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is full, the water surface is 100 ft. above the ground. How many gallons does the tank hold?1 What is the vertical pressure on the bottom? What is the pressure on the cylindrical side? What is the pressure per square foot at the bottom of the feed pipe?
8. What is the pressure upon a gate 2 ft. square in the side of a dam when the water extends 21 ft. above the top of the gate?
9. A circular window 56 cm. in diameter is set in the side of a box with its center 129 cm. below the surface of the water. What is the pressure upon it?
10. What is the pressure per square inch at the spigot of a water system which is fed from a source 127 ft. higher than the spigot? At what point would the pressure be half as great? Why?
11. An hydraulic elevator with a plunger 8 in. in diameter is connected with water works having a head of 186 ft. What is the lifting power of the water upon the plunger in the elevator well? If the elevator, with plunger, weighs 1700 lb. more than its counterweight, how many people averaging 150 lb. each, can it carry?
12. A submarine boat goes to the depth of 50 ft. in the sea. What is the pressure on a hatchway cover 2 ft. square?
13. What is the pressure upon a section 20 ft. long of the wall of a canal in which the water is 12 ft. deep? What is it that determines the upon the side of a dam?
15. What is the pressure per square foot upon the body of a diver who is working at a depth of 32 ft. in sea water, sea water being 1.025 times as heavy as fresh water.
16. What is the pressure per square inch of the water striking the blades of a turbine water wheel when the head of water is 48 ft.?
1 Vol. of cylinder 231 cu. in.
14. The two leaves of a lock gate are each 12 ft. long and 15 ft. high. What is the pressure on each when the lock is filled with water?
Tr2 x H. Vol. of sphere = πTM3. 1 gal.
17. Where is the center of pressure on the leaves of the lock gate mentioned in problem 14?
18. A box 1 m. long, 80 cm. wide, and 60 cm. deep is filled with water. Find the pressure on the bottom, sides, and ends in kilograms.
19. An air-tight wooden box 60 cm. long, 50 cm. wide, and 40 cm. deep is weighted so that it sinks to the bottom of a pond 6 m. deep. Compute the pressure with which the water tends to crush the box.
148. The Principle of Archimedes.- Demonstrations. a strong thread to a stone, suspend the stone from a spring scale, and note its weight. Weigh again, letting the stone hang in a beaker of water, and the scale will be found to read less. Why?
Suspend from one side of a balance a short brass tube A (Fig. 131), and from a hook in the closed bottom of this tube suspend a solid cylinder B, which will just fill the tube. Put weights upon the other Immerse B in water, and Fill A with water, and the
scale pan until the beam is horizontal. the equilibrium will be destroyed. equilibrium will be restored.
We learn from the above, both that a body appears to lose weight when it is immersed in a liquid, and that the amount of this apparent loss is exactly the weight of the water displaced. The fact that a submerged body seems to weigh less in a liquid than in the air was observed by the Greek philosopher Archimedes in the third century B.C. He not only observed the apparent loss of weight, but discovered the
exact law governing it, hence the law is called the Principle of Archimedes. It may be stated as follows:
A body immersed in a liquid is buoyed up by a force equal to the weight of the displaced liquid.
This tendency of a liquid to lift a submerged body is called its buoyancy, and depends in amount upon the density of the liquid and the size of the body.
Since weight is the measure of the mutual attraction between the earth and the body weighed, there can be no real loss of weight, when a body is submerged in water. If, however, we suspend the body by means of a spring scale and weigh it in the air and then weigh it again in water, there will be a decrease indicated on the scale, and it is this decrease that is often called loss of weight.
If a body, a cube for instance, is immersed in a liquid, the horizontal pressure acting upon any side will be exactly counterbalanced by the pressure upon the opposite side. The downward pressure upon the upper surface A will be equal to the weight of a column of water having for its base the area of A, and for its height the depth of A below the surface of the water. The upward pressure upon the lower surface B will be equal to the weight of a column of water having for its base the area of B, and for its height the depth of B below the surface of the water. The difference between these pressures is the buoyancy of the liquid, and is equal to the weight of a quantity of the liquid that has the same
volume as the submerged cube. This conclusion is verified by the result of experiment.
149. Floating Bodies. When a body is placed in a liquid, the position it finally takes will depend upon the relative densities of the body and the liquid. If a stone or a drop of mercury is placed in water, it will sink, since it is heavier than the water. If a drop of olive oil is placed in a mixture of alcohol and water, of the same density as itself, it will remain wherever it is placed. If a piece of wood is placed in water, it will rise to the surface and float. The Principle of Archimedes applies to each of these cases, however, and we may write this Law of Floating Bodies :
A floating body displaces a volume of liquid that has the same weight as the floating body.
Demonstration. Make a bar of pine wood 25 cm. long and 1 cm. square. Bore a hole in one end and run in molten lead. Divide off one side of the bar into centimeters. Cover the bar with melted paraffin, melting it into the pores of the wood over a flame. Float the bar upright in a tall jar of pure water (Fig. 133); then, since 1 cc. of water weighs 1 g., the reading of the height at which it floats will give the approximate weight of the bar in grams.
150. Density. The quantity of matter, or the mass, in a unit volume measures the density of a substance. If a piece of lead, for example, has a mass of 45.4 g. and a volume of 4 cc., then the density of this lead equals 45.4 ÷ 4 = 11.35 g. per cubic centimeter. The general expression is