Question

Three identical particles each of mass m are placed at the three corners of an equilateral triangle
"". Find the gravitational force exerted on one body due to the other two.​

Answer 1

Answer:

solved

Explanation:

let side length of triangle = L

F1 = F2 = F = Gm²/L²

Resultant force = √2F²(1 - cos 60) = √F² = F = Gm²/L²

Answer 2

Dear student,

Given :-

• mass of each object = m
• angle between two sides = 60°
• length of side = a

To find :-

• the gravitational force exerted on one body due to the other two

Solution :-

Consider the gravitational force to be applied at any of the body out of the 3 on any of the one. So, currently the gravitational force is being applied at one mass m by two other masses m both at a distance of a at an angle 60°.​

From the concept of vector and finding a resultant vector of two vectors acting at an angle theta, we know that, two vectors acting at angle 60° with a force of 'F' have resultant vector of √3 F.

So, for force₁ applied by mass m₁;

        f₁ = Gm²/a²

Similarly, for force₂ applied by mass m₂;

       f₂ = Gm²/a²

Their resultant is given by;

    Fnet = √3 F

    Fnet = √3 × Gm²/a²

    Fnet = √3 Gm²/a²