Four blocks and two springs are arranged as shown in $F i g . 6.213$ the system at rest, determine the acceleration of all the loads immediately 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.

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Solution

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. For the equilibrium of load $m_{3}$ , we have
$m_{3} g + T_{2} = T_{0}$ .........(i)
For $m_{1}, m_{2} g + T_{1} = T_{0}$
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)
$T_{2} + m_{3} g - T_{0} = m_{3} a_{2}$ ........(v)
and $T_{2} - m_{34} g = m_{4} a_{4}$ ......(vi)
Solving above equations, we get
$a_{1} = a_{2} = a_{3} = 0$
and $a_{4} = \frac{(m_{3} + m_{4} - m_{1} m_{2}) g}{m_{4}}$

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Four blocks and two springs are arranged as shown in figure. The system is at rest. Determine the acceleration of all the loads immediately after the lower thread keeping the system in equilibrium has been cut. Assume that the thread and pulleys are weightless and there is no friction at the point of suspension.