Question

Three particles of mass $m_1=1\text{kg}$, $m_2=2\text{kg}$, $m_3=3\text{kg}$ are placed at the corners $A,B,C$ of an equilateral triangle, respectively. The edge length of the triangle is $1\text{m}$. Find the distance of $\text{COM}$ of the system of particles from corner $A$.

• A. $\frac{\sqrt{11}}{9}\text{m}$
• B. $\frac{\sqrt{19}}{6}\text{m}$
• C. $\frac{\sqrt{15}}{7}\text{m}$
• D. $\frac{\sqrt{14}}{6}\text{m}$

Answer 1

The correct option is B $\frac{\sqrt{19}}{6}\text{m}$


Since masses $m_1,m_2,m_3$ are placed at corners $A,B,C$ respectively, the coordinates of the masses as per the axes chosen and taking origin at A$(0,0)$, are given by:
$m_1\Rightarrow (0\text{m},0\text{m})$
$m_2\Rightarrow (\frac{1}{2}\text{m},\frac{\sqrt{3}}{2}\text{m})$
$m_3\Rightarrow (1\text{m},0\text{m})$

Applying the formula for $\text{COM}$ of system:
$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$x_{\text{CM}}=\frac{1\times 0+2\times \left(0.5\right)+3\times 1}{1+2+3}=\frac{2}{3}\text{m}$
Similarly,
$y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$y_{\text{CM}}=\frac{1\times 0+2\times \left(\frac{\sqrt{3}}{2}\right)+3\times 0}{1+2+3}=\frac{\sqrt{3}}{6}\text{m}$

Distance of $\text{COM}$ of system from corner $A$ will be:
$d=\sqrt{(x_{\text{CM}}{)}^2+(y_{\text{CM}}{)}^2}$
$d=\sqrt{{\left(\frac{2}{3}\right)}^2+\left(\frac{\sqrt{3}}{6}\right){)}^2}$
$\therefore d=\frac{\sqrt{19}}{6}\text{m}$