Question

A block of mass $m$ rests on a horizontal table with coefficient of static friction $\mu$. What minimum force must be applied on the block to drag it on the table?

• A. $\frac{\mu }{\sqrt{1+{\mu }^2}}mg$
• B. $\frac{\mu -1}{\mu +1}mg$
• C. $\frac{\mu }{\sqrt{1-{\mu }^2}}mg$
• D. $\mu mg$

Answer 1

The correct option is A $\frac{\mu }{\sqrt{1+{\mu }^2}}mg$


$N+Fsin\theta =mg\Rightarrow N=mg-Fsin\theta$

$Fcos\theta =\mu (mg-Fsin\theta )$
$\Rightarrow F=\frac{\mu mg}{cos\theta +\mu sin\theta }$ ___(1)

$F$ is minimum when $cos\theta +\mu sin\theta$ is maximum.
Hence $Y=cos\theta +\mu sin\theta$
$\frac{dy}{d\theta }=0\Rightarrow \mu =\tan \theta$___(2)

From (1) and (2)
$F_{min}=\frac{\mu }{\sqrt{{\mu }^2+1}}mg$
as $cos\theta =\frac{1}{\sqrt{{\mu }^2+1}}$
and $sin\theta =\frac{\mu }{\sqrt{{\mu }^2+1}}$