Chapter 9 Solids and Fluids - Croom Physics · Chapter 9 Solids and Fluids ... Solids and Fluids...
Transcript of Chapter 9 Solids and Fluids - Croom Physics · Chapter 9 Solids and Fluids ... Solids and Fluids...
Chapter 9
Solids and Fluids
Quick Quizzes
1. (c). The mass that you have of each element is: ( ) ( )3 3 3 319.3 10 kg/m 1 m 19.3 10 kggold gold goldm Vρ= = × = ×
(
) ( )3 3 310.5 10 kg/m 2 m 21.0 10 ksilver silver silverm Vρ= = × = ×
(3 g
) ( )3 3 32.70 10 kg/m 6 m 16.aluminum aluminum aluminumm Vρ= = × 310 kg= ×2
2. (a). At a fixed depth, the pressure in a fluid is directly proportional to the density of the fluid. Since ethyl alcohol is less dense than water, the pressure is smaller than P when the glass is filled with alcohol.
3. (c). For a fixed pressure, the height of the fluid in a barometer is inversely proportional to the density of the fluid. Of the fluids listed in the selection, ethyl alcohol is the least dense.
4. (b). The blood pressure measured at the calf would be larger than that measured at the arm. If we imagine the vascular system of the body to be a vessel containing a liquid (blood), the pressure in the liquid will increase with depth. The blood at the calf is deeper in the liquid than that at the arm and is at a higher pressure.
Blood pressures are normally taken at the arm because that is approximately the same height as the heart. If blood pressures at the calf were used as a standard, adjustments would need to be made for the height of the person, and the blood pressure would be different if the person were lying down.
5. (c). The level of floating of a ship is unaffected by the atmospheric pressure. The buoyant force results from the pressure differential in the fluid. On a high-pressure day, the pressure at all points in the water is higher than on a low-pressure day. Because water is almost incompressible, however, the rate of change of pressure with depth is the same, resulting in no change in the buoyant force.
6. (b). Since both lead and iron are denser than water, both objects will be fully submerged and (since they have the same dimensions) will displace equal volumes of water. Hence, the buoyant forces acting on the two objects will be equal.
7. (a). When there is a moving air stream in the region between the balloons, the pressure in this region will be less than on the opposite sides of the balloons where the air is not moving. The pressure differential will cause the balloons to move toward each other. This is demonstration of Bernoulli’s principle in action.
331
332 CHAPTER 9
Answers to Even Numbered Conceptual Questions
2. We approximate the thickness of the atmosphere by using 0P P ghρ= + with P at the top of the atmosphere and P at sea level. This gives an approximation of
0 0=1 atm=
( )( )5
40 10 Pa0
1 kg m 1P P
hgρ−
3 1 2
0~ 1
0 m s−
= = m ~ 1h or
Because both the density of the air,
0 km
ρ , and the acceleration of gravity, g, decrease with altitude, the actual thickness of the atmosphere will be greater than our estimate.
4. Both must have the same strength. The force on the back of each dam is the average pressure of the water times the area of the dam. If both reservoirs are equally deep, the force is the same.
6. The external pressure exerted on the chest by the water makes it difficult to expand the chest cavity and take a breath while under water. Thus, a snorkel will not work in deep water.
8. A fan driven by the motor removes air and hence decreases the pressure inside the cleaner. The greater air pressure outside the cleaner pushes air in through the nozzle toward this region of lower pressure. This inward rush of air pushes or carries the dirt along with it.
10. The larger the density of a fluid, the higher does an object float in it. Thus, an object will float lower in low density alcohol.
12. The water level on the side of the glass stays the same. The floating ice cube displaces its own weight of liquid water, and so does the liquid water into which it melts.
14. A breeze from any direction speeds up to go over the mound, and the air pressure drops at this opening. Air then flows through the burrow from the lower to the upper entrance.
16. No. The somewhat lighter barge will float higher in the water.
18. Although the area and pressures is the same at the base of each vessel, the volumes of the vessels differ, so the weight of the liquid in each vessel increases as its volume increases.
Solids and Fluids 333
Answers to Even Numbered Problems
2. (a) (b) 43.14 10 N× 46.28 10 N×
4. 81.65 10 Pa×
6. 22 N directed down the page in the figure
8. 67.5 10 Pa×
10. (a) 2.5 mm (b) 0.75 mm (c) 36.9 10 kg×
12. The stress is , so the arm should survive. 75.6 10 Pa×
14. 41.9 10 N×
16. 61.2 10 Pa×
18. (a) (b) 30.053 8 m− 3 31.09 10 kg m× (c) Yes, in most practical circumstances.
20. (a) 65.1 N (b) 275 N
22. 10.5 m; no, some alcohol and water evaporate.
24. 2.3 lb
26. 0.611 kg
28. 10.7% of the volume is exposed
30. (a) 1 017.9 N, 1 029.7 N (b) 86.2 N (c) 11.8 N for both
32. 4 21.28 10 m×
34. 16.5 cm
36. (a) 3 38.57 10 kg m× (b) 3714 kg m
38. 78 kg
40. 154 in s
42. (a) 11.0 m s (b) 42.64 10 Pa×
44. 24.4 10 Pa−×
334 CHAPTER 9
46. (a) 17.7 m s (b) 1.73 mm
48. (a) 15.1 MPa (b) 2.95 m s (c) 4.34 kPa
50. 347 m s
52. 27.32 10 N m−×
54. 5.6 m
56. 0.694 mm
58. 0.12 N
60. 1.5 m s
62. 51.5 10 Pa×
64. 455 kPa
66. 8.0 cm s
68. 10 29.5 10 m s−×
70. 3 31.02 10 kg m×
72. (b) 41.25 10 Pa×
74. (a) 10.3 m (b) zero
78. 1.9 m
80. (a) 1.25 cm (b) 13.8 m s
82. 0.72 mm
84. (b) 26 kN
86. (a) 18.3 mm (b) 14.3 mm (c) 8.56 mm
88. 1.71 cm
90. 0.86 mm
Solids and Fluids 335
Problem Solutions
9.1 The elastic limit is the maximum stress, F A where F is the tension in the wire, that the wire can withstand and still return to its original length when released. Thus, if the wire is to experience a tension equal to the weight of the performer without exceeding the elastic limit, the minimum cross-sectional area is
2min
min 4mgD F
Aelastic limit elastic limit
π= = =
( )
and the minimum acceptable diameter is
( ) ( )
( )2
3min 8
4 70 kg 9.8 m s41.3 10 m 1.3 mm
5.0 10 Pamg
Delastic limitπ π
−= = = × =×
9.2 (a) ( ) ( )23 8 25.00 10 m 4.00 10 N m 3.14 10 NA stress π − = × × × 4F = ⋅
(b) The area over which the shear occurs is equal to the circumference of the hole times the thickness of the plate. Thus,
( )
( ) ( )3 3
4 2
2
2 5.00 10 m 5.00 10 m
1.57 10 m
A r tπ
π − −
−
=
= × ×
= ×
So, ( )( )4 21.57 10 m 4.F A stress −= ⋅ = × 8 2 400 10 N m 6.28 10 N× = ×
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9.3 F
StressA
= , where ( )0.30F weig= ht and 2A rπ=
Thus, ( )
( )610 Pa×23
0.30 480 N
5.0 10 m−= =
×1.8Stress
π
336 CHAPTER 9
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V , so
9.4 As a liquid, the water occupied some volume V . As ice, the water would occupy volume if it was not compressed and forced to occupy the original volume. Consider the pressure change required to squeeze ice back into volume V . Then,
l
1.090 lV
9 and V V= ∆l
0 1.0 0.090l lV = −
9 82
0N10 1.65 10 Pa 600 atm
m 1 09l
l
V − × = × 0
2.00V ∆
= − .090.V V
1≈P B∆ = −
9.5 Using ( )
0F LY
A L∆= with
2
4d
Aπ
= and F mg= , we get
( )( ) ( )
( ) ( )
2
822
4 90 kg 9.80 m s 5010 Pa
1.0 10 m 1.6 mY
π −
×
m3.5= = ×
)
9.6 From (
0F LY
A L∆= , the tension needed to stretch the wire
by 0.10 mm is
( ) ( )( )
( ) ( ) ( )( )
2
0 0
210 3 3
-2
4
18 10 Pa 0.22 10 m 0.10 10 m22 N
4 3.1 10 m
Y d LY A LF
L L
π
π − −
∆∆= =
× × ×= =
×
cos30 cos30 0x xR F F F
The tension in the wire exerts a force of magnitude F on the tooth in each direction along the length of the wire as shown in the above sketch. The resultant force exerted on the tooth has an x-component of = Σ = − ° + ° =
sin 30 sin 30 22 Ny y F F F= − ° − ° = − = −, and a y-component of
.
Thus, the resultant force is
R F= Σ
22 N directed down the page in the diagram=R .
9.7 From ( )0L LFstres 0sY
A L L = = ∆ ∆
( )(
, the maximum compression the femur can withstand
is
) ( ) ( )6
0 39
160 10 Pa 0.50 m4.4 10 m 4.4 mm
18 10 Pa
stress LL
Y−
×∆ = = = × =
×.
Solids and Fluids 337
9.8 The shear modulus is given by ( )
shear stress stressS
shear strain x h= =
∆.
Hence, the stress is
( )10 63
5.0 m1.5 10 Pa 7.5 10 Pa
10 10 mx
stress Sh∆ = = × = × ×
9.9 From the defining equation for the shear modulus, we find the displacement, , as
x∆
( ) ( ) ( )( ) ( )
3 4 2
26 2
5
5.0 10 m 20 N 10 cm1 m3.0 10 Pa 14 cm
2.4 10 m 0.024 mm
h F A h Fx
S S A
−
−
× ⋅∆ = = = ⋅ ×
= × =
9.10 (a) When at rest, the tension in the cable equals the weight of the 800-kg object,
. Thus, from 37.84 10 N×( )
0F LY
A L=
∆, the initial elongation of the cable is
( ) ( )
( ) ( )3
34 2 10
7.48 10 N 25.0 m2.45 10 m= 2.5 mm
4.00 10 m 20 10 Pa−
−
×= = ×
× ×0F L
LA Y⋅
∆ =⋅
(b) When the load is accelerating upward, Newton’s second law gives
yF mg ma− = , or ( )yF m g a= + (1)
If 23.0 m sy = +800 kg an= d m a , the elongation of the cable will be
( )( ) ( )
( )( )2
310
3.0 m s 25.0 m3.2 10 m=3.2 mm
20 10 Pa−
= ×× ×
04 2
800 kg 9.80
4.00 10 mF L
LA Y −
+⋅∆ = =
⋅
(
Thus, the increase in the elongation has been ) ( ) 3.20initialincrease L L= ∆ − ∆ = mm 2.45 mm 0.75 mm− =
338 CHAPTER 9
(c) From the definition of the tensile stress, stress F A= , the maximum tension the cable can withstand is ( ) ( )( )4 2 4
max max 4.00 10 m 8.8 10 NF A stress −= ⋅ = × = ×82.2 10 Pa×
Then, equation (1) above gives the mass of the maximum load as
( )
43max
max 2
8.8 10 N6.9 10 kg
9.8+3.0 m sF
mg a
×= = = ×
+.
9.11 The tension and cross-sectional area are constant through the entire length of the rod, and the total elongation is the sum of that of the aluminum section and that of the copper section.
( ) ( ) ( ) ( )0 00 0Cu CuAl Al
Al CuAl Cu Al Cu
rod
F L LF L LFL L L
AY AY A Y Y
∆ = ∆ + ∆ = + = +
2
where A rπ= 30.20 cm 2.0 10 m−= = ×
(
with r . Thus,
)
( )
3
2 103
10 N 1.3 m7.0 10 Pa10 m−
×
×2
10
5.8 2.6 m1.9 10 m 1.9 cm
11 10 Pa2.0π− + = × = × ×
rodL∆ =
9.12 The acceleration of the forearm has magnitude
3
3 23
km 10 m 1 h80
h 1 km 3 600 s4.4 10 m/s
5.0 10 sv
at −
∆
= = = ×∆ ×
The compression force exerted on the arm is F ma= and the compressional stress on the bone material is
( ) ( )
( )3 2
72 4 2 2
3.0 kg 4.4 10 m s0 Pa
2.4 cm 10 m 1 cmF
StressA −
×= = 5.6 1= ×
Since the stress is less than the allowed maximum, the arm should survive .
Solids and Fluids 339
9.13 Applying Newton’s second law to the dancer gives ( ), or F mg ma F m g a− = = +
where F is the normal force exerted on the dancer by the floor, and a is the upward acceleration (if any) the dancer is given.
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(a) When 0a = , then ( ) ( )2 50.0 kg 9.80m s 490 NF mg= = = , and the pressure is
5-4 2
490 N1.88 10 Pa
26.0 10 mP
FA
= = = ××
(b) When 24.00 m s
(
a = + , the normal force is
) ( )2 50.0 kg 13.8m s 690 NF = = and 5-4 2
690 N2.65 10 Pa
26.0 10 mP = = ×
×.
9.14 Let the weight of the car be W. Then, each tire supports 4
W,
and the gauge pressure is 4
F WP
A A= = .
Thus, ( ) ( )2 5 44 4 0.024 m 2.0 10 Pa 1.9 10 NW AP= = × = ×
9.15 Neglecting surface tension effects, the pressure inside the bubble is the same as the pressure in the alcohol at this distance below its surface. This is
( )0 3 2 5
kg m 1 at1.10 atm 806 9.80 4.0 m
m s 1.013 10alcoholP P ghρ = + = + ×
m Pa
1.4 atmP =
9.16 The total downward force is the combined weight of the man and chair. This force is distributed over an area equal to 2 times the cross-sectional area of a leg. Hence, the pressure is
( )
( )( )( )
( )2
622 2
75 kg 9.80 m s1.2 10 Pa
2 2 1.0 10 mman chairm m gF
PA rπ π −
+= = = = ×
×
340 CHAPTER 9
9.17 The volume of concrete in a pillar of height h and cross-sectional area A is V , and
its weight is
Ah=
( )( )45.0 10 N mgF Ah= × 3 . The pressure at the base of the pillar is then
( )( ) ( )
4 34 3
5.0 10 N m5.0 10 N mg AhF
P hA A
×= = = ×
max 1.7 10P = ×
Thus, if the maximum acceptable pressure is, , the maximum allowable height is
7 Pa
7max
max 4 3
1.7 10 Pa5.0 10 N m 5.0 10 N m
Ph
×= = =
× ×2
4 3 3.4 10 m×
9.18 (a) From the definition of bulk modulus, ( )0B P V V= −∆ ∆ , the change in volume of
the 1. of seawater will be
300 m
( ) ( )( )3 8 53
10
1.00 m 1.13 10 Pa 1.013 10 Pa0.053 8 m
0.210 10 Pawater
V P × − ×∆= − = −
×0VB
∆ = −
(b) The quantity of seawater that had volume V at the surface has a mass of 1 030 kg. Thus, the density of this water at the ocean floor is
30 1.00 m=
( )3 3
30
1 030 kgm
1.00 0.053 8 mm mV V V
ρ = = =+ ∆ −
1.09 10 kg= ×
(c) Considering the small fractional change in volume (about 5%) an enormous change in pressure generated, we conclude that
it is a good approximation to think of water as incompressible
9.19 The density of the solution is 31.02 1.02 10 kg mwaterρ ρ= = × 3 . The gauge pressure of the fluid at the level of the needle must equal the gauge pressure in the vein, so
, and
41.33 10 PagaugeP ghρ= = ×
( ) ( )4
3 3 2
1.33 10 Pa1.33 m
1.02 10 kg m 9.80 m sgaugeP
hgρ
×= = =
×
Solids and Fluids 341
9.20 (a) Suppose the “vacuum cleaner” functions as a high–vacuum pump and produces “zero” pressure inside the hose. The air below the brick will then exert a net upward force of
( ) ( ) ( )22 5 21.013 10 Pa 1.43 10 m 65.1 NhoseF PA P rπ π − = = = × × =
(b) The octopus can pull the bottom away from the top shell with a force that could be no larger than ( )
( ) ( )
0
25 23 2
kg m1.013 10 Pa 1 030 9.80 32.3 m 1.43 10 m
m s
F PA P gh Aρ
π −
= = +
= × + ×
or 275 NF =
9.21 The excess water pressure (over air pressure) acting on the wall is
( ) ( )( )3 3 2 4avav
0 2.40 m10 kg m 9.80 m s 1.18 10 Pa
2gaugeP ghρ+ = = = ×
(
Hence, the net inward, horizontal force exerted on the wall by the water is
) ( ) ( )( )4 5
av1.18 10 Pa 9.60 m 2.40 m 2.71 10 N 271 kNgaugeF P A= = × = × =
9.22 If we assume a vacuum exists inside the tube above the wine column, the pressure at the base of the tube (that is, at the level of the wine in the open container) is
0atmoP gh ghρ ρ= + = . Thus,
( ) ( )5
3 2
1.013 10 Pa10.5 m
984 kg m 9.80 m satmoP
hgρ
×= = =
Some alcohol and water will evaporate, degrading the vacuum above the column.
342 CHAPTER 9
9.23 We first find the absolute pressure at the interface between oil and water.
( )( )( )
1 0
5 3 21.013 10 Pa 700 kg m 9.80 m s 0.300 m 1.03 10 Pa
oil oilP P gh
5
ρ= +
= × + = ×
2 1 water waterP P gh
This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use ρ= + , or
( ) ( ) ( )5 3 3 2 52 1.03 10 Pa 10 kg m 9.80 m s 0.200 m 1.05 10 PaP = × + = ×
9.24 First, use Pascal’s principle, 1 1 2 2F A F A= , to find the force piston 1 will exert on the handle when a 500-lb force pushes downward on piston 2.
( )( )
( )
2 211 1
1 2 22 22 2
2
2
44
0.25 in500 lb 14 lb
1.5 in
dA dF F F
A d dππ
= = =
= =
0
22
F
Now, consider an axis perpendicular to the page, passing through the left end of the jack handle. τΣ = yields ( )( ) ( )14 lb 2.0 in 12 in 0F+ − ⋅ = , or 2.3 lbF =
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9.25 Pascal’s principle, 1 1 2 2F A=F A , gives
( )2
2aster
1.8 cm44 N 12.4 N
6.4 cmbrake cylinder
brake pedalm cylinder
AF F
A
= =
(
= .
This is the normal force exerted on the brake shoe. The frictional force is )0.50 12.4 N 6.2 Nkf nµ= = = ,
and the torque is ( ) ( )6.2 N 0.34 m 2.1 N mdrumf rτ = ⋅ = = ⋅
Solids and Fluids 343
9.26 Since the frog floats, the buoyant force must equal the weight of the frog. Then, from Archimedes’ principle, the weight of the displaced fluid equals the weight of the frog.
Hence, ( )fluid frogV g m gρ = , or
( )3
3
g 4 6.00 cm15 611 g 0.611 kg
cm 2 3π = =
1.3frog fluidm Vρ= =
9.27 The boat sinks until the weight of the additional water displaced equals the weight of the truck. Thus,
( )
( )( )( )3 23 2
kg m10 4.00 m 6.00 m 4.00 10 m 9.80 ,
m s
truck waterW V gρ
−
= ∆
= ×
or 39.41 10 N 9.41 kNtruckW = × =
9.28 When the iceberg floats, the weight of the displaced water must equal the weight of the ice. Thus,
( ) ( )water water ice iceV g Vρ ρ= g , or water ice
ice water
VV
ρρ
=
The volume of the displaced water equals the volume of ice submerged, so the volume of ice exposed is exposed ice submerged ice waterV V= −V V and the fraction of the ice exposed is
V= −
3
3 0.10=920 kg m
11 030 kg m
ce
ater
= −1 1exposed water i
ice ice w
V VV V
ρρ
= − = − 7 , or 10.7%
9.29 (a) From Archimedes’s principle, the granite continent will sink down into the peridotite layer until the weight of the displaced peridotite equals the weight of the continent. Thus, at equilibrium,
( ) ( )g pAt g Ad gρ ρ =
or g pt dρ ρ=
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344 CHAPTER 9
(b) If the continent sinks 5.0 km below the surface of the peridotite, then , and the result of part (a) gives the first approximation of the thickness of the continent as
5.0 kmd =
( )3 3
p
3 3g
3.3 10 kg m5.0 km 5.9 km
2.8 10 kg mt d
ρρ
×= = = ×
9.30 Note: We deliberately violate the rules of significant figures in this problem to illustrate a point.
(a) The absolute pressure at the level of the top of the block is
( )
0
5 3 23 2
5
kg m1.0130 10 Pa 10 9.80 5.00 10 m
m s
1.0179 10 Pa
top water topP P ghρ
−
= +
= × + ×
= ×
and that at the level of the bottom of the block is
( )2
0
5 33 2
5
kg m1.0130 10 Pa 10 9.80 17.0 10 m
m s
1.0297 10 Pa
bottom water bottomP P ghρ
−
= +
= × + ×
= ×
(
Thus, the downward force exerted on the top by the water is )( )251.0179 10 Pa 0.100 m 1017.9 Ntop topF P A= = × =
(
and the upward force the water exerts on the bottom of the block is ) ( )251.0297 10 Pa 0.100 m 1029.7 Nbot botF P A= = × =
(b) The scale reading equals the tension, T, in the cord supporting the block. Since the block is in equilibrium, 0y bot topF T F F mgΣ = + − − = , or
( ) ( ) ( )2s 1029.7 1017.9 N= − −10.0 kg 9.80 m 86.2 NT =
Solids and Fluids 345
(c) From Archimedes’s principle, the buoyant force on the block equals the weight of the displaced water. Thus, ( )
( ) ( ) ( ) ( )23 3 210 kg m 0.100 m 0.120 m 9.80 m s 11.8 N
water blockB V gρ=
= =
From part (a), ( )1 029.7 1 017.9 N=11.8 Nbot topF F− = − , which is the same as the
buoyant force found above.
9.31 Constant velocity means that the submersible is in equilibrium under the gravitational force, the upward buoyant force, and the upward resistance force:
0y yF maΣ = =
( )41.20 10 kg− × 1100 N 0sea waterm g gVρ+ + + =
where m is the mass of the added sea water and V is the sphere’s volume.
Thus, ( )3 43 2
kg 1100 N4 1.50 m 1.20 10 kgm 3 9.80 m s
π + − × 31.03 10m= ×
or 32.67 10m = × kg
9.32 By Archimedes’s principle, the weight of the fifty planes is equal to the weight of a horizontal slice of water 11.0 cm thick and circumscribed by the water line, or
( ) ( ) ( ) ( )3 250 1 030 k m 11.0 10 mplane waterm g B g V g Aρ −= ∆ = ∆ = ×
(
Thus, )
( ) ( )4 2
3 2
50 29 000 kg1.28 10 m
1 030 k m 11.0 10 m
gA
g −= = ×
×
Note that the free-fall acceleration cancels in this calculation and does not affect the answer.
346 CHAPTER 9
9.33 The balloon is in equilibrium under the action of three forces. These are the buoyant force, B, the total weight, W, of the balloon and the helium, and the tension T in the string. Hence, ( ) 0y balloon heliumF B m m g TΣ = − + − = , or ( )balloon heliumT B m m g= − +
The buoyant force is
( )air balloonB V gρ= helium heliumm and balloonVρ= , where 34
3balloon
rV
π=
Thus, ( )
( ) ( ) ( ) ( )( )
3
3 2
43
4 kg m 9.80 m s 0. 0 kg 9.80 m s
3
balloon
rT g m g
π
π
−
3500 m 0.012− 21.29 0.181
air heliumρ ρ= −
= −
or 5.57 NT =
9.34 At equilibrium, Σ = so the spring force is
0y springF B F mg− − =
( )springF B mwater blockmg Vρ = − = − g
where -3 310 mblock ρ×3
5.00 kg7.69
650 kg mwood
m= = =V .
Thus, ( )( ) ( )3 3 -3 310 kg m 7.69 10 m 5.00springF = × 2 kg 9.80 m s 26.4 N − =
The elongation of the spring is then
26.4 N
0.165 m 16.5 cm160 N m
springFx
k∆ = = = =
9.35 (a) The buoyant force is the difference between the weight in air and the apparent weight when immersed in the alcohol, or 300 N 200 N 100 NB = − = . But, from Archimedes’s principle, this is also the weight of the displaced alcohol, so
( )alcoholB Vρ= g . Since the sample is fully submerged, the volume of the displaced
alcohol is the same as the volume of the sample. This volume is
( ) ( )2 31.46 10 m−= ×
3 2
100 N700 kg m 9.80 m salcohol
BV
gρ= =
Solids and Fluids 347
(b) The mass of the sample is 2
300 N30.6 kg
9.80 m sweight in air
mg
= = = ,
and its density is 3 32 3
30.6 kg2.10 10 kg m
1.46 10 mmV
ρ −= = = ××
9.36 The difference between the weight in air and the apparent weight when immersed is the buoyant force exerted on the object by the fluid.
(a) The mass of the object is 2
300 N30.6 kg
9.80 m sweight in air
g= = =m . The buoyant force
when immersed in water is the weight of a volume of water equal to the volume of the object, or ( )w wB ρ= V g . Thus, the volume of the object is
( )3 3
2
265 N3.57 10 m
9.80 m sw
w
BV
gρ−
3 3
300 N10 kg m
−= = = × ,
and its density is 3 33 3
30.6 kg8.57 10 kg m
3.57 10 mobject
mV
ρ −= = = ××
( )
(b) The buoyant force when immersed in oil is equal to the weight of a volume of oil. Hence, 33.57 10 mV −= × 3 ( )oil oilB Vρ= g , or the density of the oil is
( ) ( )3
3 3 2
300 N 275 N714 kg m
3.57 10 m 9.80 m soil
oil
BV g
ρ−
−= = =
×
9.37 The volume of the shell is 3
3 34 4 0.200 m4.19 10 m
3 3 2π π − = = = ×
3V r , so the mass of
alcohol filling the shell is ( ) ( )3 3 3m 4.19 10 m 3.38 kg−× =806 kga am Vρ= =
(
,
and the mass of the filled shell is )0.shell am m m= + = 400 3.38 kg 3.78 kg+ =
348 CHAPTER 9
The buoyant force exerted on the shell by the water is ( ) ( )( )( )3 3 3 3 210 kg m 4.19 10 m 9.80 m s 41.1 NwaterB V gρ −= = × = ,
and the upward acceleration is
2 241.1 N9.80 m s 1.07 m s
3.78 kgy
y
F B mg Ba g
m m m
Σ −= = = − = − =
9.38 When the mattress is fully submerged, the buoyant force exerted by the water (and hence the total weight that can be supported) is
( )( ) ( )( )( ) ( )3 3 2 210 kg m 2.0 m 0.50 m 0.080 m 9.80 m s 7.8 10 N
waterB V gρ=
= = ×
Thus, the total mass that can be supported is
2
2
7.8 10 N80 kg
9.80 m stotal
Bm
g×
= = =
The addition mass that can be placed on the mattress is then 80 kg 2.0 kg 78 kgadditional total mattressm m m= − = − =
9.39 The volume of the iron block is
4 33 3
2.00 kg2.54 10 m
7.86 10 kg miron
iron
mV
ρ−= = = ×
×
(
,
and the buoyant force exerted on the iron by the oil is ) ( )( )( )3 4 3 2916 kg m 2.54 10 m 9.80 m s 2.28 NoilB V gρ −= = × =
0yFΣ =
Applying to the iron block gives the support force exerted by the upper scale
(and hence the reading on that scale) as 19.6 N 2.28 N 17.3 Nupper ironF m g B= − = − =
Solids and Fluids 349
From Newton’s third law, the iron exerts force B downward on the oil (and hence the beaker). Applying to the system consisting of the beaker and the oil gives
0yFΣ =
( ) 0lower erF B m g− − =
(
oil beakm + The support force exerted by the lower scale (and the lower scale reading) is then ) ( ) ( )228 N 2.00 1.00 kg 9.80 m s 31.7 N + + = 2.lower akerF B m g= + =oil bem +
9.40 The volume flow rate of an incompressible fluid through a conduit is V t Av∆ ∆ = , where A is the cross-sectional area of the conduit and v is the average velocity of the fluid. Thus, for the specified pipe,
( )
( ) ( )( ) ( )
3
2
20.0 gal 231 in 1 gal154 in s
1.00 in 30.0 sV
vA t∆
= = =∆
9.41 (a) The volume flow rate is Av , and the mass flow rate is ( ) ( ) ( )3 21.0 g cm 2 40 cm s 80 g sAvρ = =.0 cm
(b) From the equation of continuity, the speed in the capillaries is
( )2
3 2
2.0 cm40 cm s
3.0 10 cmaorta
capiliaries aortacapillaries
Av v
A
= = ×
,
or 22.7 10 cm s 0.27 mm scapiliaries
−= × =v
9.42 (a) From the equation of continuity, the flow speed in the second pipe is
( )2
12 1 2
2
10.0 cm2.75 m s 11.0 m s
2.50 cmA
v vA
= = =
350 CHAPTER 9
(b) Using Bernoulli’s equation and choosing 0y = along the centerline of the pipes gives
( )
( ) ( ) ( )
2 22 1 1 2
2 25 3 3
12
11.20 10 Pa 1.65 10 kg m m s 11.0 m s
2
P P v vρ= + −
2.75 = × + × −
or 42 2.64 10 Pa= ×P
9.43 From Bernoulli’s equation, choosing 0y = at the level of the syringe and needle,
2 22 2 1
1 12 2
P v P 1vρ ρ+ = + ,
so the flow speed in the needle is ( )1 22 P P
ρ−2
2 1v v= +
In this situation,
( ) 41 2 1 1 5 2
2.00 N8.00 10 Pa
50 10 matmo gaugeP P P P P −− = − = = = ×
×
1 0v ≈
1 2.FA
=
Thus, assuming , ( )4
2 3 3
.00 10 Pa0 1
10 kg mv
×=
×
2 8
1.00= + 2.6 m s
9.44 We apply Bernoulli’s equation, ignoring the very small change in vertical position, to
obtain ( ) ( )22 2 21 2 2 1 1 1
1 12
2 2P P v v v v v2
1
32
ρ ρ ρ − = − = − = , or
( ) ( )23 2 231.29 kg m 15 10 m s 4.4 10 Pa
2P − −∆ = × = ×
Solids and Fluids 351
9.45 First, consider the path from the viewpoint of projectile motion to find the speed at
which the water emerges from the tank. From 20
12yy v t a t∆ = + y with , we find the
time of flight as
0 0yv =
( ) ( )2
2 2 1.00 m0.452 s
9.80 m sy
yt
a
∆ −= = =
−
From the horizontal motion, the speed of the water coming out of the hole is
2 0
0.600 m1.33 m s
0.452 sx
xv v
t∆
= = = =
1 2 atmoP P P= = 1v
We now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the hole. With and 0≈ , this gives
21 2 2
12
gy v gyρ ρ ρ= + , or
( )( )
222
1 2 2
1.33 m s9.0
2 2 9.80 m sv
h y yg
= − = = = 210 m 9.00 cm− =0 ×
9.46 (a) Apply Bernoulli’s equation with point 1 at the open top of the tank and point 2 at the opening of the hole. Then, 1 2 atmoP P P= = and we assume v . This gives 1 0≈
22 2
12
v gy gρ ρ ρ+ =
(
1y , or
) ( ) ( )20 m s 16.0 m2 1 22 2 9.8 17.7 m sv g y y= − = =
(b) The area of the hole is found from
3 3
6 22
2
2.50 10 m min 1 min2.35 10 m
17.7 m s 60 sflow rate
Av
−−× = = = ×
The diameter is then
( )6 2
322
4 2.35 10 m41.73 10 m 1.73 mm
Ad
π π
−−
×= = = × =
352 CHAPTER 9
9.47 First, determine the flow speed inside the larger portions from
( )
4 3
1 221
1.80 10 m s0.367 m s
2.50 10 m 4
flow ratev
A π
−
−
×= = =
×
The absolute pressure inside the large section on the left is P P1 0 gh1ρ= + , where is the height of the water in the leftmost standpipe. The absolute pressure in the constriction is
1h
2 0 2P P ghρ= +
(
, so ) ( )1 2 1 2 5.00 cmP P g h h gρ ρ− = − = The flow speed inside the constriction is found from Bernoulli’s equation with 1 2y y= .
This gives ( ) ( )2 2 22 1 1 2 1 1 2
22v v P P v g h h
ρ= + − = + − , or
( ) ( ) ( )20.367 m s 2= + 210 m 1−×2 9.80 m sv 5.00 .06 m s=
The cross-sectional area of the constriction is then
4 3
2
10 m s1.06 m s
flA
−× 4 210 m−×
2
1.80ow ratev
= = 1.71= ,
and the diameter is
( )4 210 m
1.47π π
−×= 2
2 10 m 1.47 cmd −× =24 1.714 A
= =
9.48 (a) For minimum pressure, we assume the flow is very slow. Then, Bernoulli’s equation gives
2 21 1=
2 2river rim
P v gy P v gyρ ρ ρ ρ + + + +
(
) ( )0 1 atm 0river rim rivermin
P g yρ+ = + + − y
or, ( ) ( )3 2
m80 2096 m 564 m
m s −
5 3 kg1.013 10 Pa 10 9.river min
P = × +
(
.
) ( )5 71.013 10 1.50 10 Pa=1river min
P = × + × 7.51 10 Pa 15.1 MPa× =
Solids and Fluids 353
(b) The volume flow rate is 2
=4d
flow rate Av vπ
= . Thus, the velocity in
the pipe is ( ) ( )
( )
3
22
4 4500 m d4 1 d2.95 m s
86 400 s0.150 m
flow ratev
dπ π
= = =
(c) We imagine the pressure being applied to stationary water at river level, so Bernoulli’s equation becomes
( ) 210 1 atm
2river rim river rimP g y yρ ρ + = + − +
( )
v , or
( )
( )
22 3 kg1 1
10 2.95 2 2 m
4.34 kPa
river river rim rivermin min
river min
P P v P
P
ρ = + = + 3
= +
ms
The additional pressure required to achieve the desired flow rate is 4.34 kPaP∆ =
9.49 (a) For upward flight of a water–drop projectile from geyser vent to fountain–top, ( )2 2
0 2y y yv v a y= + ∆ , with 0yv = when maxy y∆ = ∆ , gives
( ) ( ) ( )0 0 2y yv a= − 2y = − − 2max
9.8 m s 40.0 m 28.0 m s∆ =
(b) Because of the low density of air and the small change in altitude, atmospheric pressure at the fountain top will be considered equal to that at the geyser vent. Bernoulli’s equation, with v 0top = , then gives
( )210
2 vent top ventyv g yρ ρ= + −
(
, or
) ( ) ( )22 0 m s 40.0 m 28.0 m svent top ventv g y y= − =2 9.8=
354 CHAPTER 9
(c) Between the chamber and the geyser vent, Bernoulli’s equation with 0chamberv ≈ yields
( ) 210
2atm vent ventchamberP gy P v gyρ ρ+ + = + + ρ , or
( )
( ) ( )
2
2
33 2
12
28.0 m skg m10 9.80 175 m 2.11 MPa
m 2 s
atm vent vent chamberP P v g y yρ − = + −
= +
20.8 atmospheresgauge atmoP P P= − =
=
or
9.50 The assumption of incompressibility is surely unrealistic, but allows an estimate of the
speed. From 2 21 1 1 2 2
1 12 2
v gy P v gyρ ρ ρ ρ+ + = + + 2P , with 2 0iy y= = and inside the
passenger compartment, we have
0v ≈
( )3 22
1 atm 1.20 kg m 0 1.00
2v+ +
( )
0.287 atm+0+0= ,
or 5 2
347
= = 2 3
2 1.00 atm 0.287 atm 1.013 10 N m m s
1.20 kg m 1 atmv
− ×
9.51 (a) Choosing point 1 at the top of the tank and point 2 at the exit from the tube, Bernoulli’s equation with v1 0≈ gives
( )22 1 2
12
v P P gρ ρ= − +
1 2 1 atmP P= =
( )1 2y y−
1 2y y
But, , and h− = . Thus, 2 2=v g h
Solids and Fluids 355
(b) Use Bernoulli’s equation with point 1 at the top of the tank and point 2 at the highest point in the tube. This gives
( ) ( ) ( )2 22 1 2 1 2
12atmg y y P P v vρ ρ− = − + −
2 1 0v v= = 2 1 maxy y y− =
When the siphon ceases to work, the fluid will be at rest at point 2, so and . Thus, P P2 matm gy axρ= + .
Since the minimum value of is 0, 2P 0max
atmP Py
g gρ ρ= =
9.52 Because there are two edges (the inside and outside of the ring) we have,
( )-2
2-2
= = 2( )
1.61 10 N= = 7.32 10 N m
4 4 1.75 10 m
total
F FL circumference
Fr
γ
π π−×
= ××
9.53 From Σ = , the balance reading is found to be 0yF T mg F− − =y yT mg F= + where yF is
the vertical component of the surface tension force. Since this is a two-sided surface, the surface tension force is ( )2F γ L= and its vertical component is ( )2 cyF osLγ φ= where φ
is the contact angle. Thus, T m 2 cosg Lγ φ= + . when 0.40 NT = 0 φ = °⇒ 2 0.40mg L Nγ+ = (1) when 0.39 NT = 180 φ = °⇒ 2 0.39mg L Nγ− = (2) Subtracting equation (2) from (1) gives
( )28.3 10 N m−×
2
N 0.39 N.0 10 m−
− −= = =
×0.40 N
4γ
0.3L
9 N 0.404 3
.
9.54 The height the blood can rise is given by
( )
( )( )( )2 2 6
2 0.058 N m cos02 cos5.6 m
1 050 kg m 9.80 m s 2.0 10 mh
grγ φρ −
°= = =
×
356 CHAPTER 9
9.55 From 2 cos
grh
γ φρ
= , the surface tension is
( )( )( )( )2 3 2 42
2cos
2.1 10 m 1 080 kg m 9.80 m s 5.0 10 m5.6 10 N m
2cos0
h grργ
φ
− −−
=
× ×= =
°×
9.56 From 2 cos
grh
γ φρ
= , the radius of the capillary tube is
( )
( )( )( )4
3 2 2
2 0.088 N m cos02 cos3.47 10 m
1 035 kg m 9.80 m s 5.00 10 mr
ghγ φρ
−−
°= = = ×
×
(
The diameter is then )4 42 2 3.47 10 m 6.94 10 m= 0.694 mmd r − −= = × = ×
9.57 From the definition of the coefficient of viscosity, F LAv
η = , the required force is
( ) ( )( ) ( )3 2
3
1.79 10 N s m 0.800 m 1.20 m 0.50 m s8.6 N
0.10 10 mAv
FL
η −
−
× ⋅ = = =×
9.58 From the definition of the coefficient of viscosity, F LAv
η = , the required force is
( ) ( )( ) ( )3 2
3
1 500 10 N s m 0.010 m 0.040 m 0.30 m s0.12 N
1.5 10 mAv
FL
η −
−
× ⋅ = = =×
Solids and Fluids 357
9.59 Poiseuille’s law gives ( ) 4
1 2 8
P P Rflow rate
L
πη
−= , and 2 atmP P= in this case. Thus, the
desired gauge pressure is
( ) ( )( )( )
( )2 5
1 44
8 0.12 N s m 10 m s8
0.50atm
L flow rateP P
R
ηπ π
−⋅ ×− = =
3
2
50 m 8.6
10 m−×,
or 6
1 2.1 10 Pa 2.1 MPaatmP P− = × =
9.60 From Poiseuille’s law, the flow rate in the artery is
( ) ( ) ( )
( )( )
4345 3
3 2 2
400 Pa 2.6 10 m 3.2 10 m s
8 8 2.7 10 N s m 8.4 10 mP R
flow rateL
ππη
−−
− −
×∆= = = ×
× ⋅ ×
Thus, the flow speed is ( )
5 3
2-3
3.2 10 m s1.5 m s
2.6 10 m
flow ratev
A π
−×= = =
×
9.61 If a particle is still in suspension after 1 hour, its terminal velocity must be less than
( ) 5max
cm 1 h 1 m5.0 1.4 10 m s
h 3 600 s 100 cmtv − = =
× .
Thus, from ( )22
9t f
r gv ρ ρ
η= −
( )( )
, we find the maximum radius of the particle:
( )( )( ) ( )
max9
2
9 1.00 10 N
2 9.80 m s
t
f
v
g
η
ρ ρ=
−max
3 2 56
2 3
s m 1.4 10 m s2.8 10 m 2.8 m
1 800 1 000 kg m
r
µ− −
−× ⋅ ×
= = −
× =
358 CHAPTER 9
9.62 From Poiseuille’s law, the excess pressure required to produce a given volume flow rate of fluid with viscosity η through a tube of radius R and length L is
( )
4
8 L VP
R
ηπ∆ ∆
∆ =t
If the mass flow rate is ( ) -31.0 10 kg sm t∆ ∆ = × , the volume flow rate of the water is
-3
6 33 3
1.0 10 kg s1.0 10 m s
10 kg mm tV
t ρ−∆ ∆ ×∆
= = = ×∆ ×
(
1.0
and the required excess pressure is
) ( ) ( )( )
3 2 6 35
43
8 1.0 10 Pa s 3.0 10 m 1.0 10 m s1.5 10 Pa
0.15 10 mP
π
− − −
−
× ⋅ × ×∆ = = ×
×
9.63 With the IV bag elevated 1.0 m above the needle, the pressure difference across the needle is ( ) ( ) ( )3 3 2 31.0 10 kg m 9.8 m s 1.0 m 9.8 10 PaP ghρ∆ = = × = ×
and the desired flow rate is
( )( )
3 3 6 37 3
500 cm 1 m 10 cm2.8 10 m s
30 min 60 s 1 minVt
−∆= = ×
∆
Poiseuille’s law then gives the required diameter of the needle as
( )( )
( ) ( ) ( )( )
1 43 2 7
3
10 m 2.8 10 m s
10 PaD R
− − −1 48 1.0 10 Pa s 2.58
2 2 29.8
L V t
P
ηπ π
3 × × × ⋅∆ ∆= = =
∆ ×
or 44.1 10 m 0.41 mm−= × =D
Solids and Fluids 359
9.64 We write Bernoulli’s equation as 2 21 1
2 2out out out in in inP v gy P v gρ ρ ρ ρ+ + = + + y
or ( ) ( )2 212gauge in out out in outP P P v v g yρ= − = − + iny −
Approximating the speed of the fluid inside the tank as
,
we find
0inv ≈
( ) ( ) ( ) ( )23 3 211.00 10 kg m 30.0 m s 0 9.80 m s 0.500 m
2gaugeP = × − +
or 54.55 10 Pa 455 kPa= × =gaugeP
������
������
9.65 The Reynolds number is
( )( )( )3 2
3 2
1 050 kg m 0.55 m s 2.0 10 m
2.7 10 N s mvd
RNρη
−
−
×= =
× ⋅= 34.3 10×
In this region (RN > 3 000), the flow is turbulent .
9.66 From the definition of the Reynolds number, the maximum flow speed for streamlined (or laminar) flow in this pipe is
( ) ( )( )
( )( )3 2
maxmax 3 2
1.0 10 N s m 2 0000.080 m s 8.0 cm s
1 000 kg m 2.5 10 m
RNv
d
ηρ
−
−
× ⋅⋅= = = =
×
9.67 The observed diffusion rate is 14
158.0 10 kg5.3 10 kg s
15 s
−−×
= × . Then, from Fick’s law, the
difference in concentration levels is found to be
( )
( ) ( )( ) ( )
2 1
153 3
10 2 4 2
5.3 10 kg s 0.10 m1.8 10 kg m
5.0 10 m s 6.0 10 m
Diffusion rate LC C
DA
−−
− −
− =
×= =
× ××
360 CHAPTER 9
9.68 Fick’s law gives the diffusion coefficient as ( ) Diffusion rate
DA C L
=⋅ ∆
, where C L∆ is the
concentration gradient.
Thus, ( ) ( )15
10 24 2 2 4
5.7 10 kg s9.5 10 m s
2.0 10 m 3.0 10 kg mD
−−
− −
×= =
× ⋅ ××
9.69 Stokes’s law gives the viscosity of the air as
( ) ( )13
5 26 4
3.0 10 N1.4 10 N s m
6 6 2.5 10 m 4.5 10 m sFrv
ηπ π
−−
− −
×= = = × ⋅
× ×
9.70 Using (22
9)t f
r gv ρ ρ
η−= , the density of the droplet is found to be 2
92
tf
vr gη
ρ ρ= + . Thus, if
30.500 10 m2
r −= = ×d
and 21.10 10 m st−= ×v when falling through 20 °C water
1.00 10 N s mη = × ⋅ , the density of the oil is
( ) ( )
3 2 2
− −
3 323 4 2
9 1.00 10 N s m 1.10 10 m skg1 000 1.02 10 kg m
m 2 5.00 10 m 9.80 m sρ
−
× ⋅ ×= + = ×
×
( )3 2−
( )( )
9.71 (a) Both iron and aluminum are denser than water, so both blocks will be fully submerged. Since the two blocks have the same volume, they displace equal amounts of water and the buoyant forces acting on the two blocks are equal.
(b) Since the block is held in equilibrium, the force diagram at the right shows that
The buoyant force B is the same for the two blocks, so the spring scale reading is
0 yF T mgΣ = ⇒ = −
T
B
largest for the iron block which has a higher
density, and hence weight, than the aluminum block.
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��
������� ��
Solids and Fluids 361
(c) The buoyant force in each case is
( ) ( )( )( )3 3 3 2 3water 1.0 10 kg m 0.20 m 9.8 m s 2.0 10 NB V gρ= = × = ×
(
For the iron block: ) ( ) ( ) ( )3 3 3 2
iron iron 7.86 10 kg m 0.20 m 9.8 m sT V g B Bρ= − = × −
or 4 3 3iron 1.5 10 N 2.0 10 N 13 10 NT = × − × = ×
(
For the aluminum block: ) ( ) ( ) ( )3 3 3 2
aluminum aluminum 2.70 10 kg m 0.20 m 9.8 m sT V g Bρ= − = × B−
or 3 3 3aluminum 5.2 10 N 2.0 10 N 3.3 10 NT = × − × = ×
9.72 (a) Starting with P P0 ghρ= + , we choose the reference level at the level of the heart, so
0 HP P= . The pressure at the feet, a depth Hh below the reference level in the pool of blood in the body is F HP P Hghρ= + . The pressure difference between feet and heart
is then F H Hghρ− =P P .
(b) Using the result of part (a),
( ) ( ) ( )3 3 2 41.06 10 kg m 9.80 m s 1.20 m 1.25 10 PaF HP P− = × = ×
9.73 The cross-sectional area of the aorta is 21 1 4A dπ= and that of a single capillary is
22 4cA dπ= . If the circulatory system has N such capillaries, the total cross-sectional area
carrying blood from the aorta is
22
2 4c
N dA NA
π= = From the equation of continuity,
12 1
2
vA A
v
= , or
2 22 11
24 4N d dv
vπ π
= ,
which gives
2 22
72 6
1.0 m s 0.50 10 m2.5 10
1.0 10 m s 10 10 m
−
− −
×= = = × × ×
1 1
2 2
v dN
v d
362 CHAPTER 9
9.74 (a) We imagine that a superhero is capable of producing a perfect vacuum above the water in the straw. Then 0P P ghρ= + , with the reference level at the water surface inside the straw and P being atmospheric pressure on the water in the cup outside the straw, gives the maximum height of the water in the straw as
( ) ( )5 2
max 3 3 2
3 10 N m010.3 m
kg m 9.80 m satm atm
water water
P Ph
g gρ ρ×−
= = =1.01
1.00 10=
×
(b) The moon has no atmosphere so 0atmP = , which yields max 0h =
9.75 (a) ( )22 H O160 mm of H O 160 mmP gρ= =
( )33 2
kg m10 9.80 0.160 m
m s = =
1.5
( )
7 kPa
35
1 atm1.57 10 Pa
1.013 10 PaP = × = ×
2 2H O H O H
21.55 10−× atm
The pressure is g HgP gh ghρ ρ= = , so
( )2
2
H OHg H O
Hg
10 kg m13.6 10 kg
h hρρ
= = ×
3 3
3 3 160m
mm 11.8 mm of Hg=
(b) The fluid level in the tap should rise.
(c) Blockage of flow of the cerebrospinal fluid.
9.76 When the rod floats, the weight of the displaced fluid equals the weight of the rod, or
0f displaced rodgV gVρ ρ= . But, assuming a cylindrical rod, V . The volume of fluid
displaced is the same as the volume of the rod that is submerged, or
2rod rπ= L
( )2displaced r Lπ= −V .
Thus,
h
( )20h g r Lρ π − =
2 f g r Lρ π , which reduces to 0fρ ρ= LL h−
9.77 When the balloon floats, the weight of the displaced air equals the combined weights of the filled balloon and its load. Thus, air balloon balloon helium balloon loadgV m g gV m gρ ρ= + + ,
or ( )
3 33
600 kg 4 000 kg4.14 10 m
1.29-0.179 kg mballoon load
balloonair helium
m mV
ρ ρ++
= = = ×−
Solids and Fluids 363
9.78 When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the filled balloon plus the weight of string that is above ground level. If m and L are the total mass and length of the string, the mass of string that is above ground
level is
s
s
hm
L
. Thus,
air balloon balloon helium balloon s
hgV m g gV m g
Lρ ρ = + +
(
,
which reduces to )air helium balloon balloon
s
V mh L
m
ρ ρ − −=
.
This yields
( ) ( )
( )31.29 3 0.25 kg
2.0 m 1.9 mh − − = =
3 3 kg m 0.179 kg m 4 0.40 m
0.050 kg
π
9.79 In the sketches below, is the scale reading when only the sinker is submerged and is the reading when both sinker and block are submerged.
1 200 NT =0 N2 14T =
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Applying the first condition of equilibrium to the entire system shown in each case gives With only sinker submerged: 10 y b s sF T W W BΣ = ⇒ = + −
20
With both block and sinker submerged: y b s s bF T W W B BΣ = ⇒
b w
= + − −
1 2aterB Vg T T
Thus, the buoyant force on the submerged block is ρ= = − , or the volume
of the wooden block is ( ) ( )1 2 waterT gρ= −V T .
364 CHAPTER 9
The mass of the wooden block is: ( )50.0 Nbm W g g= = so the density of the wood is
( )( ) ( )
( ) ( ) ( )3 33
1 21 2
50.0 N 1.00 10 kg m50.0 N 50.0 N833 kg m
200 N 140 Nwater
woodwater
gmV T TT T g
ρρ
ρ×
= = = = =− −−
9.80
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(a) Consider the pressure at points A and B in part (b) of the figure by applying
0 fP P ghρ+= . Looking at the left tube gives ( ) A atm waterP P g L hρ+= − , and looking at
the tube on the right, B atm oilP P gLρ+= . Pascal’s principle says that B AP P= . Therefore, ( )atm oilP g atL P+ +m water g L hρ ρ= − , giving
( )3
31 5.001 0 m
l
ter
h L
= −
750 kg m00 kg
1 oi
wa
ρρ
= − cm 1.25 cm=
Solids and Fluids 365
(b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bernoulli’s equation to points A and B. This gives
2 21 12 2A air A air A B air B air BP v gy P v gyρ ρ ρ ρ+ + = + +
, , and 0A B A By y v v v= = =
Since , this reduces to
212B A airP P vρ− = (1)
Now use 0 fP P ghρ+= to find the pressure at points C and D, both at the level of
the oil–water interface in the right tube. From the left tube, C A waterP P gLρ+= , and from the right tube, D B oilP P gLρ+= . Pascal’s principle says that D CP P= , and equating these two gives B oil A wP gL P ater gLρ ρ+ = + , or ( ) B A water oilP P gLρ ρ− = − (2) Combining equations (1) and (2) yields
( )
( )( )( )2 200 10 m13.8
−− ×= =
2
2 9.80 m
v =
1 000 750
water oil
air
gLρ ρρ−
m s 5.
1.29s
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9.81 Consider the diagram and apply Bernoulli’s equation to points A and B, taking at the level of point B, and recognizing that v . This gives
0y =0A ≈
( )2
0 sin
1 0
2B w B
h L
P v
ρ θ
ρ
+ +
= + +
A wP g −
Recognize that A BP P= = atmP
( )
since both points are open to the atmosphere. Thus, we obtain
( ) ( )22 2 9.80 m s 10.0 m 2.00 m sin 30.0 13.3 m s = − ° = sh= − inv g L θB
366 CHAPTER 9
Now the problem reduces to one of projectile motion with 0 sin 30.0 6.64 m sy Bv v= ° =
0, and yv y= =
(
At the top of the arc, .
Then,
maxy
)2 20 2y y yv v a y= + ∆ gives ( ) ( )( )2 2
maxs 2 9.80 m s 0y0 6.64 m= + − − ,
or max 2.25 m y = above the level of point B
9.82 The increase in pressure as the ball sinks is wP ghρ∆ = , where h is the depth at the ocean bottom.
From the definition of bulk modulus, ( )i
PB
V V∆
= −∆
, the change in volume will be
( ) ( )i wV P V B gh Vρ∆ =− ∆ =− i B . Since the volume of a sphere is 343
V rπ
= , this result
may be written as ( ) ( )3 34 43 3
w 3f i
ghr r r
B
ρi
π π− = − , which reduces to
( )3 3 1 wf i
ghr r
B
ρ = −
.
Thus, the final diameter is
( ) ( )1 3 1 3
2 2 1 1w wf f i i
gh ghD r r D
B B
ρ ρ = = − = −
and the decrease in the diameter has been
( )
( )( )( )( )
1 3
1 33 2 3
10
1 1
1 030 kg m 9.80 m s 10.0 10 m3.00 m 1 1
14 10 Pa
0.72 mm
wi f i
ghD D D D
B
D
ρ ∆ = − = − −
× = − − ×
∆ =
Solids and Fluids 367
9.83 While the ball is submerged, the buoyant force acting on it is ( )wVρ=B . The upward acceleration of the ball while under water is
g
( )( ) ( )
3
33
41
3
1 000 kg m 40.10 m 1 9.80 m s
1.0 kg 3
y wy
F B mga r g
m m mπρ
π
Σ − = = = −
= − =
2 231 m s
Thus, when the ball reaches the surface, the square of its speed is ( ) ( )( )2 2 2 2
0 2 0 2 31 m s 2.0 m 125 my y yv v a y= + ∆ = + = 2s
When the ball leaves the water, it becomes a projectile with initial upward speed of
0 125 m syv = and acceleration of 29.80 m sya g= − = − . Then, ( )2 20 2y y ya y= + ∆v v gives
the maximum height above the surface as
( )2 2
max 2
0 125 m s6.4 m
2 -9.80 m sy
−= =
9.84 (a) The gauge pressure on the surface of one of the hemispheres is the same at all points, and the inward force exerted on each small element of surface is directed along the radius of the hemisphere. To separate the hemispheres, the force applied along the axis must overcome the vector sum of all these small elements of force. This sum is equal to the force the gauge pressure exerts on a circular area, 2A Rπ= , which is the projection of the hemispherical surface onto a plane perpendicular to the axis. Therefore, the required force is
( ) 2gauge oF P A P P Rπ= = −
(b) If P P , the necessary force is
0.10 and 0.30 mo R= =
( )
( ) ( )
20 00.10
0.90 1.013 10 Pa
F P P Rπ
π
= −
= × 25 40.30 m 2.6 10 N= 26 kN= ×
368 CHAPTER 9
9.85 The weight of the soap bar is equal to the buoyant force when it floats in water alone, or ( )1.5 cmbar ww Aρ= ⋅
total oil waterB B B= +
(
g ,
where A is the surface area of either the top or bottom of the rectangular bar. When both oil and water are present, the weight of the floating bar equals the total buoyant force, . Thus, ) [ ] ( )1.5 cmw o 2.0 cmil wA g Aρ ρ x g A xρ g ⋅ + ⋅ −⋅ = ,
which reduces to ( )2.0 cm 1.5 cmw oilρ ρ
= −
wxρ−
0.60oil w
Since ρ ρ= , this gives ( )12.0 cm 1.5 cm
1 0.60 x = − = −
1.3 cm
9.86 Let s = be the length of one edge of the ice cube. Then, the area of one face of the cube is and its volume is . We shall ignore any buoyant force exerted by the air in comparison to those exerted by the more dense fluids in all cases.
20.0 mm
3s
2s
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(a) Since the ice cube is floating, the weight of the displaced water must equal the weight of the cube. Thus, ( )2 3
w a ices x g s gρ ρ⋅ = , or
( )3
3
m20.0 mm
g m
917 kg18.3 mm
1 000 kice
aw
x sρρ
= = =
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(b) Here, the sum of the buoyant forces exerted by the alcohol and the water must equal the weight of the floating ice cube. This gives ( )2 25.00 mmal w b ices g s x gρ ρ + ⋅ =
(
3s gρ ,
or )
( )( ) ( )( )
5.00 mm
917 20.0 mm 806 5.00 mm14.3 mm
1 000
ice alb
w
sx
ρ ρρ
⋅ −=
−= =
Solids and Fluids 369
(c) Again the sum of the buoyant forces exerted by the alcohol and the water must equal the weight of the floating ice cube, so ( )2 2
al c w c ices x g s s x g s gρ ρ ρ ⋅ + − = 3 .
This gives ( )1 000 9171 000 806
w icec
w al
ρ ρρ ρ
−−= = − −
20.0 mm 8.56 mm
=
x s
9.87 A water droplet emerging from one of the holes becomes a projectile with 0 00 and y xv v v= = .
The time for this droplet to fall distance h to the
floor is found from 20
12y yt a t+y v∆ = to be
2h
tg
=
The horizontal range is 2
t v= =
1 2R R=
hg
R v .
If the two streams hit the floor at the same spot, it is necessary that , or
1 22 2h hg g1 2v v=
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With h , this reduces to
1 25.00 cm and 12.0 cmh= =
21 2 2
1
12.0 cm5.00 cm
hv v v
h= = , or 1 2 2.40v v=
3 0v
(1)
Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water). The pressure is atmospheric pressure at both points and, if the tank is large in comparison to the size of the holes, ≈ . Thus, we obtain
21 1
12atm atmP v gh Pρ ρ+ + = 3h0+ gρ+ , or ( )2
1 32v g h h= − 1 . (2)
Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and point 3 gives
22 2
12atm atmP v gh Pρ ρ+ + = 3h0+ gρ+ , or ( )2
2 32v g h h= − 2 . (3)
370 CHAPTER 9
Square equation (1) and substitute from equations (2) and (3) to obtain ( ) ( )3 1 3 22 2.40 2g h h g h h − = −
3h
Solving for yields
( )2 1
3
2.40 12.0 cm 5.00 cm017.0 cm
1.40 1.40h h
h−−
= = =2.4
,
so the surface of the water in the tank is 17.0 cm above floor level .
9.88 Since the block is floating, the total buoyant force must equal the weight of the block. Thus,
( ) [ ]
( )
4.00 cm
4.00 cm
oil water
wood
A x g A x g
A g
ρ ρ
ρ
− + ⋅
=
where A is the surface area of the top or bottom of the rectangular block. Solving for the distance x gives
( ) ( )960 9304.00 cm 4.00 cm 1.71 cm
1 000 930wood oil
water oil
xρ ρρ ρ
− −= = − −
=
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9.89 In order for the object to float fully submerged in the fluid, its average density must be the same as that of the fluid. Therefore, we must add ethanol to the water until the density of the mixture is 3900 kg m 0.900 g cm= 3 . The mass of the mixture will be
( )30.900 g cmM V Vρ= = , where V is the total volume of the mixture.
The mass of water in the mixture is ( ) ( )3 31.00 g cm 500 cm 500 gw w wm Vρ= = = ,
and the mass of ethanol added is ( )30.806 g cme e e em V Vρ= =
e
where V is the volume of ethanol added.
Solids and Fluids 371
The total mass is ( )3500 g+ 0.806 g cmw eM m m V= + =
3500 cm +w eV= + =
e
e
and the total volume is V V Substituting these into
V
( )30.900 g cmM = V from above gives
( ) ( ) ( )3 3500 g+ 0.806 g cm 0.900 g cm= 3500 cm +e eV V
Solving for the volume of the added ethanol yields
( ) ( )
( )
3 3
3
.900 g cm 500 cm53
0.806 g cm= =
−3
500 g 02 cm
0.900eV−
9.90 When the section of walkway moves downward distance L∆ , the cable is stretched distance ∆ and the column is compressed distance L L∆ . The tension force required to stretch the cable and the compression force required to compress the column this distance is
steel cable
cable
Y A LF
L∆
cable = and Al columcolumn
colum
Y AL
n
n
LF
∆=
Combined, these forces support the weight of the walkway section:
or 8 500 Ncolumn gF F F+ = =cable 8 500 Nsteel cabl column
cable column
Y A A Le L AlYL L
∆ ∆+ =
giving 8 500 N
steel cable Al column
cable column
Y A Y AL L
+L∆ =
The cross-sectional area of the cable is ( )222 1.27 10 m
cable
DA
π −
= =4 4
×π
and the area of aluminum in the cross section of the column is
( ) ( ) ( )2 2
2 22 2 0.161 4 m
4 4 4 4outer innerouter inner
D DD D πππ π 0.162 4 mcableA
−− = − = =
372 CHAPTER 9
Thus, the downward displacement of the walkway will be
( ) ( )( )
( ) ( ) (( )
)2 22 1010 2
8 500 N
7.0 10 Pa 0.162 4 m 0.161 4 m20 10 Pa 1.27 10 m
4 5.75 m 4 3.25 m
Lππ −
∆ = × −× × +
or 48.6 10 m 0.86 mmL −∆ = × =