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Gravitational force question

  1. Jun 28, 2008 #1
    In the figure below, two spheres of mass m and a third sphere mass M form an equilateral triangle, and a
    fourth sphere of mass m4 is at the center of the triangle. The net gravitational force on m4 from the three
    other spheres is zero; what is M in terms of m.
  2. jcsd
  3. Jun 28, 2008 #2


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    Welcome to PF!

    Hi zila24! Welcome to PF! :smile:

    Show us how far you've got, and where you're stuck, and then we'll know how to help you. :smile:
  4. Jun 28, 2008 #3
    i divided the triangle into a right triangle and found that if each side is R, for the right triangle the base would be 1/2 r and the length would be radical 3 over 2 and since the mass m4 is in the center the lengths above and below it would be radical 3 over 4
    the forces of both m in the x direction would cancel each other, so we need to find out the force in the y direction, this is what im having a problem with.

    ifound the force between M and m4 (which is only in the y direction to be)- GMm4 / (3/16)d^2
  5. Jun 28, 2008 #4
    i found the force for m y to be Gmm cos 49 / ( 7/16)d^2
  6. Jun 28, 2008 #5
    but when i added them together to figure out the mass relationship, the answer was not right it should be M=m
  7. Jun 28, 2008 #6


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    Hi zila24! :smile:

    Nooo … your geometry is wrong …

    it's in the middle of the triangle, not the middle of that line. :frown:

    Hint: concentrate on the small right-angled triangle at the bottom! :wink:
  8. Jun 28, 2008 #7
    Oh ok so now i got the hypotenuse to be .577 and my calculation for the force to be
    Gm(m4)cos 30 / (.577)^2

    but it should be sin of 30 i dont get why =/
  9. Jun 28, 2008 #8
    o wait is it cos of 60?

    but how do i know the length between m4 and M
  10. Jun 28, 2008 #9


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    Yes, (√3)/6 = 0.577 is correct.

    But you didn't need to calculate it, did you, since it's the same distance to all three masses? :smile:
    Because components always use the cos of the angle, and 30º isn't the angle. :smile:
  11. Jun 28, 2008 #10
    ok i can solve the question now thank u for your help =) ... but i just had one last question.. i just wanted to know how you got the .577
  12. Jun 28, 2008 #11


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    Now I'm confused … I thought you got the .577? :confused:

    There's various ways of doing it.

    One is to use that small right-angled triangle I mentioned, another is that the centre of mass is always at the one-third point. :smile:
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