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Particle Acceleration(No clue how to do)

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data
    An 8.50kg point mass and a 14.0kg point mass are held in place 50.0cm apart. A particle of mass M is released from a point between the two masses 17.0cm from the 8.50kg mass along the line connecting the two fixed masses.

    Find the magnitude of the acceleration of the particle.

    2. Relevant equations
    I have a feeling i'm suppos to get the Sum of the Forces in the x and y direction.

    3. The attempt at a solution
    Tried something from the book, but the book uses angles, while angles aren't mentioned here.
    8.50 kg-------.17m--------.5m-----------------14.0kg

    Please help. Just tell me where to begin and I'll do the work. THanks.
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2

    Chi Meson

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    Are these particles outside the gravitational effects of the Earth? If not, then the particle accelerates at 9.81 m/s^2.

    Assuming that the Earth's g-field is ignored, then you need to find the two forces between M and the other two particles. No angles, since all three particles are lines up. So, get the Law of gravitation formula, convert distances to meters, and find those two forces.
  4. Nov 7, 2007 #3
    Fg = G*m1*m2 / r^2
    So I would do (6.673x10^(-11)) * 8.50 * 14.0 / .5^2 = 3.176x10^-8
    Which would be Fg of m1m2 right? Would I then proceed to make G*M*(m1+m2) / r^2 and solve for M?
    Last edited: Nov 7, 2007
  5. Nov 7, 2007 #4

    Chi Meson

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    Not exactly. right formula, but the first two masses (.5 m apart) are fixed. the mass M in between will feel a force from each of the other two. Distance to one is given, figure out the other. Calculate the two forces. Determine if they will be in the same direction on mass M or opposite direction. Determine net force.
  6. Nov 7, 2007 #5
    Ok. But would I still use the 3.176x10^8 that was calculated using m1 and m2?
    Cause then I would assume I would do
    Fg = G*M*8.5 / .17^2
    Fg = G*M*14.0 / .33^2
  7. Nov 7, 2007 #6

    Chi Meson

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    No. If I understand the question correctly, you want the acceleration of particle M. The force between the two other particles are NOT forces on particle M. Much the same way that the force of attraction between the sun and the earth have nothing to do with the force of the sun on you and the force of the earth on you.
  8. Nov 7, 2007 #7
    Yea, i'm looking for the acceleration of particle M. The formula for that would be a= F/M
    I know i'm suppose to get the F and mass of M to get the acceleration.
  9. Nov 7, 2007 #8
    I know how to get the Fg if it was just between m1 and m2 lol Based of an example in my book. But adding in a third unknown mass with distance is throwing me off.
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