Question

1. A 10-lb cylinder is held in equilibrium by the belt and wall. There is no slipping at the wall and between the belt and the cylinder is 0.25. • The minimum force P must be applied to the belt for equilibrium = 140 16 o The tensile force along the belt between the point B and the support ring at the wall • The normal force acting on the cylinder at A- lb (negative if toward wall) • The friction force at A= lb (negative if going down)

Answer 1

Ans FBD - - - - - NA - WA Elk He - MB To se 0.25 (78730) T = 1:1398 P (+EMATO PX 0.2 + 10x01) - T (030 (011+01110930) -t gin30 (0.1 Sin30) =0 01 2P = -1 +0.0866T+0.075 T +0.025T=0 0.2 p = -1+0.1966 T lll) POZ from 12 & 12). 1p = 78.8 ub ] . T = 1:1399 X7808 Ta 89,8217 ub Scanned with CamScanner

I For = 0 --Ng+ T. Fin30 =0 NA = .89082 x 1 / 2 NA = 44.9/1 Mb 7 { Fy=0 Tuzo & FA -p-w =0 89.82x60430+ FA-78.8-1050 FA = 11.013 lb .? (Filmax = MS NA = 0.25X441911 = 11:22 lb - FA = 11.013 jb Scanned with CamScanner