the other by some multiple of two right angles; therefore 18. One asymptote of an hyperbola lies in the surface of a fluid: find the depth of the centre of pressure of the area included between the immersed asymptote, the curve, and two given horizontal lines in the plane of the hyperbola. Let BB'C'C (fig. 10) be the included area. Draw PM, horizontally, equidistantly from BB', CC'. Take any two strips PM, P'M', of equal breadths, and equidistant from PM. Then, 7 denoting the breadth, Pressure on PM = 7.PM.OM.sin a = 17 (a2+b2) sin a = pressure on P'M'. Hence PM, P'M', balance about PM. Similarly for all like pairs of strips. Hence the centre of pressure of BB'C' C lies in the line PM. 19. A cone is totally immersed in a fluid, the depth of the centre of its base being given. Prove that, P, P', P'", being the resultant pressures on its convex surface, when the sines of the inclination of its axis to the horizon are s, s', s", respectively, P2 (s′ — s′′) + P12 (s′′ − s) + P′′2 (s — s′) = 0. = Let R the resultant pressure on the whole surface of the cone, the base included; P= the resultant pressure on the convex surface, when the axis is inclined at an angle a to the horizon; B = the pressure on the base; h = the altitude of the cone; the depth of the centre of its base; r = the radius k of its base; σ = the density of the fluid. = Multiplying these three equations in order by s' s-s', and adding, we have = 0. 20. Light emanating from a luminous circular disk, placed horizontally on the ceiling of a room, passes through a rectangular aperture in the floor: ascertain the form and area of the luminous patch on the floor of the room below. Shew that neither the shape nor the area of the patch will be affected by any movement of the disk along the ceiling. Let 0 (fig. 11) be the centre of the disk, M any point in its circumference. Through P, any point in a side of the aperture ABCD, draw OPO' to meet the floor of the lower room in O'. Draw MP and produce it to M', a point in the floor. With O' as centre and radius O'M' describe a circle on the floor. This circle will be the area illuminated by the rays which pass through the point P. Again, lines drawn from 0 through A, B, C, D, to meet the floor will form a rectangle A'B'C'D' on the floor. The form of the patch is therefore such as represented in (fig. 12). Let AB = a, BC = b, and let h, h', denote the heights of the higher and lower rooms. = πr'2 + a'b' + 2 (a' + b') r' 1 · 7 = {πr2h'2 + ab (h+h')2 + 2rh' (h + h') (a+b)}. h2 This result shews that the form and area of the patch are independent of the position of the disk on the ceiling of the upper room. 21. If c,, C, C, be the lengths of the meridian shadows of C2 three equal vertical gnomons, on the same day, at three different places on the same meridian, prove that the latitudes λ, λ λ of the places are connected together by the equation 17 27 the altitude of the gnomon, 8 when on the meridian. = tan (sun's zenith distance when on the meridian), Hence λ-λ2 -1 C2 C -1 3 + x = (cc) cot (λ, -λ) .......... Multiplying (1), (2), (3), in order by c1 (c-c2), c. (c2 — c1), adding, and observing that TUESDAY, Jan. 17, 1854. 9 to 12. m 1. IF C denote generally the number of combinations of m m things s together, and C be taken to denote unity for all values of m; prove that, if 0 n-1 n-2 n-r+1 n-r n-3 then S+S+S+S+...+ S=1′′+1+2′′ +3′′-1 +...+ (n−1)3+n2+(n+1)'. whence the triangle formed by producing IS, HJ, is isosceles, and therefore, CS, CH, being equal, the vertex of the triangle must lie in BC produced. Since the angles SIY', HJZ, are equal respectively to the angles SPY, HPZ, they can never be zero, and therefore SI, HJ, can never be perpendicular to the major axis. Thus the point of intersection of IS, JH, BC, can never move off to an infinite distance from C. 8. From any point T, (fig. 5), two tangents are drawn to a given ellipse, the points of contact being Q, Q': CQ, CQ, QQ', CT, are joined; V is the intersection of QQ, CT. Prove that the area of the rectilinear triangle QCQ' varies inversely as But CP2 CV CT.CV-CV2 CV. TV. = |