4 .. tan, 0 = 37 which gives 0, and therefore the complement of , or the inclination sought 684. Letr be the radius of the sphere; then the pressure on the base is Par? ХrEmr3. and the depth of its centre of gravity (Vince's Flux. p. 119,) To 2 the pressure upon the curved surface is 685. Let » be the radius of the interior; then the surface is 4ars and the depth of the centre of gravity is r. Therefore if s be the specific gravity of the flaid, the internal pressure of P = 47° XrXs= 4r's. Also the weight of the fluid is V pase -a 686. Letr be the radius of the inner sphere; then if y be the radius of any section parallel to the base, and a — * (a being the altitude of the fluid,) its depth in the fluid, the pressure upon its circumference I surface is measured by (Vince, Prop. VII.) 2ny (a - x) and resolving this, the pressure vertically upon this section is y 27 2ny (a – x) x (*_xo) (a-x) Hence the whole vertical pressure is 5 ( 2 – x®) (a-x) ds = 2# / (38 —x") (a—x) dx ) ) SN*) a :: P=anxa (12–20) + (12 – 2*)}+rar sin."x + c Let x =u; then P = 0, and C= – {ama. V(92–a*) + . (r® – ) +raw sin.--a} Hence when x = 0; the whole vertical pressure of the fluid whose altitude is a (S being the specific gravity of the fluid) is p3 XS = 1 1 27 P= 1 21 3 Q= 3 3 raf sin.-'a} x S. Again, if S' denote the specific gravity of the vessel, its solid part being in magnitude 2 (r+t) 276 .(1 + 71 – 7) 3 3 21 W = 3 its weight is {(r + t )– pix} x S'; {(r + 1)* which, by the question, being equated to Q, will give S , as required. 687. The surface pressed is the same in both cases ; the pressures will be as the depths of the centres of gravity. Now the distance of the centre of gravity of the surface from the vertex is Sæd (surf.) altitude. surf. 2 3 Hence the pressures are as 2 1 or as 2 to 1. 3 688. Let a be the axis of the paraboloid, and B the radius of its base, s its specific gravity, and s' that of the fluid ; then the part immersed is (see Vince, Prop. XVI.) Hence, if x denote the depth of the axis immersed, and p the parameter of the generating parabola, we have ту°х p.ro s' 2 2 Again, the distance of the centre of gravity of the whole solid, and of the part immersed from the vertex, are respectively (Vince's Fluxions, p. 117.) a and a Hence the distance between these two centres is 2 a X (1 3 Again, the area of the section made by the surface of the water and if du denote an element of this area, and x denote the distance of this particle from the diameter about which the solid will revolve (if at all); then by Poisson's Mech. vol. II. p. 416, the body will be stavle, unstable, or in a state of indifference, according as fx®du – dxQ is positive, negative, or zero, Sz?du being taken through the whole extent of the area apa V Now if generally R be the radius of this section, it is easily found that ᎡᏎ- 4 Hence the equilibrium is stable, unstable, or indifferent, according as R** - dxQ S which will shew the state of the equilibrium when p, a and are given in numbers. When the paraboloid tends to fall, a must be increased to a', so that P (1 = 0 3 1 689. Let ABC (Fig. 104) be a vertical section, parallel to the base of the horizin, and the ZB being immersed, let DE be the surface of the water, and suppose E the given point which meets the water; also if AC, DE, be bisected in F, H, and BF, BH be trisected in G,g; G, g, will be the centres of gravity of the whole section ABC, and of the part immersed DBE. Now the condition of equilibrium of the prism is evidently the same for the prism as for this section; it is therefore that Gg be vertical or I DE. Hence FG : Hg :: BF: 1 HB 3 3 :: BF : HB :: Gg is parallel to FH and FH is also I DE. Moreover DE is bisected in H, Therefore FD = FE, which gives the required position. Since in the same fluid the volume of the part immersed is constant, the area of DBE is constant, and the locus of the points H, G is an hyperbola (see Problem 44, vol. II.); to find whose equation, let L FBC = a BC = a, } BE = X |