с a с a с с hence (hAz) = ac. 16 Again, integrating from y = 0 to y = (ce? – xe?)t, and from x = 0 to x = a, we have *(hC) = [[zdxdy = S SCA? – x? – y*)* dx dy - a* – ) = c Sin (a? – xo) dx = štałc. (hCx) = SSxzdx dy 4 dx dy = Επαc. Hence, R denoting the magnitude of the resultant pressure, we have, since by symmetry (HB) = S(hA), R=gp {fa?c* + $faa*c*}* = dgpac {8c° + n’a?}!! Also, from the equation Σ (hA) - Σ (%C) = Σ (h A2) - Σ (%Cu), we have ž.jac – T. ana*c = i traci - To nac, .(1). In like manner, from the equation yΣ (%C) - 3. Σ (%B) = Σ (%Cy) - Σ (hΒα), as is evident from symmetry, we should obtain 8may - 16cz - 37 (a' - c). ... .(2) Again, it is manifest from symmetry that (hB) = (hA), (hBx) = (hAy): T or we have hence from the equation .. (3). Any one of the equations (1), (2), (3), is deducible from the other two, any two of them being the equations to the direction of the resultant pressure. 3. A space bounded by the three coordinate planes and the surface y b is filled with an elastic fluid without weight: to investigate the equations to the direction of the resultant pressure. Since the unit of pressure will be constant, the three equations for the direction of the resultant reduce themselves to yΣ (C) - ΖΣ (B) = Σ(Cy) - Σ (B2). .......(1), ΖΣ (Α) - Σ(C) = Σ(Α2) - Σ (Ca)........(2), :(2), αΣ (Β) - yΣ(A) = Σ(Βα) - Σ(Ay)........(3). Now, integrating first from y = 0 to y = y, and then from z = 0 to z = C, (A) = Ssdy dz = Sy dz z 1 - 2 + ch с dz = zbc, = rs1 5. ) dz yzdz and Σ(Az) = - Bobc?. The values of the other analogous quantities are obvious from symmetry. Hence, from (1), 7. jab - z .ca = Bab? – Xoaca, (4). Similarly, from (2) and (3), c(z – c) = a (x - ļa)........ (5), (6). or Any one of the three equations (4), (5), (6), may be deduced from the other two, and therefore any two of them determine the direction of the resultant pressure. 4. A light vessel in the form of a segment of an ellipsoid, the ends of which are sections made by a principal plane and a plane parallel to it, stands with its larger end on a horizontal plane: to determine how much fluid may be poured into it without its raising the vessel and forcing its way out. Let the equation to the ellipsoid be the base of the segment being principal plane coinciding with the plane X, y. The semi-axes of a section of the ellipsoid made by a plane parallel to the plane xy and at a distance z from it, will be equal to 61 a and therefore the area of this section will be equal to a From this it is plain that, h being the depth of the fluid, the volume of the fluid will be equal to h3 1 dz = πα (και 3c But the volume of a cylinder on the base tab of the ellipsoidal segment and of an altitude h, is equal to mabh. Hence the vertical pressure of the fluid on the vessel will be equal to the weight of a mass of the fluid, the volume of which is equal to παύλ - παο (και παbh3 – ( h 3c 3c and therefore, if W denote the weight of the vessel and p the density of the fluid, the greatest value of h, in order that the vessel may not be raised upwards, is given by the equation 3 or 3cW h apgab 5. A thin hemispherical bowl is placed in vacuum and filled with fluid, the surface of which is subjected to a pressure equal to that of the atmosphere. If the bowl be divided into two equal parts by a vertical plane through the centre, and be kept together by a string in the diameter perpendicular to the plane of section, to find the tension of the string, the two halves of the bowl being supposed to be joined together at the lowest point. Let AOB (fig. 9) be the dividing plane, C being the centre of the bowl, and O its lowest point. Let P be any point in the area of the bowl: join CP, and let OPQ be a quadrant of a great circle of the sphere: join CQ. Let r = the radius of the bowl, LACQ = 0, LPCQ = 0, and h = the altitude of a column of the fluid of which the pressure is equal to that of the atmosphere. Then, T denoting the tension of the string, the moment of T about a tangent line at O parallel to AB, must be equal to the moment of the pressure of the fluid upon the surface AOBQ about the same line. Now the component of the pressure upon an elemental area r cos 0 dp.rdo at P, at right angles to the plane AOB, will act through the point C, and will be equal to gp (h + r sin 6).r cos Odo.rdo.cos 0 sin - gpr (h + r sin 6) cos O sin pdo do; and therefore the moment of the whole pressure about the tangent line at O will be equal to ST*(+r 0 h = = 0 Hence, Tr being the moment of T about the line at 0, we see that T = ågpr2 (37h + 4r). 6. A hemisphere, filled with fluid, is divided into four unconnected portions by two vertical planes through its axis at right angles to each other; the parts are just kept together by ; four strings fastened at the centre and at four points in the surface: to find the position and tension of one of the strings. Let the equation to the hemisphere be x? + y + z = px, 2 the planes of 2, 2, and y, z, being the two dividing planes. Then, T being the tension of the string in the octant of space + X, + Y, +2, we shall have, p denoting the density of the fluid, T = ¿gpro (n® + 8), the equations to the direction of T being + 7. A sphere is just filled with a homogeneous fluid ; to determine the resultant of the pressures upon either of the hemispheres into which it is divided by a vertical plane. If the equation to the sphere be x + y2 + x2 = r?, the plane of yz being the dividing plane, and the axis of z extending vertically downwards; then, p being the density of the fluid and R the resultant pressure, R = gptors, and the equations to the direction of R will be y = 0, 4x. 8. To find the form of an open vessel full of fluid, supposed to be a surface of revolution with its axis vertical, that the whole horizontal pressure exerted upon it by the fluid may be the greatest possible ; the altitude and volume of the vessel being given. 32 a |