Question

A mass $m$ is supported as shown in the figure by ideal strings connected to a rigid wall and to a mass $3m$ at rest on a fixed horizontal surface. The string connected to larger mass is horizontal, that connected to smaller mass is vertical and the one connected to wall makes an angle ${60}^o$ with horizontal. Then the minimum coefficient of static friction between the larger mass and the horizontal surface that permits the system to remain in equilibrium in the situation shown is

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{3\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

Answer: (B)

$\frac{1}{3\sqrt{3}}$

At the instant $3m$ is about to slip, fbd is as shown:
$\therefore 3\mu mg$ $=Tcos{60}^0.....\left(1\right)$
and $mg=Tsin{60}^0.....\left(2\right)$
$\therefore \mu =\frac{1}{3\sqrt{3}}$