Question

Three particles of masses 1.0 kg and 2.0 kg and 3.0 kg are placed at the corners A,B and C respectively of an equilateral triangle ABC of edge 1 m.Locate the centre of mass of the system.

Answer 1

$m_1$=1 kg,

$m_2$=2 kg,

$m_3$=3 kg

$x_1=0$,

$x_2=1,x_3=\frac{1}{2}$

$y_1=0,y_2=0,y_3=\frac{\sqrt{3}}{2}$

The position of centre of mass is

$C.M=\frac{(m_1{x}_2+m_2{x}_2+m_3{x}_3)}{m_1+m_2+m_3},\frac{(m_1{y}_2+m_2{y}_2+m_2{y}_3)}{m_1+m_2+m_3}$

$=(\frac{(1\times 0)+(2\times 1)+(3\times \frac{1}{2})}{1+2+3},\frac{(1\times 0)+(2\times 0)+(3\times \frac{\sqrt{3}}{2})}{1+2+3})$

$=(\frac{7}{12},\frac{\sqrt{3}}{4})$ from the point B.