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a uniform chain of mass M and length L is lying on a horizontal table with one third of its length hanging over the edge of the table if the chain is in limiting equilibrium what is the coefficient of friction for the contact between the table and chain |
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Answer» 0.5Explanation:MASS of CHAIN on table = (2/3)*MFriction force = UMG = u(2/3)*MgThis will be equal to mg of HANGING partMg/3 = u(2/3)*Mgu = 1/2 = 0.5 |
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