Question

Three particles a b and c each of mass m are lying at the corners of equilateral triangle of side l if the particle is released keeping the particles b and c fixed the magnitude of instantaneous acceleration of a east

Answer 1

The gravitational force from Point A to point B:

F₁ = G(m)(m)/L²

= Gm²/L²

The gravitational force from Point A to point C:

F₂ = G(m)(m)/L²

= Gm²/L²

Now we will look at the magnitudes of F₁ and F₂

Let F₁ =  F₂ = F

Angle is 60° which is between F₁  and F₂

Now we will calculate the net force on the particle A:

F base net = √F² + F² + 2(F) (F) cos 60°

= √3F²

= √3F

Now we will calculate the acceleration for the particle A:

a = F base net/ m

= √3F/m

= √3/m (Gm²/L²)

= √3Gm/ L²

Answer 2

here is the answer ..