A uniform chain of $$6$$ m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is $$0.5$$, the maximum length of the chain hanging from the table is ______ m.

Let the total length of the chain be $$L = 6$$ m and the coefficient of static friction be $$\mu = 0.5$$.

Let $$x$$ be the length of the chain hanging over the edge. Then $$(L - x) = (6 - x)$$ is the length on the table.

Let the mass per unit length of the chain be $$\lambda = \frac{m}{L}$$.

Weight of the hanging part (pulling the chain down):

$$W_{hang} = \lambda x g = \frac{m x g}{6}$$

Normal force on the table part:

$$N = \lambda (6 - x) g = \frac{m(6 - x)g}{6}$$

Maximum static friction force:

$$f = \mu N = 0.5 \times \frac{m(6 - x)g}{6}$$

At the maximum hanging length, the system is on the verge of sliding, so:

$$W_{hang} = f$$

$$\frac{m x g}{6} = 0.5 \times \frac{m(6 - x)g}{6}$$

Cancelling common terms:

$$x = 0.5(6 - x)$$

$$x = 3 - 0.5x$$

$$1.5x = 3$$

$$x = 2 \text{ m}$$

The maximum length of the chain hanging from the table is 2 m.