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Math Help - Cant solve these simoltaneous equations

  1. #1
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    Cant solve these simoltaneous equations

    Fr = m alpha (L/2) cos(beta)

    N - mg = - m alpha (L/2) sin(beta)

    Fr cos(beta) - N sin(beta) = 2 Io alpha / L

    ================================
    The problem is a dynamics problem for an engineering class im in and i've gotten the three equations (which is suppose to be the hard part) that I have to simoltaneously solve. Knowns in the above equation include 'm', 'g', and beta and everything must be in terms of these variables.

    Also Io = (mL^2)/12

    I'm trying to solve for Fr and N in terms of known quantities but i keep getting stuck with some messy algebra and wrong answers. Any help would be greatly appreciated.
    Last edited by fogel1497; April 22nd 2010 at 08:59 AM. Reason: for clarity
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  2. #2
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    Hi

    The first equation gives Fr = f(alpha)
    The second equation gives N = g(alpha)

    Substitute into the third equation
    f(alpha) cos(beta) - g(alpha) sin(beta) = 2 Io alpha / L

    This leads to alpha = 3g sin(beta) / L

    Substitute into the first equation Fr = 3mg sin(beta) cos(beta) / 2
    and into the second equation
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi
    The first equation gives Fr = f(alpha)
    What are you calling f here? Where are the terms L, m, and beta going? I'm just a little confused because in the context here there is no variable defined as 'f'.

    Quote Originally Posted by running-gag View Post
    Hi
    The second equation gives N = g(alpha)
    When you bring the term 'mg' to the other side your left with:
    N = - m alpha (L/2) sin(beta) + mg
    I don't see how this condenses into N = g(alpha)

    Once again thanks for your help! I'm just trying to understand the algebra thats going on here.
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  4. #4
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    Quote Originally Posted by fogel1497 View Post
    What are you calling f here? Where are the terms L, m, and beta going? I'm just a little confused because in the context here there is no variable defined as 'f'.
    I was talking about a function of alpha

    F_r = \alpha \frac{mL}{2}\cos \beta

    N = mg - \alpha \frac{mL}{2}\sin \beta

    Substitute into the third equation

    \alpha \frac{mL}{2}\cos^2 \beta - mg \sin \beta + \alpha \frac{mL}{2}\sin^2 \beta = \alpha \frac{mL}{6}

    And since \alpha \frac{mL}{2}\cos^2 \beta + \alpha  \frac{mL}{2}\sin^2 \beta = \alpha \frac{mL}{2} (\cos^2 \beta + \sin^2 \beta) = \alpha \frac{mL}{2}

    It comes
    \alpha \frac{mL}{2} - mg \sin \beta = \alpha \frac{mL}{6}

    \alpha \frac{mL}{3} = mg \sin \beta

    \alpha = \frac{3g}{L} \sin \beta
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