Question

A pushing force making an angle $\theta$ with the horizontal. is applied on a block of weight $W$ placed on a horizontal table. If the angle of friction is $\varphi ,$ the magnitude of force required to move the body is equal to

• A. $\frac{W\cos \varphi }{\cos (\theta -\varphi )}$
• B. $\frac{W\sin \varphi }{\cos (\theta +\varphi )}$
• C. $\frac{W\tan \varphi }{\sin (\theta -\varphi )}$
• D. $\frac{W\sin \theta }{\tan (\theta -\varphi )}$

Answer 1

Answer: (B)

The various forces acting on the block are shown figure.

Here, $\mu =\tan \varphi =\frac{f}{R}$
or $f=R\tan \varphi ...(i)$
The condition for the block just to move is
$F\cos \theta =f=R\tan \varphi ...(ii)$
and $F\sin \theta +W=R$
of $R=W+F\sin \theta ...(iii)$
From equations $(ii)$ and $(iii)$, we get
$F\cos \theta =(W+F\sin \theta )\tan \varphi =W\tan \varphi +F\sin \theta \tan \varphi$
or $F\cos \theta -F\sin \theta \frac{\sin \varphi }{\cos \varphi }=\frac{W\sin \varphi }{\cos \varphi }$
or $F(\cos \theta \cos \varphi -\sin \theta \sin \varphi )=W\sin \varphi$
or $F\cos (\theta +\varphi )=W\sin \varphi$ or $F=\frac{W\sin \varphi }{\cos (\theta +\varphi )}$