Also (see Vince's Fluxions) AG = 4 a, and AT = x, y and nG = (a + x) sin, V = sin. Vxy' + cos. V. Since ng, gG are in the same straight line. Hence (a + yo sy) sin. V cos. V=y (3) p But since 2T = _ V, and y = 2x tan. Ì = 2x tan. V, we easily get (yo = px) sin. Ve P and 2y v (p* + 4y") Moreover y – x)} . 2 P 4 Hence equations (2) and (3) become SV g°= 3 (p° + 4yo) (2) sp y = 94 (f« - 1 .. (3) P which being equated, and the resulting equation resolved according to y, will give y, and therefore the position of the tangent PT, &c. &c. 653. Let 2r be the diameter of the sphere, s the specific gravity of the sphere, and s' that of the fluid ; then since the volume of that segment of the sphere, which is cut off by a plane I to the part of the diameter x, is (Vince) (rxo - 123) ihat part of it which corresponds to x = {(21) = är, is 9783 Now, when the air is admitted, let P, Q, be the parts in the upper and lower fluids; then (Vince's Hyd. p. 35,) P:Q :: s' :s - 0.00122 0.00122 4973 X S which being put = = (180 - ) will give x the depth required. If the fluid be common water s'= 1. 654. Let , be the exterior sphere, x that of the interior one, and suppose s, s' the specific gravities of iron and water respectively. Then since the body loses just its weight, or the weight of the quantity of water it displaces, we have 476 655. Let s, s', s", denote the given specific gravities of the bodies and of the fluid; also let Q, Q' be the magnitudes of the bodies; then the absolute weights are Qs, Q's', and they lose in the fluids the weights Qs", Q's", consequently by the question, we have Qs Qs" = Q's – Q's" Ś si" .. 656. Let s denote the specific gravity of the solid ; then if P, Q be the parts in the upper and lower fluids, we have (Vince, p. 35), 657. The pressure on the top of the vessel is the same as it would be on the other side of it, if a.column of fluid equal in height to the tube, were supported by it. Hence, the pressure required is a? x ma = m 23 a being the side of the cube. 658. The pressure of the fluid against any section parallel to the horizon oc depth of that section in the fluid. Hence the thickness of the cylinder must increase proportionally with the depth, and the exterior form of the vessel will be that of the frustum of a cone. 659. Let s be the specific gravity required, and P, Q the magnitudes of the pårts immersed in the upper and lower fluids ; then (Vince, p. 33) P:Q:: 9-sis – :P+Q:Q:: 4 : 3 - 3 4Q ..SE + 3 P+Q But the volume of a paraboloid whose axis is a and radius of e base B is (! its circumscribing cylinder) πρα 2 тра? ? 660. Let M, M' be the magnitudes of the bodies, s, s' their specific gravities, and S, S' the specific gravities of water and air; then the absolute weights of the bodies are Ms, M's'; and since a body, when weighed in a fluid, loses in weight that of the fluid it displaces, :: 6 = Ms — MS, 2 = Ms MS 2 S- S S M' = 7 5-S 78 - 4S' 3 the specific gravities required. 661. Letr be the radius of the base of the cylinder, x the height of the fluid, and y the height of the section; then the press sure on the base is 7072 X X; the pressure on the upper surface (Vince, Prop. VII.) 2ar (- y) x y 2 - and P = and that on the lower surface is 2rry x y + x - y); and these pressures are all equal by the question. :: rx = (x - y) = 2y (x or rx = 2* – 2xy + y* = 2xy – Y Hence y = x(1 - :: x = 2r, and y = 2r (1 – 12). 662. If a be a side of the squarè, the pressures on the upper and lower halves are (Vince, Prop. VII.) a* a? Х (1 - x) = a + ::P:P :: j :}:: 1:3. 663. Let x be the specific gravity required'; then by Vince, Prop. XXI, and the question, we have 1 -1 ::6 x :3 - a 664. Let M, M' be the magnitudes of the bodies, s, s' their specific gravities, and S the specific gravity of water; then their absolute weights are 14 = Ms, 8 = M's' also the weights lost in the water are MS, M'S. |