JEE Main 2026 — Physics Mechanics

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

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

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

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

The distance of the $2\text{ kg}$2 kg and $3\text{ kg}$3 kg masses from the altitude is $10 \sin(60^\circ) = 5\sqrt{3}\text{ m}$10sin(60∘)=53​ m.

The coordinates of the masses with respect to $p$p are:

$m_1 = 15\text{ kg}$m1​=15 kg at $(0, 2.5)$(0,2.5)

$m_2 = 2\text{ kg}$m2​=2 kg at $(-5\sqrt{3}, -2.5)$(−53​,−2.5)

$m_3 = 3\text{ kg}$m3​=3 kg at $(5\sqrt{3}, -2.5)$(53​,−2.5)

The x-coordinate of the center of mass is:

$$x_{cm} = \dfrac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \dfrac{15(0) + 2(-5\sqrt{3}) + 3(5\sqrt{3})}{15 + 2 + 3} = \dfrac{5\sqrt{3}}{20} = \dfrac{\sqrt{3}}{4}$$xcm​=m1​+m2​+m3​m1​x1​+m2​x2​+m3​x3​​=15+2+315(0)+2(−53​)+3(53​)​=2053​​=43​​

The y-coordinate of the center of mass is:

$$y_{cm} = \dfrac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \dfrac{15(2.5) + 2(-2.5) + 3(-2.5)}{20} = \dfrac{37.5 - 12.5}{20} = \dfrac{25}{20} = 1.25$$ycm​=m1​+m2​+m3​m1​y1​+m2​y2​+m3​y3​​=2015(2.5)+2(−2.5)+3(−2.5)​=2037.5−12.5​=2025​=1.25

The position of the center of mass is $\left(\dfrac{\sqrt{3}}{4}, 1.25\right)$(43​​,1.25).

Answer: $\left(\dfrac{\sqrt{3}}{4}, 1.25\right)$(43​​,1.25)