Question

Three particles of masses 1.0 kg , 2.0 kg and 3.0 kg are placed at the corners A , B and C respectively of an equilateral triangle ABC of edge 1 m . Locate the centre of mass of the system.

Answer 1

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ANSWER::

Given:-

m₁=1 kg   m₂=2 kg   m₃= 3kg
x₁=0         x₂=1         x₃= 1/2
y₁=0         y₂=0         y₃=√3/2

So , the position of centre of mass is =
 [(m₁x₁+m₂x₂+m₃x₃)/(m₁+m₂+m₃) , (m₁y₁+m₂y₂+m₃y₃)/(m₁+m₂+m₃)]

=[(1x0+2x1+3x1/2)/(1+2+3) , (1x0+2x0+3x√3/2)/(1+2+3)]

=(7/12,3√3/12) from point B.

Hope it helps!

Answer 2

Answer:

Explanation:

Given:-

m₁=1 kg   m₂=2 kg   m₃= 3kg

x₁=0         x₂=1         x₃= 1/2

y₁=0         y₂=0         y₃=√3/2

So , the position of centre of mass is =

 [(m₁x₁+m₂x₂+m₃x₃)/(m₁+m₂+m₃) , (m₁y₁+m₂y₂+m₃y₃)/(m₁+m₂+m₃)]

=[(1x0+2x1+3x1/2)/(1+2+3) , (1x0+2x0+3x√3/2)/(1+2+3)]

=(7/12,3√3/12) from point B