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Homework Help: Gravitation problem

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Two identical point masses, each of mass M, always remain separated by a distance of 2R. A third mass m, is then placed a distance x along the perpendicular bisector of the original two masses. (Picture attached.) Show that the gravitational force on the third mass is directed inward along the perpendicular bisector and has a magnititude of :

    [itex] F = \frac{2GMmx}{(x^2+R^2)^\frac{3}{2}} [/itex]

    2. Relevant equations

    [itex] F=G\frac{m1m2}{r^2}[/itex]

    3. The attempt at a solution

    First I figured out the magnitude of the force of gravitation between one of the two mass Ms and the third mass m. To find the distance between them I used the Pythagorean theorem where:

    [itex] r^2 = x^2 + R^2[/itex]

    plugging this and the two masses into the gravitation formula:

    [itex] F = G\frac{Mm}{x^2+R^2} [/itex]

    That is the magnitude of the force of gravitation between each M and m. To find the magnitude and direction of the total force I should find the vector sum of these two forces? This is where I end up lost. The Pythagorean theorem wont work here. Can anyone help point me in the right direction? I know that after i find the magnitude of the resultant force I can use arctan to find the direction.

    Attached Files:

  2. jcsd
  3. Apr 4, 2014 #2


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    Hint: use vectors for the attractive force applied by each mass M on the smaller mass 'm'. Find the resultant of these two attractive forces on 'm'.
  4. Apr 4, 2014 #3
    You need to resolve the force into components in the x- and y directions. This will involve sines and cosines in right triangles. Geometrically, what is the cosine of the angle between the force and the bisector?

  5. Apr 5, 2014 #4
    so if i take the force between one of the two Ms and m and break it into components, its y component would be:

    [itex] \frac{GMmSin(theta)}{x^2+R^2}[/itex]

    and its x component would be:

    [itex] \frac{GMmCos(theta)}{X^2+R^2} [/itex]

    then i do the same of the other force and add the components of the two forces so that the y components would cancel out and become 0 (same magnitude except one is above and one below the x axis) and the net x component would be

    [itex] \frac{2GMmCos(theta)}{X^2+R^2} [/itex]

    Then the resultant vector would be the radical of that squared? but that doesnt leave me with the answer im supposed to be getting. Im still missing something =[ am I at least heading in the right direction? Im so lost lol
  6. Apr 6, 2014 #5
    Yes. You're heading in the right direction. Now, what is cosθ in terms of the sides of a right triangle? (Hint: there was a right triangle involved when you used the Pythagorean theorem).
  7. Apr 6, 2014 #6
    ahh i see!

    [itex] Cos(theta) = \frac{adj}{hyp} = \frac{x}{(R^2+x^2)^\frac{1}{2}} [/itex]

    this would make the magnitude of the force

    [itex] \frac{2GMmx}{(x^2+R^2)^\frac{3}{2}} [/itex]

    thank you! :)

    I've proved the magnitude, is showing the y components cancel out enough to prove the direction? or do i have to use arctan (y component/ x component) ?
  8. Apr 7, 2014 #7
    Sure. You can see by symmetry that the y components cancel out.
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