Let the hanging length be $x$ m

Total length of chain $= 6$ m.

So, length on the table $= 6-x$ m.

Forces acting on the chain

Since the chain is uniform, let its mass per unit length be $\lambda$.

• Weight of hanging part pulls the chain downward: $F_{\text{pull}} = \lambda x g$

• The part lying on the table experiences friction.

  Normal reaction on the portion on table: $N = \lambda (6-x) g$

  Maximum static friction: $f_{\max} = \mu_s N = 0.5\,\lambda (6-x) g$

Condition for maximum hanging length at rest

For the chain to be just on the verge of slipping, $\lambda x g = 0.5\,\lambda (6-x) g$

Cancel $\lambda g$ from both sides: $x = 0.5(6-x)$

Solve for $x$

$x = 3 - 0.5x$ $1.5x = 3$ $x = 2$

Final answer

The maximum length of the chain hanging from the table is $\boxed{2\text{ m}}$