127x Filetype PDF File size 0.06 MB Source: athena.ecs.csus.edu
COSMOS: Complete Online Solutions Manual Organization System Chapter 16, Solution 1. ∑MN=°1.5m sin60 294.3N 0.75m cos60° ()()()() AB 2 =°30 kg 4 m/s sin60 0.75 m ()()() () NB =144.96 N ∑FN=F=30kg 4m/s2 xB () F = 24.96 N 22 (a) RN=+F=295.36N AA or RA = 295 N 85.2° ! 1 294.3 α ==tan 85.2° 24.96 and B =145.0 N ! (b) F 24.96 µ == =0.08481 NA 294.3 or µ = 0.0848 ! 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.
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