COSMOS Chapter 3 Solution 62 - Edubirdie

University:
Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2024
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1
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diamundgalx45p

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 62. From the solution of Problem 3.53: λ AD = 0.8i − 0.6k TBH = ( 375 N ) i + ( 750 N ) j − ( 750 N ) k; TBH = 1125 N M AD = −180 N ⋅ m Only the perpendicular component of TBH contributes to the moment of TBH about line AD. The parallel component of TBH will be used to find the perpendicular component. Have ( TBH )Parallel = λ AD ⋅ TBH = [ 0.8i − 0.6k ] ⋅ ( 375 N ) i + ( 750 N ) j − ( 750 N ) k  = ( 300 + 450 ) N = 750 N Since TBH = ( TBH )Perpendicular + ( TBH )Parallel Then (TBH )Perpendicular = ( TBH )2 − ( TBH )2Parallel = (1125 N )2 − ( 750 N )2 = 838.53 N and M AD = ( TBH )Perpendicular d 180 N ⋅ m = ( 838.53 N ) d d = 0.21466 m or . d = 215 mm

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