A uniform chain of length $L$ has one of its end attached to the wall at point A, while  $\frac{3L}{4}$ of the length of the chain is lying on table as shown in figure. The minimum co-efficient of friction between table and chain so that chain remains in equilibrium is 

The vertical component of the force at A must balance the weight of chain that is hanging, so

 $F\cos 37°=\frac{\lambda L}{4}g\Rightarrow F=\frac{5\lambda L}{16}g$
(where $\lambda$  is the mass/length of the chain).
The horizontal component of the force will act on the chain which must be balanced by the tension at the point where the chain touches the table.

$F\sin 37°=T=f\le \mu N$.

here, $N=\frac{3L\lambda }{4}g$
$\Rightarrow \mu \ge \frac{1}{4}\Rightarrow {\mu }_{\min }=\frac{1}{4}$
$F\cos 37°=\frac{\lambda L}{4}g$