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Force with three masses

  1. Sep 25, 2010 #1
    I can't find what I'm doing wrong in this problem:

    Equation: f= (Gm1m)/r^2

    F1= (G*(8.5))/(.2)^2 = 1.417 x 10^-8
    F2= (G*(12))/.5^2 = 3.2016x10^-9

    Fnet= -1.417 x 10^-8 + 3.2016x10^-9 = -1.1x10^-8m

    a= 1.1x10^-8, and this is wrong and I can't figure out where I messed up at.
    Last edited: Sep 25, 2010
  2. jcsd
  3. Sep 25, 2010 #2
    Well firstly, on F2, it should be 0.3^2 not 0.5^2 as the point is released 20cm from the 8.50kg mass which makes it 30cm from the 12.0kg mass.

    Secondly, force is measured in Newtons (kg*m / s^2). Not metres (m) as you have it. You don't know the mass of particle m so you can't use it. You need the following.

    g = GM / r^2 = (m^3 / kg*s^2)(kg) / m^2 = m/s^2

    G = 6.67x10^-11m^3 / kg*s^2
    M1 = 8.5kg
    M2 = 12.0kg
    r1 = 0.2
    r2 = 0.3

    g1 = GM1/r1^2
    g2 = GM2/r2^2

    g = g1 - g2

    Last edited: Sep 25, 2010
  4. Sep 25, 2010 #3

    oh I didn't think of that, thanks Jared
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