In the shown arrangement, both pulleys and the string are massless and all the surfaces are frictionless. Find the acceleration of the wedge.

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Solution

F.B.D. of the system is shown in the fig.
So, according to the fig.
Let the acceleration of the wedge of mass $m_{1}$ be $a$
So, acceleration of mass $m_{3}$ will be $a$ in the vertical direction.
Writing equilibrium forces in the horizontal direction for $m_{1} a n d m_{2}$ will be respectively
$T - N = m_{1} a ⟶ (e q u a t i o n 1) N = m_{2} a ⟶ (e q u a t i o n 2)$
Now writing equation of forces in the vertical direction for $m_{2} {a n d m}_{3}$ we have
$T - m_{2} g = m_{2} a ⟶ (e q u a t i o n 3) m_{3} g - T = m_{3} a ⟶⟶ (e q u a t i o n 4) s o l v i n g t h i s w e h a v e a = \frac{(m_{3} - m_{2}) g}{(m_{2} + m_{3})}$
which is the acceleration of the wedge.

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