Question

A uniform rope lies on a table with some portion hanging. The rope begins to slide when the length of the hanging part is 25% of the entire length. The coefficient of friction between the rope and table is:
$\begin{align}
  & (A)0.33 \\
 & (B)0.25 \\
 & (C)0.50 \\
 & (D)0.20 \\
\end{align}$

Answer 1

Hint: In this problem, some portion of rope is left hanging from the table. Now, there are two forces on this rope which are balanced, first the force of gravity and secondly tension due to the remaining part of the rope on the table. The leftover part of rope on the table will experience friction due to the normal reaction of this part only and we shall use this fact to calculate the coefficient of friction between the rope and table.

Complete answer:
Let the total length of the rope be denoted by ‘L’ and let the mass per unit length of the rope be given by ‘m’. Then, we can write the mass of the hanging rope (say ${{M}_{H}}$) as:
$\begin{align}
  & \Rightarrow {{M}_{H}}=\dfrac{25}{100}\times L\times m \\
 & \Rightarrow {{M}_{H}}=0.25mL \\
\end{align}$
Thus, the weight due to this mass will be equal to:
$\begin{align}
  & \Rightarrow W={{M}_{H}}g \\
 & \Rightarrow W=0.25mgL \\
\end{align}$
Now, the portion of rope left on the table will experience a frictional force due to its own weight. This frictional force will be equal to:
$\Rightarrow f=\mu {{M}_{T}}g$
Where, ${{M}_{T}}$ is the mass of the rope on the table and it is equal to:
$\begin{align}
  & \Rightarrow {{M}_{T}}=\dfrac{75}{100}\times L\times m \\
 & \Rightarrow {{M}_{T}}=0.75mL \\
\end{align}$
Therefore, the frictional force is:
$\Rightarrow f=0.75\mu mgL$
Now, this frictional force will act as tension for the part of rope hanging. Thus, it will balance the weight of hanging rope. Therefore, on equating these two forces, we get:
$\begin{align}
  & \Rightarrow 0.75\mu mgL=0.25mgL \\
 & \Rightarrow 0.75\mu =0.25 \\
 & \therefore \mu =0.33 \\
\end{align}$
Hence, the value of coefficient of friction between the rope on the table and the table comes out to be 0.33 .

Hence, option (A) is the correct option.

Note:
Since the whole system is at rest, therefore we can say that this is the static coefficient of friction. If the system were moving with some speed then, then it would have been called dynamic coefficient of friction. Also, in real life the dynamic coefficient of friction is always slightly less than the static coefficient of friction.