Question

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.

• A. = 0 = 0 = 0 upwards
• B. = 0 = 0 = 0 upwards
• C. = g downloads = g downloads = g upwards upwards
• D. = g uploads = g uploads = g downwards downwards

Answer 1

The correct option is B

= 0

= 0

= 0

upwards

Free body drawing

Before cutting

$K{x}_1=m_{2}g$ -------------(1)

$T_1$ = $m_1$g + k$x_1$

$T_1$ = ($m_1$ + $m_2$)g (from eq (1)) ...(ii)

From eq (ii)

$T_1$ = k$x_2$ + $m_3$g

k$x_2$ = ($m_1$ + $m_2$ - $m_3$)g ...(iii)

$T_2$ + $m_4$g = k$x_2$

From eg (iii) $T_2$ = ($m_1$ + $m_2$ - $m_3$ - $m_4$)g ...(iv)

Now when the lower string is cut it slogs and $T_2$ becomes 0

So now there is a greater force in upward direction so the block will start going up. Lets see

k$x_2$ - $m_4$g = $m_4$$a_4$

From (iii) ($m_1$ + $m_2$ - $m_3$ - $m_4$g = $m_4$ $a_4$)

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

Spring force is not instantaneously affected by the cutting of down string (as spring will still have the elongation) so on $m_3$ downward pull is maintained and hence the Tension in the upward string will be same.

This means $m_3$ will remain in equilibrium.

So $a_3$ = 0

Similarly,

Nothing is changed just after cutting the string

$\Rightarrow$ $a_2$ = $a_1$ = 0