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Law of gravitation

  1. Nov 19, 2006 #1


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    law of gravitation- is this better?

    Could someone check this please?

    Four identical masses of mass 600 kg each are placed at the corners of a square whose side lengths are 14.0cm (.14m).

    What is the magnitude of the net gravitational force on one of the masses, due to the other three?

    It's asking for the absolute value of the force.

    Ok so the equation to use is (GMm)/r^2

    And superposition of forces should be used, correct?

    This is what I am trying: for the mass in the upper left hand corner
    G= 6.67 x 10^-11

    (G*600*600)/(.14^2)= 1.23 x 10^-3 for the masses beside and below

    (G*600*600)/(.07^2)= 4.90 x 10^-3 for the mass diagonal from the one used
    (.07 because the attraction would decrease by half, right?)

    So adding these together:
    (2)(1.23 x 10^-3) + 4.90 x 10^-3= 7.35 x 10^-3

    (1.23 x 10^-3)+(-1.23 x 10^-3)+ 4.90 x 10^-3 = 4.90 x 10^-3
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 19, 2006 #2


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    You have to break them up into x and y axis components, add those, then use Pythagoras to add them back up.
  4. Nov 19, 2006 #3


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    I'm not sure of your reasoning here.

    I would calculate the new distance by pythagoras and use that.

    You're right - the attaraction DOES half, but you've used that fact to half your distance, squared your distance and substitued it into the equation which has given you a GREATER force for that diagonal one... which can't be right!

    Also remember when you add, you're adding vectors.
  5. Nov 19, 2006 #4


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    Okay let's try again:

    (G*600*600)/(.14^2)= 1.23 x 10^-3 for the masses beside and below

    use cos(0) and sin(0)
    x-component: 1.23 x 10^-3 y-component: 0
    use cos(90) and sin(90)
    x-comp: 0 y-comp: 1.23 x 10^-3

    (G*600*600)/[(.14^2)+(.14^2)]= 6.13 x 10^-4 for the mass diagonal from the one used
    (I saw something like this denominator in an example I found)
    use cosine and sine of 45 degrees
    x and y-comp: 4.33 x 10^-4

    So adding these together:
    F(x)=0 + 1.23 x 10^-3 + 4.33 x 10^-4 = 1.66 x 10^-3
    F(y)=4.33 x 10^-4 + 1.23 x 10^-3

    using pythagorean theorem:
    2.35 x 10^-3
    Last edited: Nov 19, 2006
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