COSMOS Chapter 6 Solution 127

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 127. (a) FBD E: ΣFx = 0: FDE cos α −160 N = 0, where α = tan −1 FBD Joint D: β = tan −1 FDE = 160 N cosα 38.8 mm = 22.823°, so FDE = 173.59 N C 92.2 mm 37.5 mm − 18.8 mm = 25.732° 38.8 mm FCD cos ( 45° − 25.732° ) ΣFx = 0: − (173.59 N ) cos ( 25.732° + 22.823° ) = 0 FCD = 121.717 N C FBD BC: ΣM B = 0: ( 0.045 m )(121.717 N ) sin 45° − M M = 3.873 N ⋅ m, = 0, M = 3.87 N ⋅ m (b) FBD E: ΣFx = 0: FDE cos α −160 N = 0, where α = tan −1 FDE = 160 N cos α 20 mm + 35.4 mm − 26.4 mm = 16.8584° 92.2 mm + 27.8 mm − 24.3 mm so FDE = 167.185 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 = tan −1 27.8 mm − 24.3 mm = 7.5520° 26.4 mm γ = tan −1 20 mm + 35.4 mm − 26.4 mm = 42.363° 24.3 mm + 7.5 mm ΣFy′ = 0: (167.185 N ) sin (16.8584° + 42.363°) − FCD sin ( 90° − 7.5520° − 42.363° ) = 0, FCD = 223.07 N C FBD BC: δ = tan −1 35.4 mm = 51.857° 27.8 mm ΣM B = 0: M − ( 0.045 m )( 223.07 N ) sin ( 90° − 51.857° − 7.5520° ) = 0 M = 5.12 N ⋅ m ! 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.