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Homework Help: Gravitational net force of zero

  1. Apr 22, 2005 #1
    A 200 kg mass and a 500kg mass are separated by 0.4m. At what position can a 50kg mass be placed so as to experience a net force of zero (other than infinity)?

    here's what I did:

    I need both forces to equal each other.


    did I do this correctly? was there a faster way?
    Last edited: Apr 22, 2005
  2. jcsd
  3. Apr 22, 2005 #2
    Your question isnt clear,are you trying to balance the entire system? How many dimensions here, 1 or 2? Do you want just the 50kg to experience no force? It looks like your talking about gravitational forces among the 3, right?
  4. Apr 22, 2005 #3
    well, I copied the book. What I got was that I had to place the 50kg mass in between the 200kg and 500kg masses so that the 50kg mass experiences a net force of zero.
  5. Apr 22, 2005 #4
    then you want the pull from left to equal the pull from the right. Though I didnt check your math, the method is right and your answer seems logical.
  6. Apr 22, 2005 #5
    Are the CORES of the masses seperated by 0.4m or just surface to surface? Anyway if it is the cores, don't bother with two variables. Say one will be x, the other 0.4-x.










    It would have been faster if you knew this formula!
    Last edited: Apr 22, 2005
  7. Apr 22, 2005 #6
    Since the radius of each mass isnt provided, you are forced to assume that it is core to core.
  8. Apr 22, 2005 #7
    In this case my guess was good :)
  9. Apr 23, 2005 #8
    Theres a mistake in the algebra somewhere there. The answer should have + 1, not -1.
  10. Apr 23, 2005 #9


    0.4/x =1 + ((M1^0.5)/M2^0.5)
    x = 0.4 / ( 1+ ((M1^0.5)/M2^0.5))
  11. Apr 23, 2005 #10

    Doc Al

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    Staff: Mentor

    So far, so good. (Assuming these masses can be treated as particles.) Simplifying and rewriting this in standard form for a quadratic:
    [tex]15y^2 -20y + 4 = 0[/tex]
    Which yields two solutions, only one of which is relevant to this problem.
    I don't know why you wrote it this way.
    Perfectly correct.

    It would have been slightly faster to immediately take the (positive) square root (as Werg22 showed), which eliminates the need to solve a quadratic:
    [tex]\frac{200}{x^2}=\frac{500}{y^2} \; \Longrightarrow \; \frac{2}{x^2}=\frac{5}{y^2} [/tex]
  12. Apr 23, 2005 #11
    Sorry guys! It is actually because I forgot the 1 was an individual term... I tought it was part of the parenthesis when it is not.
    Last edited: Apr 23, 2005
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