Calculating Net Gravitational Force in Equilateral Triangle Configuration

[QuantumKnight]

Homework Statement

Three point particles are fixed in place in the xy plane. The three partiles sit on the
corners of an equilateral triangle with sides of length a = 2.50 mm. Particle 1 has a mass m1 = 12.0 kg,
particle 2 has a mass m2 = 18.0 kg, and particle 3 has a mass m3 = 15.0 kg.

1. What is the magnitude and direction of the net gravitational force exerted on particle 1 by the other
two particles?[/B]

Homework Equations

G((m₁m₂)/a²)

and

G((m₁m₃)/a²)

The Attempt at a Solution

I'm trying to use the above equation and separate them by its x and y components. But I feel that because its an equilateral triangle I need to somehow make it into a right triangles so that the radius from mass 1 of both particles will be 30 degrees from mass 1 if I make mass 1 the origin.

How do I attack this problem?

 

[NascentOxygen]

You have 2 forces acting on particle #1. You know these 2 forces act with an angle of how many degrees between them?

Draw the diagram.

 

[lightgrav]

probably easiest to add the components ... it is the same thing as splitting the equilateral into 2 right triangles.
put mass 1 at the tip of the A , there's only 1 component.

 

[QuantumKnight]

You have 2 forces acting on particle #1. You know these 2 forces act with an angle of how many degrees between them?

Draw the diagram.

Sorry, I forgot to mention I did. I drew what forces acted on mass 1 as well.

 

[QuantumKnight]

Where I am drawing a mental blank is if I divide it into two right trianges I get that a sides are the hypotenuse and the opposite angle of mass 1 is !/2a but the how do I find the adjacent side to mass 1?

 

[NascentOxygen]

Doesn't Pythagoras help you with right triangles?