The position of center of mass of three masses \(2\) kg, \(3\) kg and \(15\) kg placed with respect to mid point (\(p\)) of normal bisector, as shown in the figure is _______.

Let the midpoint \(p\) of the normal bisector be the origin \((0,0)\).

The triangle is isosceles with two sides of \(10\text{ m}\) and an included angle of \(120^\circ\). The altitude from the \(15\text{ kg}\) mass bisects the angle into two \(60^\circ\) angles.

The length of the altitude is \(10 \cos(60^\circ) = 5\text{ m}\).

Since \(p\) is the midpoint of this altitude, the \(15\text{ kg}\) mass is at a distance of \(2.5\text{ m}\) above \(p\), and the base is \(2.5\text{ m}\) below \(p\).

The distance of the \(2\text{ kg}\) and \(3\text{ kg}\) masses from the altitude is \(10 \sin(60^\circ) = 5\sqrt{3}\text{ m}\).

The coordinates of the masses with respect to \(p\) are:
\(m_1 = 15\text{ kg}\) at \((0, 2.5)\)
\(m_2 = 2\text{ kg}\) at \((-5\sqrt{3}, -2.5)\)
\(m_3 = 3\text{ kg}\) at \((5\sqrt{3}, -2.5)\)

The x-coordinate of the center of mass is:
\(x_{cm} = \dfrac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \dfrac{15(0) + 2(-5\sqrt{3}) + 3(5\sqrt{3})}{15 + 2 + 3} = \dfrac{5\sqrt{3}}{20} = \dfrac{\sqrt{3}}{4}\)

The y-coordinate of the center of mass is:
\(y_{cm} = \dfrac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \dfrac{15(2.5) + 2(-2.5) + 3(-2.5)}{20} = \dfrac{37.5 - 12.5}{20} = \dfrac{25}{20} = 1.25\)

The position of the center of mass is \(\left(\dfrac{\sqrt{3}}{4}, 1.25\right)\).

Answer: \(\left(\dfrac{\sqrt{3}}{4}, 1.25\right)\)