Mechanics of Liquids

Chapter III. Mechanics or Liquids.


(1) The Gravity Pressure of Liquids

35. Pressure of Liquids against Surfaces.—The sight of a great ship, perhaps built of iron and floating on water, causes one to wonder at the force that supports it. This same force is noticed when one pushes a light body, as a cork, under water. It is quite evident in such a case that a force exists sufficient to overcome the weight of the cork so that it tends to rise to the surface. Even the weight of our bodies is so far supported by water that many persons can float.
Fig. 17.—Water forces the card against the chimney.
The following experiment provides a means of testing this force:
If an empty can is pushed down into water, we feel at once the force of the liquid acting against the object and tending to push it upward. It may be noticed also that so long as the can is not completely submerged the deeper the can is pushed into the water the greater is the upward force exerted by the liquid.
We may test this action in various ways: a simple way is to take a cylindrical lamp chimney, press a card against its lower end and place it in the water in a vertical position. The force of the water will hold the card firmly against the end of the chimney. (See Fig. 17.) The amount of force may be tested by dropping shot into the tube until the card drops off. At greater depths more shot will be required, showing that the force of the water increases with the depth. Or one may pour water into the chimney. It will then be[Pg 37] found that the card does not drop until the level of the water inside the chimney is the same as on the outside. That is, before the card will fall off, the water must stand as high within the chimney as without no matter to what depth the lower end of the chimney is thrust below the surface of the water.
36. Law of Liquid Pressure.—As there is twice as much water or shot in the chimney when it is filled to a depth of 10 cm. as there is when it is filled to a depth of 5 cm. the force of the water upward on the bottom must be twice as great at a depth of 10 cm. as at a depth of 5 cm. Since this reasoning will hold good for a comparison of forces at any two depths, we have the law: "The pressure exerted by a liquid is directly proportional to the depth."
The amount of this force may be computed as follows: First, the card stays on the end of the tube until the weight of water from above equals the force of the water from below, and second, the card remains until the water is at the same height inside the tube as it is outside. Now if we find the weight of water at a given depth in the tube, we can determine the force of the water from below. If for instance the chimney has an area of cross-section of 12 sq. cm. and is filled with water to a depth of 10 cm., the volume of the water contained will be 120 ccm. This volume of water will weigh 120 g. This represents then, not only the weight of the water in the tube, but also the force of the water against the bottom. In a similar way one may measure the force of water against any horizontal surface.
37. Force and Pressure.—We should now distinguish between force and pressure. Pressure refers to the force acting against unit area, while force refers to the action against the whole surface. Thus for example, the atmospheric pressure is often given as 15 pounds to the square inch or as one kilogram to the square centimeter. On[Pg 38] the other hand, the air may exert a force of more than 300 pounds upon each side of the hand of a man; or a large ship may be supported by the force of thousands of tons exerted by water against the bottom of the ship.
In the illustration, given in Art. 36, the upward force of the water against the end of the tube at a depth of 10 cm. is computed as 120 grams. The pressure at the same depth will be 10 grams per sq. cm. What will be the pressure at a depth of 20 cm.? at a depth of 50 cm.? of 100 cm.? Compare these answers with the law of liquid pressure in Art. 36.
38. Density.—If other liquids, as alcohol, mercury, etc., were in the jar, the chimney would need filling to the same level outside, with the same liquid, before the card would fall off. This brings in a factor that was not considered before, that of the mass[B] of a cubic centimeter of the liquid. This is called the density of the liquid. Alcohol has a density of 0.8 g. per cubic centimeter, mercury of 13.6 g. per cubic centimeter, while water has a density of 1 g. per cubic centimeter.
39. Liquid Force against Any Surface.—To find the force exerted by a liquid against a surface we must take into consideration the area of the surface, and the height and the density of the liquid above the surface. The following law, and the formula representing it, which concisely expresses the principle by which the force exerted by a liquid against any surface may be computed, should be memorized:
The force which a liquid exerts against any surface, equals the area of the surface, times its average depth below the surface of the liquid, times the weight of unit volume of the liquid.
Or, expressed by a formula, F = Ahd. In this formula,[Pg 39] "F" stands for the force which a liquid exerts against any surface, "A" the area of the surface, "h," for the average depth (or height) of the liquid pressing on the surface, and "d", for the weight of unit volume of the liquid. This is the first illustration in this text, of the use of a formula to represent a law. Observe how accurately and concisely the law is expressed by the formula. When the formula is employed, however, we should keep in mind the law expressed by it.
We must remember that a liquid presses not only downward and upward but sideways as well, as we see when water spurts out of a hole in the side of a vessel. Experiments have shown that at a point the pressure in a fluid is the same in all directions, hence the rule given above may be applied to the pressure of a liquid against the side of a tank, or boat, or other object, provided we are accurate in determining the average depth of the liquid; The following example illustrates the use of the law.
For Example: If the English system is used, the area of the surface should be expressed in square feet, the depth in feet and the weight of the liquid in pounds per cubic foot. One cubic foot of water weighs 62.4 lbs.
Suppose that a box 3 ft. square and 4 ft. deep is full of water. What force will be exerted by the water against the bottom and a side?
From the law given above, the force of a liquid against a surface equals the product of the area of the surface, the depth of the liquid and its weight per unit volume, or using the formula, F = Ahd. To compute the downward force against the bottom we have the area, 9, depth, 4, and the weight 62.4 lbs. per cubic foot. 9 × 4 × 62.4 lbs. = 2246.4 lbs. To compute the force against a side, the area is 12, the average depth of water on the side is 2, the weight 62.4, 12 × 2 × 62.4 lbs. = 1497.6 lbs.

Important Topics

1. Liquids exert pressure; the greater the depth the greater the pressure.
[Pg 40]
2. Difference between force and pressure.
3. Rules for finding upward and horizontal force exerted by a liquid. F = Ahd.
4. Weight, mass, density.


1. What is the density of water?
2. What force is pressing upward against the bottom of a flat boat, if it is 60 ft. long, 15 ft. wide and sinks to a depth of 2 ft. in the water? What is the weight of the boat?
3. If a loaded ship sinks in the water to an average depth of 20 ft., the area of the bottom being 6000 sq. ft., what is the upward force of the water? What is the weight of the ship?
4. If this ship sinks only 10 ft. when empty, what is the weight of the ship alone? What was the weight of the cargo in Problem 3?
5. What is the liquid force against one side of an aquarium 10 ft. long, 4 ft. deep and full of water?
6. What is the liquid force on one side of a liter cube full of water? Full of alcohol? Full of mercury? What force is pressing on the bottom in each case?
7. What depth of water will produce a pressure of 1 g. per square centimeter? 10 g. per square centimeter? 1000 g. per square centimeter?
8. What depth of water will produce a pressure of 1 lb. per square inch? 10 lbs. per square inch? 100 lbs. per square inch?
9. What will be the force against a vertical dam-breast 30 meters long, the depth of the water being 10 meters?
10. A trap door with an area of 100 sq. dcm. is set in the bottom of a tank containing water 5 meters deep. What force does the water exert against the trap door?
11. What is the force on the bottom of a conical tank, filled with water, the bottom of which is 3 meters in diameter, the depth 1.5 meters?
12. If alcohol, density 0.8 were used in problem 11, what would be the force? What would be the depth of alcohol to have the same force on the bottom as in problem 11?
13. What is the pressure in pounds per square inch at a depth of 1 mile in sea water, density 1.026 grams per cc.?
[Pg 41]
14. Find the force on the sides and bottom of a rectangular cistern filled with water, 20 ft. long, 10 ft. wide, and 10 ft. deep?
15. Find the force on the bottom of a water tank 14 ft. in diameter when the water is 15 ft. deep, when full of water.
16. Find the force on one side of a cistern 8 ft. deep and 10 ft. square, when full of water.
17. Find the force on a vertical dam 300 ft. long and 10 ft. high, when full of water.
18. Find the pressure at the bottom of the dam in question 17.
19. Why are dams made thicker at the bottom than at the top?
20. A ship draws 26 ft. of water, i.e., its keel is 26 ft. under water. What is the liquid force against a square foot surface of the keel? Find the pressure on the bottom.

(2) Transmission of Liquid Pressure

40. Pascal's Principle.—Liquids exert pressure not only due to their own weight, but when confined, may be made to transmit pressure to considerable distances. This is a matter of common knowledge wherever a system of waterworks with connections to houses is found, as in cities. The transmission of liquid pressure has a number of important applications. The principle underlying each of these was first discovered by Pascal, a French scientist of the seventeenth century. Pascal's Principle, as it is called, may be illustrated as follows:
Suppose a vessel of the shape shown in Fig. 18, the upper part of which we may assume has an area of 1 sq. cm., is filled with water up to the level AB. A pressure will be exerted upon each square centimeter of area depending upon the depth. Suppose that the height of AB above CD is 10 cm., then the force upon 1 sq. cm. of CD is 10 g., or if the area of CD is 16 sq. cm., it receives a force of 160 g.
Fig. 18.—The force increases with the depth.
[Pg 42]
If now a cubic centimeter of water be poured upon AB it will raise the level 1 cm., or the head of water exerting pressure upon CD becomes 11 cm., or the total force in CD is 16×11 g., i.e., each square centimeter of CD receives an additional force of 1 g. Hence the force exerted on a unit area at AB is transmitted to every unit area within the vessel.
The usual form in which this law is expressed is as follows: Pressure applied to any part of a confined liquid is transmitted unchanged, in all directions, and adds the same force to all equal surfaces in contact with the liquid.
Fig. 19.—The force is proportional to the area.
The importance of this principle, as Pascal himself pointed out, lies in the fact that by its aid we are able to exert a great force upon a large area by applying a small force upon a small area of a confined liquid, both areas being in contact with the same liquid. Thus in Fig. 19 if the area of the surface CD is 2000 times the area of the surface AB, then 1 lb. applied to the liquid on AB will exert or sustain a force of 2000 lbs. on CD.
41. Hydraulic Press.—An important application of Pascal's principle is the hydraulic press. See Fig. 20. It is used for many purposes where great force is required, as in pressing paper or cloth, extracting oil from seeds, lifting heavy objects, etc. Many high school pupils have been seated in a hydraulic chair used by a dentist or barber. This chair is a modified hydraulic press.
[Pg 43]
Fig. 20.—Cross-section of a hydraulic press.
The hydraulic press contains two movable pistons, P and p (see Fig. 20). The larger of these, P, has a cross-sectional area that may be 100 or 1000 times that of the smaller. The smaller one is moved up and down by a lever; on each upstroke, liquid is drawn in from a reservoir, while each down-stroke forces some of the liquid into the space about the large piston. Valves at V and prevent the return of the liquid. If the area of P is 1,000 times that of p, then the force exerted by P is 1000 times the force employed in moving p. On the other hand, since the liquid moved by the small piston is distributed over the area of the large one, the latter will move only 1/1000 as far as does the small piston. The relation between the motions of the two pistons and the forces exerted by them may be stated concisely as follows: The motions of the two pistons of the hydraulic press are inversely proportional to the forces exerted by them. The cross-sectional areas of the two pistons are, on the other hand, directly proportional to the forces exerted by them.
An application of Pascal's principle often employed in[Pg 44] cities is the hydraulic elevator. In this device a long plunger or piston extends downward from the elevator car into a cylinder sunk into the earth, sometimes to a depth of 300 ft. Water forced into this cylinder pushes the piston upward and when the water is released from the cylinder the piston descends.
Fig. 21 represents another form of hydraulic elevator, where the cylinder and piston are at one side of the elevator shaft. In this type, to raise the elevator, water is admitted to the cylinder pushing the piston downward.
42. Artesian Wells.—Sometimes a porous stratum containing water in the earth's crust is inclined. Then if there are impervious strata (see Fig. 22), both above and below the water-bearing one, and the latter comes to the surface so that rain may fill it, a well sunk to the water-bearing stratum at a point where it is below the surface will usually give an artesian well, that is, one in which the water rises to or above the surface. Many are found in the United States.
Fig. 21.—A hydraulic freight elevator.
Fig. 22.—Conditions producing an artesian well.
Fig. 23.—A standpipe.
43. Standpipes and Air Cushions.—Many who have[Pg 45] lived in cities where water is pumped into houses under pressure know that the water pressure is changed when several faucets are opened at the same time. Again, if several persons are using a hose for sprinkling, the pressure may be lessened so as to be insufficient to force the water above the first floor. In order to allow for these changes some flexibility or spring must be introduced somewhere[Pg 46] into the water-pipe system. Water is nearly incompressible and if no means were employed to take care of the pressure changes, the sudden stopping and starting of the flow would cause serious jars and start leaks in the pipes. Two common devices for controlling sudden changes in the water pressure are the standpipe and the air cushion.
The standpipe is simply a large vertical tube connected to the water mains from which and into which water readily flows. When many faucets are opened the water lowers; when most faucets are closed the water rises, giving a simple automatic control of the surplus water and a supply of water for a short time during a shut-down of the pumps. Standpipes are often used in towns and small cities. Fig. 23 represents the standpipe at Jerome, Idaho.
The air cushion (Fig. 24) is a metal pipe or dome filled with air attached to a water pipe where sudden changes in pressure are to be controlled. At many faucets in a city water system such an air cushion is employed. It contains air; this, unlike water, is easily compressible and the confined air when the tap is suddenly closed receives and checks gradually the rush of water in the pipe. Even with an air cushion, the "pound" of the water in the pipe when a tap is suddenly closed is often heard. If air cushions were not provided, the "water hammer" would frequently crack or break the pipes.
Fig. 24.—The short pipe above the faucet contains air forming an air cushion.

Important Topics

1. Pascal's law.
2. Hydraulic press.
3. Artesian wells.
4. Standpipes and air cushions.
[Pg 47]


1. Where have you seen an air cushion? Describe it and its use.
2. Where have you seen an hydraulic press? Why and how used?
3. Where have you seen hydraulic elevators? What moves them?
4. Where do you know of liquids under pressure? Three examples.
5. What is the pressure in water at a depth of 1500 cm. Express in grams per square centimeter and in kilograms per square centimeter.
6. What head[C] of water is required to give a pressure of 200 g. per square centimeter? 2 kg. per square centimeter?
7. What pressure will be produced by a "head" of water of 20 meters?
8. If 1728 cu. in. of water are placed in a vertical tube 1 sq. in. in cross section to what height would the water rise? It would give how many feet of head?
9. What would the water in problem 8 weigh? What pressure would it produce at the bottom, in pounds per square inch? From this, compute how many feet of "head" of water will produce a pressure of 1 lb. per square inch.
10. Using the result in problem 9, what "head" of water will produce a pressure of 10 lbs. per square inch? 100 lbs. per square inch?
11. From the result in 9, 100 ft. of "head" of water will produce what pressure? 1000 ft. of "head?"
12. If the diameter of the pump piston in a hydraulic press is 2 cm. and that of the press piston 50 cm. what will be the force against the latter if the former is pushed down with a force of 40 kg.?

(3) Archimedes' Principle

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[Pg 48] 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.
[Pg 49]
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.
[Pg 50]
(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.

Important Topics

1. Archimedes' Principle.
2. Law of floating bodies.
3. The applications of Archimedes' Principle are to determine (a) the weight of a floating body; (b) the volume of an immersed solid, and (c) the density of a body.


1. Look up the story of Archimedes and the crown. Write a brief account of it.
2. Why is it easier for a fat man to float in water than for a lean one?
[Pg 51]
3. A fish weighing 1 lb. is placed in a pail full of water. Will the pail and contents weigh more than before adding the fish? Why?
4. Why can a large stone be lifted more easily while under water than when on the land?
5. Why does the air bubble in a spirit level move as one end of the instrument is raised or lowered?
6. Why does a dead fish always float?
7. A ship is built for use in fresh water. What will be the effect on its water line when passing into the ocean?
8. Why can small bugs walk on water while large animals cannot?
9. If an object weighing 62.4 lbs. just floats in water, what weight of water does it displace? What volume of water is displaced? What is the volume of the body?
10. What is the volume of a man who just floats in water if he weighs 124.8 lbs.? If he weighs 187.2 lbs.?
11. An object weighing 500 g. just floats in water. What is its volume? How much water does a floating block of wood displace if it weighs 125 lbs.? 125 g.? 2 kg.? 2000 kg.?
12. A flat boat 10 × 40 ft. in size will sink how much in the water when 10 horses each weighing 1250 lbs. are placed on board?
13. A ship 900 ft. long and 80 ft. average width sinks to an average depth of 25 ft. when empty and 40 ft. when loaded. What is the weight of the ship and of its load?
14. Will a 1000 cc. block sink or float in water if it weighs 800 g.? If it weighs 1200 g.? Explain.
15. If a 1000 cc. block of metal weighing 1200 g. is placed in the water in mid ocean what will become of it?
16. Prove Archimedes' Principle by use of the principles of liquid pressure.
17. An irregular stone, density 2.5 g. per ccm. displaces 2 cu. ft. of water. What is its weight? Its apparent weight in water?
18. Will the depth to which a vessel sinks in water change as she sails from Lake Ontario into the Atlantic Ocean? Why?
19. If the density of sea water is 1.0269 g. per cubic centimeter and that of ice 0.918 g. per ccm., what portion of an iceberg is above water?
20. In drawing water from a well by means of a bucket, why is less force used when it is under water than when entirely above?
[Pg 52]
21. A stone which weighs 300 lbs. can be lifted under water with a force of 150 lbs. What is the volume of the stone?
22. The average density of the human body is 1.07 grams per c.c. How much water will a man who weighs 150 lbs. displace when diving? How much when floating?

(4) Density and Specific Gravity

48. Density.—The density of a substance is often used as a test of its purity. Archimedes in testing King Hiero's crown to find out if it were made of pure gold determined first its density. It is by such tests that the purity of milk, of alcohol, of gold, and a great variety of substances is often determined.
Knowledge of methods of finding density is of value to everyone and should be included in the education of every student. The density of a substance is the mass of unit volume of the substance. In the metric system, for example, the density of a substance is the mass in grams per 1 ccm. Taking water, 1 ccm. weighs 1 gr. or its density is therefore 1 g. to the cubic centimeter. A cubic centimeter of aluminium weighs 2.7 g. Its density therefore is 2.7 g. per ccm.
49. Specific Gravity.Specific gravity is the ratio of the weight of any volume of a substance to the weight of an equal volume of water. Its meaning is not quite the same as that of density, since specific gravity is always a ratio, i.e., an abstract number, as 2.7. Density of a substance is a concrete number, as 2.7 grams per ccm. In the metric system the density of water is one gram per cubic centimeter, therefore we have:
Density (g. per ccm.) = (numerically) specific gravity.
In the English system, the density of water is 62.4 pounds per cubic foot, therefore in this system we have:
Density (lbs. per cu. ft.) = (numerically) 62.4 × sp. gr.
[Pg 53]
50. Methods for Finding Density and Specific Gravity
(a) Regular Solids.—Solids of regular shapes such as cubes, spheres, etc., whose volumes may be readily found by measurement, may be weighed. The mass divided by the volume gives the density, or D = Mμ/v.
(b) Irregular Solids.—with these the volume cannot be found by measurement but may be obtained by Archimedes' Principle. Weigh the solid first in the air and then in water. The apparent loss of weight equals the weight of the equal volume of water displaced. From this the volume may be found. And then the density equals mass/volume; the specific gravity =
wt. in air / wt. of equal volume of water = wt. in air / ((wt. in air) - (wt. in water))
(c) Solids Lighter than Water.—This will require a sinker to hold the body under water. Weigh the solid in air (w). Weigh the sinker in water (s). Attach the sinker to the solid and weigh both in water (w´). The specific gravity equals
(wt. of solid in air)/(loss in wt. of solid in water) or w/((w + s) - w´)
The apparent loss of weight of the solid is equal to the sum of its weight in air plus the weight of the sinker in water, less the combined weight of both in water.
(d) The Density of a Liquid by a Hydrometer.—One may also easily find the density of any liquid by Archimedes' Principle. If one takes the rod described in Art. 46, and places it in water, the number of cubic centimeters of water it displaces indicates its weight in grams. On placing the rod in another liquid in which it floats, it will of course displace its own weight and the height to which the liquid rises on the scale gives the volume. By dividing[Pg 54] the weight of the rod as shown by its position in water by the volume of the liquid displaced we obtain the density of the liquid. Commercial hydrometers for testing the density of milk, alcohol and other liquids are made of glass of the form shown in Fig. 28. The long narrow stem permits small differences in volume to be noticed, hence they are more accurate than the rod described in the preceding paragraph. For convenience this rod contains a paper scale, so that when the height of the liquid on the stem is noted, the density is read at once.
Fig. 28.—A hydrometer used to find the density of a liquid.
Density of Liquids by Loss of Weight. Weigh a piece of glass in air (Wa), in water (Ww), and in the liquid to be tested (Wl).
Then (Wa - Ww)gives the weight of the water displaced.
And (Wa - Wl) gives the weight of the liquid displaced.
Hence, (Wa - Wl)/(Wa - Ww) equals the specific gravity of the liquid.

Important Topics

1. Definitions of density and specific gravity.
2. Methods of finding density: (a) regular solids; (b) irregular solids; (c) solids lighter than water; (d) liquids by hydrometer; (e) liquids by loss of weight.


Note.—Consider that 1 cu. ft. of water weighs 62.4 lbs. Consider that 1 ccm. of water weighs 1 g.
1. What is meant by the statement that a block of wood has a specific gravity of 0.6?
2. Considering that the density of the human body is the same as that of water, what is the volume of a 125-lb. boy? Of a 250-lb. man? Of a 62.4-lb. boy? What is the volume of your body?
3. How is the weight of large ships found? Give an example.
[Pg 55]
4. Mention three cases where determinations of density are important.
5. A body weighs 40 g. in air, 15 g. in water, 5 g. in an acid. Find (a) the density of the body; (b) its volume; (c) density of the acid.
6. If the specific gravity of a horse is 1, what is the volume of a horse weighing 500 kg.? Of one weighing 1248 lbs.?
7. A weighted wooden box sinks to a depth of 20 cm. in water and 24 cm. in alcohol, and to a depth of 18 cm. in brine. What is the density of the alcohol and of the brine?
8. A glass stopper weighs in the air 25 g., in water 15 g., in oil 18 g. Find the density and volume of the stopper. Find the density of the oil.
9. What would a cubic foot of wood weigh if the specific gravity were 0.5.?
10. The specific gravity of aluminum is 2.7. Find the weight of a cubic foot of it.
11. A block of wood weighs 40 g. A piece of lead appears to weigh 70 g. in water. Both together appear to weigh 60 g. in water. Find the density of the wood.
12. A stone weighs 30 g. in air, 22 g. in water, and 20 g. in salt water. Find the density of the salt water.
13. Will iron sink in mercury? Why?
14. A submarine boat weighing 200 tons must have what volume in order to float?
15. Find the weight of 2 cu. ft. of copper from its density.
16. What is the weight in water of a mass whose specific gravity is 3.3 and whose weight is 50 kg.?
17. A block of granite weighs 1656 lbs.; its volume is 10 cu. ft., what is its density?
18. If the specific gravity of hard coal is 1.75 how would you determine how many tons of coal a bin would hold?
19. A hollow copper ball weighs 2 kg. What must be its volume to enable it to just float in water?
20. A mass having a volume of 100 ccm. and a specific gravity of 2.67 is fastened to 200 ccm. of wood, specific gravity 0.55. What will the combination weigh in water?
21. A block weighing 4 oz. in air is tied to a sinker which appears to weigh 14 oz. in water. Both together appear to weigh 6 oz. in water. What is the specific gravity of the block?