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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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