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.
AIEEE 2012, Difficult
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Let the force $F$ is applied at an angle $\theta$ with the horizontal.
For horizontal equilibrium,
$F \cos \theta=\mu R$
For vertical equilibrium,
$R+F \sin \theta=m g$
$\text { or }, R=m g-F \sin \theta$
Substituting this value of $R$ in eq. $(i),$
we get
$F \cos =\mu(m g-F \sin \theta)$
$=\mu m g-\mu F \sin \theta$
or, $F(\cos \theta+\mu \sin \theta)=\mu m g$
or, $F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$
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, $F_{\text {min }}$
$=\frac{\mu w}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu w}{\sqrt{1+\mu^2}}$
For horizontal equilibrium,
$F \cos \theta=\mu R$
For vertical equilibrium,
$R+F \sin \theta=m g$
$\text { or }, R=m g-F \sin \theta$
Substituting this value of $R$ in eq. $(i),$
we get
$F \cos =\mu(m g-F \sin \theta)$
$=\mu m g-\mu F \sin \theta$
or, $F(\cos \theta+\mu \sin \theta)=\mu m g$
or, $F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$
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, $F_{\text {min }}$
$=\frac{\mu w}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu w}{\sqrt{1+\mu^2}}$
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