windsofchange
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- Feb 11, 2019
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Question: A uniform solid hemisphere of weight W rests in limiting equilibrium with its curved surface on a rough inclined plane. The plane face is kept horizontal by a weight P attached to a point on its rim. By resolving horizontally, and by taking moments about the point of contact between the hemisphere and the inclined plane, show that the coefficient of friction is given by P/sqrt(W(W+2P)).
I resolved horizontally and the equations which came out were R (Reaction force) = (W+P)cos(x) and uR = (W+P)sin(x), where x is the angle of inclination of the plane. I found that the coefficient of friction is tan(x) but I don't know where to go from here.
Please respond I don't know what else to do.
I resolved horizontally and the equations which came out were R (Reaction force) = (W+P)cos(x) and uR = (W+P)sin(x), where x is the angle of inclination of the plane. I found that the coefficient of friction is tan(x) but I don't know where to go from here.
Please respond I don't know what else to do.
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