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Net Gravitational Force

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data

    The masses and coordinates of three particles are as follows: 20.0 kg, (0.50, 1.00) m; 432.0 kg, (-1.00, -1.00) m; 71.0 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.

    2. Relevant equations

    Fgrav=(Gm*m)/r2

    3. The attempt at a solution
    I did this equation from the 20 kg sphere to all the other ones, and did that force times cos and sin. but I am not getting the right answer.
     
  2. jcsd
  3. Mar 10, 2010 #2

    rl.bhat

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    Show your calculations.
     
  4. Mar 10, 2010 #3
    for the 20 kg to the 71 kg: F = (6.67E-11Nm^2/kg^2*20kg*71kg)/(.5m^2)=.000000379 (and this should be negative since its in the negative y direction and there is no x force)

    for the 20kg to the 432 kg: F =(6.67E-11Nm^2/kg^2*20kg*432kg)/(sqrt(2)m^2)=.000000288
    (negative because below x-axis) fx= .000000288cos(45)=.000000204=Fy

    for the 20kg to the 20 kg: F =(6.67E-11Nm^2/kg^2*20kg*20kg)/(sqrt(1.25)m^2)=.000000019
    Fx= .000000021cos(63.43)=.000000009 and Fy=.000000021sin(63.43)=.000000019

    then I added them together: Fx= .000000009+(-.000000204)=-.000000195
    Fy= .000000019+(-.000000379)+(-000000204)=-.000000564

    (-.000000195)^2 + (-.000000564)^2= 3.56121E-13
    sqrt(3.56121E-13)=5.97E-7
     
  5. Mar 10, 2010 #4
    I know I'm doing something wrong...but I don't know what it is.
     
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