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Find the magnitude of the acceleration of the particle.

  1. Nov 28, 2007 #1
    An 8.50 kg point mass and a 13.5 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 19.0 cm from the 8.50 kg mass along the line connecting the two fixed masses.

    Find the magnitude of the acceleration of the particle.



    so Fg =((6.67*10^-11)*8.50*13.5)/(.25^2)

    F=ma since particle close to 8.5kg
    a= 1.2246*10^-7/8.50
    =1.440*10^-8 m/s^2

    why this answer wrong please help me
  2. jcsd
  3. Nov 28, 2007 #2
    You seem to be trying to calculate the force that exists between the 8.5 and the 13.5 kg masses (although the distance between them is 0.5 not .25m). But this question is really about mass m.

    mass m is being pulled by two different forces...one from the 8.5 kg mass and the other from the 13.5 kg mass. It may help to draw a free body diagram of mass m and then write a net force equation for it. It is this net force that is responsible for the acceleration.
    Last edited: Nov 28, 2007
  4. Nov 28, 2007 #3
    so Fg =((6.67*10^-11)*8.50*13.5)/(.5^2) = 3.6018*^-9 N

    Fnet = F2+F1= F(m2+m1)

    so a= Fg/(m2+m1) ???
  5. Apr 8, 2008 #4
    bump, is this equation right

    a= Fg/(m2+m1)
  6. Apr 8, 2008 #5
    I figured it out

    you take ((G)(8.5)(m))/(.19m)^2
    =1.57E-8 m

    then you take ((G)(13.5)(m))/(.31m)^2
    =9.37E-9 m

    now just subtract the bottom one from the top to get your acceleration.
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