Question

Three particles $\text{A, B}$ & $\text{C}$ of masses $4m,2m$ & $3m$ are placed on the vertices of a equilateral triangle of length $2L$ as shown in the figure.

Find out the perpendicular distance of centre of mass from the side $\text{AB}$.

• A. $\frac{L}{\sqrt{3}}$
• B. $\sqrt{3}L$
• C. $\frac{\sqrt{3}L}{2}$
• D. $\frac{\sqrt{3}L}{4}$

Answer 1

The correct option is A $\frac{L}{\sqrt{3}}$

Let us consider the $x-y$ plane and vertex of equilateral triangle at origin and side AB along $x$ axis. The coordinates of the vertices of the triangle will be as shown in figure.

From the figure, we can conclude that the perpendicular distance of $COM$ from side $AB$ is $y$ co-ordinate of $COM$.
therefore, $y_{com}=\frac{y_A{m}_A+y_B{m}_B+y_C{m}_C}{4m+2m+3m}$
$=\frac{0+0+\sqrt{3}L\times 3m}{9m}=\frac{3\sqrt{3}L}{9}=\frac{L}{\sqrt{3}}$
$⟹$Required distance is $\frac{L}{\sqrt{3}}$