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# M2 Equilibrium Question Watch

1. I have a diagram that looks like this and I need to prove that the coefficient of friction is less than or equal to tan(alpha). So far I have taken moments about A to get the normal contact force R at P:
(3l)*R = 2lcos(alpha)W
R=(2/3)cos(alpha)W

When I try to get the coefficient of friction I said that Ff=Wsin(alpha) when resolving parallel to the plane but when equating that to the coefficient of friction I get u = 3/2tan(alpha) not just tan(alpha)

Can anyone help?
2. (Original post by Bobrey)
I have a diagram that looks like this and I need to prove that the coefficient of friction is less than or equal to tan(alpha). So far I have taken moments about A to get the normal contact force R at P:
(3l)*R = 2lcos(alpha)W
R=(2/3)cos(alpha)W

When I try to get the coefficient of friction I said that Ff=Wsin(alpha) when resolving parallel to the plane but when equating that to the coefficient of friction I get u = 3/2tan(alpha) not just tan(alpha)

Can anyone help?
There is a vertical reaction at the bottom end of the rod, and it will have a component parallel to the rod.

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