Question

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 at an angle of $30^{\circ}$ to the horizontal is

• A. $25\, N$
• B. $100\, N$
• C. $50\, N$
• D. $\frac{50}{\sqrt{3}} N$

Answer 1

Answer: (C)

Let $P$ force is acting at an angle $30^{\circ}$ with the horizontal.
For the condition of motion $F=\mu R$
$P \cos 30^{\circ}=\mu\left(m g-P \sin 30^{\circ}\right)$
$\Rightarrow P \frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}\left(100-P \frac{1}{2}\right)$
$\Rightarrow \frac{3 P}{2}=\left(100-\frac{P}{2}\right)$
$\Rightarrow 2 P=100$
$\Rightarrow P=50\, N$