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

2024
Views:

74

Pages:

2
Author:

gaarmaakbn9h

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 128. (a) FBD BC: ( 0.045 m ) FCD sin 45° − 6.00 N ⋅ m = 0 ΣM B = 0: FCD = 188.562 N C FBD Joint D: α = tan −1 38.8 mm = 22.823° 92.2 mm β = tan −1 37.5 mm − 18.8 mm = 25.732° 38.8 mm ΣFx′ = 0: (188.562 N ) cos ( 45° − 25.732°) − FDE cos ( 22.823° + 25.732° ) = 0, FDE = 268.92 N C FBD E: ( 268.92 N ) cos ( 22.823°) − P = 0, ΣFx = 0: P = 248 N (b) FBD BC: δ = tan −1 35.4 mm = 51.857° 27.8 mm β = tan −1 27.8 mm − 24.3 mm = 7.5520° 26.4 mm ΣM B = 0: ! ( 0.045 m ) FCD sin ( 90° − 51.857° − 7.5510°) − 6.00 N ⋅ m = 0 FCD = 262.00 N C 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 FBD Joint D: β = 7.5520°, α = tan −1 γ = tan −1 20 mm + 35.4 mm − 26.4 mm = 42.363° 24.3 mm + 7.5 mm 20 mm + 35.4 mm − 26.4 mm = 16.8584° 92.2 mm + 27.8 mm − 24.3 mm ΣFy′ = 0: FDE sin (16.8584° + 42.363° ) − ( 262.00 N ) sin ( 90° − 7.5520° − 42.363° ) = 0, FDE = 196.366 N C FBD E: ΣFx = 0: (196.366 N )( cos16.8584°) − P = 0, P = 187.9 N 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. !