)2 = 0.74 kg
(0.83Hz)2 − (0.60 Hz)2
11. We assume that the spring is stretched some distance y0 while the A θ Fs
rod is in equilibrium and horizontal. Calculate the net torque
about point A while the object is in equilibrium, with clockwise
torques as positive.
∑τ = Mg ( 1 l ) − Fsl = 1 Mgl − ky0l =0 Mg
2 2
Now consider the rod being displaced an additional distance y
below the horizontal, so that the rod makes a small angle of θ as shown in the free-body diagram.
Again write the net torque about point A. If the angle is small, then there has been no appreciable
horizontal displacement of the rod.
∑τ = Mg ( 1 l ) − Fsl = 1 Mgl −k(y+ y0 ) l = Iα = 1 Ml 2 d 2θ
2 2 3 dt 2
Include the equilibrium condition, and the approximation that y = l sinθ ≈ l θ.
1 Mgl − kyl − ky0l = 1 Ml 2 d 2θ → 1 Mgl − kyl − 1 Mgl = 1 Ml 2 d 2θ →
2 3 dt 2 2 2 3 dt 2
−kl 2θ = 1 Ml 2 d 2θ → d 2θ + 3k θ = 0
3 dt 2 dt 2 M
This is the equation for simple harmonic motion, corresponding to Eq. 14-3, with ω2 = 3k .
M
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446
Chapter 14 Oscillations
ω2 = 4π 2 f 2 = 3k → f= 1 3k
M 2π M
12. (a) We find the effective spring constant from the mass and the frequency of oscillation.
f = 1 k →
2π m
k = 4π 2mf 2 = 4π 2 (0.055kg) (3.0 Hz)2 = 19.54 N m ≈ 20 N m (2 sig fig)
(b) Since the objects are the same size and shape, we anticipate that the spring constant is the same.
f = 1 k1 19.54 N m = 1.4 Hz
2π m = 2π 0.25 kg
13. (a) For A, the amplitude is AA = 2.5 m . For B, the amplitude is AB = 3.5 m .
(b) For A, the frequency is 1 cycle every 4.0 seconds, so fA = 0.25 Hz . For B, the frequency is 1
cycle every 2.0 seconds, so fB = 0.50 Hz .
(c) For C, the period is TA = 4.0s . For B, the period is TB = 2.0s
(d) Object A has a displacement of 0 when t = 0 , so it is a sine function.
xA = AAsin (2π fAt) → xA = (2.5m) sin ( 1 πt)
2
Object B has a maximum displacement when t = 0 , so it is a cosine function.
xB = ABcos (2π fBt) → xB = (3.5m) cos (π t)
14. Eq. 14-4 is x = Acos (ωt + φ ) .
(a) If x (0) = − A, then − A = Acosφ → φ = cos−1 (−1) → φ = π .
(b) If x (0) = 0 , then 0 = Acosφ → φ = cos−1 (0) → φ = ± 1 π .
2
(c) If x (0) = A , then A = Acosφ → φ = cos−1 (1) → φ = 0 .
(d) If x (0) = 1 A , then 1 A= A cos φ → φ = cos−1 ( 1 ) → φ = ± 1 π .
2 2 2 3
(e) If x (0) = − 1 A, then − 1 A= Acosφ → φ = cos−1 (− 1 ) → φ = ± 2 π .
2 2 2 3
( )(f)
If x (0) = A 2 , then A 2 = Acosφ → φ = cos−1 1 → φ = ± 1 π .
2 4
The ambiguity in the answers is due to not knowing the direction of motion at t = 0.
15. We assume that downward is the positive direction of motion. For this motion, we have
k = 305 N m , A = 0.280 m, m = 0.260 kg, and ω = k m = (305 N m) 0.260 kg = 34.250 rad s .
(a) Since the mass has a zero displacement and a positive velocity at t = 0, the equation is a sine
function.
y (t ) = (0.280 m)sin[(34.3rad s)t]
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447
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(b) The period of oscillation is given by T = 2π = 2π s = 0.18345s . The spring will have
ω 34.25 rad
its maximum extension at times given by the following.
tmax T + nT = 4.59×10−2 s + n (0.183 s) , n = 0,1, 2,
=4
The spring will have its minimum extension at times given by the following.
tmin = 3T + nT = 1.38×10−1s + n (0.183 s), n = 0,1, 2,
4
16. (a) From the graph, the period is 0.69 s. The period and the mass can be used to find the spring
constant.
T = 2π m → k = 4π 2 m = 4π 2 0.0095 kg = 0.7877 N m≈ 0.79 N m
k T2
( 0.69 s)2
(b) From the graph, the amplitude is 0.82 cm. The phase constant can be found from the initial
conditions.
x = Acos ⎝⎜⎛ 2π t + φ ⎠⎟⎞ = (0.82 cm) cos ⎝⎜⎛ 2π t + φ ⎠⎟⎞
T 0.69
x (0) = (0.82 cm) cosφ = 0.43cm → φ = cos−1 0.43 = ±1.02 rad
0.82
Because the graph is shifted to the RIGHT from the 0-phase cosine, the phase constant must be
subtracted.
x= ( 0.82 cm) cos ⎜⎛⎝ 2π t − 1.0 ⎠⎟⎞ or (0.82 cm) cos (9.1t − 1.0)
0.69
17. (a) The period and frequency are found from the angular frequency.
ω = 2π f → f = 1 ω = 1 5π = 5 Hz T = 1 = 1.6 s
2π 2π 4 8 f
(b) The velocity is the derivative of the position.
x = (3.8 m) cos ⎛ 5π t + π ⎞ v = dx = − (3.8m) ⎛ 5π ⎞ sin ⎛ 5π t + π ⎞
⎝⎜ 4 6 ⎠⎟ dt ⎝⎜ 4 ⎠⎟ ⎝⎜ 4 6 ⎟⎠
x (0) = (3.8 m) cos ⎛ π ⎞ = 3.3 m v (0) = − ( 3.8 m) ⎛ 5π ⎞ sin ⎛ π ⎞ = −7.5 m s
⎜⎝ 6 ⎠⎟ ⎜⎝ 4 ⎠⎟ ⎜⎝ 6 ⎠⎟
(c) The acceleration is the derivative of the velocity.
v = − ( 3.8 m) ⎛ 5π ⎞ sin ⎛ 5π t + π ⎞ a = dv = − (3.8 m) ⎛ 5π ⎞2 cos ⎛ 5π t + π ⎞
⎜⎝ 4 ⎠⎟ ⎜⎝ 4 6 ⎠⎟ dt ⎝⎜ 4 ⎟⎠ ⎜⎝ 4 6 ⎠⎟
v ( 2.0) = − (3.8 m) ⎛ 5π ⎞ sin ⎛ 5π ( 2.0) + π ⎞ = −13 m s
⎝⎜ 4 ⎟⎠ ⎜⎝ 4 6 ⎠⎟
a (2.0) = − ( 3.8 m) ⎝⎜⎛ 5π ⎞2 cos ⎛⎜⎝ 5π (2.0) + π ⎟⎠⎞ = 29 m s2
4 ⎟⎠ 4 6
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