Angle of friction $\theta=\tan ^{-1}(\mu)$

Or           $\theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$

Suppose the body is dragged by a force $F$ acting at an angle $\alpha$ with horizontal. Then,

$\mathrm{N}=\mathrm{mg}-\mathrm{F} \sin \alpha$

and $\mathrm{F} \cos \alpha=\mu \mathrm{N}=\mu(\mathrm{mg}-\mathrm{F} \sin \alpha)$

$\therefore \quad \mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \alpha+\mu \sin \alpha} \quad \cdots(i)$

$F$ is minimum when denominator is maximum or, $\frac{\mathrm{d}}{\mathrm{d} \alpha}(\cos \alpha+\mu \sin \alpha)=0$

or $\quad-\sin \alpha+\mu \cos \alpha=0$

or $\quad \tan \alpha=\mu=\frac{1}{\sqrt{3}}$

$\alpha=30^{\circ},$ the angle of friction $\theta$

At $\alpha=30^{\circ},$ force needed is minimum.

Substituting the values in Eq. $(i)$ we have,

$\mathrm{F}_{\min }=\frac{\left(\frac{1}{\sqrt{3}}\right)(25)(\mathrm{g})}{(\sqrt{3} / 2)+(1 / \sqrt{3})(1 / 2)}$

$=12.5 \mathrm{g}$

$=12.5 \mathrm{kgf}$