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

2023
Views:

459

Pages:

2
Author:

Tyshawn Shaffer

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 152. FBD Gear A: looking from C (a) M A = 15 lb ⋅ ft ΣM A = 0: rA = 4 in. M A − P rA = 0 P= M A 180 lb ⋅ in. = rA 4 in. P = 45 lb FBD Gear B: looking from F ΣM B = 0: M 0 − rB P = 0 M 0 = rB P = ( 2.5 in.)( 45 lb ) = 112.5 lb ⋅ in. M 0 = 112.5 lb ⋅ in. i ! FBD ABC: looking down (b) ( 2 in.)( 45 lb ) − ( 5 in.) C ΣM B = 0: ΣFz = 0: FBD BEG: =0 45 lb − B + 18 lb = 0 By analogy, using FBD DEF ΣFz = 0: E = 63 lb k C = 18 lb k B = −63 lb k F = 18 lb k Gz + 63 lb − 63 lb = 0 Gz = 0 ΣFy = 0 Gy = 0 ΣM G = 0 M G − ( 6.5 in.)( 63 lb ) = 0 M G = ( 410 lb ⋅ in.) i ! 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. COSMOS: Complete Online Solutions Manual Organization System FBD CFH: ΣF = 0: Hz = H y = 0 ΣM H = 0 M H = − ( 6.5 in.)(18 lb ) = −117 lb ⋅ in. M G = − (117.0 lb ⋅ in.) i ! 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.