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A block of mass m rests on a horizontal table with a co-efficient of static friction $\mu .$ What minimum force must be applied on the block to drag it on the table?
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Verified Answer
The correct answer is:
$\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}$
Hint:
$F \sin \theta+N=m g \quad N=m g-F \sin \theta$
$F \cos \theta=\mu(m g-F \sin \theta) \quad F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$,
for $F_{\min }, \frac{d}{d \theta}[\cos \theta+\mu \sin \theta]=0$
$\therefore \tan \theta=\mu$
$\therefore F_{\min }=\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}$
$F \sin \theta+N=m g \quad N=m g-F \sin \theta$
$F \cos \theta=\mu(m g-F \sin \theta) \quad F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$,
for $F_{\min }, \frac{d}{d \theta}[\cos \theta+\mu \sin \theta]=0$
$\therefore \tan \theta=\mu$
$\therefore F_{\min }=\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}$
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