Question

For the situation shown in the figure, pulleys are light and smooth. Pulleys $C$ and $D$ are fixed, but $A$ and $B$ are movable. Find the value of $m$ (in $kg$ ) if $m$ is in equilibrium. (take, $m_1=2kg$ and $m_2=1kg$ )

Answer 1

Answer: 2

For $m_2$,
$T-m_{2}g=m_2(2a)$ ...(i)
Free body diagram (FBD) of $m_2$

For $m_1{m}_{1}g-2m_2=m_1(a)$ ...(ii)
Free body diagram (FBD) of $m_1$

From Eqs. (i) and (ii), we get
$m_{1}g-2m_{2}g=4m_{2}a+m_{1}a$
$\Rightarrow a=0m/s^2$
Or $T=m_{2}g=10N$
For $m,2T=mg$
$\Rightarrow m=\frac{20}{10}=2kg$
Free body diagram of $m$
$mg=2T$
$\Rightarrow m=\frac{2\times 10}{10}=2kg$