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, m₁ = 2 kg and m₂ =1 kg)

Let tension  of $m_2$is T and tension of  $m_1$ is 2T

Let acceleration of $m_1$is a and acceleration of  $m_2$ is 2a

$For m_2$

$$\begin{gathered} T-m_{2}g=m_2\left(2a\right) \\ \Rightarrow T-1\left(10\right)=1\left(2a\right)......\left(i\right) \end{gathered}$$

Free body diagram (FBD) of m₂

$$\begin{gathered} \mathbf{\text{ For }}{\mathit{m}}_{\mathbf{1}}: m_{1}g-2T=m_1\left(a\right) \\ \Rightarrow 2\left(10\right)-2T=2a ......\left(ii\right) \end{gathered}$$

Free body diagram (FBD) of m₁

From Eqs. (i)2 and (ii), we get

$\begin{array}{l}2T-20=4a\\ \Rightarrow 20-2T=2a\\ \\ \Rightarrow a=0\mathrm{m}/{\mathrm{s}}^2\\ \text{ Or substituting in (1)}\\ \Rightarrow T-10=2\left(0\right)\\ \Rightarrow T=10N\\ \end{array}$

Free body diagram of m

$\mathbf{\text{For }}\mathit{m},2T=mg\Rightarrow 2\left(10\right)=m\left(10\right)\Rightarrow m=2\mathrm{kg}$