A block of weight $W$ rests on a horizontal floor with coefficient of static friction $\mu$. It is desired…

1. $\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$
2. $\theta=\tan ^{-1}\left(\frac{1}{\mu}\right), F=\frac{\mu W}{\sqrt{1+\mu^2}}$
3. $\theta=0, F=\mu W$
4. $\theta=\tan ^{-1}\left(\frac{\mu}{1+\mu}\right), F=\frac{\mu W}{1+\mu}$

Solution

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, $R=m g-F \sin \theta$ Substituting this value of $R$ in eq. (i), we get $ \begin{aligned} & 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{aligned} $ 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}} $

Asked in: JEE Main 2012 (19 May Online)

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