Example

Question:

Find the centre of mass of three particles at the vertices of an equilateral triangle.
The masses of the particles are 100 g, 150 g, and 200 g respectively.
Each side of the equilateral triangle is 0.5 m long.

Solution:

With the \(x\)- and \(y\)-axes chosen as shown:
The coordinates of points \(O\), \(A\), and \(B\) forming the equilateral triangle are respectively \((0,0)\), \((0.5,0)\), and \((0.25,\,0.25\sqrt{3})\).
Let the masses 100 g, 150 g, and 200 g be located at \(O\), \(A\), and \(B\) respectively.

The center of mass \(X, Y\) are:

\( X = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} = \frac{100 \times 0 + 150 \times 0.5 + 200 \times 0.25}{100 + 150 + 200} \)
\( \ \ \ = \frac{75 + 50}{450} = \frac{125}{450} = \frac{5}{18}~{\rm m} \)

\( Y = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} = \frac{100 \times 0 + 150 \times 0 + 200 \times 0.25\sqrt{3}}{450} \)
\( \ \ \ = \frac{50\sqrt{3}}{450} = \frac{\sqrt{3}}{9}~{\rm m} \)

Therefore, the centre of mass is at \( \left(\frac{5}{18},\; \frac{\sqrt{3}}{9} \right) \) meters.
Note: The centre of mass is not the geometric centre of the triangle since the masses are different at the three vertices.