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

2024
Views:

161

Pages:

2
Author:

pravitate8aj10

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 141. Free-Body Diagram: (a) The force acting at E on the free-body diagram of rod AB is perpendicular to AB and CD. Letting λ E = direction cosines for force E, λE = rB/ A × k rB/ A × k  − ( 32 in.) i + ( 24 in.) j − ( 40 in.) k  × k = 2 2 ( 32 ) + ( 24 ) in. = 0.6i + 0.8 j Also, W = − (10 lb ) j B = Bk E = E ( 0.6i + 0.8 j) From free-body diagram of rod AB ΣM A = 0: rG/ A × W + rE/ A × E + rB/ A × B = 0 i j k i j k i j k ∴ −16 12 −20 (10 lb ) + −24 18 −30 E + −32 24 −40 B = 0 0 −1 0 0.6 0.8 0 0 0 1 ( −20i + 16k )(10 lb ) + ( 24i − 18 j − 30k ) E + ( 24i + 32 j) B = 0 From k-coefficient 160 − 30 E = 0 ∴ E = 5.3333 lb and E = 5.3333 lb ( 0.6i + 0.8 j) E = ( 3.20 lb ) i + ( 4.27 lb ) j or (b) From j-coefficient −18 ( 5.3333 lb ) + 32 B = 0 ∴ B = 3.00 lb or B = ( 3.00 lb ) k COSMOS: Complete Online Solutions Manual Organization System From free-body diagram of rod AB ΣF = 0: A + W + E + B = 0 Ax i + Ay j + Az k − (10 lb ) j + ( 3.20 lb ) i + ( 4.27 lb ) j + ( 3.00 lb ) k = 0 From i-coefficient Ax + 3.20 lb = 0 ∴ Ax = −3.20 lb j-coefficient Ay − 10 lb + 4.27 lb = 0 ∴ Ay = 5.73 lb k-coefficient Az + 3.00 lb = 0 ∴ Az = −3.00 lb Therefore A = − ( 3.20 lb ) i + ( 5.73 lb ) j − ( 3.00 lb ) k