Question

A block of mass $m=2\text{kg}$ is resting on a rough inclined plane of inclination ${30}^{\circ }$ as shown in figure. The coefficient of friction between the block and the plane is $\mu =0.5$. What minimum force $F$ should be applied on the block as shown in figure, so that the block does not slip on the plane?( $g=10{\text{ms}}^{-2})$

• A. Zero
• B. $6.24\text{N}$
• C. $2.68\text{N}$
• D. $4.34\text{N}$

Answer 1

The correct option is C $2.68\text{N}$

Since, $\text{mg sin}{30}^{\circ }>\mu \text{mg cos}{30}^{\circ }$
The block has a tnedency to slip downwards, Let $F$ be the minimum force applied on it, so that it does not slip.
Then,
$N=F+\text{mg cos}{30}^{\circ }$
$\therefore \text{mg sin}{30}^{\circ }=\mu N=\mu (F+\text{mg cos}{30}^{\circ }$
$F=\frac{\text{mg sin}{30}^{\circ }}{\mu }-\text{mg cos}{30}^{\circ }$
$F=\frac{\left(2\right)\left(10\right)\left(\frac{1}{2}\right)}{0.5}-\left(2\right)\left(10\right)\left(\frac{\sqrt{3}}{2}\right)$
$F=20-17.32$
$F=2.68\text{N}$