Three equal masses of 3 kg each are placed at the vertices of an equilateral triangle of side 1m. The gravitation force in newton, acting on the mass of 4 kg kept at centroid is

1. Positioning of Masses

The vertices of the equilateral triangle are A, B, and C, each having a mass m = 3 kg. The centroid G of the triangle is equidistant from all three vertices.

2. Distance from Centroid to Vertices

The distance from the centroid to any vertex of an equilateral triangle is given by:

r = side length / sqrt(3) = 1 / sqrt(3) m.

3. Gravitational Force Due to One Mass

The gravitational force between the 4 kg mass at G and one 3 kg mass at a vertex is given by Newton's law of gravitation:

F = G * m1 * m2 / r^2

Where:

• G = 6.67 × 10-11 N·m2/kg2
• m1 = 4 kg, m2 = 3 kg
• r = 1 / sqrt(3) m

Substituting the values:

F = (6.67 × 10-11)(4)(3) / (1 / sqrt(3))2

Simplify:

F = (6.67 × 10-11)(12) / (1/3) = (6.67 × 10-11)(12)(3) = 2.4 × 10-9 N

4. Net Gravitational Force at Centroid

The gravitational forces due to the three masses are equal in magnitude and directed toward their respective vertices. Since the centroid is equidistant from the three masses, the forces form a symmetric system. Their vector sum cancels out due to symmetry.

The net gravitational force acting on the 4 kg mass at the centroid is zero.