9–3. Locate the distance x to the center of gravity of the homogeneous rod bent into the parabolic shape. If the rod has a weight per unit length of 0.5 lb>ft, determine the reactions at the fixed support O. y 0.5 ft SOLUTION O dL = 2dx 2 + dy 2 dy = x dx 1 ' x = L x dL = L dL L0 x2dx2 + x2 dx2 1 L0 2dx2 + x2 dx2 Let x = tan u dx = sec2 u du p 4 x = L0 tan u21 + tan2 u sec2 ud u p 4 L0 21 + tan2 u sec 2 udu p C sec3 u D 04 3 = p C sec u2tan u + 12 {ln | sec u + tan u|} D 04 ' x = 0.531 ft Ans. Also, dL = 2dx2 + dy 2 L = LB 1 + a dy 2 b dx dx 1 = L0 21 + x 2 dx = 1.148 ft L ' xdL = 1 L0 x 21 + x 2 dx = 0.6095 x = 0.6095 = 0.531 ft 1.148 Ans. + ©F = 0; : x Ox = 0 + c ©Fy = 0; Oy - 0.5(1.148) = 0 Oy = 0.574 lb a + ©MO = 0; Ans. Ans. MO - 0.5(1.148)(0.531) = 0 MO = 0.305 lb # ft Ans. y 0.5x2 x 1 ft 9–4. Locate the distance y to the center of gravity of the homogeneous rod bent into the parabolic shape. y 0.5 ft SOLUTION O dL = 2dx2 + dy2 L = LA 1 + a dy 2 b dx dx 1 = L0 21 + x 2 dx = 1.148 ft L ' ydL = 1 L0 0.5x2 21 + x2 dx = 0.2101 ft y = 0.2101 = 0.183 ft 1.148 Ans. y 0.5x2 x 1 ft 9–7. Determine the area and the centroid x of the parabolic area. y h SOLUTION y Differential Element:The area element parallel to the x axis shown shaded in Fig. a will be considered. The area of the element is dA = x dy = a h1>2 y 1>2 dy x a ' ' Centroid: The centroid of the element is located at x = y 1>2 and y = y. = 2 2h1>2 Area: Integrating, h A = LA LA = dA LA h ' xdA x = dA = L0 ¢ a 1>2 2h a L0 h 1>2 y1>2 ≤ ¢ 2 ah 3 y 1>2 dy = a 1>2 h h 2a A y3>2 B 2 = 1>2 3h y 1>2 dy ≤ 0 2 ah 3 Ans. a2 y2 h a ¢ ≤` y dy 2h 2 0 3 L0 2h = = = a Ans. 2 2 8 ah ah 3 3 h 2 h x2 –– a2 x a 9–11. y Locate the centroid x of the area. h y ⫽ —2 x2 b SOLUTION h dA = y dx ' x = x x b x = LA h 3 x dx 2 b L0 ' x dA = LA dA b h 2 x dx 2 b L0 B b h 4 b x R 4b2 0 = = b h B 2 x3 R 3b 0 3 b 4 Ans. 9–12. y Locate the centroid y of the shaded area. h y ⫽ —2 x2 b SOLUTION dA = y dx y ' y = 2 x b y = LA = dA 2 h 4 x dx 4 2b L0 ' y dA LA h b h 2 x dx 2 L0 b B 2 b b h x5 R 10b4 0 = = b h B 2 x3 R 3b 0 3 h 10 Ans. *9–28. y Locate the centroid x of the area. 2 –– SOLUTION y x3 Differential Element: The differential element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is 1m 3 –– dA = (y1 - y2) dx = (x 2>3 - x 3>2 y x2 ) dx x ' Centroid: The centroid of the element is located at x = x. 1m 1 3 2 1 (x2>3 - x3>2) dx = ¢ x5>3 - x5>2 ≤ ` = m2 Ans. 5 5 5 0 1m A = dA = LA L0 We have 1m ' x dA x = LA L0 = dA LA 3 8 ¢ x8>3 = 1m x(x2>3 - x3>2) dx 1 5 = 1 5 2 7>2 1m x ≤` 7 0 = L0 25 m 56 (x5>3 - x5>2) dx 1 5 Ans. 9–29. y Locate the centroid y of the area. 2 –– SOLUTION y x3 Differential Element: The differential element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is 1m 3 –– dA = (y1 - y2) dx = (x 2>3 - x 3>2 y x2 ) dx x 1 1 ' Centroid: The centroid of the element is located at y = (y1 + y2) = (x2>3 + x3>2). 2 2 1 3 2 1 (x2>3 - x3>2) dx = ¢ x5>3 - x5>2 ≤ ` = m2 5 5 5 0 1m A = dA = LA L0 Ans. We have 1m ' y dA LA y = L0 = dA LA 1 2>3 (x + x3>2)(x2>3 - x3>2) dx 2 L0 = 1 5 1m 1 4>3 (x - x3) dx 2 1 5 x4 1 m 1 3 7>3 B x - R` 2 7 4 0 = = 1 5 25 m 56 Ans. 1m 9–39. Locate the centroid y of the paraboloid. z z2 = 4y 4m y SOLUTION Volume and Moment Arm: The volume of the thin disk differential element is ' dV = pz2dy = p14y2dy and its centroid y = y. Centroid: Applying Eq. 9–3 and performing the integration, we have y = LV 4m ' ydV LV = dV L0 y3p14y2dy4 4m p14y2dy L0 4p = y3 3 4m 2 4m y 4p 2 0 0 = 2.67 m Ans. 4m 9–40. Locate the center of gravity of the volume. The material is homogeneous. z 2m SOLUTION y 2 = 2z Volume and Moment Arm: The volume of the thin disk differential element is ' dV = py2dz = p12z2dz = 2pzdz and its centroid z = z. Centroid: Due to symmetry about z axis ' ' x = y = 0 Ans. Applying Eq. 9–3 and performing the integration, we have ' Lv z = Lv 2p = 2m ' z dV = dV 2m L0 z3 3 2m 2 2m z 2p 2 z12pzdz2 L0 0 0 = 2pzdz 4 m 3 Ans. 2m y 9–46. z The hemisphere of radius r is made from a stack of very thin plates such that the density varies with height r = kz, where k is a constant. Determine its mass and the distance to the center of mass G. G r SOLUTION y Mass and Moment Arm: The density of the material is r = kz. The mass of the thin disk differential element is dm = rdV = rpy2dz = kz3p(r2 - z2) dz4 and its ' centroid z = z. Evaluating the integrals, we have r m = Lm dm = L0 kz3p(r2 - z2) dz4 = pk ¢ z4 r r2z2 pkr4 - ≤` = 2 4 0 4 Ans. r Lm ' z dm = L0 z5kz3p(r2 - z2) dz46 = pk ¢ r2z3 z5 r 2pkr5 - ≤` = 3 5 0 15 Centroid: Applying Eq. 9–3, we have z = _ z Lm ' z dm = Lm dm 2pkr5>15 pkr4>4 = 8 r 15 Ans. x
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