HW8-Solution

HW8-Solution - HW8 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW8 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 4, Solution 37 Similar triangles: ABE and ACD AE AD BE ; CD 0.15 m 0.5 m BE ; BE 0.25 m 0.075 m (a) MA 0: Tx (0.25 m) 0.075 Tx (0.5 m) (400 N)(0.1 m) (400 N)(0.4 m) 0.35 Tx 0 1400 N Ty 0.075 (1400 N) 0.35 300 N 0 1100 N T 1432 lb (b) Fy 0: A 300 N 400 N 400 N A A 1100 N (c) Fx 0: C 1400 N 0 C 1400 N C 1400 N Chapter 4, Solution 42 Free-Body Diagram: Fy 0: T (80 N) cos 30 T 0 69.282 N T 69.3 N MC 0: (69.282 N) cos 30 (0.2 m) (80 N)(0.2 m) ( A sin 30 )(0.4 m) 0 A MA 0: (69.282 N) cos 30 (0.2 m) (80 N)(0.6 m) (C sin 30 )(0.4 m) 0 140.000 N A 140.0 N 30.0° C 180.000 N C 180.0 N 30.0° Chapter 4, Solution 53 Free-Body Diagram: (a) From free-body diagram of rod AB MC 0: P(l cos ) P(l sin ) M 0 or sin cos M Pl (b) For sin M cos 150 lb in., P 20 lb, and l 150 lb in. 5 1.25 (20 lb)(6 in.) 4 6 in. Using identity sin sin 2 (1 sin cos 2 2 1 1.25 1.25 sin 1.5625 2.5sin sin 2 (1 sin 2 )1/ 2 ) 1/ 2 1 sin 2sin 2 2.5sin 2 0.5625 0 Using quadratic formula sin ( 2.5) 2.5 4 1.75 0.29428 17.1144 (625) 4(2)(0.5625) 2(2) or sin 0.95572 and sin 72.886 and or 17.11 and 72.9 ...
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This note was uploaded on 11/07/2010 for the course ME 211 taught by Professor Staff during the Fall '05 term at Cal Poly.

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