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Newton must respect the law of lever!

  1. Jul 24, 2004 #1
    Newton must respect the law of lever as we all must.
    The law of lever is the 1st law ever in physics and every thing
    must agree with it.
    \documentclass{article}
    \begin{document}
    \section{Newton must respect the law of lever}
    \hspace{1 cm}I know most of you will object me as I'm the first
    with such approach. But, I do have a point so please let me make
    it. Assume that body A and body B have masses $M_a$ and $M_b$,
    such that $M_a>M_b$. They are initially at distance R and without
    initial push they start to fall on each other only because their
    mutual gravity pull. We view only the part of the interaction from
    the release until the collision. The gravity force for each body
    according to Newton is given with:
    \begin{equation}
    F=G\frac{M_a M_b}{R^2}
    \end{equation}
    Newton's gravity law respects his third law, so:
    \begin{equation}\label{Newton3}
    F_a=-F_b
    \end{equation}
    Newton's second law claims that force equals to mass times
    acceleration so \ref{Newton3} will turn into:
    \begin{equation}
    a_a M_a = -a_b M_b
    \end{equation}
    This means that $a_a < a_b$ because $M_a > M_b$ so within same
    time body A will pass smaller distance from body B i.e.
    \begin{equation}\label{condit}
    \Delta X_a<\Delta X_b
    \end{equation}
    The work done by some force while making displacement is given
    with:
    \begin{equation}\label{Work_done}
    W=\int_{X1}^{X2} Fdx
    \end{equation}
    In our system we have only two forces making displacement - only
    two works done and they must cancel because of conservation of
    energy for such a closed system given with:
    \begin{equation}\label{Cosevr_energy}
    \sum W_i=0
    \end{equation}
    From \ref{Cosevr_energy} we have $W_a=-W_b$. Widen up it should
    give:
    \begin{center}
    $\int_{X_{1a}}^{X_{2a}}F_a dx=-\int_{X_{1b}}^{X_{2b}}F_b dx$

    $\int_{X_{1a}}^{X_{2a}}Fdx=\int_{X_{1b}}^{X_{2b}}Fdx$

    $F(X_{2a}-X_{1a})=F(X_{2b}-X_{1b})$

    \end{center}
    So it must be:
    \begin{equation}\label{must}
    F\Delta X_a=F\Delta X_b
    \end{equation}
    But it is not so because of \ref{condit}. By the way the equation
    \ref{must} is very similar to the law of lever only though the
    forces must have inverse the ratio of their distances or:
    \begin{equation}\label{Lever}
    F_a D_a = D_b F_b
    \end{equation}
    Therefore, the conflict between Newton's gravity and conservation
    of energy is due to Newton's disobedience for the law of lever.
    \end{document}

    tex doesn't work that well. Just copy pase into your TeX editor
     
  2. jcsd
  3. Jul 24, 2004 #2
    If the work done on the bodies have opposite sign, why do they both gain kinetic energy? Last time I checked, work was positive when force and displacement are pointing in the same direction.
     
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