Question

Figure-4.10 shows a fixed wedge on which two blocks of masses 2 kg and 3 kg are placed on its smooth inclined surfaces. When the two blocks are released from rest, find the acceleration of centre of mass of the two blocks.

Answer 1

The acceleration of the center of mass of the two blocks is $\frac{7g}{10}$.

Answer 2

1. Calculate individual accelerations down the smooth inclines: $a_1 = g \sin(30^\circ) = g/2$ and $a_2 = g \sin(60^\circ) = \sqrt{3}g/2$.
2. Define a horizontal (x) and vertical (y) coordinate system. Assume inclines slope downwards to the right, so acceleration vectors are at angles $-30^\circ$ and $-60^\circ$ with the horizontal.
3. Express accelerations as vectors: $\vec{a}_1 = \frac{g}{2}(\cos(-30^\circ)\hat{i} + \sin(-30^\circ)\hat{j})$ and $\vec{a}_2 = \frac{\sqrt{3}g}{2}(\cos(-60^\circ)\hat{i} + \sin(-60^\circ)\hat{j})$.
4. Calculate the acceleration of the center of mass using $\vec{a}_{CM} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}$.
5. Compute the magnitude of $\vec{a}_{CM}$.