COSMOS Chapter 3 Solution 66 | Michigan State University - Edubirdie

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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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marthamanayay9qr16

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 66. From the solution of Prob. 3.55: λ AD = 0.99228 i + 0.124035 j FEF = ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k FEF = 24.3 kN M AD = − 24.916 kN ⋅ m Only the perpendicular component of FEF contributes to the moment of FEF about edge AD. The parallel component of FEF will be used to find the perpendicular component. Have ( FEF )Parallel = λ AD ⋅ FEF = [ 0.99228 i + 0.124035 j] ⋅ ( 2.7 kN ) i + (10.8 kN ) j + ( 21.6 kN ) k  = 4.0187 kN Since FEF = ( FEF )Perpendicular + ( FEF )Parallel Then ( FEF )Perpendicular = ( FEF )2 − ( FEF )2Parallel = ( 24.3)2 − ( 4.0187 )2 = 23.965 kN and M AD = ( FEF )Perpendicular d 24.916 kN ⋅ m = ( 23.965 kN ) d d =1.039683m . or d = 1.040 m

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