Here, both block and wedge are moving.
Consider the acceleration of the block with respect to the wedge is
a1 and the acceleration of the wedge is
a2.
Given, mass of the wedge, M = 16 kg
and mass of block, m = 8 kg
Let’s draw the free body diagram of the wedge,
In the x-directions,
N cos60º =
Ma2 N(0.5)=16×a2 ⇒
N=32a2 Now, draw the free body diagram of the block with respect to thewedge.
Along the perpendicular to the inclined plane,
N=8gcos30°−8a2sin30º 32a2=4√3g−4a2 36a2=4√3g ⇒
a2=g Along the inclined plane,
mgsin30°+ma2cos30°=ma1 8g×+8×g×=8a1 ⇒
a1= ∴ The acceleration of the block with respect to the wedge is
.