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point nor slide along its base, the weight of a cubic foot of granite being 160 lbs., and the coefficient of friction between it and the soil being .6?

SOLUTION.

First, to find the thickness necessary to prevent rotation outwards. Denote the height of the wall, by h, and suppose the water to extend from the bottom to the top. Denote the thickness, by t, and the length of the wall, or dam, by l. The weight of the wall in pounds, will be equal to

lht × 160;

and this being exerted through its centre of gravity, the moment of the weight with respect to the outer edge, as an axis, will be equal to

tlh × 160 = 80lht.

The pressure of the water against the inner face, in pounds, is equal to

lh2 × 62.5 = lh2 × 31.25.

This pressure is applied at the centre of pressure, which is (Example 1) at a distance from the bottom of the wall equal to }; hence, its moment with respect to the outer edge of the wall, is equal to

Th3 x 10.4166.

The pressure of the water tends to produce rotation outwards, and the weight of the wall acts to prevent this rotation. In order that these forces may be in equilibrium, their moments must be equal; or

80lht=lh3 x 10.4166.

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Next, to find the thickness necessary to prevent sliding along the base. The entire force of friction due to the weight of the wall, is equal to

160/ht X.6 = 961ht;

and in order that the wall may not slide, this must be equal to the pressure exerted horizontally against the wall. Hence,

96/ht31.25/h2.

Whence, we find,

t = .325h.

If the wall is made thick enough to prevent rotation, it will be secure against sliding.

6. What must be the thickness of a rectangular dam 15 feet high, the weight of the material being 140 lbs. to the cubic foot, that, when the water rises to the top, the structure may be just on the point of overturning?

Ans. 5.7 ft.

7. The staves of a cylindrical cistern filled with water, are held together by a single hoop. Where must the hoop be situated?

Ans. At a distance from the bottom equal to one-third of the height of the cistern.

8. Required the pressure of the sea on the cork of an empty bottle, when sunk to the depth of 600 feet, the diameter of the cork being of an inch, and a cubic foot of sea water being estimated to weigh 64 lbs.? Ans. 134 lbs.

Buoyant Effort of Fluids.

A

Fig. 139.

157. Let A represent any solid body suspended in a heavy fluid. Conceive this solid to be divided into vertical prisms, whose horizontal sections are infinitely small. Any one of these prisms will be pressed downward by a force equal to the weight of a column of fluid, whose base (Art. 155) is equal to the horizontal section of the filament, and whose altitude is the distance of its upper surface from the surface of the fluid; it will be pressed upward by a force equal to the weight of a column of fluid having the same base and an altitude equal to the distance of the lower base of the filament from the surface of the fluid. The resultant of these two pressures is a force exerted vertically upwards, and is equal to the weight of a column of fluid, equal in bulk to that of the filament and having its point of application at the centre of gravity of the volume of the filament. This being true for each filament of the body, and the lateral pressures being such as to destroy each other's effects, it follows, that the resultant of all the pressures upon the body will be a vertical force exerted upwards, whose intensity is equal to the weight of a portion of the fluid, whose volume is equal to that of the solid, and the point of application of which is the centre of gravity of the volume of the displaced fluid. This upward pressure is called the buoyant effort of the fluid, and its point of application is called the centre of buoyancy. The line of direction of the buoyant effort, in any position of the body, is called a line of support. That line of support which passes through the centre of gravity of a body, is called the line of rest.

Floating Bodies.

158. A body wholly or partially immersed in a heavy fluid, is urged downwards by its weight applied at its centre of gravity, and upwards, by the buoyant effort of the fluid applied at the centre of buoyancy.

The body can only be in equilibrium when the line through the centre of gravity of the body, and the centre of buoyancy, is vertical; in other words, when the line of rest is vertical. When the weight of the body exceeds the buoyant effort, the body will sink to the bottom; when they are just equal, it will remain in equilibrium, wherever placed in the fluid. When the buoyant effort is greater than the weight, it will rise to the surface, and after a few oscillations, will come to a state of rest, in such a position, that the weight of the displaced fluid is equal to that of the body, when it is said to float. The upper surface of the fluid is then called the plane of floatation, and its intersection with the surface of the body, the line of floatation.

If a floating body be slightly disturbed from its position of equilibrium, the centres of gravity and buoyancy will no longer

be in the same vertical line. Let DE represent the plane of floatation, G the centre of gravity of the body (Fig. 141), GH its line of rest, and the centre of buoyancy in the disturbed position of the body.

If the line of support CB, intersects the line of rest in M, above G, as in Fig. 141, the buoy

D

E

M

G

Fig. 140.

H

B

M

D

E

G

K

W

Fig. 141.

ant effort and the weight will conspire to restore the body to its position of equilibrium; in this case, the equilibrium must be stable.

If the point M falls below G, as in Fig. 142, the buoyant effort and the weight will conspire to overturn the body; in this case, the body must, before being disturbed, have been in a state of unstable equilibrium.

D

B

H

E

L

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If the centre of buoyancy and centre of gravity are

D

E

M

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Fig. 148.

always on the same vertical, the point M will coincide with G (Fig. 143), and the body will be in a state of indifferent equilibrium. The limiting position of the point M, or of the intersection of the lines of rest and of support, obtained by disturbing the floating body through an infinitely small angle, is called the metacentre of the body. Hence,

If the metacentre is above the centre of gravity of the body, it will be in a state of stable equilibrium, the line of rest being vertical; if it is below the centre of gravity, the body will be in unstable equilibrium; if the two points coincide, the body will be in indifferent equilibrium.

The stability of the floating body will be the greater, as the metacentre is higher above the centre of gravity. This condition is practically fulfilled in loading ships, or other floating bodies, by stowing the heavier objects nearest the bottom of the vessel.

Specific Gravity.

159. The specific gravity of a body is its relative weight; that is, it is the number of times the body is heavier than an equivalent volume of some other body taken as a standard.

The numerical value of the specific gravity of any body, is the quotient obtained by dividing the weight of any volume of the body by that of an equivalent volume of the standard.

For solids and liquids, water is generally taken as the standard, and, since this liquid is of different densities at different temperatures, it becomes necessary to assume also a standard temperature. Most writers have taken 60° Fahrenheit as this standard. Some, however, have taken 38°75 Fah., for the reason that experiment has shown that water has its maximum density at this temperature. We shall adopt the latter standard, remarking that specific

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