Question

A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is $0.5$, what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

• A. $L/3$
• B. $L/2$
• C. $2L/3$
• D. $3L/4$

Answer 1

Answer: (C)

Let x be length of the chain that lies on the table.
Mass per unit length of the chain = $\frac{M}{L}$
Mass of length x of the chain = $\frac{M}{L}x$
Mass of the length $(L-x)$ of hanging chain = $\frac{M}{L}(L-x)$
At equilibrium, friction force between table and chain
= weight of hanging part of chain
$\mu \left(\frac{M}{L}x\right)g=\frac{M}{L}\left(L-x\right)g$
$0.5x=L-x;1.5x=L$
$\therefore x=\frac{2L}{3}$