Center of Mass Position of Three Masses System

To find the position of the center of mass of the given three-mass system, we first determine the geometric properties of the triangle formed by the masses.

Step 1: Geometry of the System

• Mass $m_1 = 2\text{ kg}$ is located at the bottom-left vertex.
• Mass $m_2 = 3\text{ kg}$ is located at the bottom-right vertex.
• Mass $m_3 = 15\text{ kg}$ is located at the top vertex.

From the figure:

• The triangle is isosceles with equal side lengths $l = 10\text{ m}$.
• The apex angle at the top vertex is $120^\circ$.

The length of the normal bisector (altitude $h$) from the top vertex to the base is: $h = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5\text{ m}$

The half-length of the base ($d$) is: $d = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\text{ m}$

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Step 2: Coordinates with respect to Point $p$

Point $p$ is the midpoint of the normal bisector (altitude). Since the total height of the altitude is $5\text{ m}$, the point $p$ is located at a vertical distance of $2.5\text{ m}$ from both the top vertex and the midpoint of the base.

Taking $p(0,0)$ as the origin, with the $x$-axis parallel to the base (pointing right) and the $y$-axis along the normal bisector (pointing upwards):

• Position of $m_3 = 15\text{ kg}$: $(x_3, y_3) = (0, 2.5)\text{ m}$
• Position of $m_1 = 2\text{ kg}$: $(x_1, y_1) = (-5\sqrt{3}, -2.5)\text{ m}$
• Position of $m_2 = 3\text{ kg}$: $(x_2, y_2) = (5\sqrt{3}, -2.5)\text{ m}$

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Step 3: Calculation of Center of Mass

The total mass of the system is: $M = m_1 + m_2 + m_3 = 2 + 3 + 15 = 20\text{ kg}$

1. $X$-coordinate of Center of Mass ($X_{\text{cm}}$): $X_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}$ $X_{\text{cm}} = \frac{2(-5\sqrt{3}) + 3(5\sqrt{3}) + 15(0)}{20}$ $X_{\text{cm}} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{5\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$

2. $Y$-coordinate of Center of Mass ($Y_{\text{cm}}$): $Y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}$ $Y_{\text{cm}} = \frac{2(-2.5) + 3(-2.5) + 15(2.5)}{20}$ $Y_{\text{cm}} = \frac{-5 - 7.5 + 37.5}{20} = \frac{25}{20} = 1.25$

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Conclusion

The position of the center of mass with respect to point $p$ is: $\left(\frac{\sqrt{3}}{4}, 1.25\right)$

This matches Option A.