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Question
A heavy uniform chain lies on a horizontal table. If the coefficient of friction between the chain and the table is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is
Options
20%
25%
30%
40%
MCQ
Solution
20%
Explanation:
Let `M/L` be the mass per unit length of chain. Let the length of chain that hangs is L', so the length of chain that rests on table is L – L'. Thus, mass of the chain. That hangs and that rests are `M/L L`' and `M/L (L - L^')` respectively.
Let the gravitational force on the hanging chain be balanced by the frictional force from the chain on the table.
∴ f = m'g
∴ μmg = m'g
∴ `μ M/L (L - L^') = M/L (L^')`
∴ `0.25 (L - L^') = L^'`
∴ `L^'/L = 0.25/1.25` = 0.2 = 20%
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Friction
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