Question

A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the table surface is $0.25$, then find the maximum $\%$ of the length of the chain that can hang over one edge of the table.

• A. $20\%$
• B. $25\%$
• C. $35\%$
• D. $15\%$

Answer 1

The correct option is A $20\%$

$\text{Step 1: Free body diagram[Refer Fig.]}$
If $M$ is the mass of the chain and $L$ is the length of the chain.

Let $x$ length is lying on the horizontal table.

$\text{Step 2: Mass of various parts of chain}$

Now, mass per unit length of the chain$\left(\lambda \right)=\frac{M}{L}$

$\therefore$ Mass of the length $L-x$, hanged part$=\frac{M}{L}(L-x)$
And, Mass of the length $x$ lying on table $=\frac{M}{L}x$

$\text{Step 3: Condition for maximum length that can hang}$

As the friction force from the table is stopping the chain from falling down.

So, To get the maximum $\%$ of the length of the chain hanging, frictional force should be maximum.

$f_{\text{max}}=\mu N$

From FBD

$N=\frac{Mgx}{L}$

$\therefore f_{\text{max}}=\mu \frac{M}{L}xg$ $....\left(1\right)$

Weight of hanging chain

$W=\frac{M}{L}(L-x)g$

For system to be stable, friction should balance the weight of hanging chain

$\therefore f_{\text{max}}=W$

$\Rightarrow f_{\text{max}}$ $=\frac{M}{L}(L-x)g$ $....\left(2\right)$

$\text{Step 3: Equation Solving}$

From equation $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \mu \frac{M}{L}xg=\frac{M}{L}(L-x)g$

$\Rightarrow L-x=\frac{x}{4}$

$\Rightarrow$ $x=\frac{4}{5}L$

Hence, Hanging part is $\frac{L}{5}$ which is $20\%$ of the total length $L$

Hence Option $A$ is correct.