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

2020
Views:

433

Pages:

2
Author:

teespagmaxrtf

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 123. Member FBDs: θ = tan −1 = tan −1 ( BC + CD ) sin 60° AF + ( AB + BC − CD ) cos 60° ( 2.5 in. + 1 in.) sin 60° 4.5 in. + (1.5 in. + 2.5 in. − 1 in.) cos 60° θ = 26.802° From FBD CDE: ΣM C = 0: ( 7.5 in.)( 20 lb ) − (1 in.) FDF cos ( 30° − 26.802°) = 0, FDF = 150.234 lb C ΣFx = 0: (150.234 lb ) cos ( 26.802°) − ( 20 lb ) sin 60° − Cx = 0, Cx = 116.774 lb ΣFy = 0: (150.234 lb ) sin 26.802° − ( 20 lb ) cos 60° − C y =0 C y = 57.742 lb From FBD ABC: ΣM A = 0: (1.5 in.) sin 60° Bx + (1.5 in.) cos 60° By − ( 4 in.) sin 60°  (116.774 lb ) + ( 4 in.) cos 60° ( 57.742 lb ) = 0, 3 Bx + By = 385.37 lb 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. (1) COSMOS: Complete Online Solutions Manual Organization System From FBD BFG: ΣM G = 0: − (1.5 in.) (150.234 lb ) sin 26.802° + (1.5 in.) cos 30°  (150.234 lb ) cos 26.802° − (1.5 in.) cos 30° Bx − 6 in. + (1.5 in.) sin 30°  B y = 0, Bx = 243.32 lb, Solving (1) and (2): ΣFx = 0: ΣFy = 0: 3Bx − 9 B y = 96.775 lb By = − 36.075 lb 243.32 lb − (150.234 lb ) cos 26.802° − Gx = 0, G x = 109.226 lb G y = 31.667 lb G y − (150.234 lb ) sin 26.802° + 36.075 lb = 0, On G, G = 113.7 lb 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. 16.17° ! (2)