Question

A block of mass $2\text{kg}$ rests on a rough inclined plane making an angle of ${30}^{\circ }$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is

• A. $9.8\text{N}$
• B. $9.8\sqrt{3}\text{N}$
• C. $4.9\sqrt{3}\text{N}$
• D. $19.6\text{N}$

Answer 1

The correct option is A $9.8\text{N}$

Angle of repose ${\theta }_r={\tan }^{-1}\left({\mu }_s\right)={\tan }^{-1}\left(0.7\right)$
or $\text{tan}{\theta }_r=0.7$
Angle of plane is $\theta ={30}^{\circ },\text{tan}\theta =\text{tan}{30}^{\circ }=0.577$

Since $\text{tan}\theta <\text{tan}{\theta }_r$
$\Rightarrow \theta <{\theta }_r$
$\therefore$Block will not slide
$\Rightarrow f=\text{mg sin}\theta \ne \mu \text{mg cos}\theta$
or $f=2\times 9.8\times \text{sin}{30}^{\circ }=9.8\text{N}$