75. (a) We note that FN= mgin this situation, so fs,max= μsmg= (0.52)(11 kg)(9.8 m/s2) = 56 N. Consequently, the horizontal force GFneeded to initiate motion must be (at minimum) slightly more than 56 N. (b) Analyzing vertical forces when GFis at nonzero θyields ,maxsin (sin ).NssFFmgfmgF+=¡=−Now, the horizontal component of GFneeded to initiate motion must be (at minimum)
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.