COSMOS Chapter 6 Solution 151 | Answer Key - Edubirdie

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

2022
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348

Pages:

1
Author:

Lily Callahan

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 151. \ FBD Central Gear: By symmetry: F1 = F2 = F3 = F ΣM A = 0: 3 ( rA F ) − 15 N ⋅ m = 0, ΣM C = 0: rB ( F − F4 ) = 0, F = 5 N⋅m rA FBD Gear C: ΣFx′ = 0: C x′ = 0 ΣFy′ = 0: C y′ − 2 F = 0, F4 = F C y′ = 2 F Gears B and D are analogous, each having a central force of 2F FBD Spider: ΣM A = 0: 75 N ⋅ m − 3 ( rA + rB ) 2 F = 0 25 N ⋅ m − ( rA + rB ) 10 N⋅m = 0 rA rA + rB r = 2.5 = 1 + B , rA rA FBD Outer gear: Since rA = 24 mm, ΣM A = 0: rB = 1.5 rA rB = 36.0 mm ! (a) 3 ( rA + 2rB ) F − M E = 0 3 ( 24 mm + 72 mm ) (b) 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. 5 N⋅m −ME = 0 24 mm M E = 60.0 N ⋅ m !

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