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

2021
Views:

306

Pages:

1
Author:

Kundulu8mbj

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 49. ( ) WD = (18 kg ) 9.81 m/s 2 = 176.58 N Fs = kx = ( 3.5 kN/m )( 0.1 m ) = 0.35 kN = 350 N φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.0362° FBD Lever: ΣM C = 0: ( 0.3 m )( 350 N ) − ( 0.4 m )(176.58 N ) − ( 0.525 m ) RA cos 24.036° − ( 0.05 m ) RA sin 24.036° = 0 RA = 68.758 N ΣFx = 0: ΣFy = 0: C x − ( 68.758 N ) sin 24.036° = 0, Cx = 28.0 N C y − 350 N − 176.58 N + ( 68.758 N ) cos 24.036° = 0 C y = 464 N FBD Wedge: P 68.758 N = sin 38.072° sin 75.964° (a) (b) P = 43.7 N C x = 28.0 N C y = 464 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.