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Superposition Gravataional Forces

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known data
    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 Pa and Pb is .5m. The mass of Pa is 6.00g and the mass of Pb 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

    2. Relevant equations

    Need the gravatational force equation Fg = Gm1m2/r2

    3. 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

    Fnet on particle A cos(-163.8) = Fgravity AB cos(150) + Fgravity ACCOS(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.
  2. jcsd
  3. Dec 4, 2008 #2


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    Homework Helper

    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.
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