Question 188 of 210

A block of weight $W$ rests on a horizontal floor with coefficient of static friction $\mu$. It is desired to make the block move by applying minimum amount of force. The angle $\theta$ from the horizontal at which the force should be applied and magnitude of the force $F$ are respectively.

A$\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$B$\theta=\tan ^{-1}\left(\frac{1}{\mu}\right), F=\frac{\mu W}{\sqrt{1+\mu^2}}$C$\theta=0, F=\mu W$D$\theta=\tan ^{-1}\left(\frac{\mu}{1+\mu}\right), F=\frac{\mu W}{1+\mu}$

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Answer: (A) $\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$

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, $F_{\text {min }}$
$$=\frac{\mu w}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu w}{\sqrt{1+\mu^2}}$$