Universal Gravitation Problem

In summary, three point particles with masses of 12.0 kg, 18.0 kg, and 15.0 kg are fixed in an equilateral triangle with sides of length 2.50 mm in the xy plane. The net gravitational force exerted on particle 1 by the other two particles can be calculated using the equation G((m1m2)/a^2) and G((m1m3)/a^2). To find the components of this force, the triangle can be divided into two right triangles. Pythagoras' theorem can then be used to find the adjacent side of mass 1.
  • #1

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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
NascentOxygen said:
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.
 
  • #5
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?
 

Attachments

  • hw1.pdf
    168.8 KB · Views: 252
  • #6
Doesn't Pythagoras help you with right triangles?
 

Suggested for: Universal Gravitation Problem

Back
Top