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HW16 - PROBLEM 5.90 The composite body shown is formed by...

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Unformatted text preview: PROBLEM 5.90 The composite body shown is formed by removing a hemisphere of radius r from a cylinder of radius R and height 2R. Determine (a) the y coordinate of the centroid when r : 3R/4, (b) the ratio r/R for which 57 : —L2R. SOLUTION Note, for the axes shown V 7 9V V 1 (nR3)(2R) : 277R; —R —2;rR‘ 2 7 2 m J —:r 1711“, Q 3 8 4 / ® J 4 X E 27: RtL 72x R‘—L G) 3 l 4 4 )7 zyy R — g1 Thcn = = _ q 1 1 ):V R, _ 3 I. r=—R: )2— (a) 4 ) or )7 = —1.118R 4 (b) y=+2Rz —1.2R=— 4 , or [Lj — 3.2[L +1l6 = 0 R R Solving numerically i : 08844 R PROBLEM 5.106 A Window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the conesponding area. M = 292.2 mm 3;? in = iv] = w = 212.2 mm )7" : J7V1= 80 + yw : so + LX500) = 398.3 mm 72' * (EMS—00) = 318.31nm zlv A“ = AVI = \$500)2 : 196 350 mm2 AW : g(500)(680) : 534 071mm2 A, mm2 )7, mm 4:: mm iA, mm3 fA,1mn3 1 (80)(500) : 40 000 40 250 1.6 x 106 10 x 105 11 I— 196 350 292.2 212.2 57.4 x106 41.67 ><106 111 (80)(680) : 54 400 40 500 0.2176 x 106 27,2 x 106 IV 534 071 398.3 318.3 212.7 x 106 170 ><106 V (80)(500) = 40 000 40 250 l 1.6 ><106 10 x 106 VI 196 350 292.2 212.2 57_4 ><106 41.67 x 106 2 1.061><106 332.9 x 106 300.5 x106_‘ Now, symmetry implies X z 340 mm 4 and 172A = EiA: 17(1061x106 mmz) = 332.9 x 106 mm3 or 17 = 314 mm‘ 22A : 22A: Z(1.061x105 m2) = 300.5 x106 mm3 or 2:283mm4 PROBLEM 5.125 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R. SOLUTION First note that symmetry implies with two planes parallel to the xy plane. Now Now Symmetry implies dA = (7rr)(Rd6?) _ 2r Yet. ' ’i r = Rsin t9 dA = 7:122 sin eds) fEL : —Esin6 7f o NI: A = ﬁnRz sint9d6 = ﬂR2[—cos¢9] = IrR2 Wad/1 = EZL[—%sin 0)(7zR2 sin Ode) = _2R3|:g7 sirIZHT 2 4 0 = _£R3 2 7 _ 7 _ — 2 _ _£ 3 yA— beLdA. y(7rR )— 2R 2 : E = 0 4 The element of area dA of the shell shown is obtained by cutting the shell ...
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