Four blocks and two springs are arranged as shown in fig. The system at rest, determine the accelration of all the loads immedialtely after the 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.

Explanation

For the equilibrium of system of loads,
`(m_(1)+m_(2)) gt (m_(3)+m_(4))`
The force in the left spring.
Let `T_(2)` is the force inthe right spring. For the equlilbrium of load `m_(3)`, we have
`m_(3)g+T_(2)=T_(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 (i), we get
`m_(3)g+T_(2)=(m_(1)g+m_(2)g)`
or `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)` ..(iii)
`m_(2)g-T_(1)=m_(2)a_(2)` ...(iv)
`T_(2)+m_(3)g-T_(0)=m_(3)a_(3)`..(v)
and `T_(2)-m_(4)g=m_(4)a_(4)` ...(vi)
Solving above equation, we get
`a_(1)=a_(2)=a_(3)=0`
and `a_(4) = ((m_(1)+m_(2)-m_(3)-m_(4))g)/(m_(4))`.