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

2019
Views:

478

Pages:

1
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 6, Solution 103. FBD BC: Note that only one of the forces B y , B′y exists. ( ) M C B y or B′y − ( 0.15 m ) ΣM C = 0:  2  FFG =  M B or B′y  0.15 m  C y   ( ( So we seek a maximum of M C By or B′y 1 FFG = 0 2 ) (1) ) ( ) M C ( B′y ) = ( 0.175 m ) B′y M C By = ( 0.125 m ) By where FBD stool: w = ( 56 kg ) ( 9.81 N/kg ) = 549.36 N consider contact at A′, A y = 0, either ΣM A′ = ( 0.425 m ) By − ( 0.225 m ) w =0 2 ( ) M C By = ( 0.03309 m ) w By = 0.26471 w, or ( M A′ ) = ( 0.475 m ) B′y − ( 0.225 m ) (i) w =0 2 ( ) B′y = ( 0.23684 ) w, M C B′y = ( 0.04145 m ) w (ii) consider contact at A, A′y = 0, either ΣM A = 0: ( 0.35 m ) By − ( 0.15 m ) w =0 2 ( ) By = 0.21429 w, M C By = ( 0.02678 m ) w or ΣM A = 0: B′y = 0.1875 w, ( 0.40 m ) B′y − ( 0.15 m ) (iii) w =0 2 ( ) M C B′y = ( 0.03281 m ) w (iv) (a) The maximum is (ii), with contact at A′ and B′, and !  2  (b) FFG max =   ( 0.04145 m ) ( 549.36 N ) = 214.69 N,  0.15 m  FFG max = 215 N T ! 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.