COSMOS Chapter 6 Solution 149 | Answer Key - Edubirdie

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

2020
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436

Pages:

1
Author:

Nicole Wang

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 149. FBD boom: θ = tan −1 Note: ( 3.2sin 24° − 1) m ( 3.2cos 24° − 0.6 ) m θ = 44.73° (a) ΣM A = 0: ( 6.4 m ) cos 24° ( 2.3544 kN ) − ( 3.2 m ) cos 24° B sin 44.73° + ( 3.2 m ) sin 24° B cos 44.73° = 0 B = 12.153 kN B = 12.15 kN 44.7° W ΣFx = 0: Ax − (12.153 kN ) cos 44.73° = 0 A x = 8.633 kN (b) ΣFy = 0: − 2.3544 kN + (12.153 kN ) sin 44.73° − Ay = 0 A y = 6.198 kN On boom: On carriage: Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. A = 10.63 kN A = 10.63 kN 35.7° 35.7° W

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