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Positioning a Third Mass So That It Experiences No Net Gravitational Force

  1. Nov 10, 2011 #1
    The problem statement, all variables and given/known data

    M1 is a spherical mass (47.4 kg) at the origin. M2 is also a spherical mass (19.9 kg) and is located on the x-axis at x = 69.6 m. At what value of x would a third mass with a 10.5 kg mass experience no net gravitational force due to M1 and M2?


    The attempt at a solution

    https://mail.google.com/mail/?ui=2&ik=2ce981a95c&view=att&th=1338feddbcf78489&attid=0.1&disp=inline&realattid=f_guuf8isl0&zw

    I think I'm on the right track, but I'm not really sure. And I'm not 100% sure how to simplify what I have. Any advice would be much appreciated.
     
  2. jcsd
  3. Nov 10, 2011 #2

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    Hi Becca93! :smile:

    If your third mass is at position x, what would be the force with which M1 attracts the third mass?
    And what would be the force with which M2 attracts the third mass?

    (I'm missing a relevant equation here...)
     
  4. Nov 10, 2011 #3
    After I got the answer incorrect the first time, loncapa gave me the hint "The force due to M1 must be of the same magnitude as the force due to M2, use Newton's law of gravitational attraction. "

    This leads me to believe that the third mass must be between the two larger masses in order to experience no net gravitational force. Also because it talks about being on the x-axis and asks for the value of x. I'm just not sure how to go about solving for x, if you know what I mean.

    Due to the hint, I think I'm supposed to set it as equal, but I'm really not sure.
     
  5. Nov 10, 2011 #4

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    Yes, you're supposed to set it as equal.
    And before you can solve for x, you first need to set up the equation...
    Can you set up the equation?
     
  6. Nov 10, 2011 #5
    I don't know if the picture in the post above came out or not, but the attached photo is as far as I got equation-wise.
     

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  7. Nov 10, 2011 #6

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    I can see the photo now, but it was not attached to any of your previous posts.

    Good! So you have the equation.
    You should also substitute the numbers of the masses.

    What you have is a so called quadratic equation.
    You are supposed to bring the x2 on the right hand side of the equal sign to the left hand side and merge it with the x2 that's already there.
    Do you know how to do that?
     
  8. Nov 10, 2011 #7
    Yeah, I was just getting some pretty ridiculous numbers. I'm trying again now to see if I get a better answer.

    (I'm reattaching the photo that should have been attached up above so you can see where I'm coming from, hah)
     

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  9. Nov 10, 2011 #8

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    Aha! That's where you're coming from! :)
     
  10. Nov 10, 2011 #9
    Third attempt, again incorrect.

    Can anyone point out where I'm going wrong?



    Edit: @Serena, that diagram isn't in the actual question. One of the physics help guys at uni said that's how it should go. If there's another way, I'm all ears!
     

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  11. Nov 10, 2011 #10

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    Well, this is the way to go!

    Why would you say the numbers are ridiculous?
    They look about right, although you appear to have switched the masses around, giving you an answer that is too close to the wrong mass.
     
  12. Nov 10, 2011 #11
    I think it was the weird quadratic throwing me off.

    But where did I mix up the masses?

    This is how I got to where I got with the formula, by the way.
     

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  13. Nov 10, 2011 #12

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    My bad, you did not mix up the masses.
    And your calculation appears correct, but there must be a calculation error somewhere.
    If I calculate it (my way) I get 2 solutions for x, which are different from what you got.
    The correct solution in the middle would be a little to the side of the smaller mass.

    In your last picture you seem to have dropped a square symbol btw.
     
  14. Nov 10, 2011 #13
    Oh, whups. I fixed it, and it's squared when I plugged the numbers in.

    Do you mind if I ask where your answer starts to vary from mine? I've gone over mine a number of times now but I keep getting 30.8 as the answer, even though I know it's incorrect and I'm running out of time. If you don't want to say, I understand, I'm just getting a little frustrated, haha.
     
  15. Nov 10, 2011 #14

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    Well, I usually keep the symbols until I have a nice formula, before substituting the numbers.
    That makes it easier to spot and correct mistakes.

    I got the equation:
    [tex](M_1 - M_2) x^2 - 2 L M_1 x + M_1 L^2 = 0[/tex]

    Solving it yields:
    [tex]x = {{2 L M_1 \pm \sqrt{(2 L M_1)^2 - 4 (M_1 - M_2) (M_1 L^2)}} \over {2(M_1 - M_2)}}[/tex]
    And I simplified it a bit afterward.



    Well, I wouldn't mind giving you my solution, but apparently you're doing some online test.
    PF rules forbid me from giving full solutions in such cases...
    But I guess you can substitute the numbers can't you?
     
  16. Nov 10, 2011 #15
    It's a loncapa assignment, actually. My prof posts my weeks work to do and I have until the deadline to get my questions done. All my other questions are completed correctly, and I still have an hour and a half, it's just this one question that's messing me up, haha.

    I understand not wanting/not being able to give answers, don't worry, it's perfectly fine. I'll try out your equation and see if I get a better answer. But do you mind if I ask what is 'L' in the context of the equations?

    Edit: If you can't answer that, it's perfectly okay! I don't want to get you in hot water or anything, I'm just curious.
     
  17. Nov 10, 2011 #16

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    Oh, that should be plenty of time!
    Thanks for explaining. I often wonder how these online assignments work.


    L is the distance between masses M1 and M2.
    L = 69.6 m


    Edit: mind the factor 2 in the denominator that I added at second instance.
     
  18. Nov 10, 2011 #17

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    Btw, I started from:
    [tex]{G M_1 \over x^2} = {G M_2 \over (L-x)^2}[/tex]
     
  19. Nov 10, 2011 #18
    Since you seemed curious: for my online assignments, I generally get a week to 10 days to complete them. If I get stuck, I can go to my prof or the math help center or other students etc and work through the questions. I get 10 tries to get a long answer question. If If I get one wrong 10 times, or if I don't get something answered by the deadline, I get locked out from submitting any more answers. The answers stored in the computer by the prof are released 7 hours after it closes. This particular one was composed of 27 long answer questions, and this is my very last one.

    Anyway, thank you, I'm plugging in my numbers now. I'll let you know how it goes.
     
  20. Nov 10, 2011 #19
    Which value did you use for M1? I took 47.4 as M1, plugged it into the second equation, and ended up with x = 199.6 m and x = 40.3 m, neither of which were correct.

    Did I mangle the formula somehow?
     
  21. Nov 10, 2011 #20

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    Thanks. It clarifies things for me and makes me understand your and other people's problems better.


    Yes, M1=47.4 kg.

    Still, I got slightly different values for x, so I suspect you made a typo while calculating.
     
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