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Gravitational Force

  1. Nov 10, 2007 #1
    The problem statement, all variables and given/known data
    Two identical spheres weighing at 100kg are suspended from massless cables at a length of 100m. The top of the cables are seperated at exactly 1m apart. What is the distance from the centers of the spheres?

    The attempt at a solution
    I tried using equilibrium to solve for it.
    Where Tcosx=mg

    However, I am still missing the angle created by the force of attraction between the two spheres. Any ideas on how else I can approach this?? Thanks
  2. jcsd
  3. Nov 10, 2007 #2


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    F is gravitational force, yes? I don't see where the 0.5 is coming from. But you can get F from a formula. And for distance between the spheres just use 1m. The deflection won't change the distance enough to matter. In that case you are all set to go. You have two equations and two unknowns, x and T. You can solve for both. You don't really need T. So how do you get x? x is the angle, right?
    Last edited: Nov 10, 2007
  4. Nov 10, 2007 #3
    http://img155.imageshack.us/img155/8247/72591671ee1.jpg [Broken]

    Despite the large angles, this is how I interpreted the question. I got 0.5F because I am only taking the force on one side. But since it's the attraction force between the two spheres, I'm not sure if we need to take it as two different forces.

    I am having trouble finding x, the angle. But this approach may be wrong in the first place.
    Last edited by a moderator: May 3, 2017
  5. Nov 10, 2007 #4


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    You are farther along than you think. The force on each sphere is given by Newton's law. There is no 0.5. Try dividing the equations by each other, you can cancel out the T.
  6. Nov 10, 2007 #5
    Okay, well that still leaves with me the two unknown variables.


    If i substitute F with GMm/R^2, I still have two unknown variables, R and x.

    The distance between the radius should be really close to 1, but is there really no mathematical calculation to solve for this?
  7. Nov 10, 2007 #6


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    The difference between R and 1m is so small you can ignore it. You could write down an exact equation in terms x, but that would be much more difficult to solve. We don't want that, do we?
  8. Nov 10, 2007 #7
    Solving x gives another unknown variable.


    x= sin^-1[(1-R)/200]

    Now i can't possible solve for x. I also can't use tan or cos to solve because i don't have the exact vertical distance.
  9. Nov 10, 2007 #8


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    I've told you twice to ignore the difference between R and 1m. The exact equation is Gm^2/(1m-200m*sin(x))^2=mg*tan(x) using your expression for R. If you try to expand that out it's cubic in sin(x) and there is still a cos(x) hanging around. So you'd need to square again. So you could write a sixth degree polynomial in sin(x). But you can't solve that. Ignore the difference between R and 1m. That's three times. Even if you ignore it the answer is exceedingly accurate.
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