Physics - Circular Motion Question with Solution | TestHub

Let the force $F$ is applied at an angle $\theta$ with the horizontal.
For horizontal equilibrium, $F\cos \theta =\mu R$ For vertical equilibrium, $R+F\sin \theta =\mathrm{mg}$ or, $R=mg-F\sin \theta$ Substituting this value of $R$ in eq. (i), we get $$\begin{array}{rl}& F\cos \theta =\mu (\mathrm{mg}-F\sin \theta )\\ & =\mu \mathrm{mg}-\mu F\sin \theta \\ & \text{ or, }F(\cos \theta +\mu \sin \theta )=\mu \mathrm{mg}\\ & \text{ or, }\mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \theta +\mu \sin \theta }\end{array}$$ For $F$ to be minimum, the denominator $(\cos \theta +\mu \sin \theta )$ should be maximum. $$\therefore \frac{d}{d\theta }(\cos \theta +\mu \sin \theta )=0$$ or, $-\sin \theta +\mu \cos \theta =0$ or, $\tan \theta =\mu$ or, $\theta ={\tan }^{-1}(\mu )$ Then, $\sin \theta =\frac{\mu }{\sqrt{1+{\mu }^2}}$ and $$\cos \theta =\frac{1}{\sqrt{1+{\mu }^2}}$$ Hence, $F_{\text{min }}$ $$=\frac{\mu w}{\frac{1}{\sqrt{1+{\mu }^2}}+\frac{{\mu }^2}{\sqrt{1+{\mu }^2}}}=\frac{\mu w}{\sqrt{1+{\mu }^2}}$$