Three balls, of mass m, 5m, and 3m, are at the vertices of an equilateral triangle with side lengths equal to L. What is the magnitude of the force on mass m from the two other masses, in terms of G, m, and L

F_{total}= 8.25 ,      θ’= 194º

Explanation:

To solve this problem we must use the law of universal gravitation and vectorly add the forces

         F =

Let us call the mass m with the subscript 1, the mass 3m with the subscript 3 and the mass 5m with the subscript 5, the total force on particle 1 is

          F_total = F₁₅ + F₁₃

The bold are vectors, in the attachment we can see a diagram of the angles and the forces, the distance between the masses is

           r = L

let's find the force between m1 and m5

           F₁₅ = G m₁ m₅ / r²

           F₁₅ = G m 5m / L²

           F₁₅ = G 5m² / L²

this force is on the line that joins the two masses, let's use trigonometry to decompose this force

           cos 30 = F₁₅ₓ / F₁₅

           sin 30 =

           F₁₅ₓ = F₁₅ cos 30

            = F₁₅ sin 30

equally with the force between mass 1 and mass 3

            F₁₃ = -G 3 m² / L²

            F₁₃ₓ = F₁₃ cos 30

            F_{13y} = F₁₃ sin 30

to find the total force we can add each component independently, see attached

X axis

           F_total x = -F₁₅ₓ + F₁₃ₓ

           F_total x = -G 5m2 / L² + G 3m² / L²

           F_total x = - G 2m² / L²

Y axis  

           F_total y = - F_{15y} - F_{13y}

           F_total y = - G 5m² / L² - G 3 m² / L²

           F_toal y = - G 8 m² / L²

We can give the result in two ways

1) F_total = - G m ^ 2 / L² (2 i ^ + 8 j ^)

2) in the form of module and angle.

Let's use the Pythagorean theorem

        =

       F_{total}= 8.25

with trigonometry

        tan θ =

        tan θ = 8/2

        θ = tan⁻¹ 4

        θ = 76º

if we measure this angle from the positive side of the x axis in a counterclockwise direction

       θ’= 270 -76

       θ’= 194º

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