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March 4, 2015

March 4, 2015

Posted by **James** on Thursday, June 21, 2012 at 2:48pm.

- Physics -
**ajayb**, Friday, June 22, 2012 at 3:39amAssume 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

- Physics -
**James**, Friday, June 22, 2012 at 10:57amThank you so much! I've been trying to figure this one out for a few hours.

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