altitudes be but equal, the bases always sustain the same pressure. And as that pressure, in the regular upright vessel, is the whole column of the fluid, which is as the base and altitude; therefore the pressure in all figures is in that same ratio. Corol. 1. Hence, when the heights are equal, the pressures are as the bases. And when the bases are equal, the pressure is as the height. But when both the heights and bases are equal, the pressures are equal in all, though their contents be ever so different. Corol. 2. The pressure on the base of any vessel is the same as on that of a cylinder, of an equal base and height. Corol. 3. If there be an inverted syphon, or bent tube, Abc, containing two different fluids co, ABD, that balance each other, or rest in equilibrio; then their heights in the two legs, AE, cd, above the point of meeting, will be reciprocally as their densities. For if they do not meet at the bottom, the part Bo balances the part BE, and therefore the part co balances the part AE; that is, the weight of cd is equal to the weight of AE. And as the surface at D is the same, where they act against each other, therefore AE : CD : : density of cd : density of Ar. So, if cd be water, and AE quicksilver, which is near 14 times heavier ; then cd will be = 14AF. ; that is, if AE be 1 inch, cd will be 14 inches; if AE be 2 inches, cd will be 28 inches ; and so on. 250. PRor. If a body be immersed in a fluid of the same density or specific gravity; it will rest in any place where it is put. But a body of greater density will sink ; and one of a less density will rise to the top, and float. The body being of the same density, or of the same weight with the like bulk of the fluid, will press the fluid under it, just as much as if its space were filled with the fluid itself. The pressure then all around it will be the same as if the fluid were in its place ; consequently there is no force, neither upward nor down. ward, to put the body out of its place. And therefore it will remain wherever it is put. But if the body be lighter; its pressure downward will be less than before ; and less thah he water upward at the same depth ; therefore the greater 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. 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. 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 bulk, lose, in the same fluid, weights which are reciprocally as the specific gravities of the bodies, or directly as their bulks. Corol. 3. The whole weight of a body which will float in a fluid, is equal to the weight of as much of the fluid, as the immersed part of the body displaces 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. 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 their magnitudes. 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 i. of the fluid ; this therefore is to its weight in the air, as the specific gravity o the fluid is to that of the 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 s the specific gravity of the fluid; then w — w; w :: s : s, which proportion will give either of those specific gravities, the one from the other, "Thus s = wo s, the specific gravity of the body; and s = *:: 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. 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 = weight of the heavier body in air, h = weight of the same in water, L = weight of the lighter body in air, l = weight of the same in water, c = weight of the compound in air, c = weight of the same in water, w = the secific gravity of water. Then, 1st, (II – h) s = IIw, From which equations may be 2d, (L — l) s = Lw, found any of the above quantities, 3d, (c. — c) f = cu, in terms of the rest. s its spec. gravity; s its spec. gravity; f its spec. gravity; - dividing the absolute weight of the body by its loss in water, and multiplying by the specific gravity of water. But if the body L be lighter than water ; then l will be negative, and we must divide by L + 1 instead of L – l, and to find l we must have recourse to the compound mass c; and because, from the 4th and 5th equations, L – l = c – c – Ltd. H - h, therefore s *(T) = (HEW); that is, divide, the absolute weight of the light body, by the difference be. tween the losses in water, of the compound and heavier body, and multiply by the specific gravity of water. Or thus, - S/L & = Fif as found 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. * =#=;o-#=#9. the quantities of the two ingredients H and L, in the com: pound c. And so for any other demand. Prop. To find the specific gravity of a body. 251. 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, art. 250, s = , where B is the weight of the body out of Buy 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, So is the specific gravity of water, To the specific gravity of the body.” ExAMPLE. If a piece of stone weigh 10b, but in water only 6;lb, required its specific gravity, that of water being 1000 1 Ans. 3077. 252. CAse II.--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, So is the specific gravity of water, To the specific gravity of the body. That is, the specific gravity is s = (C - C) - (FA)# =y by cor. 6, art. 250. 6lb. in water; required the specific gravity of the elm ? Ans. 600. 253. Case III.-For a fluid of any sort.—Take a piece of a body of known specific gravity; weigh it both in and out of the fluid, finding the loss of weight by taking the difference of the two ; then say, • In the Lectures on Natural Philosophy, in the Royal Mil. Academy, Coates's Hydrostatic steelyard is employed for this purpose. It is an improvement upon the one described in Gregory's Mathematics for “Practical Men, As the whole or absolute weight, That is, the spec. grav. w = *:: s, by cor. 6, art. 250. ExAMPLE. A piece of cast iron weighed 34.61 ounces in a fluid, and 40 ounces out of it; of what specific gravity is that fluid 7 Ans. 1000. 254. Pitop. To find the quantities of two ingredients in a given compound. . Take the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound and each ingredient; and multiply each specific gravity by the difference of the other two. Then say, by proportion, As the greatest product, Is to the whole weight of the compound, So is each of the other two products, To the weights of the two ingredients. Y (f- ose = the one, and L = (s—f)s y (s – s)f (s—s) the other, by cor. 6, art. 250. ExAMPLE. A composition of 112b. being made of tin and copper, whose specific gravity is found to be 8784; required the quantity of each ingredient, the specific gravity of tin being 7320, and that of copper 9000 ! Answer, there is 100lb. of copper and consequently 12lb. of tin. } in the composition. SCHOLIUM. 255. The specific gravities of several sorts of matter, as found from experiments, are expressed by the numbers annexed to their names in the following Tables. TABLES OF SPECIFIC GRAVITIES. solids. Platina - - 20,722 | Rhodium - - 11,000 Gold, pure, hammered 19,362 Virgin Silver - 10,744 Guinea of George III. 17,629. Shilling of George III. 10,534 Tungsten - - 17,600 Bismuth, molten - 9,822 Mercury, at 32° Fahr. 13,598 Copper, wiredrawn 8,878 Lead - - 11,352. Red Copper, molten 8,788 Palladium - - 11,300 Molybdena - - 8,611 |