Laws of Motion Part 1

Here, both block and wedge are moving. Consider the acceleration of the block with respect to the wedge is $a_1$ and the acceleration of the wedge is $a_2$. 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 = $Ma_2$ $N(0.5)=16 \times a_{2}$ ⇒ $N=32 a_{2}$ Now, draw the free body diagram of the block with respect to thewedge. Along the perpendicular to the inclined plane, $N = 8g \cos 30^{\circ} - 8a_2 \sin 30^{\circ}$ $32 a_{2}=4 \sqrt{3} g - 4 a_{2}$ $36 a_{2}=4 \sqrt{3} g$ ⇒ $a_{2}=\frac{\sqrt{3}}{9} g$ Along the inclined plane, $m g \sin 30^{\circ} + m a_{2} \cos 30^{\circ} = m a_{1}$ $8 g \times \frac{1}{2} + 8 \times \frac{\sqrt{3}}{9} g \times \frac{\sqrt{3}}{2} = 8 a_{1}$ ⇒ $a_{1}=\frac{2 g}{3}$ ∴ The acceleration of the block with respect to the wedge is $\frac{2g}{3}$.