Question

A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is $\mu .$ the block is to be pulled by applying a force to it. what minimum force is needed to slide the block ? In which direction should this force act?

Answer 1

Let P be th eforce applied to it an angle $\theta$. From the free body diagram
$R+Psin\theta -mg=0$
$\Rightarrow R=-P\sin \theta +mg\mu R=P\cos \theta$
From equation (i)
$\mu (mg-P\sin \theta )=-P\cos \theta \Rightarrow \mu mg=\mu P\sin \theta -P\cos \theta \Rightarrow P=\frac{\mu mg}{\mu \sin \theta +\cos \theta }$
Applied force P should be minimum, when $\mu \sin \theta +\cos \theta$ is maximum.
Again $\left(\mu \sin \theta \right)$
is maximum when its derivative is zero
$\frac{d}{d\theta }(\mu \sin \theta +\cos \theta )=0\Rightarrow \mu \cos \theta -\sin \theta =0\Rightarrow \theta =ta{n}^{-1}\mu So,P=\frac{\mu mg}{\mu \sin \theta +\cos \theta }=\frac{\frac{\mu mg}{\cos \theta }}{\frac{\mu \sin \theta }{\cos \theta }+\frac{\cos \theta }{\cos \theta }}$

(Dividing Numerator and denominator by \cos \theta), we get,
$P=\frac{\mu mg\sec \theta }{1+\mu \tan \theta }=\frac{\mu mg\sec \theta }{1+ta{n}^2\theta }=\frac{\mu mg}{\sec \theta }=\frac{\mu mg}{\sqrt{(1+{\tan }^2\theta )}}=\frac{\mu mg}{(1+{\mu }^2)}$
Hence minimum force is $\frac{\mu mg}{\sqrt{(1+{\mu }^2)}}$ at an angle $\theta ={\tan }^{-1}\mu$