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Net Gravitational Force

  1. Nov 16, 2009 #1
    The masses and coordinates of three particles are as follows: 20.0 kg, (0.50, 1.00) m; 309.0 kg, (-1.00, -1.00) m; 18.0 kg, (0.00, -0.50) m. What is the gravitational force on a 20.0 kg sphere located at the origin due to the other spheres, magnitude and direction? Give the direction as an angle in degrees counter clockwise with respect to the the + x-axis.

    I have split the forces into x and y components using the F = (Gm1m2)/r^2 equation.
    So for my x-forces I get:
    F = (309)(20)G/2 * (1/sqrt(2)) ----> this is a negative force pulling down
    F = (20)(20)G/1.25 * (.5/sqrt(1.25))

    y-forces:
    F = (309)(20)G/2 * (1/sqrt(2)) ----> this is a negative force pulling down
    F = (18)(20)G/.25 ----> this is a negative force pulling down
    F = (20)(20)G/1.25 * (1/sqrt(1.25))

    Then I took the sum of the Xs squared and the sum of the Ys squared and then the square root to find the magnitude of the resultant vector. I got 2.612E-7 N.
    Then for the degrees I took the inverse tan ( sum of the Ys/sum of the Xs) +180 deg to get it in the third quadrant for a result of 282.2 deg. Somewhere I've done something wrong because I'm getting this incorrect, where did I make a mistake?
     
  2. jcsd
  3. Nov 16, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi singinglupine! Welcome to PF! :smile:

    (have a square-root: √ :wink:)
    Sorry, but I don't understand what formula you've applied here. :redface:

    Use mMr/r3 :smile:
     
  4. Nov 16, 2009 #3
    Re: Welcome to PF!

    Thanks for the welcome :) I used the law of universal gravitation where
    F = GmM/r2 and G is the gravitational force constant. The parts with something else multiplied by the force are the x, cos(θ) and y, sin(θ) components. I used pythagorean theorem to figure out the length of the hypotenuse, so that cos would be adjacent/hypotenuse and sin opposite/hypotenuse.
     
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