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Factoring difficulties

  1. Jan 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Objects with masses of 200kg and 500kg are separated by 0.4m a)Find the net gravitational force exerted by these objects on a 50kg object placed midway between them. b) At what position (other than infinitely remote ones) can the 50 kg object be placed so as to experience a net force of zero?

    I have solved a), everything below is for part b)

    I'm solving for a problem right now and I'm having difficulties in factoring. =P

    Trying out Latex, it may take a while.

    2. Relevant equations
    [tex]0=\frac{Gm_{1}m_{3}}{-r_{1}{}^2} + \frac{Gm_{2}m_{3}}{r_{2}{}^2}[/tex]


    3. The attempt at a solution
    [tex]0=\frac{Gm_{1}m_{3}}{-r_{1}{}^2} + \frac{Gm_{2}m_{3}}{r_{2}{}^2}[/tex]

    [tex]0=Gm_{3}(\frac{m_1}{-r_{1}{}^2} + \frac{m_2}{r_{2}{}^2})[/tex]

    [tex]0=\frac{m_1}{-r_{1}{}^2} + \frac{m_2}{r_{2}{}^2}[/tex]

    [tex]0=m_{1}r_{2}{}^2 - m_{2}r_{1}{}^2[/tex]

    [tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex]

    Since there are two unknown variables, you need a second equation:

    [tex]0.4m=r_{1} + r_{2}[/tex]

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~


    I just don't know how to factor:
    [tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex] to make it in [tex](r_{1}+r_{2})(r_{1} - r_{2})[/tex] form. Can someone show me the simple factoring that I've forgotten? xP
     
    Last edited: Jan 5, 2007
  2. jcsd
  3. Jan 5, 2007 #2
    By the way, are my previous factoring steps correct? Thanks!
     
  4. Jan 5, 2007 #3
    I don't think you need to factor it, just do the following:

    (hold on...)
     
    Last edited: Jan 5, 2007
  5. Jan 5, 2007 #4
    Well I think it would be easier if you factored it and then added/subtracted the two equations to find one unknown variable right?

    Do I just substitute the equation you found with r_1 into the other equation?
     
  6. Jan 5, 2007 #5
    This is what I do when I get these kinds of equations: I plug them into my graphing calculator, graph them, and see where y = 0 to find the roots. Call me lazy, but it works :P So you can plug in the equation

    [tex]y=(200)(0.4 - x)^2 - (500)x^2[/tex]

    and look for where y = 0.
     
    Last edited: Jan 5, 2007
  7. Jan 5, 2007 #6
    Good idea! xD I've never tried that before with any physics equations. Hmm, mabye that's what I should do when I get stuck with these kinds of equations. Btw, they allow graphing calculators on the physics subject test lol! hehe. Thanks for your advice and help!

    While I try that, if anyone would be willing to factor:

    [tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex]

    into the (x+y)(x-y) form, it would be greatly appreciated! :smile:
     
  8. Jan 5, 2007 #7

    Hurkyl

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    Science Advisor
    Gold Member

    You know how to factor x² - y², so all you have to do is to rewrite what you actually have in that form.
     
  9. Jan 5, 2007 #8
    How do you factor 200x² - 500y²?? Seriously, I'm uncertain of how to factor that lol.
     
  10. Jan 5, 2007 #9
    Ahh, now I've got it. The only slightly difficult part comes with the coefficients. Take the square roots of them, so that when they are multiplied together you get them back.

    [tex]
    0 = 100(2r_1^2 - 5r_2^2)
    [/tex]

    [tex]
    0 = (\sqrt{2}r_1 + \sqrt{5}r_2)(\sqrt{2}r_1 - \sqrt{5}r_2)
    [/tex]

    [tex]
    0 = (\sqrt{2}(0.4 - r_2) + \sqrt{5}r_2)(\sqrt{2}(0.4 - r_2) - \sqrt{5}r_2)
    [/tex]

    [tex]
    0 = (0.4 \sqrt{2} + r_2 (\sqrt{5} - \sqrt{2}))(0.4 \sqrt{2} - r_2 (\sqrt{2} + \sqrt{5}))
    [/tex]

    Now you can find the roots of this equation.
     
    Last edited: Jan 5, 2007
  11. Jan 5, 2007 #10
    [tex]x^{2}-y^{2}=(x+y)(x-y)[/tex]

    Don't take my word for it, expand it so that you understand.
     
  12. Jan 5, 2007 #11
    Just to demonstrate as maybe the coefficients are screwing you up...

    [tex]200x^{2}-500x^{2}[/tex]
    [tex]=100(2x^2-5y^2)[/tex] (might help if there is a 0 on the other hand, makes factoring simpler, although of course unnecessary)
    [tex]=100(\sqrt{2}x+\sqrt{5}y)(\sqrt{2}x-\sqrt{5}y)[/tex]

    in this case the "[tex]x^{2}[/tex]" in [tex]x^{2}-y^{2}=(x+y)(x-y)[/tex] is [tex]2x^{2}[/tex] and because
    [tex]\sqrt{xy}=\sqrt{x}*\sqrt{y}[/tex]


    [tex]\sqrt{2x^{2}}=\sqrt{2}*\sqrt{x^{2}}=\sqrt{2}x[/tex]
     
    Last edited: Jan 6, 2007
  13. Jan 5, 2007 #12

    Hurkyl

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    Gold Member

    This is a good skill to get used to. If you have an expression like

    200 r12 - 500 r22

    and you think you want to apply an identity like

    x2 - y2 = (x + y) (x - y)

    then you don't need to be creative: you can simply solve for what you need to do. Set

    x2 = 200 r12
    y2 = 500 r22

    and then if you can solve for x and y, that shows you how you can apply your identity to your expression.
     
  14. Jan 6, 2007 #13
    I see now.. Yeah the coefficients were screwing me up =P Thanks for all your guys' help! :smile:
     
  15. Jan 6, 2007 #14
    I still don't get the answer that is posted in the back of the book with the factored equations and the second one. I don't know what I'm doing wrong. The two answers I got for r_2 is -0.6883m and -.55m... The answer for b) is Between the two objects and 0.245m from the 500kg object.
     
  16. Jan 6, 2007 #15
    :frown:
    I still am unable to solve b) At what position (other than infinitely remote ones) can the 50 kg object be placed so as to experience a net force of zero?

    I have no idea what I'm doing wrong. I have came up with 2 equations that should relate to what is being asked. I have factored one of them so I can plug the other in to find one of the lengths.. My answer is not correct however.

    The answer for b) is Between the two objects and 0.245m from the 500kg object. he two answers I got for r_2 is -0.6883m and -.55m... :frown: Help Please? Thanks.
     
  17. Jan 7, 2007 #16

    Dick

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    Homework Helper

    Check your arithmetic. The root of the second factor is positive.
     
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