Question

Three point masses $1\text{kg},1.5\text{kg},2.5\text{kg}$ are placed at the vertices of a triangle with sides $3\text{cm},4\text{cm}$ and $5\text{cm}$ as shown in the figure. The location of the centre of mass with respect to the $1\text{kg}$ mass is:

• A. $0.6\text{cm}$ to the right of $1\text{kg}$ and $2\text{cm}$ above $1\text{kg}$ mass
• B. $0.9\text{cm}$ to the right of $1\text{kg}$ and $2\text{cm}$ above $1\text{kg}$ mass
• C. $0.9\text{cm}$ to the left of $1\text{kg}$ and $2\text{cm}$ above $1\text{kg}$ mass
• D. $0.9\text{cm}$ to the right of $1\text{kg}$ and $1.5\text{cm}$ above $1\text{kg}$ mass

Answer 1

The correct option is B $0.9\text{cm}$ to the right of $1\text{kg}$ and $2\text{cm}$ above $1\text{kg}$ mass

Assume $1\text{kg}$ mass as origin.


As we know,
$X_{\text{cm}}=\frac{m_1\times x_1+m_2\times x_2+m_3\times x_3}{m_1+m_2+m_3}$
$\Rightarrow X_{\text{cm}}=\frac{1\times 0+1.5\times 3+2.5\times 0}{5}=0.9\text{cm}$
and,
$Y_{\text{cm}}=\frac{m_1\times y_1+m_2\times y_2+m_3\times y_3}{m_1+m_2+m_3}$
$\Rightarrow Y_{\text{cm}}=\frac{1\times 0+1.5\times 0+2.5\times 4}{5}=2\text{cm}$
Hence, center of mass of the system will be $0.9\text{cm}$ to the right of $1\text{kg}$ and $2\text{cm}$ above $1\text{kg}$ mass.