Posted by James on Thursday, June 21, 2012 at 2:48pm.
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
Thank you so much! I've been trying to figure this one out for a few hours.
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