Dr. Hayel Shehadeh Ch. 6 Homework Solution SCI 101 1st Semester 2007

Chapter 6: Fluid Mechanics

Solutions to Chapter 6 Exercises

2. The concept of pressure is being demonstrated. He is careful that the pieces are small and numerous so that his weight is applied over a large area of contact. Then the sharp glass provides insufficient pressure to cut the feet.

5. A person lying on a waterbed experiences less bodyweight pressure because more of the body is in contact with the supporting surface. The greater area reduces the support pressure.

8. The use of a water-filled garden hose as an elevation indicator is a practical example of water seeking its own level. The water surface at one end of the hose will be at the same elevation above sea level as the water surface at the other end of the hose.

10. Buoyant force is the result of differences in pressure; if there are no pressure differences, there is no buoyant force. This can be illustrated by the following example: A Ping-Pong ball pushed beneath the surface of water will normally float back to the surface when released. If the container of water is in free fall, however, a submerged Ping-Pong ball will fall with the container and make no attempt to reach the surface. In this case there is no buoyant force acting on the ball because there are no pressure differences—the local effects of gravity are absent.

13. Heavy objects may or may not sink, depending on their densities (a heavy log floats while a small rock sinks, or a boat floats while a paper clip sinks, for example). People who say that heavy objects sink really mean that dense objects sink. Be careful to distinguish between how heavy an object is and how dense it is.

17. The balloon will sink to the bottom because its density increases with depth. The balloon is compressible, so the increase in water pressure beneath the surface compresses it and reduces its volume, thereby increasing its density. Density is further increased as it sinks to regions of greater pressure and compression. This sinking is understood also from a buoyant force point of view. As its volume is reduced by increasing pressure as it descends, the amount of water it displaces becomes less. The result is a decrease in the buoyant force that initially was sufficient to barely keep it afloat.

20. Ice cubes will float lower in a mixed drink because the mixture of alcohol and water is less dense than water. In a less dense liquid a greater volume of liquid must be displaced to equal the weight of the floating ice. In pure alcohol, the volume of alcohol equal to that of the ice cubes weighs less than the ice cubes, and buoyancy is less than weight and ice cubes will sink. Submerged ice cubes in a cocktail indicate that it is predominantly alcohol.

24. A Ping-Pong ball in water in a zero-g environment would experience no buoyant force. This is because buoyancy depends on a pressure difference on different sides of a submerged body. In this weightless state, no pressure difference would exist because no water pressure exists.

34. The height would be less. The weight of the column balances the weight of an equal-area column of air. The denser liquid would need less height to have the same weight as the mercury column.

35. Drinking through a straw is slightly more difficult atop a mountain. This is because the reduced atmospheric pressure is less effective in pushing soda up into the straw.

40. The end supporting the punctured balloon tips upwards as it is lightened by the amount of air that escapes. There is also a loss of buoyant force on the punctured balloon, but that loss of upward force is less than the loss of downward force, since the density of air in the balloon before puncturing was greater than the density of surrounding air.

49. The troughs are partially shielded from the wind, so the air moves faster over the crests than in the troughs. Pressure is therefore lower at the top of the crests than down below in the troughs. The greater pressure in the troughs pushes the water into even higher crests.

Solutions to Chapter 6 Problems

1. A 5-kg ball weighs 49 N, so the pressure is 49 N/cm2 = 490 kPa.

2. (Density = m/V = 6 kg/1 liter = 6 kg/liter. (Since there are 1000 liters in 1 cubic meter, density may be expressed in units kg/m3. Density = 6 kg/1 liter  1000 liter/m3 = 6000 kg/m3, six times the density of water.)

3. Pressure = weight density  depth = 9800 N/m3 220 m = 2,160,000 N/m2 = 2160 kPa.

4. (a) The volume of the extra water displaced will weigh as much as the 400-kg horse. And the volume of extra water displaced will also equal the area of the barge times the extra depth. That is, V = Ah, where A is the horizontal area of the barge;

Then h = .

Now A = 5m  2m = 10 m2; to find the volume V of barge pushed into the water by the horse’s weight, which equals the volume of water displaced, we know that

density = . Or from this, V = = = 0.4 m3.

So h = = = 0.04 m, which is 4 cm deeper.

(b) If each horse will push the barge 4 cm deeper, the question becomes: How many
4-cm increments will make 15 cm? 15/4 = 3.75, so 3 horses can be carried without sinking. 4 horses will sink the barge.

5. From Table 6.1 the density of gold is 19.3 g/cm3. Your gold has a mass of 1000 grams, so = 19.3 g/cm3. Solving for V, V = = 51.8 cm3.

8. The displaced water, with a volume 90 percent of the vacationer’s volume, weighs the same as the vacationer (to provide a buoyant force equal to his weight). Therefore his density is 90 percent of the water’s density. Vacationer’s density = (0.90)(1,025 kg/m3) = 923 kg/m3.

10. To decrease the pressure ten-fold, back to its original value, in a fixed volume, 90% of the molecules must escape, leaving one-tenth of the original number.

12. If the atmosphere were composed of pure water vapor, the atmosphere would condense to a depth of 10.3 m. Since the atmosphere is composed of gases that have less density in the liquid state, their liquid depths would be more than 10.3 m, about 12 m. (A nice reminder of how thin and fragile our atmosphere really is.)

14. On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is

(a) The weight of the displaced air must be the same as the weight supported, since the total force (gravity plus buoyancy) is zero. The displaced air weighs 20,000 N. (b) Since weight = mg, the mass of the displaced air is m = W/g = (20,000 N)/(10 m/s2) = 2,000 kg. Since density is mass/volume, the volume of the displaced air is vol = mass/density = (2,000 kg)/(1.2 kg/m3) = 1,700 m3 (same answer to two figures if g = 9.8 m/s2 is used).

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