Sliding of a Uniform Hanging Chain – Rankers Physics

Friction: Practice Problem & Solution

A uniform chain of length \( L \) is placed on a rough horizontal table with some part hanging below the table. If the length of the hanging part becomes \( \frac{2L}{5} \) then the chain starts sliding on the table. The co-efficient of friction between the chain and the table is:
\( \frac{2}{5} \)
\( \frac{3}{5} \)
\( \frac{1}{3} \)
\( \frac{2}{3} \)

Solution Explained:

To solve this problem, we apply the core principles of Friction. Understanding the underlying formula is key to arriving at the correct answer below:

The hanging part of length \( \frac{2L}{5} \) exerts a pulling force of \( \frac{2}{5} Mg \). The part on the table of length \( \frac{3L}{5} \) experiences a maximum friction force of \( \mu \frac{3}{5} Mg \). Equating the forces at the verge of sliding gives \( \mu = \frac{2}{3} \).

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