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

2020
Views:

71

Pages:

1
Author:

imophulebf10

COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 60. FBD Cylinder: ( ) W = ( 80 kg ) 9.81 m/s 2 = 784.8 N ΣM G = 0: FA − FB = 0, ΣM D = 0: dN B − dN A + rW = 0, so N A > NB, FA = FB (1) N A = NB + W 3 (2) FA max > FB max ∴ slip impends first at B. FB = µ s N B = 0.25 N B ΣM A = 0: ( r cos 30° ) N B − ( r sin 30° )W − r (1 + sin 30° )( 0.25 N B ) = 0 r = note d = tan 30 N B = 1.01828W = 799.15 N 3r FB = 0.25 N B = 199.786 N From (2) above, N A = 799.15 N + 784.8 N = 1252.25 N 3 From (1), FA = FB = 199.786 N ΣFy = 0: FBD Wedge: NC − (1252.25 N ) cos10° + 199.786 N sin10° = 0 NC = 1198.53 N Impending slip FC = µ s NC = 0.25 (1198.53 N ) = 299.63 N ΣFx = 0: P − 299.63 N − (199.786 N ) cos10° − (1252.25 N ) sin10° = 0 P = 714 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. 20.0°