A girl of mass mg = 43 kg is pulling a sled up a slippery slope. The coefficient

Question

A girl of mass mg = 43 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.162; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a amax = 0.025 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is = 6°.

1.) What is the maximum value of the force of static friction on the girl?

2.) If the sled is being pulled up the slope with a=amax, what is the mass of the sled?

Explanation / Answer

(I) N = M g cos 6 + T sin 6 where N is normal force on girl and M = 43 (II) u N - T - M g sin 6 = M a where u = .162 and a = .025 acceleration of girl N = [M (a + g sin 6) + T] u rearranging (II) u (M g cos 6 + T sin 6) = M (a + g sin 6) + T combining (I) and (II) T = [u M g cos 6 - M (a + g sin 6) / (1 - u sin 6) T = [.162 * 43 * 9.8 * cos 6 - 43 (.025 + 9.8 sin 6)] / (1 - .162 sin 6) = 23.2 N 1) F = u N = u( M g cos 6 + 23.2 sin 6) using (I) F = .162 ( 43 * 9.8 cos 6 + 23.2 sin 6) = 68.3 N the frictional force 2) T - m g sin 6 = m a m = T / (a + g sin 6) = 23.2 / (.025 + 9.8 * sin 6) = 22.1 kg

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