Question

A uniform rigid pole of length L and mass M is to be supported from a vertical wall in a horizontal position, as shown in the figure. The pole is not attached directly to the wall, so the coefficient of static friction, μs,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 a distance D directly above the point where the pole contacts the wall. Determine the minimum value of μs, as a function of L and D, that will keep the pole horizontal and not allow its end to slide down the wall. (Use any variable or symbol stated above as necessary.)

Answer 1

Assume the rope makes an angle theta with horizontal and tension in the rope is T. Let the normal reaction at the pole and wall contact point is N (in horizontal dir.)

T*Cos theta = N ....(1)
T*Sin theta + mu*N = Mg ........(2)
So, Tan theta = (M*g-mu*N)/N
or, Tan theta = M*g/N - mu
But Tan theta = D/L
So, M*g/N - mu = D/L
or, mu = M*g/N - D/L ......(3)

Now take moment of the forces about the point of contact:
T*(Sin theta)*L = M*g*L/2
or T*(Sin theta)= M*g/2 .....(4)
From (1)& (4) Tan theta = (M*g/2)/N
or, N = (M*g/2)/(D/L)
= M*g*L/2*D ........(5)
From (3) & (5)
mu = M*g*2D/M*g*L - D/L
= 2D/L - D/L
= D/L

Answer 2

Thank you so much! I've been trying to figure this one out for a few hours.