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General Problem

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Four 9.5 kg spheres are located at the corners of a square of side 0.60 m. Calculate the magnitude of the total gravitational force exerted on one sphere by the other three.

    2. Relevant equations

    F=G(mass of #1)(mass of #2)/r^2
    F=G(mass of #1)(mass of #3)/r^2
    F=G(mass of #1)(mass of #2)/r^2
    For #3 take Fsin(45)=x value
    For #3 x value=y value

    x and y components added
    F=squareroot(x^2+y^2)


    3. The attempt at a solution
    r=0.6m
    m=9.5kg
    F#2x=1.625e-8
    F#2y=0
    F#3x=0
    F#3y=-1.672e-8
    F#4x=1.182e-8
    F#4y=-1.182e-8
    FTotal= 4.037e-8

    Correct answer in book for FTotal: 3.2e-8
     
  2. jcsd
  3. Oct 5, 2008 #2

    hage567

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    Can you show how you got

    F#2x=1.625e-8
    F#4x=1.182e-8
    F#4y=-1.182e-8
     
  4. Oct 5, 2008 #3
    RE:
    Can you show how you got

    F#2x=1.625e-8
    F#4x=1.182e-8
    F#4y=-1.182e-8

    Begin Reply:
    F#2x=(G)(9.5)(9.5)/(0.6^2)
    =1.672e-8
    Which was a typo, thanks for pointing that out.
    F=#4x=[(G)(9.5)(9.5)/(0.6^2)]sin(45)=
    =1.182e-8
    F=#4y=[(G)(9.5)(9.5)/(0.6^2)]cos(45)=
    =1.182e-8
     
  5. Oct 5, 2008 #4

    hage567

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    The distance between the two masses diagonally across from each other will NOT be 0.6 m. 0.6 m is the length of the sides of the square.
     
  6. Oct 5, 2008 #5
    Wow thank you so much! I can't believe I missed that. I plugged in my values again and I got 3.13e-8, which I attribute to rounding. Thanks again!
     
  7. Oct 5, 2008 #6

    hage567

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    You're welcome.
     
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