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, $m_{1}=2\, kg$ and $m_{2}=1\, kg$ )

Answer 1

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

For $m_{1} m_{1} g-2 m_{2}=m_{1}(a)$ ...(ii)
Free body diagram (FBD) of $m_{1}$

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