HW4-Solution

HW4-Solution - First note 2 2 (17.2 in.) (7.62 in.) 18.8123...

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HW4 - Solution Terry Cooke: Cal Poly San Luis Obispo
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Chapter 3, Solution 3 First resolve the 4-lb force into components P and Q , where (16 N)sin 28 7.5115 N Q Then / (0.17 m)(7.5115 N) B A B M r Q 1.277 N m or 1.277 N m B M
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Chapter 3, Solution 8 ( a ) We have / (4 in.)(200 lb) 800 lb in. B C B N M r F or 800 lb in. B M ( b ) By definition / sin 10 (180 70 ) 120 B A B M r P     Then 800 lb in. (18 in.) sin120 P or 51.3 lb P ( c ) For P to be minimum, it must be perpendicular to the line joining Points A and B . Thus, P must be directed as shown. Thus min / B A B M dP d r or min 800 lb in. (18 in.) P or min 44.4 lb P min 44.4 lb P 20°
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Chapter 3, Solution 13
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Unformatted text preview: First note 2 2 (17.2 in.) (7.62 in.) 18.8123 in. CB d Then 17.2 in. cos 18.8123 in. 7.62 in. sin 18.8123 in. and ( cos ) ( sin ) 125 lb [(17.2 in.) (7.62 in.) ] 18.8123 in. CB CB CB F F F i j i j Now / A B A CB M r F where / (20.5 in.) (4.38 in.) B A r i j Then 125 lb [(20.5 in.) (4.38 in.) ] (17.2 7.62 ) 18.8123 in. A M i j i j (1538.53 lb in.) (128.2 lb ft) k k or 128.2 lb ft A M...
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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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HW4-Solution - First note 2 2 (17.2 in.) (7.62 in.) 18.8123...

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