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Unformatted text preview: nd law applied to the vertical direction yields 0 = P + N mg N = mg P = 136 N The maximum static friction is f s,max = s N = 54.4 N The magnitude of horizontal force F is smaller than the maximum static friction. Therefore, the static friction keeps the block from sliding. Newtons 2 nd law applied to the horizontal direction yields the friction: 0 = F f f = 50.0 N b) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 80.0 N. The maximum static friction is f s,max = s N = s (mg P)= 46.4 N This is smaller than the applied force. Hence the static friction cannot hold the block and the block starts to slide. The friction exerted on the block is kinetic friction. The magnitude of the friction is f = k N = k (mg P)= 23.2 N...
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- Spring '08