the depth; the tangent at the vertex of the segment lies in the surface of the fluid: to find the depth of the centre of pressure. If h represent the length of the axis of the parabolic segment, the required depth will be equal to zh. 5. A segment of a parabola, cut off by a double ordinate to the axis, is immersed in a fluid, the density of which varies as the depth; the base of the segment lies in the surface of the fluid: to find the centre of pressure. If h denote the length of the axis of the segment, and 7 the altitude of the centre of pressure above the vertex, ã = 1h. CHAPTER III. RESULTANT PRESSURES OF FLUIDS ON CURVED SURFACES OF IMMERSED SOLIDS. = LET K be any indefinitely small element of the surface of a solid at a depth h below the surface of a homogeneous incompressible fluid, and let p be the density of the fluid. Let A, B, C, be the areas of the projections of K upon the coordiA nate planes yz, zx, xy, respectively, the axes of coordinates being supposed to be rectangular. Let X, Y, Z, be the components of the resultant pressure, and , y, z, the coordinates of any point in its direction. Then, the limits of the summation being defined by the boundary of the surface, X = gp (hA), Y = gp3 (hB), Z = gp (hC), Y = (%), Ζ = (), Yü - Xy = gp {E(hBx) – 3 (h. Ay)} = N: from the first three of these equations we can determine the magnitude of the resultant; and, eliminating X, Y, Z, we have for the equations to its direction, any two of the three equations yΣ (%C) - Σ (%B) = Σ (%Cy) - Σ (hΒα), αΣ (%B) - yΣ (MA) = Σ (hΒα) - Σ(hAy). In these formulæ, whenever X acts in the negative instead of the positive direction, we must replace A by - A: analogous remarks are applicable to the actions of Y and Z. = Multiplying L, M, N, by X, Y, Z, respectively, we see that LX + MY +NZ = 0, which is the condition necessary that the three equations defining the direction of the resultant force may coexist ; in other words, this relation is the condition that the pressures on the elements of the surface may be all reducible to a single resultant. If the fluid be a heterogeneous fluid, then, p denoting the unit of pressure at K, we must replace the six preceding equations for the determination of the resultant pressure by the six following: X = Σ(pA), Y = Σ(pΒ), Ζ = Σ(pC'); αΣ (pΒ) - yΣ (DA) = Σ(p Βα) - Σ(pAy): dp = gpdh: if the fluid be elastic, then, g denoting the force of gravity, k a constant quantity dependent upon the elasticity of the fluid, and C an arbitrary constant which may be determined when simultaneous values of h and p are given, gh p = Ciek. k If g be very small compared with or the weight of the fluid be inconsiderable in comparison with its elastic force, then p= C very nearly, or the unit of pressure is the same at every point. Let S represent the magnitude of any area immersed in a fluid ; through the periphery of S conceive an indefinite number of vertical lines to be drawn so as to trace out on the horizontal plane coinciding with the surface of the fluid a curvilinear figure including an area of magnitude S". Also let V denote the volume included between these vertical lines and the areas S, S'. Then, in accordance with the equations Z= (PC), p = g Sp dh, we see that the vertical pressure upon S is equal to the weight of the volume V of the fluid. 1. A vessel, bounded by two vertical planes at right angles to each other, and a paraboloid of revolution, the axis of which coincides with the intersection of the planes, is filled with fluid as far as the focus : to determine the magnitude and direction of the resultant pressure. Let the vertex of the paraboloid be taken as the origin of coordinates, and its axis as the axis of z, the plane of xz being supposed to bisect the right angle contained between the two vertical planes. It is evident that, the surface being symmetrical with regard to the plane of xz, the resultant pressure must lie within the plane xz. Hence it will be necessary to take only the equations X = gp (hA), Z=-gp(hC), z (hA) + ū (hC) = (hA2) + (hCx). Let 4m be the latus rectum of the paraboloid ; then h= m - 2, and therefore, the integrations being performed from y=-(2mz) (2mz)', and from z = 0 to 2 = m, (hA) = S/(m – z) dy dz to y dz 2 (2m)(met - z*) dx = $ mov2, " 2 (h42) = $5(m – 2) z.dy dz = 2 (2m) ["(m2* --') dz = 3 m• v2. * m 0 S Again, r being the distance of the small area C from the origin of coordinates, and 0 the inclination of r to the axis of x, (hC) = S/(m – 2) rd0dr the integration being performed from r = 0 to g = 2m, and from 0 = -17 to 0 = 17 : hence (hC) = m*sd0 = mo : 2. A hemispheroidal bowl is filled with fluid; to find the magnitude and direction of the pressure upon a quarter of the surface bounded by two planes passing through the axis, which is vertical. Let the axis of the bowl be taken as the axis of 2, and the intersections of the two vertical planes with the plane bounded by the rim of the bowl, the axes of x and y. Then a and c being the semi-axes of the generating ellipse, the equation to the surface will be xx2 1. с Hence, integrating from z = 0 to z = (a? - y)t, and from – ?}, a |