The vessel will be a cone, and, if h be its altitude and its volume, the radius of its base will be equal to 303\ πή a 9. A vessel in the form of a paraboloid of revolution is partly filled with fluid, and then inverted upon a horizontal plane: having given the weight of the vessel, to find the altitude of the greatest volume of fluid which can be contained without running out at the rim of the vessel. If W denote the weight of the vessel, l the latus rectum of the paraboloid, and x the required altitude, 2W ngpl" 10. A hollow vessel in the form of a tetrahedron is filled with a known fluid, and placed in an inverted position on a horizontal plane: to find the least weight of the vessel that the fluid may not escape from under it. Least weight of the vessel = twice the weight of the fluid. X = CHAPTER IV. EQUILIBRIUM OF SOLID BODIES FLOATING FREELY. a In order that a solid, partially or totally immersed in a fluid, may float in equilibrium, it is sufficient and necessary that the resultant pressure of the fluid upon it act in the vertical line passing through its centre of gravity and be equal to its weight. If the fluid be incompressible (or even if it be elastic, provided that the solid be totally immersed), these conditions are equivalent to the two following conditions, viz. that the weight of the fluid displaced be equal to that of the solid, and that the line joining the centres of gravity of the solid and fluid displaced be vertical. The discovery of the principles of the equilibrium of floating bodies is due to Archimedes; his investigations on this subject were published in his treatise entitled Tepi Tôv oxovuévwy, the original of which is no longer extant. The Latin translation by Tartaglia of this remarkable work, by which our knowledge of the hydrostatical discoveries of Archimedes has been preserved, is entitled De iis quæ vehuntur in aquâ ; an edition of which by Commandine was published at Bologna in the year 1565. SECTION 1. 1. To find the positions in which a square lamina ABCD (fig. 10) may float in a fluid with one angle A out of the fluid, the plane of the lamina being vertical. Let PQ be the line of floatation : let AP = 2, AQ = y. Join AC and bisect it in G. Bisect PQ in F, join AF, and take HF = {AF. Then G and H will be the centres of gravity X E of ABCD and APQ respectively. Join GH and draw FE parallel to HG. Then, in order that the lamina may be at rest, G and the centre of gravity of PBCDQ must be in the same vertical line: but H, G, and the centre of gravity of PBCDQ are in a straight line; hence GH and EF must be vertical lines. Join PE, QE; these two lines will be equal, because PF= QF and PFE = L QFE. Let a denote the length of a side of the square, p the density of the lamina, and o of the fluid. Then, the weight of the displaced fluid being equal to that of the lamina, we must have 6.PBCDQ = p.a', or hence (x - y). (x + y - ya) = 0, which equation gives us either x - y = 0.. 2 + y -a = 0...... Taking (1) and (2) we have .(2), (3). or x2 2-3(1-9)==y' 1 ay which gives one position of equilibrium, provided that l be less than 1 and greater than . and and therefore 23 < 1, 23 a Thus there will be two other positions of equilibrium, provided that x and y be both possible and positive quantities less than a. This will be the case if the two inequalites hold good, viz. р 32 23 24 that is, if be greater than and less than 32 32 Bossut: Traité d'Hydrodynamique, tom. I. p. 165. 2. A given triangular lamina ABC (fig. 11), all the sides of which are unequal, floats in a fluid with one angle (A) immersed; to determine its positions of equilibrium, the plane of the lamina being vertical. Let D be the middle point of BC, and E that of B'C', B'C' being the line of floatation. Join AD, AE, DE, BD, C'D. , , , Then, G being the centre of gravity of the lamina and H of the fluid displaced, AG = {AD, AH = {AE, and therefore AG: AD:: AH: AE; which shews that GH is parallel to DE. But GH must be vertical and therefore at right angles to B'C': hence DE is at right angles to B'C'. Since DEB = 1 DEC", and BE L C'E, it follows that DB' = DC'. . Let AB = x, AC' = Y, AD = l, AC = b, AB = a, . B'AD = 1, L C'AD =u: also let p = the density of the solid, and o = that of the fluid. Then, the weight of BAC being equal to that of the fluid displaced by B'AC', we have jxy o sin 2 B'AC" = jabp sin L BAC, = = or 2lx cos , 2 or O xy = pab.... ..(1). Again B'D2 = x + 72 C'D2 = y + 1 – 2ly cos ; and therefore, B'D being equal to C'D, 2? + – 21x cos X = y + 12 – 2ly cos M, I – ? From the construction of these two curves it is easily seen that they can intersect each other either once only or three times only in the positive quadrant: there may therefore be one and there may be three positions of equilibrium. Should all the points of intersection of the hyperbolæ be such that x is greater than a or y than b, the lamina will not rest in any position with only one angle immersed. Cor. 1. If the triangle be isosceles, A being the vertical angle, cos X = cos y, and then, instead of the equation (2), we shall have (x - y) {x + y - 21 cos ^} = 0, 21 cos N, 21 cos X = ja, and therefore, instead of the equation (2), we have either of the two following, y = 0, or x + y = ja. From (1) and the former of these we get xe - La, ) or X which gives one position of equilibrium. |