the whole area of the spherical envelope, viz. 4na’, by the invariable unit of pressure of the fluid, viz. B. 4. A fluid being supposed to consist of an indefinite number of particles at finite intervals, and the repulsive force between two neighbouring particles being supposed to vary inversely as the nth power of their distance, to determine the relation between the density and the elastic force of the fluid. Elastic force o (Density) *** The converse proposition is also true. Newton: Principia, Lib. 11., Sect. 5, Prop. 23, Scholium. n+2 3 act. LET K be an elemental portion of a plane area immersed in a fluid at any inclination to the horizon, and let h be the depth of K below the surface of the fluid ; then, p being the density of the fluid, the pressure on K will be equal to gphK. On every such small element of the immersed area a corresponding force will The immersed area will therefore be acted upon at right angles by an indefinite number of parallel forces all tending in the same direction; let the resultant of these forces be R, and, taking any system of coordinate axes, rectangular or oblique, in the plane of the immersed area, let ī, , be the coordinates of the point of the area at which R acts. Then ł, y, R, may be R, determined from the three equations R = gp (hK), T.R = gp (hxK), y.R = gp (hyK), the formulæ for , y, being $(hak) (hy K) y Σ (hK)' (hK) the limits of the summations being dependent upon the form of the immersed area. The point of which ż and y are the coordinates is called the Centre of Pressure. If the axes of coordinates be at right angles to each other, gp (hxK) and gps (hyK) denote the moments of the pressure of the fluid about the axis of y and x respectively. = = X = If the axes of coordinates be at right angles to each other, and that of y be in the surface of the fluid, then, 6 being the inclination of the immersed area to the vertical, h = x cos 0, K = dxdy, and therefore Σ (GK) 2 (cm ) Ý (2K) Σ (Κ) these formulæ shew that, if the immersed area be a plane lamina of uniform thickness, moveable about the axis of Y, the Centre of Pressure will coincide with the Centre of Percussion, a proposition demonstrated by Cotes, (Hydrostatical and Pneumatical Lectures, pp. 41, 42, 3rd. edit.). If we adopt the notation of the Integral Calculus, we have, putting K = dx dy, SShx dx dy SShy dx dy ; Sshdx dy' Ssh dx dy the limits of the integration depending upon the boundary of the area. If the boundary of the area be discontinuous, we must divide the area into a number of portions with continuous boundaries; we must then take the sum of the values of each of the integrals Sfx dx dy, Sla* dx dy, SSxy dx dy, for these several portions, to express its value for the whole of the immersed area. It is frequently more convenient to adopt polar formula. If the axes of coordinates be rectangular, that of y being parallel to the surface of the fluid, and the area immersed be symmetrical with regard to the axis of x, it is plain that the Centre of Pressure must lie in the axis of x, and it is therefore necessary to evaluate only the two expressions (hK) and $(heo). The idea of a Centre of Pressure on a plane surface subjected to the pressure of a fluid, was first developed by Stevin in his Beghinselen der Waaghconst, published in 1585. The existing state of analysis rendered it impossible to make any but the most elementary applications of his conception. He has actually calculated however the position the centre of pressure of a parallelogram immersed to any depth in a fluid, with one edge parallel to the surface, the plane of the parallelogram being inclined at any angle whatever to the horizon, and has laid down on correct mechanical principles a general sketch of the method of extending such a computation to any plane rectilinear figures whatever. These investigations may be seen in the French translation of Stevin's works by Albert Girard, pp. 495, 496, 497. On the discovery of the principles of the infinitesimal analysis, the determination of the centres of pressure of planes of various forms was effected without difficulty. Among the earliest writers who have treated on the problem of the centre of pressure may be mentioned Herman, Phoronomia, p. 141, anno 1716, and Cotes, Hydrostatical and Pneumatical Lectures, pp. 40, 3rd. edit. 1. To find the centre of pressure of a parallelogram immersed in a fluid, one edge of the parallelogram being in the surface. Let ABCD (fig. 4) be the parallelogram, OE being a straight line joining the middle points 0, E, of the two horizontal sides. Let OE, OB, produced indefinitely to x, y, be taken as the axes of coordinates. Let PP', pp', be two lines parallel to AB, cutting OE in two points M, m, very near to each other. Then, being the inclination of Ox to Oy, K = 2y dx. sin a, h = x sin a; and therefore, if OE = 1, a 2y a . x 0 or, y being invariable, S. 2y dx . sin a. x sin a. x = as * 24 dx sin a . x sin a, * do - 3 Sizda , 2 x2 dx x dx, 0 378 = 122.ā, 2 = 1, which shews that, if G be the centre of pressure, OG = 2GE. “Si le fond d'une eau n'est à niveau, estant parallelogramme, duquel le plus haut costé soit à fleur d'eau, et de son milieu, au milieu de son costé opposite, est menée une ligne ; le centre de gravité (du pressement de l'eau congregé contre le fond) divise ceste ligne de telle sorte, que la partie haute à la basse est en raison double.” Stevin : Euvres Mathématiques par Albert Girard, p. 495. 2. To find the centre of pressure of a portion of a parabolic area bounded by the axis of the parabola, the curve, and an ordinate at right angles to the axis, supposing the ordinate to lie in the surface of the fluid. Let a denote the portion of the axis between the foot of the ordinate and the vertex of the parabola, and b the length of the ordinate. Let the axis of x coincide with the axis of the parabola, and the axis of y with the tangent at its vertex, and let O be the inclination of the axis of the parabola to the vertical. Then, supposing the parabola to be divided into small strips, contained between consecutive ordinates, we have K = y dx, h = (a – x) cos O; and therefore a soydx . (a – x) = { "xy dx. (a – x); 2; 2 w } Again, dividing the parabolic area into elementary rectangles dx dy, we have y SD <a – x) dx dy = SS.(a – x) y dedy, osca a) y dx = t["(a – a) yʻda; or, 4m being the latus rectum of the parabola, ofica - ) (a - * ’ . 0 0 a a 0 |