Question

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

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

Answer 1

The correct option is A

$20\%$

Step 1. Given data

The coefficient of friction between the chain and the table surface is$0.25$.

Step 2. Formula used

$f_{max}=\mu N$

Where $\mu$ is the coefficient of friction and $N$ is the normal force.

Step 3. Explanation for correct option

Let $x$ be the length of the chain on the table and $\left(L-x\right)$ is the length hanging.

Let the total mass is $M$.

Therefore the mass of the chain hanging is $\frac{M\left(L-x\right)}{L}$.

Mass of the chain on the table $\frac{Mx}{L}$.

$f_{max}=\mu N=\frac{\mu Mgx}{L}$ where $g$ is the acceleration due to gravity.

Weight of the chain hanging,

$W=m\text{'}g=\frac{M\left(L-x\right)g}{L}$

For the system to be stable

$$\begin{gathered} \frac{M\left(L-x\right)g}{L}=\frac{\mu Mgx}{L} \\ \Rightarrow L-x=0.25x \\ \Rightarrow x=\frac{4}{5}L \end{gathered}$$

Therefore the hanging part is $L-x=L-\frac{4L}{5}=\frac{L}{5}$

Now we calculate the percentage:

$\frac{{\displaystyle \frac{L}{5}}}{L}\times 100=20\%$

Hence, the correct option is option A.