Question

A heavy uniform chain of length $L$ lies on the top of a horizontal table. If the coefficient of static friction between the chain and the table surface is $0.2$, then the maximum fraction of the length of the chain that can be hung over one edge of the table is:

• A. $\frac{L}{6}$
• B. $\frac{L}{3}$
• C. $\frac{L}{2}$
• D. $\frac{L}{4}$

Answer 1

The correct option is A $\frac{L}{6}$


Let mass of chain $=M$
Length of the chain $=L$
Let the length of hanging portion is $x$.
$\Rightarrow$For equilibrium of whole chain, limiting friction should balance net downward force on chain.
$\text{Limiting friction}=\text{weight of hanging part of chain}$
$\Rightarrow f_L=m_{x}g...\left(i\right)$
$\therefore$ Mass of length $x$ $m_x=\left(\frac{M}{L}\right)\times x\text{kg}$
Normal reaction on the part of chain kept on horizontal table is,
$N=(M-m_x)g$
$\because N$ balances the weight of part kept on table.
$\Rightarrow N=(M-\frac{M}{L}x)g$
$\therefore N=\frac{M}{L}(L-x)\times g$
Limiting friction acting on the chain kept on the table is,
$\Rightarrow f_L=\mu \times N$
$f_l=\mu \times \frac{M}{L}(L-x)g...\left(ii\right)$

From Eq $\left(i\right)\&\left(ii\right)$
$\Rightarrow m_{x}g=\mu \left(\frac{M}{L}\right)(L-x)g$
$\Rightarrow \left(\frac{M}{L}\right)\times x\times g=\frac{\mu M}{L}(L-x)g$
$\Rightarrow x=\mu (L-x)$
$x(1+\mu )=\mu L$
$\Rightarrow \frac{x}{L}=\frac{\mu }{1+\mu }=\frac{0.2}{1.2}$
$\therefore x=\frac{L}{6}$