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

2021
Views:

120

Pages:

1
Author:

FrextWhestembki

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 63. From the solution of Problem 3.54: λ AD = 0.8i − 0.6k TBG = − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k TBG = 1125 N M AD = − 222 N ⋅ m Only the perpendicular component of TBG contributes to the moment of TBG about line AD. The parallel component of TBG will be used to find the perpendicular component. Have ( TBG )Parallel = λ AD ⋅ TBG = [ 0.8i − 0.6k ] ⋅  − ( 500 N ) i + ( 925 N ) j − ( 400 N ) k  = ( − 400 + 240 ) N = −160 N Since TBG = ( TBG )Perpendicular + ( TBG )Parallel Then (TBG )Perpendicular = (TBG )2 − (TBG )2Parallel = (1125 N )2 − ( −160 N )2 = 1113.56 N and M AD = (TBG )Perpendicular d 222 N ⋅ m = (1113.56 N ) d d = 0.199361 m or d = 199.4 mm .

COSMOS Chapter 3 Solution 63

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