Question

Four bodies of masses $2\text{kg},3\text{kg},5\text{kg}$ and $8\text{kg}$ respectively are placed at the four corners of a square of side $2\text{m}$. Then, position of centre of mass of the system will be:

• A. $(\frac{11}{9},\frac{13}{9})$
• B. $(\frac{11}{9},\frac{7}{9})$
• C. $(\frac{13}{9},\frac{11}{9})$
• D. $(\frac{7}{5},\frac{11}{9})$

Answer 1

The correct option is A $(\frac{11}{9},\frac{13}{9})$


Let us place the masses at the corners of a square $ABCD$ of side $2\text{m}$, assuming origin ($0,0$) at $A$

Writing the $x$ and $y$ coordinates of the bodies as shown below in the table:

$\begin{array}{cccc}& \text{mass (kg)}& x\left(\text{m}\right)& y\left(\text{m}\right)\\ A& 2& 0& 0\\ B& 3& 2& 0\\ C& 5& 0& 2\\ D& 8& 2& 2\end{array}$

From the formula for $x$ and $y$ coordinates of $\text{CM}$:
$x_{\text{CM}}=\frac{m_A{x}_A+m_B{x}_B+m_C{x}_C+m_D{x}_D}{m_A+m_B+m_C+m_D}$
$=\frac{(2\times 0)+(3\times 2)+(5\times 0)+(8\times 2)}{2+3+5+8}=\frac{22}{18}$
$\therefore x_{\text{CM}}=\frac{11}{9}$

Simillarly;
$y_{\text{CM}}=\frac{m_A{y}_A+m_B{y}_B+m_C{y}_C+m_D{y}_D}{m_A+m_B+m_C+m_D}$

$=\frac{\left(0\right)+\left(0\right)+(5\times 2)+(8\times 2)}{2+3+5+8}=\frac{13}{9}$

$\therefore (x_{\text{CM}},y_{\text{CM}})=(\frac{11}{9},\frac{13}{9})$