" to the sum of all the ds. Consequently the whole pressure of the fluid on the body or surface b is equal to the weight of the bulk bg of the fluid, that is, of the column whose base is the given surface b, and its height is g the depth of the centre of gravity in the fluid. PROPOSITION LXIII. 315. The Pressure of a Fluid, on the Base of the Vessel in which it is contained, is as the Base and Perpendicular Altitude : whatever be the Figure of the Vessel that contains it. If the sides of the base be upright, so that it be a prism of a uniform width throughout; then the case is evident; for then the base supports the whole fluid, and the pressure is just equal to the weight of the fluid. But if the vessel be wider at top than bottom, then the bottom sustains or is pressed by, only the part contained within the upright lines ac, bo; because the parts Aca, BDb are supported by the sides AC, BD; and those parts have no other effect on the part aboc than keeping it in its position, by the la teral pressure against ac and bo, which does not alter its perpendicular pressure downwards. And thus the pressure on the bottom is less than the weight of the contained fluid. And if the vessel be widest at bottom; then the bottom is still pressed with a weight which is equal to that of the whole upright column aboc. For, as the parts of the fluid are in equilibrio, all the parts have an equal pressure at the same depth; so that the parts within cc and do press equally as those in cd, and therefore equally the same as if the sides of the vessel had gone upright to a and b, the defect of fluid in the parts aca and BDb being exactly compensated by the downward pressure or resistance of the sides AC and BD against the contiguous fluid. And thus the pressure on the base may be made to exceed the weight of the contained fluid, in any proportion whatever. So that, in general, be the vessels of any figure whatever, regular or irregular, upright or sloping, or variously wide and narrow in different parts, if the bases and perpendicular altitudes be but equal, the bases always sustain the same pressure. And as that pressure, in the regular upright vessel, : vessel, is the whole column of the fluid, which is as the base 316. Corol. 1. Hence, when the heights are equal, the 317. Corol. 2. The pressure on the base of any vessel is For, if they do not meet at the bottom, A9 E So, if cn be water, and AE quicksilver, which is near 14 times heavier; then co will be = 14AE; that is, if AE be 1 inch, co will be 14 inches; if Aɛ be 2 inches, co will be 28 inches; and so on. PROPOSITION LXIV. 319. If a Body be Immersed in a Fluid of the same Density or THE body, being of the same den- But But if the body be lighter; its pressure downward will be less than before, and less than the water upward at the same depth; therefore the great force will overcome the less, and push the body upward to a. And if the body be heavier than the fluid, the pressure downward will be greater than the fluid at the same depth; therefore the greater force will prevail, and carry the body down to the bottom at c. 320. Corol. 1. A body immersed in a fluid, loses as much weight, as an equal bulk of the fluid weighs. And the fluid gains the same weight. Thus, if the body be of equal density with the fluid, it loses all its weight, and so requires no force but the fluid to sustain it. If it be heavier, its weight in the water will be only the difference between its own weight and the weight of the same bulk of water; and it requires a force to sustain it just equal to that difference. But if it be lighter, it requires a force equal to the same difference of weights to keep it from rising up in the fluid. 321. Corol. 2. The weights lost, by immerging the same body in different fluids, are as the specific gravities of the fluids. And bodies of equal weight, but different bulks, lose in the same fluid, weights which are reciprocally as the specific gravities of bodies, or directly as their bulks. > 322. Corol. 3. The whole weight of a body which will float in a fluid, is equal to as much of the fluid, as the immersed part of the body takes up, when it floats. For the pressure under the floating body, is just the same as so much of the fluid as is equal to the immersed part; and therefore the weights are the same, 323. Corol. 4. Hence the magnitude of the whole body, is to the magnitude of the part immersed, as the specific gravity of the fluid, is to that of the body. For, in bodies of equal weight, the densities, or specific gravities, are reciprocally as their magnitudes, 324. Corol. 5. And because when the weight of a body taken in a fluid, is subtracted from its weight out of the fluid, the difference is the weight of an equal bulk of the fluid; this therefore is to its weight in the air, as the specific gravity of the fluid, is to that of body. Therefore, if w be the weight of a body in air, w its weight in water, or any fluid, s the specific gravity of the body, and then then w-w: w::s:s, which proportion will give either of those specific gravities, the one from the other. W W and s W s, the specific gravity of the fluid, So that the specific gravities of bodies, are as their weights in the air directly, and their loss in the same fluid inversely. 325. Corol. 6. And hence, for two bodies connected together, or mixed together into one compound, of different specific gravities, we have the following equations, denoting their weights and specific gravities, as below, viz. H C, c, 6th, + f S Then, s its spec. gravity; s its spec. gravity ; fits spec. gravity; From which equations may be found any of the above quantities, in terms of the rest. Thus, from one of the first three equations, is found the specific gravity by dividing the absolute weight of the body by its loss in water, and multiplying by the specific gravity of water, and But if the body L be lighter than water; then I will be negative, and we must divide by L+ instead of L-l, and to find 7 we must have recourse to the compound mass c i because, from the 4th and 5th equations, L-l=c-cu-h, therefore s ; that is, divide the absolute weight of the light body, by the difference between the losses in water, of the compound and heavier body, and multiply by LW (cc) (H-h) SfL as found the specific gravity of water. Or thus, s from the last equation. Also, if it were required to find the quantities of two ingredients mixed in a compound, the 4th and 6th equations would give their values as follows, viz. (s s) f 1 the quantities of the two ingredients H and D, in the compound c. And so for any other demand. PROPOSITION LXV. To find the Specific Gravity of a Body. 326. CASE 1.—When the body is heavier than water: weigh it both in water and out of water, and take the difference, which will be the weight lost in water. Then, by corol. 6, BW prop. 64, s= where B is the weight of the body out of B-b' water, b its weight in water, s its specific gravity, and w the' specific gravity of water. That is, As the weight lost in water, Is to the whole or absolute weight, EXAMPLE. If a piece of stone weigh 10lb, but in water 327. Case 11.—When the body is lighter than water, so that it will not sink : annex to it a piece of another body, heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound mass, separately, both in water and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater. Then say, by proportion, As the last remainder, Is to the weight of the light body in air, To the specific gravity of the body. That is, the specific gravity is s by cor. 6, prop. 64. LW (cc) - (H-h)' EXAMPLE. Suppose a piece of elm weighs 15lb in air; and that a piece of copper, which weighs 181b in air and 16lb in water, is affixed to it, and that the compound weighs 6lb water; required the specific gravity of the elm? Ans. 600. 328. CASE : ༈༙་ |