Question

A mass 'M' hangs by two strings making angles $\alpha$ and $\beta$ to the vertical. The strings are connected through frictionless pullies to two masses $m_1$ and $m_2$. If the system is in equilibrium, the ratio of $m_1:m_2$ is

• A. $\sin \alpha :\sin \beta$
• B. $\sin \beta :\sin \alpha$
• C. $\cos \alpha :\cos \beta$
• D. $\cos \beta :\cos \alpha$

Answer 1

Answer: (B)

$\sin \beta :\sin \alpha$

We have from equilibrium of $m_1,T_1=m_{1}g$
We have from equilibrium of $m_2,T_2=m_{2}g$
From horizontal equilibrium of M $T_{1}sin\alpha =T_{2}sin\beta$
implies $m_{1}gsin\alpha =m_{2}gsin\beta$
Hence $m_1:m_2=sin\beta :sin\alpha$