Superposition Gravataional Forces

[koab1mjr]

Homework Statement

There are three point particles in fixed positions in an xy plane. Particle A lies at the origin and particle B lies 150 degrees from positive A (reference is positive x axis). The distance between P_a and P_b is .5m. The mass of P_a is 6.00g and the mass of P_b is 12.0g. The position of the third particle C is not known but the mass is 8.00g. The net forces acting on particle A (only gravataional) magnittude is 2.77*10^-14 at an angle of -163.8 degrees. Find the x and y coordiantes of particle C

Homework Equations

Need the gravatational force equation F_g = Gm₁m₂/r²

The Attempt at a Solution

This problem is straight forward. Knowing that the force from particle B on particle A plus the coresponding force from particle C must equal the resulting force. So to solve I will break it up into x and y componenet and find the quadanents. So i am working along with the following

F_{net on particle A} cos(-163.8) = F_{gravity AB} cos(150) + F_{gravity AC}COS(theta)

So I subtract the first term on the right from the one on the left. And here is where things go to hell. I have some x component of a force. on the left. I have on the right two unknowns I do not know the distance between A and C, and I do not know x position of particle see which i want to know. I am stuck on how to proceed. I am sure this becomes a pythagorean thing but not sure. Any help much appreciated.

 

[rl.bhat]

If you draw the forces in a co-ordinate system, you can see that the particle c lies in the fourth quadrant. Now take the X and Y components of F(AB), F(AC) and F(net)
X-component of F(net) = X-component of F(AB) +X-component of F(AC). Similarly for Y-components. From these you can find the angle made by F(AC) with X -axis, and the rest of the things.