Question

Three particles of masses $1\text{kg},2\text{kg}$ and $3\text{kg}$ are situated at the corners of an equilateral triangle of side $b$. The $(x,y)$ coordinates respectively for the centre of mass of the system of particles will be:

• A. $[\frac{7b}{12},\frac{\sqrt{3}b}{12}]$
• B. $[\frac{3\sqrt{3}b}{12},\frac{7b}{12}]$
• C. $[\frac{7b}{12},\frac{3\sqrt{3}b}{12}]$
• D. $[\frac{7b}{12},\frac{3\sqrt{3}b}{4}]$

Answer 1

The correct option is C $[\frac{7b}{12},\frac{3\sqrt{3}b}{12}]$

The coordinates of points A, B and C are
$A\to (0,0)$
$B\to (b,0)$
$C\to (\frac{b}{2},b\sin {60}^{\circ })=(\frac{b}{2},\frac{\sqrt{3}b}{2})$


Applying the formula for coordinates of $COM$ of system:
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$x_{\text{CM}}=\frac{(1\times 0)+(2\times b)+(3\times \frac{b}{2})}{1+2+3}=\frac{7b}{12}$
Simillarly,
$y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$y_{\text{CM}}=\frac{(1\times 0)+(2\times 0)+(3\times \frac{\sqrt{3}b}{2})}{1+2+3}=\frac{3\sqrt{3}b}{12}$
So, the coordinates of the centre of mass are $\left[\begin{array}{c}\frac{7b}{12},\frac{3\sqrt{3}b}{12}\end{array}\right]$