COSMOS Chapter 3 Solution 64 | Michigan State University - Edubirdie

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

2019
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253

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1
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Kundulu8mbj

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 64. From the solution of Problem 3.59: λ DI = 0.97014 i − 0.24254 j TEF = ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k TEF = 29.7 lb M DI = 233.39 lb ⋅ ft Only the perpendicular component of TEF contributes to the moment of TEF about line DI. The parallel component of TEF will be used to find the perpendicular component. Have ( TEF )Parallel = λ DI ⋅ TEF = [ 0.97014 i − 0.24254 j] ⋅ ( 5.4 lb ) i − (16.2 lb ) j + ( 24.3 lb ) k  = ( 5.2388 + 3.9291) = 9.1679 lb Since TEF = ( TEF )Perpendicular + ( TEF )Parallel Then (TEF )Perpendicular = ( TEF )2 − ( TEF )2Parallel = ( 29.7 )2 − ( 9.1679 )2 = 28.250 lb and M DI = ( TEF )Perpendicular d 233.39 lb ⋅ ft = ( 28.250 lb ) d d = 8.2616 ft or d = 8.26 ft .

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