That these results may be admissible, x and y must be both possible, both positive, and both less than a. Hence we must have 9 > 164, and 9 16 < 1, > (0-100) '< 9 8 that is, must be less than P and greater than 16 3. A given triangular lamina ABC, all the sides of which are unequal, floats in a fluid with two angles B and C immersed; to determine its position of equilibrium, the plane of the lamina being vertical. Adhering to the diagram and notation of the preceding problem, it is plain that, BB'C'C being in the fluid and B'AC' being in vacuum, BBCCxo = BAC.p, whence o (ab – xy) = pab, oxy = (0-p) ab. Also the centres of gravity of BB'C'C and BAC, and therefore of B'AC", must be in the same vertical line. or Thus the solution of this problem coincides in form with that of the preceding, op replacing p throughout: and therefore we know that, supposing the angles B and C to be both immersed, there may be one and there may be three positions of equilibrium. Cor. 1. There will be therefore at most three positions of equilibrium when A alone is immersed, and three when A alone is out of the fluid: thus, varying the angles of the triangle, we see that, at most, the triangle can have only eighteen positions of equilibrium. Cor. 2. There will be three positions of equilibrium, A being without the fluid, if the triangle be equilateral, provided that 7 8 p be greater than and less than 16 16 Bossut: Traité d'Hydrodynamique, tom. I. p. 160. Poisson: Traité de Mécanique, tom. II. p. 584. 4. To find the positions of equilibrium of a parabolic lamina BAC, bounded by a double ordinate BC at right angles to its axis, which floats in a fluid with its vertex A immersed, the plane of the lamina being vertical. First, let us suppose the axis to be vertical, and let c be that portion of the axis which is within the fluid, a being the length of the axis of the parabolic area and 26 its double ordinate. Let e р be the density of the lamina and o of the fluid. Then, for equilibrium, sab.p = {c.b pa' = oc?, = which determines c, and therefore the position of equilibrium. Next, let us suppose the parabola to lie in an oblique position. Let Qq (fig. 12) denote the line of floatation, G the centre of gravity of the lamina, which will lie in the axis AD; H the centre of gravity of the fluid displaced, which will lie in a diameter VP through a point P, the tangent of which is horizontal. In order that the lamina may be at rest, HG must be vertical. 2m ; 3 = Draw PE vertical to meet AD in E, and draw PM at right angles to AD. Let the tangent at P meet the axis DA produced in the point T. Let 4 PTA = 0, and 4m = the latus rectum. Then, by the nature of a parabola, AM = m coť 0, ME and GE = PH-PV, AG=AD. Hence m cot 0 + 2m + PV = ža, or, since 62 = 4ma, 72 72 cot 0 + + {PV = {a ...........(1). 4a 2a Again, area QPq, jp.ab = jo.QV.PV.sin 0, pa’b? = 0?.QV.PV?. sin? 0 = 0.4SP.PV.sino, S being the focus ; 2m but SP 1 - cos 20 sino 0 hence płab2 = 40°. PV.m, 32 o?. PV3. p.area BAC or m ; From (2) and (3) we know PV and 0, which will have two equal values with opposite signs. The positions of equilibrium of the lamina are therefore determined. The values of 0 will be impossible if 72 Thus the parabola will sometimes have one and sometimes three positions of rest. Bossut: Traité d'Hydrodynamique, tom. I. p. 170. 5. A solid homogeneous cube floats in a fluid with only one angle immersed ; to determine its position of equilibrium. 1 Let 0A, OB, OC, (fig. 13) be three edges of the cube, O being the angle immersed. Let PQR be the plane of floatation. Let OP = a, OQ = b, OR = C, and let 2a denote the length of an edge of the cube. The equation to the plane PQR, OA, OB, OC, being taken as the axes of x, y, z, respectively, will be у X + a + 1. с = a, = a. =C Also, ğ, y, z, being the coordinates of the centre of gravity of the pyramid OPQR, and , y, z', of the centre of gravity of the cube. ä = ja, y=b, = c, 있 But the line passing through the two points (@, y 2) and (', y', 2') must be at right angles to the plane PQR; hence a (a – įa) = b (a - 16) = c(a - Àc), a (a - b) = {(u’ – 6), = c a - b = 0, a + 6: the latter of these results is incompatible with the condition that the angle 0 is the only one immersed. Hence we must have a = b, and similarly, from the second and third equations, 6 C, Thus it appears that the cube can float in only one way, provided that only one angle be immersed. or 4a C = a. Again, p being the density of the cube, and o of the fluid, p the weight of the fluid pyramid OPQR will be {abc.o.g, and that of the cube will be 8a'. p.g. Hence &abc.o. 1.0.9 = 8a'.p.gg σαbe = 48ρα”, But a = b = c; hence σα = 48ρα”. 6P 6 = a = 2a or C= In order that the problem may be possible, a, b, c, must be less than 2a; hence we must have the inequality 6р <l, 6. A hemispherical bowl of given weight (2W) floats upon a fluid with one third of its axis below the surface: if a weight 5 W be put into it, how much of its axis will be immersed when it is in its position of rest? Length of the axis immersed = { x length of the whole axis. 7. A cylindrical vessel in an upright position contains a quantity of fluid: supposing a solid cylinder of smaller radius to be placed in the fluid so as to float at rest in an upright position, to determine how much the fluid will rise in the hollow cylinder. р denote the density of the solid cylinder and o of the fluid ; let R be the radius of the internal surface of the hollow cylinder, r the radius and h the altitude of the solid cylinder. Then, x denoting the altitude through which the fluid rises above its original surface, Let p pr? X = h. ORP 8. A hollow cone, of which the vertical angle is 37, is placed with its axis vertical and vertex downwards : to find what quantity of fluid it must contain, in order that a given sphere, |