Question

A block of mass $m$ rests on a horizontal floor with which it has a coefficient of static friction $\mu$. It is desired to make the body move by applying the minimum possible force $F$. Find the magnitude of $F$ and the direction in which it has to be applied.

Answer 1

Normal force,

$N=mg-F\sin \theta$

If $F$ is minimum to move the block, then;

$f=$ frictional force (static)

$f=\mu N=\mu (mg-F\sin \theta )$

For the block to just move,

$F\cos \theta =f=\mu (mg-F\sin \theta )$

$F(\cos \theta +\mu \sin \theta )=\mu mg$

$F=\frac{\mu mg}{(\cos \theta +\mu \sin \theta )}$

If $F$ is minimum, then

let $y=\cos \theta +\mu \sin \theta$ ...........(1) should be maximum

$\frac{dy}{d\theta }=-\sin \theta +\mu \cos \theta =0$

$\sin \theta =\mu \cos \theta$

$\theta ={\tan }^{-1}\mu$

For the value of $\theta ={\tan }^{-1}\mu$, the force $F$ will be minimum,

and

$y=\cos \theta +\mu \sin \theta =\cos \theta +{\mu }^2\cos \theta =\cos \theta (1+{\mu }^2)=\sqrt{(1+{\mu }^2)}$

hence, minimum force,

$F=\frac{\mu mg}{y}$ $=\frac{\mu mg}{\sqrt{(1+{\mu }^2)}}$