Consider $F BD$ of structure.

Applying equilibrium equations,

$Av + Bv =200 N$

$A _{ H }= B _{ H}{ \ldots \text { (ii) }}$

From F BD of block B,

$B _{ H }+ F _{ B } \cos 60^{\circ}- N _{ B } \sin 60^{\circ}=0$

$N_B \cos 60^{\circ}-B_V-300+F_V \sin 60^{\circ}=0$

$F _{ B }=0.25 N _{ B }$

$B _{ H }-0.74 N _{ B }=0 \ldots \text {.(iii) }$

$- B _{ V }+0.71 N _{ B }=300$

$F _{ A }- A _{ H }=0$

$N _{ A }- A _{ V }=400$

$F _{ A }=\mu_{ A } N _{ A }$

$\therefore \mu_{ A } N _{ A }- A _{ H }=0$.

On solving above equations, we get

$N _{ A }=650 N , F _{ A }=260 N , F _{ A }=\mu_{ A } N _{ A }$

$\therefore \mu_{ A }=\frac{260}{250}=0.4$