COSMOS Chapter 3 Solution 120 | Michigan State University - Edubirdie

University:
Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
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COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 120. R = ΣF = FF + FE (a) Have FF = 50 N ( sin 60° ) j + ( cos 60° ) k  = ( 43.301 N ) j + ( 25 N ) k where FE = − ( 25 N ) k ∴ R = ( 43.301 N ) j or R = ( 43.3 N ) j M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE Have rF /C = ( 0.225 m ) i − ( 0.050 m ) j where rE/C = ( 0.450 m ) i − ( 0.325 m ) j ∴ M CR i j k i j k = 0.225 −0.050 0 N ⋅ m + 0.450 −0.325 0 N ⋅ m 0 43.301 25 0 0 −25 = ( 6.875 N ⋅ m ) i + ( 5.625 N ⋅ m ) j + ( 9.7427 N ⋅ m ) k or M CR = ( 6.88 N ⋅ m ) i + ( 5.63 N ⋅ m ) j + ( 9.74 N ⋅ m ) k (b) To determine which direction duct section CD has a tendency to turn, have R M CD = λ DC ⋅ M CR where λ DC = Then − ( 0.45 m ) i + ( 0.1 m ) j ( −0.45)2 + ( 0.1)2 = −0.97619i + 0.21693j R = ( −0.97619i + 0.21693j) ⋅ ( 6.875i + 5.625 j + 9.7427k ) N ⋅ m M CD = ( −6.7113 + 1.22023) N ⋅ m = −5.4911 N ⋅ m Since λ DC ⋅ M CR < 0, duct DC tends to rotate counterclockwise relative to elbow C as viewed from D to C. .

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