HW20-Solution

HW20-Solution - HW20 Solution Terry Cooke Cal Poly San Luis Obispo Chapter 5 Solution 8 A in.2 x in y in xA in.3 y A in.3 1 2 30 50 22500 –8353.0

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Unformatted text preview: HW20 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 5, Solution 8 A, in.2 x , in. y , in. xA, in.3 y A, in.3 1 2 30 50 1500 15 23.634 25 30 22500 –8353.0 14147.0 37500 –10602.9 26.897 2 (15) 2 353.43 1146.57 x A 14147.0 A 1146.57 y A 26897 A 1146.57 Then X Y X Y 12.34 in. 23.5 in. Chapter 5, Solution 10 Dimensions in mm A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 2 47 26 1919.51 1 94 70 3290 2 0 11.0347 0 21181 2 15.6667 23.333 51543 51543 76766 55584 5209.5 Then X xA A yA A 51543 5209.5 55584 5209.5 X 9.89 mm Y Y 10.67 mm Chapter 5, Solution 29 First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X 0 So that Then d 0.75 cos 55 m (0.75 m) 2 (0.75 d )m (1.5 m) (1.5 d )m 1 2 m cos 55 2 1 (0.75) 2 2 2 cos 55 xL 0 (2 m) 0 or (0.75 1.5 2)d (0.75)(1.5) 3 or d 0.739 m ...
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This note was uploaded on 01/13/2011 for the course ME 211 taught by Professor Staff during the Fall '05 term at Cal Poly.

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