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

2023
Views:

66

Pages:

1
Author:

behend6abs

COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 67. From the solution of Prob. 3.56: λ AD = 0.99228 i + 0.124035 j FGH = − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k FGH = 21.3 kN M AD = − 35.931 kN ⋅ m Only the perpendicular component of FGH contributes to the moment of FGH about edge AD. The parallel component of FGH will be used to find the perpendicular component. Have ( FGH )Parallel = λ AD ⋅FGH = ( 0.99228 i + 0.124035 j) ⋅  − ( 7.1 kN ) i + (14.2 kN ) j + (14.2 kN ) k  = − 5.2839 kN Since FGH = ( FGH )Perpendicular + ( FGH )Parallel Then ( FGH )Perpendicular = ( FGH )2 − ( FGH )2Parallel = ( 21.3)2 − ( 5.2839 )2 = 20.634 kN and M AD = ( FGH )Perpendicular d 35.931 kN ⋅ m = ( 20.634 kN ) d d = 1.741349m . or d = 1.741 m

COSMOS Chapter 3 Solution 67

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