ch06-p075 - 75. (a) We note that FN = mg in this situation,...

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75. (a) We note that F N = mg in this situation, so f s ,max = μ s mg = (0.52)(11 kg)(9.8 m/s 2 ) = 56 N. Consequently, the horizontal force G F needed to initiate motion must be (at minimum) slightly more than 56 N. (b) Analyzing vertical forces when G F is at nonzero θ yields ,max sin ( sin ). Ns s FF m g f m g F += ¡ =− Now, the horizontal component of G F needed 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.

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