# 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 ?

## The Answer provided to me is $\frac{\sqrt{19}}{6}$ please i want a proof to this answe , altough i m getting a different one .

Apr 23, 2017

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

$\overline{x} = \frac{\sum {m}_{i} {x}_{i}}{\sum {m}_{i}}$

$= \frac{1 \cdot \frac{\sqrt{3}}{2} + 2 \cdot 0 + 3 \cdot 0}{6} = \frac{\sqrt{3}}{12}$

$\overline{y} = \frac{\sum {m}_{i} {y}_{i}}{\sum {m}_{i}}$

$= \frac{3 \cdot \frac{1}{2} - 2 \cdot \frac{1}{2} + 1 \cdot 0}{6} = \frac{1}{12}$

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

$d = \sqrt{{\left(O A - \overline{x}\right)}^{2} + {\left(\overline{y}\right)}^{2}}$

$= \sqrt{{\left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{12}\right)}^{2} + {\left(\frac{1}{12}\right)}^{2}}$

$= \frac{\sqrt{19}}{6}$