Laws of Motion Part 1

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=

√3

9

g
Along the inclined plane,
mg‌sin‌30°+ma2‌cos‌30°=ma1
8g×

1

2

+8×

√3

9

g×

√3

2

=8a1
⇒ a1=

2g

3

∴ The acceleration of the block with respect to the wedge is

2g

3

.