and mom. press. on AOB round Oy = 1 gpa' (Sp=a — fe=0) (cos 0)2 = 1 gpa1 (2a + sin a). d, (mom. press. on PPO round 04). Sr = mom. press. on PQ round OA = gpr3 sin cos 0. Sr. Se ultimately; .. mom. press. on PP'O round OA = gp sin 0 cos 0.80 Sr. No3 = gp sin 0 cos 0 + C; and mom. press. on ROS round OA =gp sin cos 0.80 (fra fr=0)3 = gp sin 0. cos 0.80 de (mom. press. on AOR round OA) dr = mom. press. on ROS round OA a1 4 .. mom. press. on AOR round OA = 4gpa1 ↓ sin 0 cos 0 = gpa (sin 0)2+ C; and mom. press. on AOB round OA ==gpa1 (Se=a - Se=0) sin cos 0 = gpa1 (sin a)2. 122. A hemispherical bell is placed with its mouth downwards on a horizontal plane, and water is poured into the bell through a hole in its vertex; to find how high the water will rise without lifting the bell. Let BAB (fig. 67.) be a section of the bell made by a plane through its axis AC, BCB a section of the horizontal plane, PHP a section of the surface of the water. Draw BQ parallel to AC meeting HP in Q; and let p be the density of the water; W the weight of the bell. The pressure of the water on the interior of the bell, estimated vertically upwards, is equal to the weight of the superincumbent column of fluid, or the weight of a quantity of fluid of the same bulk as the solid generated by the revolution of BPQ round AC = Tgp HC3; and when gp HC W, the weight of the bell is sustained П = by the pressure of the water. 123. A hollow sphere just filled with fluid, is divided into two parts by a vertical plane through its centre; the two hemispheres are held together by ligaments at their highest and lowest points; to find the tensions of the ligaments. Let the circle APQ (fig. 68.) be the section of the sphere made by the vertical plane; P, Q, the highest and lowest points in the circle APQ, Cits centre, K its centre of pressure. Then, (Art. 19.) the pressure on each hemisphere resolved in a direction perpendicular to APQ, is equal to the pressure on the circle APQ; and it acts in a line passing through K. Hence if P, Q, be the tensions of the ligaments at P, Q respectively, P+Q=pressure on APQ; and P. PQ = (pressure on APQ) . KQ. If the radius of the sphere = a, and the density of the fluid = p, the pressure on APQ = гpπа3, 124. A rod AB (fig. 69.) of uniform thickness, suspended by a string EL, rests with one end immersed in a fluid; to find AP the portion of the rod immersed, and the tension of the string by which it is suspended. Let be the area of a section of the rod, W its weight, G its centre of gravity, p the density of the fluid. Bisect AP in F; and through F, G draw FM, GN, vertical. The resultant of the pressure of the fluid on the rod = weight of the fluid displaced = gpK. AP; and it acts in the line FM; the other forces are W acting in GN, and T in EL. Therefore, (Art. 22.) from this equation AP and therefore T may be found. 125. A ship sailing out of the sea into a river, sinks through the space b; on throwing overboard a weight P the ship rises through the space c; to find the weight of the ship. Let p, σ be the densities of fresh and salt water respectively, A the area of the plane of floatation of the ship, W its weight, V the volume of the salt water displaced by the ship; then, (Art. 22. Cor. 2.) W goV, and the volume of the fresh water displaced at first = V +b A, .. W = gp (V+b); and the volume of the fresh water displaced after the weight P is thrown overboard = V + (b −c) A, .. W - P=gp {V + (b −c) A}. Eliminating A, V, we obtain (1 - 0) b W = P. C 126. A triangular prism floats with its axis horizontal, and one edge immersed; to find its positions of equilibrium. Let RSD, AB (fig. 70.) be sections of the prism and of the plane of floatation, made by a .plane perpendicular to the axis of the prism, passing through G its centre of gravity. Let W be the weight of the prism, h the length of its axis, ρ the density of the fluid. Draw GE perpendicular to RD, GF perpendicular to SD. Take PD = & AD, QD = & BD; and bisect PQ in H. Then H is the centre of gravity of the fluid displaced; Therefore (Art. 22. Cor. 2.) GH is perpendicular to AB or PQ; and gph. AD . BD . sin D = W. and PG QG, for PH = QH, and GH is perpendicular to PQ; = .. PD2 – QD2 – 2 ED . PD + 2 FD . QD = 0; :. AD2 – BD2 – 3 ED. AD + 3FD. BD = 0; The last term of this equation is negative, and therefore one root of it is negative; but the nature of the question excludes all negative values of AD and BD. Hence, there cannot be more than three positions of equilibrium as long as the same edge is immersed. All values of AD greater than RD, and of BD greater than SD are likewise inadmissible. -127. Two equal rods RD, SD, (fig. 71.) meeting each other at right angles, float with the angle D immersed; to find their positions of equilibrium. Let G be the centre of gravity of the rods; P, Q, the middle points of the portions immersed; GR perpendicular to DR; GS perpendicular to DS; GH perpendicular to PQ; HN perpendicular to RD; 2c the sum of the lengths of the immersed portions, when the weight of the fluid displaced is equal to the weight of the rods; RD=e; PD = a; QD = b. Then, a + b = c; and the centre of gravity of the fluid displaced must be in GH, it must also be in PQ, therefore H is the centre of gravity of the fluid displaced; y The equation to PQ referred to the axes DR, DS, is + = b a the co-ordinates of G are e, e', therefore the equation to GH is b (y - e) = a (x − e); and H is the intersection of GH and PQ, .. (b2+a2)DN = ab2 + (a2 −ab) e, .. (b2 + a2) a = cb2 + (a −b) ec, ' ... {(c − a)2 + a2 } a = c (c − a)2 + (2 a − c) ec, .. 2a3 − 3ca2 + (3c − 2e) ac − (c − e) c2 = 0, 128. To find (M) the metacentre of the prism RDS (fig. 70), the prism being inclined in the plane RDS. The moment of inertia of the plane of floatation round an axis through its centre of gravity, perpendicular to RDS and the volume of the fluid displaced = h. AD. DB. sin D ; ‚'‚ (Art. 26.) §. h . AD. DB. sin D. HM = —, AB3 . h, |