Question

Two particles of masses $4\text{kg}$ & $2\text{kg}$ are located as shown in the figure. The position of the centre of mass of the given system is $(\cos {37}^{\circ }=\frac{4}{5})$

• A. $(3,4)$
• B. $\left(\frac{3}{4}\right)$
• C. $(\frac{4}{3},1)$
• D. $(1,\frac{3}{4})$

Answer 1

The correct option is C $(\frac{4}{3},1)$

From the figure,
$x_{\text{2 kg}}=5\cos {37}^{\circ }=5\times \frac{4}{5}=4\text{m}$
$y_{\text{2 kg}}=5\sin {37}^{\circ }=5\times \frac{3}{5}=3\text{m}$

For mass $2\text{k}g:(x_2,y_2)=(4\text{m},3\text{m})$
For mass $4\text{k}g:(x_1,y_1)=(0\text{m},0\text{m})$

Now
$x_{\text{com}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=\frac{4\times 0+2\times 4}{4+2}=\frac{8}{6}=\frac{4}{3}$
$y_{\text{com}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{0+3\times 2}{4+2}=\frac{6}{6}=1$
$\therefore$ Position of $\text{com}=(\frac{4}{3},1)$