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Chapter 15. Fluids

Physics, 6th Edition

Challenge Problems
*15-37. Human blood of density 1050 kg/m3 is held a distance of 60 cm above an arm of a patient to whom it is being administered. How much higher is the pressure at this position than it would be if it were held at the same level as the arm? P = ρgh = (1050 kg/m3)(9.8 m/s2)(0.6 m); P = 6.17 kPa

*15-38. A cylindrical tank 50 fthigh and 20 ft in diameter is filled with water. (a) What is the water pressure on the bottom of the tank? (b) What is the total force on the bott0m? (c) What is the pressure in a water pipe that is located 90 ft below the water level in the tank? (a) P = (62.4 lb/ft3)(50 ft) = 3120 lb/ft2 or (b) F = PA = (3120 lb/ft2)[π(10 ft)2]; (c) P = (63.4 lb/ft3)(90 ft) = 5616 lb/ft2; 21.7 lb/in.2 (Gaugepressure)

F = 9.80 x 105 lb P = 39.0 lb/in.2 (Gauge pressure)

*15-39. A block of wood weighs 16 lb in air. A lead sinker with an apparent weight of 28 lb in water, is attached to the wood, and both are submerged in water. If their combined apparent weight in water is 18 lb, find the density of the wooden block? FB = 28 lb + 16 lb – 18 lb = 26 lb; FB = DV
V= FB 26 lb = = 0.417 ft 3 ; 3 D 62.4lb/ft Dwood = 16 lb ; D = 38.4 lb/ft3 3 0.417 ft

*15-40. A 100-g block of wood has a volume of 120 cm3. Will it float in water? Gasoline? For water: FB = ρgh = (1000 kg/m3)(9.8 m/s2)(120 x 10-6 m3) = 1.176 N W = mg = (0.1 kg)(9.8 m/s2) = 0.980 N For gasoline: FB > W YES

FB = ρgh = (680 kg/m3)(9.8 m/s2)(120 x 10-6 m3) = 0.800 N FB < W NO

W = mg = (0.1 kg)(9.8 m/s2) = 0.980 N

203Chapter 15. Fluids

Physics, 6th Edition

*15-41. A vertical test tube has 3 cm of oil (0.8 g/cm3) floating on 9 cm of water. What is the pressure at the bottom of the tube? [ PT = ροgho + ρωghw ]

PT = (800 kg/m3)(9.8 m/s2)(0.03 m) + (1000 kg/m3)(9.8 m/s2)(0.09 m) PT = 235.2 Pa + 882 Pa = 1117 Pa; PT = 1.12 kPa

*15-42. What percentage of an iceberg will remain below the surface ofseawater (ρ = 1030 kg/m3)? If iceberg floats, then: FB = mig and FB = ρωgVw mig = ρiVi g; So that: therefore,

FB

ρiVi g = ρωgVw
FB
89.3%
3

Vw ρi 920 kg/m = = = 0.893 or Vi ρ w 1030 kg/m3

Thus, 89.3 percent of the iceberg remains underwater, hence the expression: “That’s just the tip of the iceberg.” *15-43. What is the smallest area of ice 30 cm thick that will support a 90-kg man? Theice is floating in fresh water (ρ = 1000 kg/m3). [ V = Ah so that A = V/h ]

The question becomes: What volume of ice will support a 90-kg person? FB = Wice + Wman; FB = ρwgVw ; Wice = ρicegVice ; Wman = mg ( gravity divides out )

ρwgVw = ρicegVice + mg;

The smallest area occurs when the ice is level with the surface and Vice = Vw

ρwVw = ρiceVw + m or (1000 kg/m3)Vw – (920 kg/m3)Vw = 90kg
Vw = 1.125 m3; Now since V = Ah, we have

Vw 1.125 m3 A= = h 0.300 m

and

A = 3.75 m2

204

Chapter 15. Fluids

Physics, 6th Edition

*15-44. A spring balance indicates a weight of 40 N when an object is hung in air. When the same object is submerged in water, the indicated weight is reduced to only 30 N. What is the density of the object? FB = ρwgVw ;
Vw =

[ FB = 40 N –30 N = 10 N ]
FB 10 N = ; ρ w g (1000 kg/m3 )(9.8 m/s 2 ) Vw = 1.02 x 10-3 m3

The volume of the object is the same as the water displaced: V = 1.02 x 10-3 m3 m= W 40 N = = 4.08 kg; g 9.8 m/s 2

ρ=

m 4.08 kg = V 1.02 x 10−3m3

ρ = 4000 kg/m3

*15-45. A thin-walled cup has a mass of 100 g and a total volume of 250 cm 3. What is the maximum number of pennies that can be placed in thecup without sinking in the water? The mass of a single penny is 3.11 g. FB = ρgV = mcg + mpg ; mp = 150 g; mp = ρV - mc; mp = (1 g/cm3)(250 cm3) - 100 g Nbr of pennies = 48

Since each penny is 3.11 g,

15-46. What is the absolute pressure at the bottom of a lake that is 30 m deep? Pabs = 1 atm + ρgh = 101,300 Pa + (1000 kg/m3)(9.8 m/s2)(30 m); P = 395 kPa

15-47. A fluid is forced out of a...