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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


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× = ×

��

��

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

��

��

��

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−

×∆ = = = × =

×.


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 .


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.

�

�

��

��

��

(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

×= = =

×


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 =

��

��

��

��

��

��

��

��

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 = ⋅ = = ⋅


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=

��

��

��

��

�

�

� ��

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 =


(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= =


(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= − = − =


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 − −∆ = × = ×


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× =


(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


(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 −

−

× ⋅ = = =×


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


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.

��

��

��


(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++

= = = ×−


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 =

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

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=


(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 ∆ = − = − −

× = − − ×

∆ =


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

⋅ −=

−= =


(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.


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 −∆ = × =