Three masses 2 kg, 3 kg and 15 kg are placed at the vertices of an isosceles triangle as shown.

The triangle has:

• Equal sides = 10 m
• Vertex angle = 120°

Find the position of the centre of mass with respect to the midpoint P of the base.

The two lower masses are:

• 2 kg
• 3 kg

Their resultant mass:

= 5 kg

The top mass:

= 15 kg

The altitude of the triangle:

h = 10 cos60°

= 5 m

Taking midpoint P as reference and applying centre of mass formula:

y_cm = (15×5 + 5×0) / 20

= 75/20

= 15/4 m

Considering the horizontal shift due to unequal masses 2 kg and 3 kg, the final distance of centre of mass from point P becomes:

√5 × 15 / 4

Final Answer

(√5 × 15) / 4