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Need some help on this question

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Two particles are located on the x axis. Particle 1 has a mass m and is at the origin. Particle 2 has a mass 2m and is at x=+L. Where on the x axis should a third particle be located so that the magnitude of the gravitational force on BOTH particles 1 and particle 2 doubles? Express your answer in terms of L.


    2. Relevant equations

    F = G(m1)(m2)/(r^2)
    I know ur suppose to us the above equation.


    3. The attempt at a solution

    I have found the F between particle 1 and 2 to be = (G)(2(m)^2)/(L^2) and the F between 2 and 1 is the same magnitude but negative. So then place particle 3 on the x axis at point X. so then now i am kind of stuck. Is the new force of particle 1 with the addition of particle 3 equal to = F btwn. particle 1 and 2 + F btwn. particle 1 and 3.

    edit: also i assigned the mass of particle 3 as m2. So the mass of particle 3 is different then of particle 1 and 2.



    Any help will be appreciated.
     
    Last edited: Aug 26, 2007
  2. jcsd
  3. Aug 26, 2007 #2
    I think you would get something like:
    [tex]{F}_{on\_A}={F}_{AC}+{F}_{AB}[/tex]
    [tex]{F}_{on\_B}={F}_{BC}+{F}_{AB}[/tex]
    [tex]{F}_{on\_A}=2\,F[/tex]
    [tex]{F}_{on\_A}={F}_{on\_B}[/tex]

    kinda that
     
  4. Aug 26, 2007 #3
    Ok yeah so u set 2F = Fac + Fab. But doesnt the Fa = Fab. The force between particle A and B (1 and 2). Rite?

    A=1
    B=2
    C=3
    rite?
     
  5. Aug 27, 2007 #4
    mmm rootX left mre more confused then before!! help!!
     
  6. Aug 27, 2007 #5
    Before:
    The gravitational force acts on 1 and 2 are the same = x

    After:
    The gravitational force acts on 1 = The gravitational force due to 3 AND 2. = y

    The gravitational force acts on 2 = The gravitational force due to 3 AND 1. = z

    Therefore :

    y = 2x AND z = 2x since the new force acting on both the particles 1 and 2 are doubled of before.

    Answer: -L plus minus (square root of 2) *L ( if i am on the right track)
     
  7. Aug 27, 2007 #6
    ahh no i dont think so. the book says the answer is .414L but i dont know how to get it ........

    any help guys
     
  8. Aug 27, 2007 #7
    yep, I got it!! :D

    so, using the information provided, you will end up with:

    F(between A and C) = F(between A and B)
    I didn't simplify further in the previous post, and above is what you get after further simplications.

    and, C can be to the right of B, or inbetween A and B, or to the left of A.

    Case I:
    C----A----B
    Case II:
    A---C-----B
    Case III:
    A----B----C

    you know mass of A=0.5 times mass of B,
    and net force on A is equal to force on B,
    and now, find thr right case, and solve the problem!
     
    Last edited: Aug 27, 2007
  9. Aug 27, 2007 #8
    think of electric charges, and that static electric forces, if you are good at them
     
  10. Aug 27, 2007 #9
    Before:
    The gravitational force acting on the 1 and 2 are the same:
    [tex] F_{initial}=\frac{Gm*2m}{L^2}[/tex]

    After: Let's assume that particle 3 is put between 1 and 2 and is +y from the origin and its mass is m3.
    Then,
    The gravitational force that acts on 1 = The one due to 2 AND 3.
    Since 2 and 3 are at the right of 1. 2 and 3 will pull 1 in the same direction which means the gravitational force that acts on 1 is the SUM of the two gravitational forces due to 2 AND 3.

    [tex] F_{1}= F\ due\ to\ 2 + F\ due\ to\ 3[/tex]
    [tex]F_{1} = \frac{Gm*2m}{L^2} + \frac{Gm*m_{3}}{y^2}[/tex]
    [tex] But\ F_{1} = 2*F_{initial}, therefore,

    \frac{Gm*2m}{L^2} \ + \frac{Gm*m_{3}}{y^2} = 2*\frac{Gm*2m}{L^2}[/tex]
    Simplify:
    [tex]m_{3}=\frac{2my^2}{L^2}[/tex]

    Repeat the same thing for F2, at the end, you will have two equations in terms of m3 and y. Then solve for y. The one above is one of them.
     
  11. Aug 27, 2007 #10
    ok yeah thx. at the end i get 2y^2 = (L-y)^2. How do i solve for y????

    edit: nvm i got it. Thx for the help!
     
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