Chapter 12355P12.13(a)Ffnxw∑=− =0Fnyg∑=−−=8005000N NTaking torques about an axis at the foot of the ladder,8004 0030 05007 5030 015 030 00mmcmafafaf.sin.in..cos.°+°−°=nwSolving the torque equation,nw=+°=40080075050030015 0268..tan..m NNmNafaf.Next substitute this value into the Fxequation to findngfnw500 N800 NAFIG. P12.13fnw==268 Nin the positive xdirection.Solving the equation Fy∑=0,ng=1300 Nin the positive ydirection.(b)In this case, the torque equation τA=∑0 gives:9 0080030 07 5050030 015 060 00.sin..sin..sin.NNmafafafafafbg°+°−°=nwornw=421 N .Since w421 N and ffngmaxµ, we find=fngmax.4210324N.P12.14(a)n∑0(1)Fnmgmg∑−=120(2)τθθAwmgLxnL∑FHGIKJ−+=2
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .