What is a Rigid body? Find the Centre of mass of 3 particles at the edge of an equilateral triangle with masses 50g, 75g and 100g respectively, each side of the triangle is 5m long.

Let the side length of the equilateral triangle be $L = 5 \, \text{m}$. We place the first particle at the origin.

• Particle 1 ($m_1 = 50 \, \text{g}$) at $(x_1, y_1) = (0, 0)$.
• Particle 2 ($m_2 = 75 \, \text{g}$) at $(x_2, y_2) = (L, 0) = (5, 0)$.
• Particle 3 ($m_3 = 100 \, \text{g}$) at $(x_3, y_3) = \left(\frac{L}{2}, \frac{L\sqrt{3}}{2}\right) = \left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right)$.

The formula for $X_{CM}$ is $\frac{\sum m_i x_i}{\sum m_i}$.

$$\sum m_i x_i = (50 \, \text{g})(0 \, \text{m}) + (75 \, \text{g})(5 \, \text{m}) + (100 \, \text{g})\left(\frac{5}{2} \, \text{m}\right)$$ $$\sum m_i x_i = 0 + 375 \, \text{g} \cdot \text{m} + 250 \, \text{g} \cdot \text{m} = 625 \, \text{g} \cdot \text{m}$$ $$X_{CM} = \frac{625 \, \text{g} \cdot \text{m}}{225 \, \text{g}} = \frac{25}{9} \, \text{m}$$

The formula for $Y_{CM}$ is $\frac{\sum m_i y_i}{\sum m_i}$.

$$\sum m_i y_i = (50 \, \text{g})(0 \, \text{m}) + (75 \, \text{g})(0 \, \text{m}) + (100 \, \text{g})\left(\frac{5\sqrt{3}}{2} \, \text{m}\right)$$ $$\sum m_i y_i = 0 + 0 + 250\sqrt{3} \, \text{g} \cdot \text{m} = 250\sqrt{3} \, \text{g} \cdot \text{m}$$ $$Y_{CM} = \frac{250\sqrt{3} \, \text{g} \cdot \text{m}}{225 \, \text{g}} = \frac{10\sqrt{3}}{9} \, \text{m}$$

The center of mass is $(X_{CM}, Y_{CM})$.

$$\boxed{\left(\frac{25}{9} \, \text{m}, \frac{10\sqrt{3}}{9} \, \text{m}\right)}$$