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44. A Body Supported by a Liquid.—Among the applications of the force exerted by a liquid upon a surface, Archimedes' Principle is one of the most important.

Most persons have noted that a body placed in water is partly or wholly supported by the force of the water upon it. A stone held by a cord and lowered into water is felt to have a part of its weight supported, while a piece of cork or wood is wholly supported and floats.

The human body is almost entirely supported in water, in fact, many people can easily float in water. It was the consideration of this fact that led the Greek philosopher Archimedes to discover and state the principle that describes the supporting of a body in a liquid.

Fig. 25.—Theoretical proof of Archimedes' principle.

45. Archimedes' Principle.—"A body immersed in a liquid is pushed up by a force equal to the weight of the liquid that it displaces." The proof for this law is simply demonstrated. Suppose a cube, abcd, is immersed in water (Fig. 25). The upward force on cd is equal to the weight of a column of water equal to cdef. (See Art. 39.) The downward force upon the top of the cube is equal to the weight of the column of water abef. Then the net upward force upon the cube, that is, the upward force upon the bottom less the downward force upon the top, or the buoyant force exerted by the liquid is exactly equal to the weight of the displaced water abcd.

46. Law of Floating Bodies.—This same reasoning may be applied to any liquid and to any body immersed to any depth below the surface of the liquid. If the body weighs more than the displaced liquid it will sink. If it weighs less than the displaced liquid it will float or rise in the water. A block of wood rises out of the water in which it floats until its own weight just equals the weight of the water it displaces. From this we have the law of floating bodies.

A floating body displaces its own weight of the liquid in which it floats.

Fig. 26.—A floating body displaces its own weight of water.

To test the law of floating bodies, take a rod of light wood 1 cm. square and 30 cm. long (Fig. 26). Bore out one end and fill the opening with lead and seal with paraffin so that the rod will float vertically when placed in water. Mark upon one side of the rod a centimeter scale, and dip the rod in hot paraffin to make it waterproof. Now find the weight of the stick in grams and note the depth to which it sinks in water in centimeters. Compute the weight of the displaced water. It will equal the weight of the rod.

47. Applications of Archimedes' Principle. There are numerous applications of Archimedes' Principle and the law of floating bodies.

(a) To Find the Weight of a Floating Body: Problem.—A boat 20 ft. long and with an average width of 6 ft. sinks to an average depth of 3 ft. in the water. Find the weight of the boat. What weight of cargo will sink it to an average depth of 5 ft.?

Solution.—The volume of the water displaced is 20 × 6 × 3 cu. ft. = 360 cu. ft. Since 1 cu. ft. of water weighs 62.4 lbs., 360 × 62.4 lbs. = 22,464 lbs., the weight of water displaced. By the law of floating bodies this is equal to the weight of the boat. When loaded the volume of water displaced is 20 ft. × 6 × 5 ft. which equal 600 cu. ft. 600 × 62.4 lbs. = 37,440 lbs. This is the weight of the water displaced when loaded. 37,440 lbs. - 22,464 lbs. = 14,976 lbs., the weight of the cargo.

(b) To Find the Volume of an Immersed Solid: Problem.—A stone weighs 187.2 lbs. in air and appears to weigh 124.8 lbs. in water. What is its volume?

Solution.—187.2 lbs. - 124.8 lbs. = 62.4 lbs., the buoyant force of the water. By Archimedes' Principle, this equals the weight of the displaced water which has a volume of 1 cu. ft. which is therefore the volume of the stone.

(c) To Find the Density of a Body: The density of a body is defined as the mass of unit volume.

We can easily find the mass of a body by weighing it, but the volume is often impossible to obtain by measurements, especially of irregular solids.

Archimedes' Principle, however, provides a method of finding the volume of a body accurately by weighing it first in air and then in water (Fig. 27), the apparent loss in weight being equal to the weight of the displaced water. One needs only to find the volume of water having the same weight as the loss of weight to find the volume of the body.

If the metric system is used, 1 ccm. of water weighs 1 g., and the volume is numerically the same as the loss of weight.

Fig. 27.—A method of weighing a body under water.