Physics - NLM Question with Solution | TestHub

$\mathrm{FBD}$ of block $\mathrm{B}$ w.r.t. wedge $\mathrm{A}$, for maximum $\text{'}a\text{'}$:

Perpendicular to wedge:
$\Sigma f_{\mathrm{y}}=(mg\mathrm{cos\theta }+ma\mathrm{sin\theta }-N)=0$
and $\Sigma f_x=mg\sin \theta +\mu N-ma\cos \theta =0$ (for maximum a)
$\Rightarrow \mathit{ }mg\mathrm{sin\theta }+\mathrm{\mu }(mg\mathrm{cos\theta }+ma\mathrm{sin\theta })-ma\mathrm{cos\theta }=0$
$\Rightarrow \mathit{ }a=\frac{(g\mathrm{sin\theta }+\mu g\mathrm{cos\theta })}{\mathrm{cos\theta }-\mu \mathrm{sin\theta }}$

for $\theta =45°$

$a=g\left(\frac{\tan 45°+\mu }{\cot 45°-\mu }\right); \mathit{ }a=g\left(\frac{1+\mu }{1-\mu }\right)$