Question

A $1kg$ mass on a frictionless inclined plane is connected by a pulley to a hanging $0.5kg$ mass, as in the diagram above. Find out the angle at which system will be in equilibrium?
$\cos {30}^o=\sin {60}^o=\frac{\sqrt{3}}{2}$, $\cos {60}^o=\sin {30}^o=\frac{1}{2}$, $\cos {45}^o=\sin {45}^o=\frac{1}{\sqrt{2}}$

• A. $0^o$
• B. $-{30}^o$
• C. ${30}^o$
• D. ${45}^o$
• E. ${60}^o$

Answer 1

The correct option is C ${30}^o$

Given : $m=1kg$ $m_1=0.5kg$

Let the tension in the string be $T$.

For mass $m_1$ : $T=m_{1}g=0.5\times 10=5$ $N$

For mass $m$ : $T=mgsin\theta$

$\therefore$ $5=1\times 10$ $sin\theta$ $⟹sin\theta =0.5$ OR $\theta ={30}^o$