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Homework Help: Help with Newton's Law of Gravitation

  1. Feb 25, 2010 #1
    1. Find the magnitude and direction of the net gravitational force on Mass A due to masses B and C in Figure 6.27. Each mass is 2.00 kg.

    The figure looks like this:
    A<----------->(10 cm distance)C<-------------------------->(40 cm distance)B



    2.F(g)=G * (m(1)m(2))/(r^2)



    3.I'm thinking of two ways to do this, one would be just include all 3 masses and the final distance between A & B, which would be 6.674e^-11 * (2 *2 *2)/(.50 m^2), which is 2.14e-9 to the right.

    However, I could just do the two separately between A & C, and A & B, and just add them, or I could do A & C, and C & B, and just add them. Not too sure here, thanks for any help!
     
  2. jcsd
  3. Feb 25, 2010 #2

    Doc Al

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    Staff: Mentor

    :bugeye:

    Now you're cooking.

    Why C & B? You only care about the force on A, not between C and B.
     
  4. Feb 25, 2010 #3

    collinsmark

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    No, don't go that way. :eek: That's not the way the math works.

    There ya go! try that one. :approve:

    Yell, that last approach won't help you either. The problem statement is asking you the gravitational force on A, caused by the others. The gravitational force between C and B isn't relevant to the particular problem.

    [Edit:] Looks like Doc Al beat me to it.
     
  5. Feb 25, 2010 #4
    OK, so would it be, A & C 6.674e-11 * (2 *2)/.1^2, which would be 2.7e-8.

    Then, I would do A & B 6.674e-11 * (2 *2)/.5^2, which would be 1.07e-9.

    Add them together and it's 2.807e-8. Can someone double check this? For the second part of the problem, I used .5m as the distance between A & B.
     
  6. Feb 25, 2010 #5

    collinsmark

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    You might want want to redo you final result (i.e. plugging in the numbers), keeping better track of significant digits, But yes, I believe you are essentially correct. :cool:
     
  7. Feb 25, 2010 #6
    Yeah, I was in a bit of a hurry, lol.
     
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