Solution:
Due to perfect symmetric distribution, the three gravitational pull forces on the mass at the centroid are equal in magnitude and separated by \(120^circ\), resulting in a net vector sum of zero.
Three equal masses of \(3\text{ kg}\) each are fixed at the vertices of an equilateral triangle ABC. The gravitational force acting on mass \(2\text{ kg}\) placed at the centroid of triangle is
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