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

2021
Views:

135

Pages:

2
Author:

omiliar74be

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 124. Member FBDs: θ = tan −1 = tan −1 ( BC + CD ) sin 60° AF + AB cos 75° + ( BC − CD ) cos 60° ( 2.5 in. + 1 in.) sin 60° 4.5 in. + (1.5 in.) cos 75° + ( 2.5 in. − 1 in.)  cos 60° θ = 28.262° From FBD CDE: ΣM C = 0: ( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 28.262°) = 0 FDF = 150.069 lb ΣFx = 0: (150.069 lb ) cos 28.262° − ( 20 lb ) sin 60° − Cx = 0 Cx = 114.859 lb ΣFy = 0: (150.069 lb ) sin 28.262° − ( 20 lb ) cos 60° − C y =0 C y = 61.058 lb From FBD ABC: ΣM A = 0: (1.5 in.) sin 75° Bx − (1.5 in.) cos 75° By − (1.5 in.) sin 75° + ( 2.5 in.) sin 60° (114.859 lb ) + (1.5 in.) cos 75° + ( 2.5 in.) cos 60° ( 61.058 lb ) = 0 or 1.44889 Bx − 0.38823 B y = 315.07 lb (1) continued 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 From FBD BFG: ΣM G = 0: − (1.5 in.) (150.069 lb ) sin 28.262° + (1.5 in.) cos15° (150.069 lb ) cos 28.262°  − (1.5 in.) cos15° Bx + ( 6 in.) + (1.5 in.) sin15°  B y = 0 or 1.44889 Bx − 6.3882 B y = 84.926 lb Solving (1) and (2): Bx = 227.74 lb, (2) By = 38.358 lb ΣFx = 0: 227.74 lb − (150.069 lb ) cos 28.262° − Gx = 0, G x = 95.560 lb ΣFy = 0: G y − (150.069 lb ) sin 28.262° + 38.358 lb = 0, G y = 32.700 lb So force on G is 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. G = 101.8 lb 18.89° !