In the shown arrangement if $f_A\,,\, f_B$ and $T$ be the frictional forces on $A$ Block, $B$ Block and tension in the string respectively, then their values are
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$For\, A$

$\left(f_{\max }\right)_{A}=\mu \operatorname{mg} \cos 37^{\circ}=0.5 \times 50 \times \frac{4}{5} \Rightarrow 20 \mathrm{N}$

$For$ $B$

$\left(f_{\max }\right)_{B}=\mu \operatorname{mg} \cos 53^{\circ}=0.5 \times 50 \times \frac{3}{5}=15 \mathrm{N}$

$40-T-f_{B}=5 a$

$\mathrm{T}-30-\mathrm{f}_{\mathrm{A}}=5 \mathrm{a}$

$\mathrm{T}=\mathrm{f}_{\mathrm{A}}+30$

$\mathrm{f}_{\mathrm{A}}=-5 \mathrm{N}$

$\mathrm{F}_{\mathrm{Bmax}}=15$

$\mathrm{T}-\mathrm{f}_{\mathrm{B}}=40$

$\mathrm{T}=25 \mathrm{N}$

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