Question

Four point masses $2\text{kg},3\text{kg},5\text{kg}$ and $7\text{kg}$ respectively are placed at the four corners $A,B,C$ and $D$ of a square of side $4\text{m}$. The position of $COM$ will be

• A. $(\frac{48}{17},\frac{32}{17})\text{m}$
• B. $(\frac{32}{17},\frac{48}{17})\text{m}$
• C. $(2,4)$
• D. $(2,2)$

Answer 1

The correct option is B $(\frac{32}{17},\frac{48}{17})\text{m}$

Given, $m_1=2\text{kg},m_2=3\text{kg},m_3=5\text{kg},m_4=7\text{kg}$
Position of mass $m_1(x_1,y_1)=(0,0)$
Position of mass $m_2(x_2,y_2)=(4,0)$
Position of mass $m_3(x_3,y_3)=(4,4)$
Position of mass $m_4(x_4,y_4)=(0,4)$
$x-$ coordinate of $COM$: $x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$=\frac{2\times 0+3\times 4+5\times 4+7\times 0}{2+3+5+7}=\frac{32}{17}\text{m}$
$y-$ coordinate of $COM$: $y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$
$=\frac{2\times 0+3\times 0+5\times 4+7\times 4}{2+3+5+7}=\frac{48}{17}\text{m}$