bricks, or 7 bricks in this instance. There are 6 bricks on top of brick No. 1, and the pressure on the top of this brick is equal to the weight of 6 bricks, the depth of the pile between brick 1 and the top of brick 7. The pressure on top of brick 4 is 3 bricks, the number of bricks between the top of brick 4 and the top of brick 7; etc.

Suppose the size of the bricks to be 4 in. wide, 8 in. long, and 2 in. thick, and that each brick weighs 4.5 pounds. The total pressure on the surface AB is 7X 4.5

=

= 31.5 pounds, and the specific pressure is 31.5 ÷ 4 X 8 = .984375 pound per square inch. The weight of 1 cubic inch of brick is 4.5 ÷ (4X 8 X 2)=.0703125 pound. The depth of the pile is 7 X 2 14 in., and .0703125 X 14 = .984375 pound = the weight of a prism of brick 1 in. square and 14 in. high. But this value is the same as the specific pressure; hence, the specific pressure for any depth is equal to the weight of a prism of the brick whose base is the unit of area and whose height is the depth of the brick at the point considered. This also applies to fluids.

FIG. 6.

48. A cube of water measuring 1 ft. on each edge (1 cubic foot) weighs 62.4 lb. at its temperature of maximum density (4°C. or 39.2°F.); it weighs less at higher temperatures, but unless very exact results are desired, the weight of water may be taken as 62.4 pounds per cu. ft. A column of water 1 in. square and 1 ft. high evidently weighs 62.4 1448.4% pound. There÷ fore, for water at any depth, let p h = the depth in feet; then,

11

=

the specific pressure and

If the depth be taken in inches, the weight of 1 cu. in. = 62.4 ÷ 1728 30% = .036111 pound, say .0361 pound for practical purposes. Letting h' be the depth in inches, the specific pressure is

If it be desired to find the depth necessary to produce a given specific pressure, solve the above formulas for h and h', obtaining h=3&p= 2.3077p

and

(3)

For practical purposes, a depth of 2.31 ft. or 27.7 in. of water may be considered as equivalent to a pressure of 1 lb. per sq. in.

49. Pressure Due to Liquid on Submerged Surface.-In Fig. 7, let abcd be a flat plate submerged in the water contained in the tank T. It is evident that the pressure (specific) at the edge ab is less than it is at the edge dc. Let f be the center of gravity of the plate; then the specific pres

sure at f will be the same at any point along a horizontal line drawn on the plate and passing through ƒ, because every point on such a line will be at the same depth below the level of the liquid. Further, this specific pressure

will be the average pressure

a

FIG. 7

on the plate, and when multiplied by the area of the plate, the product will be the total normal pressure P on the plate. Let the depth of ƒ below the surface of the liquid, a = area of plate, and P = the total normal pressure on the plate; then,

h

=

in which w is the specific weight (weight of a unit cube) of the liquid. In this formula, if h is in feet and a is in square feet, w is the weight of a cubic foot; hence, for water

If h is in feet, a in square inches, and w = weight of 1 cu. ft.

For a curved surface, a must be equal to the projection of the surface on a plane perpendicular to the normal through the center of gravity.

Rule. The total normal pressure upon any submerged surface due to the weight of the liquid is equal to the weight of a prism of the liquid whose base is equal to the projection of the surface on a plane perpendicular to the normal at the center of gravity of the surface and whose altitude is the depth of this center of gravity below the surface of the liquid.

50. From the foregoing, it will be evident that the pressure on the bottom of a vessel has nothing to do with the shape of the vessel that contains the liquid. Thus, referring to Fig. 8, if the areas of the bottoms ab of the four vessels shown is the same, and the depth of the water is the same in all the vessels, the pressure (total pressure) on the bottoms of all the vessels will be the same, since their centers of gravity are at the same depth below the

surface of the liquid. Further, if all the vessels stand on a common flat, level surface, and the height of the liquid in each is the same, then if all are filled with the same liquid and are connected together by a pipe, the level of the liquid will still be at the same distance above the bases. This fact is strikingly shown in Fig. 9, where the vessel B has a much larger cross-section than A. If water, for example, be poured into A (or B), it will flow into B (A) through the connecting pipe, and when the pouring is stopped, it will be found that the water is at the same level in both vessels. This phenomenon is expressed in the familiar

saying "Water seeks its level." If the water level were not the same in both vessels, there would be a greater specific pressure on the bottom and at the entrance to the pipe in one vessel than in the other, and the water would flow from the place of higher pressure than to that of the lower.

51. It is to be again emphasized that the total normal pressure on any submerged surface does not depend upon the shape of the

vessel that contains the liquid. Referring to (a), Fig. 10, suppose dcbagef to be a dam, the top and bottom having the shape of a rectangle. Then the total normal pressure on the inside face of the dam depends only on the projected area of that face and the depth of the center of gravity of the face below the surface of the liquid, which is assumed to be level with the top of the dam; and it makes no difference how far back the water may extend. It also makes no difference what shape of the cross-section of the dam may be, provided the projected

area is the same; that is, the dam may be straight, as in (a), or curved, as in (b).

52. Combined Pressures.-If the upper surface of a liquid is free, the pressure at any point in the liquid is that due to the depth of the water (liquid) at that point; but if the liquid is subjected to an external pressure, the total specific pressure at any point is equal to the sum of the specific pressure due to the depth of the point below the surface of the liquid and the specific pressure due to the external pressure.

EXAMPLE. Referring to Fig. 11, A represents a cylinder filled with water, the inside dimensions being, say, 18 in. diameter and 30 inches long. A pipe B, 1⁄2 in. in diameter is connected to the cylinder at b and is filled to the point a with water. If a pressure of 16 pounds be applied to the handle C, thus pushing on a piston touching the water in the pipe at a, what will be the total pressure on the bottom of the cylinder? on the top of the cylinder? the lateral specific pressure at b? Let the vertical distance h between the top of the water in the tube and the bottom of the vessel be 8 ft., and between b and the bottom of the vessel 12 inches.

=

B

FIG. 11. per sq. in.

h

SOLUTION. The specific pressure on the bottom of the vessel due to the water in the cylinder and pipe is, by formula (2), Art. 48, since 8 ft. 96 in., p = 3 X 96 = 3.4667 lb. The specific pressure on the water at any point in the cylinder due to the push on the handle is equal to the pressure on the piston divided by the area of the piston, or

16 .7854 X .52

=

81.487 lb. per sq. in. The total specific

The specific pressure on the top due to the water in the pipe is p

=

X (96 - 30)

=

2.383 lb. per sq. in., and the total specific pressure on the top is 81.487 + 2.383 = 83.87 lb. per sq. in. The total pressure on the top is .7854 X 182 X 83.87 = 21,342 lb. Ans.

360

=

= 3.033 lb. per sq. in.

The lateral specific pressure at b is (96 - 12) due to the water only. The total specific pressure at b is 81.487 +3.033 84.52 lb. per sq. in. Ans.

=

Observe that the specific pressure due to the push on the handle is transmitted undiminished in all directions within the cylinder, while that due only to the water depends on the depth of the point considered below the level a.

EXAMPLES

(1) Referring to Fig. 2, suppose the weight on top of the piston is 1200 pounds, the diameter of the piston is 21 in., and the depth of the water under the piston is 40 in., what is (a) the total pressure on the bottom of the cylinder? (b) on the bottom of the piston? (c) the specific pressure due to the weight.

Ans.

{

(a) 1700.3 lb.
(c) 1200 lb.
(c) 3.4646-lb. per sq. in.

(2) Referring to the preceding example, what is the force (total normal pressure) tending to separate one half of the cylinder from the other half? Ans. 3516.9 lb.

(3) In Fig. 9, suppose that the diameter of A is 1.72 in. and of B 9.8 in. If a piston weighing 12 lb. rests on top of the water in B and a force of 24 lb. (including weight of small piston) is applied to a piston on top of the water in A, what force (weight) must be applied to the piston in B to keep it stationary? Ans. 767.12 lb. Suggestion. The weight of the piston in B must be subtracted from the upward pressure of the water to find the total downward force required to balance the pressure on piston in A.

(4) Neglecting the weight of the water in the last example, how far will the piston in B move when the piston in A moves 34 in.?

Ans. 0.1 in. very nearly. (5) Referring to Fig. 7, suppose abcd to be a flat rectangular plate 81⁄2 in. by 11 in., and that its center of gravity is 43 in. below the water level; what is the total normal pressure on the plate?

Ans. 145.18 lb. (6) Referring to the example of Art. 52, what will be the total upward pressure against the top of the cylinder, if the diameter of the pipe B is 2 in., the other dimensions and the pressure on the handle C being the same as before? Ans. 329,831 lb.

(7) The total difference of level between the top of the water in a reservoir and the nozzle of a fire hose is 225 ft.; what is the pressure of the water at the nozzle when the water is not flowing? Ans. 97.5 lb. per sq. in.

(8) A certain reservoir has a uniform cross-section shaped like the second illustration in Fig. 8. The bottom is a rectangle 144 ft. by 350 ft.; what is the total pressure on the bottom when the depth of the water is 75 ft., assuming that the bottom is level?