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?

• Correct answer
  A

  $\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}$
• B

  $\frac{\mu-1}{\mu+1} \mathrm{mg}$
• C

  $\frac{\mu}{\sqrt{1-\mu^{2}}} \mathrm{mg}$
• D

  $\mu \mathrm{mg}$

Solution

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 $F_{\min }, \frac{d}{d \theta}[\cos \theta+\mu \sin \theta]=0$ $\therefore \tan \theta=\mu$ $\therefore F_{\min }=\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}$

Hence, the correct answer is option: 'A'.