Question

A mass $Mkg$ is suspended by a weightless string. The horizontal force required to hold the mass at ${60}^o$ with the vertical is

• A. $Mg$
• B. $Mg$$\sqrt{3}$
• C. $Mg$$(\sqrt{3}+1)$
• D. $\frac{Mg}{\sqrt{3}}$

Answer 1

The correct option is B $Mg$$\sqrt{3}$

Let the tension in the string at equilibrium be $T$ and the force applied be $F$.

In horizontal direction : $F=T$ $\sin {60}^o$

Vertical direction : $Mg=T\cos {60}^o$

Dividing both the equations we get: $\frac{F}{Mg}=\tan {60}^o$

$\therefore$ $\frac{F}{Mg}=\sqrt{3}$ $⟹F=\sqrt{3}Mg$