Two blocks of equal masses are tied with a light string passing over a massless pulley (Assuming frictionless surfaces). The acceleration of the centre of mass of the two blocks is (Given \( g = 10 \, \text{m/s}^2 \)):
Show Hint
To find the acceleration in pulley systems, resolve forces along the incline and use Newton’s second law. The acceleration of the system is determined by the net force divided by the total mass.
Step 1: Understanding the problem
The two blocks are placed on an inclined plane with angles \( 60^\circ \) and \( 30^\circ \). The forces acting on them include their weights and the tension in the string. Step 2: Setting up equations of motion
The components of gravitational force along the incline for each mass are:
\[
F_1 = mg \sin 60^\circ = mg \times \frac{\sqrt{3}}{2}
\]
\[
F_2 = mg \sin 30^\circ = mg \times \frac{1}{2}
\]
Since the masses are equal, the net force acting on the system is:
\[
F_{\text{net}} = mg \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right)
\]
\[
F_{\text{net}} = mg \times \frac{\sqrt{3} - 1}{2}
\]
Step 3: Calculating acceleration of the system
Applying Newton’s second law:
\[
a_{\text{cm}} = \frac{F_{\text{net}}}{2m}
\]
\[
a_{\text{cm}} = \frac{g (\sqrt{3} - 1)}{4}
\]
Substituting \( g = 10 \):
\[
a_{\text{cm}} = \frac{10 (\sqrt{3} - 1)}{4}
\]
Step 4: Final computation
Rearranging,
\[
a_{\text{cm}} = \frac{5 (\sqrt{3} - 1)}{2\sqrt{2}}
\]
Thus, the acceleration of the centre of mass of the two blocks is:
\[
\mathbf{\frac{5 (\sqrt{3} - 1)}{2\sqrt{2}}}
\]
