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Fun: Gravitational Force

  1. Nov 8, 2004 #1
    Three identical masses of 650 kg each are placed on the x-axis. One mass is at -13.0 cm, one is at the origin, and one is at 42.0 cm. What is the magnitude of the net gravitational force on the mass at the origin, due to the other two masses?

    I solved this problem in the homeowrk section of the textbook and got the answer right. This is the same problem, but with different numbers. Somehow when I enter the answer in the computer it keeps on saying that I'm wrong. Please look at my work and see what went wrong. Heres my work:

    I used the formula: F_g = (Gm_1m_2)/R^2 where G= 6.67 x 10^-11

    The gravitational force from the mass at -.13 m on the mass at the origin is calculated to be: -.001668

    The gravitational force from the mass at .42 m on the mass at the origin is calculated to be: .00016

    Net gravitational force is: |F_g1 - F_g2| = .001508 N

    This answer was wrong. Thanks to anyone that is gracious enough to spend time on this problem.
     
  2. jcsd
  3. Nov 8, 2004 #2
    well, you're adding to get your answer, while your equation is showing the difference of the two forces.

    i would think that the sum would be correct, though.
     
  4. Nov 8, 2004 #3
    The sum is not correct, I tried it. The two forces are pulling in opposite directions, so I think subtracting is much more appropriate to get the "net force" on the mass at the origin.
     
  5. Nov 8, 2004 #4
    Does the computer care about sig figs? Because you're only given the mass to two sig figs, whereas you're representing your answer to 4.

    --J
     
  6. Nov 9, 2004 #5
    Teclo and Justin both of you were right! Thanks. The net force would be to the left hence negative and its to 3 sig figs. So the answer was -.00151 compared to my original: +0.001508. Thanks
     
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