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HW9-Solution

# HW9-Solution - HW9 Solution Terry Cooke Cal Poly San Luis...

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HW9 - Solution Terry Cooke: Cal Poly San Luis Obispo

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Chapter 4, Solution 55 First note T ks Where spring constant elongation of spring cos (1 cos ) cos (1 cos ) cos k s l l l kl T ( a ) From f.b.d. of collar B 0: sin 0 y F T W or (1 cos )sin 0 cos kl W or tan sin W kl ( b ) For 3 lb 6 in. 8 lb/ft 6 in. 0.5 ft 12 in./ft 3 lb tan sin 0.75 (8 lb/ft)(0.5 ft) W l k l Solving numerically, 57.957 or 58.0
Chapter 4, Solution 68 Since CD is a two-force member, the force it exerts on member ABD is directed along DC . Free-Body Diagram of ABD : (Three-Force member) The reaction at B must pass through E , where D and the 50-lb load intersect. Triangle CFD : 3 tan 0.6 5 30.964 Triangle EAD : 10tan 6 in. 6 1.5 4.5 in. AE GE AE AG Triangle EGB : 3 tan 4.5 33.690 GB GE Force triangle 50 lb sin120.964 sin33.690 sin 25.346° B D 100.155 lb 64.789 lb B D 100.2 lb B 56.3 64.8 lb C D 31.0

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Chapter 4, Solution 75 Reaction at B must pass through D . 7 in. tan 12 in. 30.256 7 in. tan 24 in. 16.26 Force triangle Law of sines 72 lb sin59.744
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HW9-Solution - HW9 Solution Terry Cooke Cal Poly San Luis...

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