Question

Three particle of masses $1kg,2kg$ and $3kg$ are placed at the vertices $A,B$ and $C$ an equilateral triangle $ABC$. If $A$ and $B$ lie at $(0,0)m$ and $(1,0)m$, the co-ordinates of their centre of mass are:

• A. $\left(\frac{\sqrt{3}}{2}m\text{and}\frac{7}{6}m\right)$
• B. $\left(\frac{7}{6}m\text{and}\frac{\sqrt{3}}{4}m\right)$
• C. $\left(\frac{7}{12}m\text{and}\frac{\sqrt{3}}{4}m\right)$
• D. $\left(\frac{7}{12}m\text{and}\frac{7}{12}m\right)$

Answer 1

The correct option is C $\left(\frac{7}{12}m\text{and}\frac{\sqrt{3}}{4}m\right)$

Three particles of masses 1 kg, 2 kg and 3 kg are at the corners of an equilateral triangle of sides 1 m. Find the coordinates of the centre of mass of the system.

Ref. image

$A=1kg$

$B=2kg$

$C=3kg$

In $\Delta ACx$

$\tan 60=\frac{Cx}{Ax}=\frac{Cx}{\frac{1}{2}}=2Cx$

Also $\tan 60=\sqrt{3}$

$\therefore Cx=\frac{\sqrt{3}}{2}$, so we get coordinates of $(\frac{1}{2},\frac{\sqrt{3}}{2})$

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=\frac{1\left(0\right)+2\left(1\right)+3\left(\frac{1}{2}\right)}{1+2+3}=\frac{7}{12}$

$y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}=\frac{1\left(0\right)+2\left(0\right)+3\left(\frac{\sqrt{3}}{2}\right)}{1+2+3}=\frac{\sqrt{3}}{4}$

$\therefore$ coordinates of corf are $(\frac{7}{12},\frac{\sqrt{3}}{4})$