Three particles of mass 1,2,3 Kg are placed at the corners A , B , C respectively of an equilateral triangle ABC of edge 1 m .Find distance of there center of masses from A ?

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The Answer provided to me is #sqrt(19)/6# please i want a proof to this answe , altough i m getting a different one .

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Answer 1 · 2017-04-23T08:27:17

The answer is correct.

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Placing the triangle in a convenient way, we take moments about the y-axis:

#bar x = (sum m_i x_i)/(sum m_i)#

#= (1 cdot sqrt3/2 + 2 cdot 0 + 3 cdot 0)/6 = sqrt3/12#

Taking moments about the x-axis:

#bar y = (sum m_i y_i)/(sum m_i)#

#= (3 cdot 1/2 - 2 cdot 1/2 + 1 cdot 0)/6 = 1/12#

Now distance #d# of the CoM from A is:

#d = sqrt((OA- bar x)^2 + (bar y)^2)#

#= sqrt((sqrt3/2- sqrt3/12)^2 + (1/12)^2)#

#= sqrt19/6#