(y = HR Sy=0) a2 = x2. HR = x2 √(2ax − x2); 5 x2)* − − ¦ a ( 2 a x − x2) 3 + 1⁄2 a2 S.√2ax − x2 ; ffy, between the proper limits, a'. fyxy = 1 xy2 + C ; = (Sy= HR — Jy=0) x y = 1 x HR2 = 1 x (2 a x − x2) ; Safyay, between the proper limits, = a'. 121. To find the center of pressure of the sector AOB (fig. 66.) having its center O in the surface of the fluid, and the radius 04 perpendicular to the surface. Oy. Let the plane of the sector meet the surface of the fluid in the arcs PP', QQ'. Let the density of the fluid = p, OA = a, AOB = a, OP=r, PQ' = dr, AOR = 0, ROS = 80. Then d, (press. on PP'O) dr = press. on PQ press, on PP0 = gp cos 0.80. fr2 = gp cos 0.80 — + C ; and press. on ROS = gp cos 0.80 (r=a - £=0) 7° .. press. on AOR = gpa3 ↓ cos 0=gpa3 sin 0 + C; and press, on AOB = } gpa3 (Se=α- Se-o) cos 0=gpa3 sin a d, (mom. press. on PP'O round Oy) dr = mom. press. on PQ round Oy = gpr3 (cos 0)2. Sr. 80 ultimately; .. mom. press. on PP'O round Oy = gp (cos 0)2 . d0 f, r2 = gp (cos 0)2 : 80 2 4 + C; ... mom. press. on AOR round Oy = 4g pa1 √ (cos and mom. press. on AOB round Oy =1&pa1 (fea — fe=0) (cos 0)2 = 1gpa1 (2a + sin a). = gp and mom. press. on ROS round OA sin cos 0.80 (fra fr-0) = gp sin e. cos 0.80 d. (mom. press. on AOR round OA) dr = mom. press. on ROS round OA = gpa' sin cos 0.80 ultimately; .. mom. press. on AOR round O4 = gpa sin cos = =gpa1 (sin 0)2 + C; and mom. press. on AOB round OA =gpa' (Sp=afe=0) sin cos 0 = gpa (sin a)2. a1 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. p 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 o 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 = 3gp HC3; π and when gp HC W, the weight of the bell is sustained 3 = 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. fluid If the radius of the sphere = a, and the density of the وم .. P+Q=gρπα, Ρ. 2α = ςρπα. α, . Ρ=ερπα, Q= ξερπα. L 4 124. 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.) T +gpк. AP = W; gpк. AP. FE = W. GE, 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 o be the densities of fresh and salt water respectively, A the area of the plane of floatation of the ship, Wits 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+bA); 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}. b P. с 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 |