A system of four masses m_1, m_2,m_3, m_4, my two springs and a fixed pulley is shown in figure. The system is kept at rest by attaching the lower thread to a rigid peg. Determine the acceleration of all the masses after the lower thread has been cut. The springs are massless, pulley is frictionless and massless.

A system of four masses m_1, m_2,m_3, m_4, my two springs and a fixed

1.	A system of four masses m_1, m_2,m_3, m_4, my two springs and a fixed pulley is shown in figure. The system is kept at rest by attaching the lower thread to a rigid peg. Determine the acceleration of all the masses after the lower thread has been cut. The springs are massless, pulley is frictionless and massless.
Answer» Solution :Clearly `(m_1 + m_2) gt (m_3 + m_4)` . Let `T_1` be the tension in left SPRING and `T_2` that in right spring and T is tension in the string passing over the pulley . Then for equilibrium of system `T_1 =m_2g , T=(m_1 + m_2)G`...........(2) and `m_3g + T_2=T` From last equation `T_2 =T-m_3g=(m_1 + m_2-m_3)g` ........ (3) After the lower THREAD is cut off Equation of motion of mass `m_1` is `m_1g + T_1-T = m_2a_3` .......... (4) Equation of motion of mass `m_2` is `m_2g -T_1 = m_2a_2` ...........(5) Equation of motion mass ` m_3` is `m_3g + T_2 -T = m_1 a_1`............. (6) Equation of motion of mass `m_4` is `m_4g -T_2 =m_4a_4` ............. (7) Using (1) , (2) and (3) we get ` a_1= a_2 = a_3 =0` ` and a_4 = (m_4g -T_2)/( m_4) =(m_4g(m_1 + m_2 - m_3)g)/( m_4) = ((m_3 + m_4 - m_1 - m_2)g)/(m_4)`