Question

Locate the centre of mass of a system of three particles of masses 1 gram 2 gram 3 gram placed at the corners of an equilateral triangle of 1m side

Answer 1

Let two of the equilateral triangle coordinates are as follow

B (-1,0) and C (1,0).  

Since all sides are same, we can easily derive that the A coordinates are (0, √(3)).

The X and Y coordinates of point A, B, and C forming the equilateral triangle are respectively (0,√(3)), (-1,0), and (1,0).  

Let the masses 1g, 2g, and 3g be located at A, B and C respectively.

(x1, y1) = (0, √(3), (x2, y2) = (-1,0) and (x3, y3) = (1,0)

m1 = 1, m2 = 2 and m3 = 3

The location of center mass can be calculated as follow

X coordinate is calculated as  

x = (m1*x1 + m2*x2 + m3*x3)/(m1 + m2 + m3)

Substituting we get,

x = (1*0 + 2*-1 + 3*1)/(1 + 2 + 3) = 1/6

Y coordinate is calculated as    

y = (m1*y1 + m2*y2 + m3*y3)/(m1 + m2 + m3)

Substituting we get,

y = (1*√(3) + 2*0 + 3*0)/(1 + 2 +3) = √(3)/6  

Thus the location for center of mass is (1/6, √(3)/6)