Question

6. Three identical particles each of mass 2 kg are placed at
the vertex of an equilateral triangle. If each side of
triangle is 10 cm then find the resultant gravitational
force acting on either one particle.​

Answer 1

Given info :Three identical particles each of mass 2 kg are placed at the vertex of an equilateral triangle. each side of triangle is 10 cm.

To find : The resultant gravitational force acting on either one particle..

solution : force between two particle is given by, F = GM²/a²

where M is mass of particles , a is side length of triangle.

now two force acting on the third particle, make 60° with each other.

so, net force, Fnet = √{F² + F² + 2F × F cos60°}

= √{F² + F² + F²} = √3F

= √3GM²/a²

here G = 6.67 × 10¯¹¹ Nm²/Kg² , M = 2kg , a = 10cm = 0.1 m

so, Fnet = (√3 × 6.67 × 10¯¹¹ × 2²)/(0.1²)

= (1.732 × 6.67 × 4) × 10^-9 N

= 46.2 × 10^-9 N

= 4.62 × 10^-8 N

Therefore the force acting on a particle by the other two is 4.62 × 10^-8 N