Question

Three particles $A,B$ and $C$ of masses $2\text{kg}$, $3\text{kg}$ and $5\text{kg}$ are placed at the vertices of a scalene triangle as shown in the figure. Find the location of $COM$ of the three particle system

• A. $(\frac{4}{5},\frac{9}{10})\text{m}$
• B. $(\frac{9}{10},\frac{4}{5})\text{m}$
• C. $(\frac{4}{5},\frac{4}{5})\text{m}$
• D. $(\frac{9}{10},\frac{9}{10})\text{m}$

Answer 1

The correct option is A $(\frac{4}{5},\frac{9}{10})\text{m}$

Given, masses of particles
$m_1=2\text{kg},m_2=3\text{kg}{m}_3=5\text{kg}$
Position of mass ${}^{\prime }{m}_1^{\prime }=(x_1,y_1)=(4,0)\text{m}$
Position of mass ${}^{\prime }{m}_2^{\prime }=(x_2,y_2)=(0,3)\text{m}$
Position of mass ${}^{\prime }{m}_3^{\prime }=(x_3,y_3)=(0,0)\text{m}$
Therefore, to find the position of COM:
$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{2\left(4\right)+3\left(0\right)+5\left(0\right)}{10}=\frac{4}{5}\text{m}$
$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{2\left(0\right)+3\left(3\right)+5\left(0\right)}{10}=\frac{9}{10}\text{m}$
$\therefore (x_{cm},y_{cm})=(\frac{4}{5},\frac{9}{10})\text{m}$