Pressure
Pressure in a Liquid
Water Dowsing
Buoyancy
Archimedes’ Principle
What Makes an Object Sink or Float
Floatation
Floating Mountains
Pascal’s Principle
Surface Tension
Capillarity
The depths of the ocean as well as the expanse of outer space are of current interest, yet liquids are seldom studied in introductory physics classes anymore. Perhaps this is because Archimedes’ Principle and the like are too far from the frontiers of present research. But because much of the physics in this chapter is more than 2000 years old is no reason that it should not be in your physics course. Liquids are a very real part of your students’ everyday world.
It is well known that falling from great heights into water has much the same effect as falling to solid ground. Less well known are the new “water saws,” with pressures of about 5500 lb/in2 used for cutting through armor-plate steel.
Regarding Figure 13.3, you may point out that the average mass of a giraffe’s heart is about 40 kilograms. That’s quite a pump.
The dedicated teacher walking on broken glass with bare feet in his classroom with Exercise 8 is Marshall Ellenstein. Marshall
In student laboratory exercises, it is more common to work with mass density than with weight density, and floating or submerged materials are more often described in units of mass rather than weight. Displaced liquid is also described in units of mass rather than weight. This is why buoyant force in this chapter is treated as “the weight of so many kilograms,” rather than “so many newtons.” The expression of buoyancy in terms of mass units should be more in keeping with what goes on in lab.
An impressive buoyancy DEMONSTRATION: Place about 8 grams of dry ice in a large (several cm) uninflated balloon. Tie the balloon. Immediately set it on a digital balance reading to the nearest milligram. As the balloon inflates (over a few minutes) the balance readout plummets at a rate of about 2 mg/sec. The scale will finally read about 2.4 grams less, assuming the balloon inflates to about 2 liters (density of air is about 1.2 g/L). I learned of this demo from my nephew John Suchocki.
Oceans tidbit: The Atlantic is getting wider, the Pacific narrower.
There is only 1 OHT for this chapter, Figure 13.22.
In the Practicing Physics book:
• Archimedes’ Principle I
• Archimedes’ Principle II
There are problems for this chapter in the Problem Solving in Conceptual Physics book.
Three activities and 1 experiment on chapter material are in the Laboratory Manual.
In the Next-Time Questions book:
• Styrofoam Cargo • Ice Cube on the Moon
• Water’s Own Level • Deuterium Ice Cube
• Fire Truck • Balance Stand
• Boat with Scrap Iron • Submerged Cube and Sphere
• Wood and Rock Float • Submerged Teabag
• Floating Block
Prerequisite to this chapter is understanding of density, covered in the previous chapter. So if you skipped Chapter 12, discuss density here. This chapter is prerequisite to the following chapter but is not prerequisite to the remaining chapters of the text.
SUGGESTED LECTURE PRESENTATION
Force versus Pressure: Begin by distinguishing between force and pressure. Illustrate with examples: Somebody pushing on your back with a force of only 1 N—with a pin! As you’re lying on the floor, a 400-N lady stands on your stomach—perched atop spike heels! Indian master lying on a bed of 1000 nails—apprentice being advised to start with one nail! The rounded corners on tables, sharp blades on cutting knives, and the absurdity of standing tall while pointing your toes downward when caught in quicksand.
Have students compare in their hands the weights of a small steel ball and a large Styrofoam ball, and after agreeing that the little ball is heavier (since density was treated in the previous chapter), weigh them and show the Styrofoam ball is heavier! Another example of pressure (on the nerve endings).
Liquid Pressure: Liquid pressure = density ´ depth: After a few words about density, you may want to derive or call attention to the derivation of this relationship in the second footnote in the chapter.
DEMONSTRATION: Pascal’s Vases (similar to Figure 13.5) and shown by Tsing Bardin in the chapter opener photo. (Dr. Bardin is a first-rate physicist who has done extensive research for IBM and now donates a day each week teaching at City College of San Francisco—a fascinating woman.) Rationalize your results in terms of the supporting forces exerted by the sloping sides of the vases. [That is, in the wide sloping vase, the water pushes against the glass, and the glass reacts by pushing against the water. So the glass supports the extra water without the pressure below increasing. For the narrow vase that slopes outward near the bottom, the water pushes up against the sloping glass. By reaction, the glass pushes down on the water, so the pressure at the bottom is as if water were present all the way to the surface.]
Discuss Figure 13.3 and why your heart gets more rest if you sleep in a prone position versus sitting up. Call attention to the fact that when swimming, the pressure one feels against the eardrums is a function of only depth—that swimming 3 meters deep in a small pool has the same effect as swimming 3 meters deep in the middle of a huge lake.
CHECK QUESTION: Would the pressure be greater swimming 3 m deep in the middle of the ocean? (Then compare the densities of fresh and salt water.)
Ask why dams are built thicker at the bottom, and after discussing Figure 13.4 sketch the top view of a couple of dams on the board and ask which design is best (Exercise 22, previous chapter). Then relate this to the shape of stone bridges (which actually need no mortar), and the arched shape of doorways in old stone structure (Figure 12.12, previous chapter), and the aqueducts shown in Figure 13.6. Another illustration is the concave ends of large wine barrels (Exercise 23, previous chapter).
Buoyant Force: Show that a consequence of pressure being depth dependent is the phenomenon of buoyancy. Sketch Figure 13.9 on the board. Follow this up with a sketch and explanation of Figure 13.14.
DEMONSTRATION: Show how an overflow can enables the measure of an object’s volume. Ask how one could measure a quarter of a cup of butter in a liquid measuring cup using this method.
DEMONSTRATION: Archimedes’ Principle, as shown in Figure 13.15.
You may find that many students who have trouble with conceptualizing buoyant force are confused about the distinction between area and volume. Be sure to make this distinction clear, (as elementary as it seems). (If you didn’t pour the contents of the spherical flask into the tall cylindrical flask of the same volume as described in the suggested lecture of the previous chapter, be sure to do so here.) Also, point out that because a liquid is incompressible (practically incompressible, as the volume of water decreases by only 50 one-millionths of its original volume for each atmosphere increase in pressure, or equivalently, for each addition 10.3 m in depth) its density is not depth dependent. The density of water near the surface is practically the same as the density far beneath the surface. You may wish to acknowledge that some variation occurs due to temperature differences. Usually a student will inquire about waterlogged objects which lie submerged yet off the bottom of the body of water. Such objects are slightly denser than the warmer surface water and not quite as dense as the cooler water at the bottom. Stress that this is unusual and that objects appreciably denser than water always sink to the bottom, regardless of the depth of the water. Scuba divers do not encounter “floating” rocks near the bottoms of deep bodies of water!
CHECK QUESTION: Two solid blocks of identical size are submerged in water. One block is lead and the other is aluminum. Upon which is the buoyant force greater? [Same, since volumes of water displaced are the same.]
After discussion, try this one:
CHECK QUESTION: Two solid blocks of identical size, one of lead and the other of wood, are put in the same water. Upon which is the buoyant force greater? [This time the BF is greater on the lead because it displaces more water than the wood that floats!]
CHECK QUESTIONS: What is the buoyant force on a ten-ton ship floating in fresh water? In salt water? In a lake of mercury? [Same BF, but different volumes displaced.]
The unit “ton” is used in several places in this text. It may be taken to mean a metric tonne, the weight of 1000 kg, or the British ton, 2000 pounds. Either interpretation is sufficient in treating the concept involved.
Flotation: Discuss boats and rafts and the change of water lines when loaded.
CHECK QUESTIONS: What is the approximate density of a fish? Of a person? What can you say of people who can’t float?
DEMONSTRATION: Cartesian diver (inverted partially filled small bottle submerged in a larger flexible plastic bottle that you squeeze to increase and decrease the weight of water in the small bottle to make it rise and fall).
Discuss the compressibility of the human body in swimming—how the density of most people a meter or two below the surface of the water is still less than the density of water, and that one need only relax and be buoyed to the surface. But that at greater depths, the greater pressure compresses one to densities greater than the density of water, and one must swim to the surface. Simply relaxing, one would sink to the bottom! Relate this to the Cartesian diver demonstration. Also state why one cannot snorkel with a tube that goes deeper than a half-meter or so.
Side point: Contrary to those old Tarzan movies, you cannot sink in quicksand. Quicksand is the name given to a mass of sand particles that are supported by circulating water rather than by each other. Its density is greater than the density of human bodies, so you can float on it. If you struggle, you’ll unfortunately succeed in digging yourself deeper in. So if you’re ever stuck in it, keep yourself still until you stop sinking (you will), and then use slow swimming motions to get yourself into a horizontal position and then roll onto the ground.
Pascal’s Principle: Begin by pushing against the wall with a meterstick and state that the stick affords a means of applying pressure to the wall—then state that the same can be done with a confined fluid. Explain how any external pressure applied to a liquid that tightly fills a volume is transmitted to all parts of the liquid equally. Discuss Figures 13.21 and 13.22. If a hydraulic press is available, crush a block of wood with it. Point out that the pressure transmitted throughout a confined liquid is pressure over and above that already in the liquid. For example, the pressure in a hydraulic system at any point is equal to the applied pressure plus the density ´ depth.
Surface Tension: In the next chapter on gases we’ll think of atoms as rigid balls that ricochet off one another. Here we think of atoms as sticky balls—capillarity. An interesting example of capillarity (not in the text) involves the tallness of trees. The cohesive forces of water explains the transport of water from roots to the top of tall trees. When a single water molecule evaporates from the cell membrane inside a leaf, it is replaced by the one immediately next to it due to the cohesive forces between water molecules. A pull is created on the column of water which is continuous from leaves to roots. Water can be lifted far higher than the 10.2 m that atmospheric pressure would serve—even to 100 m in this way, the height of the largest trees.
NEXT-TIME QUESTION: Show your class an ordinary rectangular chunk of wood with square cross-section, and ask how it will float if placed in water. [Unless it is very low density, like Styrofoam, it will float as in position b. Why?]
Solutions to Chapter 13 Exercises
1. Water.
2. Pressure would be appreciably greater by the woman, which would hurt you more.
3. A woman with spike heels exerts considerably more pressure on the ground than an elephant! Example: A 500-N woman with 1-cm2 spike heels puts half her weight on each foot, distributed (let’s say) half on her heel and half on her sole. So the pressure exerted by each heel will be (125 N/1 cm2) = 125 N/cm2. A 20,000-N elephant with 1000 cm2 feet exerting 1/4 its weight on each foot produces (5000N/1000 cm2) = 5N/cm2; about 25 times less pressure. (So a woman with spike heels will make greater dents in a new linoleum floor than an elephant will.)
4. The scale measures force, not pressure, and is calibrated to read your weight. That’s why your weight on the scale is the same whether you stand on one foot or both.
5. There is less pressure with a waterbed due to the greater contact area.
6. A sharp knife cuts better than a dull knife because it has a thinner cutting area which results in more cutting pressure for a given force.
7. More water will flow from a downstairs open faucet because of the greater pressure. Since pressure depends on depth, the downstairs faucet is effectively “deeper” than the upstairs faucet. The pressure downstairs is greater by an amount = weight density ´ depth, where the depth is the vertical distance between faucets.
8. 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.
9. Your body gets more rest when lying than when sitting or standing because when lying, the heart does not have to pump blood to the heights that correspond to standing or sitting. Blood pressure is normally greater in the lower parts of your body simply because the blood is “deeper” there. Since your upper arms are at the same level as your heart, the blood pressure in your upper arms will be the same as the blood pressure in your heart.
10. No, in orbit there are no pressure differences due to gravity.
11. Both are the same, for pressure depends on depth.
12. The water can be no deeper than the spouts, which are at the same height, so both teapots hold the same amount of liquid.
13. (a) The reservoir is elevated so as to produce suitable water pressure in the faucets that it serves. (b) The hoops are closer together at the bottom because the water pressure is greater at the bottom. Closer to the top, the water pressure is not as great, so less reinforcement is needed there.
14. Both blocks have the same volume and therefore displace the same amount of water.
15. A one-kilogram block of aluminum is larger than a one-kilogram block of lead. The aluminum therefore displaces more water.
16. A 10-N block of aluminum is larger than a 10-N block of lead. The aluminum therefore displaces more water. Only in Exercise 10 were the volumes of the block equal. In this and the preceding exercise, the aluminum block is larger. (These exercises serve only to emphasize the distinctions between volume, mass, and weight.)
17. The smaller the window area, the smaller the crushing force of water on it.
18. One, that water pressure depends on depth. The other, that the pressure due to the column of water is transmitted to all parts of the barrel.
19. From a physics point of view, the event was quite reasonable, for the force of the ocean on his finger would have been quite small. This is because the pressure on his finger has only to do with the depth of the water, specifically the distance of the leak below the sea level—not the weight of the ocean. For a numerical example, see Problem 4.
20. A typical plumbing design involves short sections of pipe bent at 45-degree angles between vertical sections two-stories long. The sewage therefore undergoes a succession of two-story falls which results in a moderate momentum upon reaching the basement level.
21. Water seeking its own level is a consequence of pressure depending on depth. In a bent U-tube full of water, for example, the water in one side of the tube tends to push water up the other side until the pressures at the same depth in each tube are equal. If the water levels were not the same, there would be more pressure at a given level in the fuller tube, which would move the water until the levels were equal.
22. 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.
23. In deep water, you are buoyed up by the water displaced and as a result, you don’t exert as much pressure against the stones on the bottom. When you are up to your neck in water, you hardly feel the bottom at all.
24. 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.
25. The diet drink is less dense than water, whereas the regular drink is denser than water. (Water with dissolved sugar is denser than pure water.) Also, the weight of the can is less than the buoyant force that would act on it if totally submerged. So it floats, where buoyant force equals the weight of the can.
26. Mercury is more dense (13.6 g/cm3) than iron. A block of iron will displace its weight and still be partially above the mercury surface. Hence it floats in mercury. In water it sinks because it cannot displace its weight.
27. Mountain ranges are very similar to icebergs: Both float in a denser medium, and extend farther down into that medium than they extend above it. Mountains, like icebergs, are bigger than they appear to be. The concept of floating mountains is isostacy—Archimedes’ principle for rocks.
28. A mostly-lead mountain would be more dense than the mantle and would sink in it. Guess where most of the iron in the world is. In the Earth’s center!
29. The force needed will be the weight of 1 L of water, which is 9.8 N. If the weight of the carton is not negligible, then the force needed would be 9.8 N minus the carton’s weight, for then the carton would be “helping” to push itself down.
30. When the ball is held beneath the surface, it displaces a greater weight of water.
31. The buoyant force on the ball beneath the surface is much greater than the force of gravity on the ball, producing a large net force and large acceleration.
32. 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.
33. While floating, BF equals the weight of the submarine. When submerged, BF equals the submarine’s weight plus the weight of water taken into its ballast tanks. Looked at another way, the submerged submarine displaces a greater weight of water than the same submarine floating.
34. The block of wood would float higher if the piece of iron is suspended below it rather than on top of it. By the law of flotation: The iron-and-wood unit displaces its combined weight and the same volume of water whether the iron is on top or the bottom. When the iron is on the top, more wood is in the water; when the iron is on the bottom, less wood is in the water. Or another explanation is that when the iron is below—submerged—buoyancy on it reduces its weight and less of the wood is pulled beneath the water line.
35. When a ship is empty its weight is least and it displaces the least water and floats highest. Carrying a load of anything increases its weight and makes it float lower. It will float as low carrying a few tons of Styrofoam as it will carrying the same number of tons of iron ore. So the ship floats lower in the water when loaded with Styrofoam than when empty. If the Styrofoam were outside the ship, below water line, then the ship would float higher as a person would with a life preserver.
36. A sinking submarine will continue to sink to the bottom so long as the density of the submarine is greater than the density of the surrounding water. If nothing is done to change the density of the submarine, it will continue to sink because the density of water is practically constant. In practice, water is sucked into or blown out of a submarine’s tanks to adjust its density to match the density of the surrounding water.
37. The water level will fall. This is because the iron will displace a greater amount of water while being supported than when submerged. A floating object displaces its weight of water, which is more than its own volume, while a submerged object displaces only its volume. (This may be illustrated in the kitchen sink with a dish floating in a dishpan full of water. Silverware in the dish takes the place of the scrap iron. Note the level of water at the side of the dishpan, and then throw the silverware overboard. The floating dish will float higher and the water level at the side of the dishpan will fall. Will the volume of the silverware displace enough water to bring the level to its starting point? No, not as long as it is denser than water.)
38. For the same reason as in the previous exercise, the water level will fall. (Try this one in your kitchen sink also. Note the water level at the side of the dishpan when a bowl floats in it. Tip the bowl so it fills and submerges, and you’ll see the water level at the side of the dishpan fall.)
39. Bouyant force will remain unchanged on the sinking rock because it displaces the same weight of water at any depth.
40. Bouyant force on a sinking swimmer will decrease as she sinks. This is because her body, unlike the rock in the previous exercise, will be compressed by the greater pressure of greater depths.
41. 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.
42. You are compressible, whereas a rock is not, so when you are submerged, the water pressure tends to squeeze in on you and reduce your volume. This increases your density. (Be careful when swimming—at shallow depths you may still be less dense than water and be buoyed to the surface without effort, but at greater depths you may be pressed to a density greater than water and you’ll have to swim to the surface.)
43. A body floats higher in denser fluid because it does not have to sink as far to displace a weight of fluid equal to its own weight. A smaller volume of the displaced denser fluid is able to match the weight of the floating body.
44. The buoyant force does not change. The buoyant force on a floating object is always equal to that object’s weight, no matter what the fluid.
45. Since both preservers are the same size, they will displace the same amount of water when submerged and be buoyed up with equal forces. Effectiveness is another story. The amount of buoyant force exerted on the heavy gravel-filled preserver is much less than its weight. If you wear it, you’ll sink. The same amount of buoyant force exerted on the lighter Styrofoam preserver is greater than its weight and it will keep you afloat. The amount of the force and the effectiveness of the force are two different things.
46. No, there does not have to actually be 14.5 N of fluid in the skull to supply a buoyant force of 14.5 N on the brain. To say that the buoyant force is 14.5 N is to say that the brain is taking up the space that 14.5 N of fluid would occupy if fluid instead of the brain were there. The amount of fluid in excess of the fluid that immediately surrounds the brain does not contribute to the buoyancy on the brain. (A ship floats the same in the middle of the ocean as it would if it were floating in a small lock just barely larger than the ship itself. As long as there is enough water to press against the hull of the ship, it will float. It is not important that the amount of water in this tight-fitting lock weigh as much as the ship—think about that, and don’t let a literal word explanation “a floating object displaces a weight of fluid equal to its own weight” and the idea it represents confuse you.)
47. 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.
48. When the ice cube melts the water level at the side of the glass is unchanged (neglecting temperature effects). To see this, suppose the ice cube to be a 5 gram cube; then while floating it will displace 5 grams of water. But when melted it becomes the same 5 grams of water. Hence the water level is unchanged. The same occurs when the ice cube with the air bubbles melts. Whether the ice cube is hollow or solid, it will displace as much water floating as it will melted. If the ice cube contains grains of heavy sand, however, upon melting, the water level at the edge of the glass will drop. This is similar to the case of the scrap iron of Exercise 38.
49. The total weight on the scale is the same either way, so the scale reading will be the same whether or not the wooden block is outside or floating in the beaker. Likewise for an iron block, where the scale reading shows the total weight of the system.
50. If water doesn’t overflow, the reading on the scale will increase by the ordinary weight of the fish. However, if the bucket is brim filled so a volume of water equal to the volume of the fish overflows, then the reading will not change. We assume here that the fish and water have the same density.
51. When the ball is submerged (but not touching the bottom of the container), it is supported partly by the buoyant force on the left and partly by the string connected to the right side. So the left pan must increase its upward force to provide the buoyant force in addition to whatever force it provided before, and the right pan’s upward force decreases by the same amount, since it now supports a ball lighter by the amount of the buoyant force. To bring the scale back to balance, the additional weight that must be put on the right side will equal twice the weight of water displaced by the submerged ball. Why twice? Half of the added weight makes up for the loss of upward force on the right, and the other half for the equal gain in upward force on the left. (If each side initially weighs 10 N and the left side gains 2 N to become 12 N, the right side loses 2 N to become 8 N. So an additional weight of 4 N, not 2 N, is required on the right side to restore balance.) Because the density of water is less than half the density of the iron ball, the restoring weight, equal to twice the buoyant force, will still be less than the weight of the ball.
52. If the gravitational field of the Earth increased, both water and fish would increase in weight and weight density by the same factor, so the fish would stay at its prior level in water.
53. Both you and the water would have half the weight density as on Earth, and you would float with the same proportion of your body above the water as on Earth. Water splashed upward with a certain initial speed would rise twice as high, since it would be experiencing only half the “gravity force.” Waves on the water surface would move more slowly than on Earth (at about 70% as fast since vwave ~ √g).
54. Because of surface tension, which tends to minimize the surface of a blob of water, its shape without gravity and other distorting forces will be a sphere—the shape with the least surface area for a given volume.
55. 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. (See the answer to Exercise 20, and Home Project 2.)
56. Part of whatever pressure you add to the water is transmitted to the hungry crocodiles, via Pascal’s principle. If the water were confined, that is, not open to the atmosphere, the crocs would receive every bit of pressure you exert. But even if you were able to slip into the pool to quietly float without exerting pressure via swimming strokes, your displacement of water raises the water level in the pool. This ever-so-slight rise, and accompanying ever-so-slight increase in pressure at the bottom of the pool, is an ever-so-welcome signal to the hungry crocodiles.
57. The strong man will be unsuccessful. He will have to push with 50 times the weight of the 10 kilograms. The hydraulic arrangement is arranged to his disadvantage. Ordinarily, the input force is applied against the smaller piston and the output force is exerted by the large piston—this arrangement is just the opposite.
58. In Figure 13.21, the increased pressure in the reservoir is a result of the applied force distributed over the input piston area. This increase in pressure is transmitted to the output piston. In Figure 13.23, however, the pressure increase is supplied by the mechanical pump, which has nothing to do with the area of fluid interface between the compressed air and the liquid.
59. When water is hot, the molecules are moving more rapidly and do not cling to one another as well as when they are slower moving, so the surface tension is less. The lesser surface tension of hot water allows it to pass more readily through small openings.
60. Surface tension accounts for the walking of water striders, needles that appear to float, and even razor blades that also appear to float. In these cases the weights of the objects are less than the restoring forces in the water surface that tends to resist stretching.
Chapter 13 Problem Solutions
1. Pressure = weight density ´ depth = 10,000 N/m3 ´ 220 m = 2,200,000 N/m2 = 2200 kPa (or for density = 9800 N/m3, pressure = 2160 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. (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.
4. First you must find the pressure. It is weight density ´ depth = (10,000 N/m3)(2 m) = 20,000 N/m2, or 20,000 Pa. Force is pressure ´ area, and 1 cm2 = 10-4 m2, so F = (20,000 N/m2)(10-4 m2) = 2 N. It would be easy for the boy to exert this force. It is about the weight of a notebook or a small box of cereal. (Note: Air pressure is not figured into this calculation because its effect in pushing down on the water from above is canceled by its effect in pushing from outside the hole against the leaking water.)
5. From Table 12.1 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.
6. Density = = =
= 4 kg/liter. And since 1 liter = 103 cm3 = 10-3 m3, density = 4,000 kg/m3.
(Or this can be reasoned as follows: The buoyant force on the object is the force needed to support 0.5 kg, so 0.5 kg of water is displaced. Since density is mass/volume, volume is mass/density, and displaced volume = (0.5 kg)/(1000 kg/m3) = 5 ´ 10-4 m3. The object’s volume is the same as the volume it displaces, so the object’s density is mass/volume =
(2 kg)/(5 ´ 10-4 m3) = 4000 kg/m3, four times the density of water.)
7. 10% of ice extends above water. So 10% of the 9-cm thick ice would float above the water line; 0.9 cm. So the ice pops up. Interestingly, when mountains erode they become lighter and similarly pop up! Hence it takes a long time for mountains to wear away.
8. (a) Weight = mg = (60 kg)(10 m/s2) = 600 N (or 588 N if 9.8 m/s2 is used).
(b) She has the same density as water, 1000 kg/m3. Since density = mass/volume, volume = mass/density, so volume = (60 kg)/(1000 kg/m3) = 0.06 m3.
(c) Buoyant force = weight of water displaced = 600 N (or 588 N). Her weight balances the buoyant force, so net force = 0.
9. 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. The relative areas are as the squares of the diameters; 62/22 = 36/4 = 9. The larger piston can lift 9 times the input force to the smaller piston.
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