Three particles of masses `m_1 = 1 kg, m_2 = 2 kg " and " m_2 = 3 kg` are placed at the corners of an equilateral triangle of side 1m as shown in Figure. Find the position of center of mass.

Explanation

The center of mass of an equilateral triangle lies at its geometrical center G. The positions of the mass `m_1, m_2` and `m_3` are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses m, and m, are easily marked as (0,0) and (1,0) respectively. To find the position of m, the Pythagoras theorem is applied. As the `Delta ` DBC is a right angle triangle,
` BC^2 = CD^2 + DB^2`
` CD^2 = BC^2 - DB^2`
` CD^2 = 1^2 - (1/2)^2 = 1 - (1/4) = 3/4`
` CD = (sqrt3)/(2)`
The position of mass `m_3` is ` ( 1/2, (sqrt3)/(2) )` " or " `(0.5 , 0.5 sqrt3) `
X coordinate of centre of mass ,
` v_(CM) = (m_1y_2 + m_2y_2 + m_3y_3)/(/(m_1 + m_2 + m_3)`
` v_(CM) = (sqrt3)/(4) m `
` therefore ` the coordinates of centre of mass G ` (x_(CM). y_(CM))` is `(7/12, (sqrt3)/(4) )`