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

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

\(\therefore \mu_{ A }=\frac{260}{250}=0.4\)