Q. A body of mass $10\ kg$ slides along a rough horizontal surface. The coefficient of friction is $1/\sqrt{3}$. Taking $g = 10\ m/s^{2}$, the least force which acts an angle of $30{^\circ}$ to the horizontal is
A) $25\ N$
B) $100\ N$
C) $50\ N$
D) $\frac{50}{\sqrt{3}}\ N$
Explanation
(c) Let $P$ force is acting at an angle $30{^\circ}$ with the horizontal
For the condition of motion $F = \mu\ R$
$P\cos{30{^\circ}} = \mu(mg - P\sin{30{^\circ})}$
$\Rightarrow P\frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left( 100 - P\frac{1}{2} \right) \Rightarrow \frac{3P}{2} = \left( 100 - \frac{P}{2} \right)$
$\Rightarrow 2P = 100\therefore P = 50\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.