Question

A uniform chain of length $L$ has one of its ends attached to the wall at point A, while $\frac{3L}{4}$ length of the chain is lying on the table as shown in figure. Then, minimum co-efficient of static friction between table and chain, so that the chain remains in equilibrium is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{4}$

FBD of the two parts of the chain are given below:


Let $\lambda =\text{mass per unit length of uniform chain}$
$\text{in kg/m}$
Now mass of the part of chain hanging from the wall,
$m_1=\frac{L}{4}\lambda \text{kg}$

Similarly for part of the chain resting on table:
mass $m_2=\frac{3L}{4}\lambda \text{kg}$

From the FBD of $m_1$,
$F\cos {37}^{\circ }=\frac{\lambda L}{4}g$ ....(1)
$F\sin {37}^{\circ }=T$

From the FBD of $m_2$,
$f=T=F\sin {37}^{\circ }$
But $f\le {\mu }_{s}N={\mu }_s\left(\frac{3}{4}\lambda Lg\right)$
i.e $F\sin {37}^{\circ }\le {\mu }_s\left(\frac{3}{4}\lambda Lg\right)$.....(2)

Substituting $F$ from (1),
$\frac{5}{16}\lambda Lg\times \frac{3}{5}\le {\mu }_s\left(\frac{3}{4}\lambda Lg\right)$
$\Rightarrow \frac{3}{16}\le {\mu }_s\times \frac{3}{4}$
$\Rightarrow {\mu }_s\ge \frac{1}{4}$
$\therefore {{\mu }_s}_{min}=\frac{1}{4}$