ch06-p014 - 14. (a) The free-body diagram for the block is...

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Unformatted text preview: 14. (a) The free-body diagram for the block is shown on the right, with F being the force applied to the block, FN the normal force of the floor on the block, mg the force of gravity, and f the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. The equations for the x and the y components of the force according to Newton’s second law are: Fx = F cos θ − f = ma Fy = F sin θ + FN − mg = 0 Now f =μkFN, and the second equation gives FN = mg – Fsinθ, which yields f = μk (mg − F sin θ ) . This expression is substituted for f in the first equation to obtain F cos θ – μk (mg – F sin θ) = ma, so the acceleration is a= F ( cos θ + μk sin θ ) − μk g . m (a) If μ s = 0.600 and μk = 0.500, then the magnitude of f has a maximum value of f s ,max = μ s FN = (0.600)(mg − 0.500mg sin 20°) = 0.497mg. On the other hand, F cos θ = 0.500mg cos 20° = 0.470mg . Therefore, F cos θ < f s ,max and the block remains stationary with a = 0 . (b) If μ s = 0.400 and μk = 0.300, then the magnitude of f has a maximum value of f s ,max = μ s FN = (0.400)(mg − 0.500mg sin 20°) = 0.332mg. In this case, F cos θ = 0.500mg cos 20° = 0.470mg > f s ,max . Therefore, the acceleration of the block is a= F ( cos θ + μk sin θ ) − μk g m = (0.500)(9.80 m/s 2 ) [ cos 20° + (0.300) sin 20°] − (0.300)(9.80 m/s 2 ) = 2.17 m/s 2 . ...
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