Question

three point of masses of 1 kg ,2 kg and 3 kg lie at (1,2 ) ,(0,-1) and (2,-3) respectively calculat co ordinates of the centre of mass of the system


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Answer 1

Answer:

Here `m_(1) = 1 kg, m_(2) = 2 kg, m_(3) = 3 kg`

`x_(1) = 1, y_(1) = 2 , x_(2) = 0, y_(2) =- 1`,

`x_(3) = 2, y_(3) =- 3`

Co-ordinates of centre of mass `(x,y)` are given

by `x = (m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3))/(m_(1) + m_(2) + m_(3))`

`= (1(1) + 2(0) + 3(2))/(1 + 2+ 3) = (7)/(6)`

`y = (m_(1)y_(1) + m_(2)y_(2) + m_(3)x_(3))/(m_(1) + m_(2) + m_(3))`

`= (1(2) + 2(-1) + 3(-3))/(1 + 2+ 3) = (-3)/(2)`

i think this is the answer you were surching