For the system at rest, determine the acceleration of all the loads immediately after lower thread keeping the system in equilibrium has been cut. Assume that the threads are weightless, the mass of the pulley is negligible small, and there is no friction at the point of suspension.

For the equilibrium of system of loads,

                               $(m_1+m_2) > (m_3+m_4)$ .

The force in the left spring, $T_1 = m_{2}g$ .

Let $T_2$ is the force in the right spring then for the equilibrium of load $m_3$ , we have

                       $m_{3}g+T_2-T_0 = 0$            …(i)

For $m_1$ ;                  $m_{1}g+T_1 = T_0$

          as                             $T_1 = m_{2}g$

          $\therefore$                $m_{1}g+m_{2}g+ = T_0$

Substituting this value in equation (i), we get

                             $m_{3}g+T_2-(m_{1}g+m_{2}g)=0$

or                                      T$T_2 = (m_1+m_2-m_3)g$            …(ii)

After cutting the lower thread, the equations of motion for the loads are;

            $m_{1}g+T_1-T_0 = m_1{a}_1$

                       $m_{2}g-T_1 = m_2{a}_2$

             $T_2+m_{3}g-T_0 = m_3{a}_3$

and        $T_2-m_{4}g = m_4{a}_4$

Solving above equations, we get

             $a_1=a_2=a_3=0$

and        $a_4=\frac{(m_3+m_4-m_1-m_2)g}{m_4}$ .