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.