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 $\phi,$ the magnitude of force required to move the body is equal to

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

Answer 1

Answer: (B)

The various forces acting on the block are shown figure.

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