<p>A uniform rigid pole of length L and mass M is to be supported froma vertical

Question

<p>A uniform rigid pole of length L and mass M is to be supported froma vertical wall in a horizontal position. The pole is not attached directly to the wall, so the coefficient of static friction between the wall and the pole provides the only vertical force on one end of the pole. The other end of the pole is supported by a light rope that is attached to the wall at a point distance D directly above the point the pole contacts the wall. Determine the minimum value of the coefficient ofstatic friction, as a function of L and D, that will keep the pole horizontal and not allow its end to slide down the wall.</p>

Explanation / Answer

Let ? be the angle of the rope to the wall. So ? = arctan(L/D) The force F in the rope must exert an upward (vertical) component on the end of the pole = Fv = F.cos? and a horizontal component = Fh = F.sin?. We also have that Fv = M.g/2 and the vertical force at the wall = Fv (by symmetry) So the ratio of vertical to horizontal forces between the wall and the pole = R = (M.g/2)/(F.sin?) Now F = Fv/cos? so R = (Mg/2)/(Fv/tan?) = Tan? Now by the definition of µs we have that µs = R = Tan? at the point of slipping Hence µs > L/D represents the minimum value of µs to prevent slipping of the pole.

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