Question

A particle of mass $m$ rests on a horizontal floor which has coefficient of static friction $\mu$. For the block to just move

• A. a minimum force $F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$ has to be applied at an angle $\theta ={\tan }^{-1}\left(\frac{1}{\mu }\right)$ with the horizontal.
• B. a minimum force $F_{min}=\mu mg$ has to be applied at an angle $\theta ={\tan }^{-1}\left(\mu \right)$ with the horizontal.
• C. a minimum force $F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$ has to be applied at an angle $\theta ={\tan }^{-1}\left(\mu \right)$ with the horizontal.
• D. a minimum force $F_{min}=\mu mg$ has to be applied at an angle $\theta ={\tan }^{-1}\left(\frac{1}{\mu }\right)$ with the horizontal.

Answer 1

The correct option is C a minimum force $F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$ has to be applied at an angle $\theta ={\tan }^{-1}\left(\mu \right)$ with the horizontal.

Let the FBD of the block be as shown in the figure, when it's just about to move.

From the FBD, we have
$F\cos \theta =f=\mu N...\left(1\right)$ and
$N+F\sin \theta =mg$
$\Rightarrow N=mg-F\sin \theta ...\left(2\right)$

On substituting (2) in (1), we get
$F\cos \theta =\mu mg-\mu F\sin \theta$
$\Rightarrow F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }...\left(3\right)$

Let $Z=\cos \theta +\mu \sin \theta$
For $F$ to be minimum, $Z$ must be maximum
$\Rightarrow \frac{dZ}{d\theta }=0$
$\Rightarrow -\sin \theta +\mu \cos \theta =0$
$\Rightarrow \tan \theta =\mu$
$\Rightarrow \sin \theta =\frac{\mu }{\sqrt{1+{\mu }^2}}$
$\&\cos \theta =\frac{1}{\sqrt{1+{\mu }^2}}$
Thus,
$F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$