Solve the following :

Let minimum force required be $\mathrm{P}$ at an angle $\theta$.

$N+P \sin \theta=m g$ (vertical equilibrium) -(i)

$P \cos \theta=\mu N$ (Body is about to move) -(ii)

From (i)

$P \cos \theta=\mu(m g-P \sin \theta)$

$P=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$

To minimize $\mathrm{P}$, we have to maximize $y=\cos \theta+\mu \sin \theta$

$\frac{d y}{d \theta}=\frac{d}{d \theta}(\cos \theta+\mu \sin \theta)$

$\frac{d y}{d \theta}=-\sin \theta+\mu \cos \theta=0$

$\tan \theta=\mu$

$\theta=\tan ^{-1} \mu$

Maximum value of

$y=\frac{1}{\sqrt{\mu^{2}+1}}+\mu \cdot \frac{\mu}{\sqrt{\mu^{2}+1}}$

$y_{\max }=\sqrt{1+\mu^{2}}$

Put in (3)

$P_{\min }=\frac{\mu m g}{\sqrt{1+\mu^{2}}}$  at $\theta=\tan ^{-1} \mu$