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Math Help - Equilibrium Independence

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    Equilibrium Independence

    Can someone show that the static equilibrium equations in two and three dimensions on any body are linearly independent.
    Last edited by ThePerfectHacker; December 27th 2006 at 10:40 AM.
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    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Can someone show that the static equilibrium equations in two and three dimensions on any body are linearly independent.
    I am unable to understand your question.
    Could you please explain it?

    Keep SMiling
    Malay
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    Quote Originally Posted by malaygoel View Post
    I am unable to understand your question.
    Could you please explain it?
    It is a linear algebra question.

    Note the equations,
    \left\{ \begin{array}{cc}x+y=2 \\ 2x+2y=4 \end{array} \right\}
    You have 2 equations, 2 varaibles.
    But no unique solution?
    Why?
    Because they are not linearly independent:
    Meaning, linear dependent:

    Let, the vectors,
    \bold{x}_1=<1,1>
    \bold{x}_2=<2,2>
    The coefficients of the system.

    Then we want to know "can be find non-zero (non-trivial) solution" to:
    k_1\bold{x}_1+k_2\bold{x}_2=\bold{0}
    Of couse, k_1=k_2=0.
    But I ask different solutions (non-trivial).
    And we can,
    2<1,1>-<2,2>=<0,0>.
    That tells us that the system has no unique solution.

    Another way, is through the determinant,
    \left| \begin{array}{cc} 1&1\\2&2 \end{array} \right|=0
    Since it is zero there cannot be a unique solution.
    ----
    Thus, I am asking that the static equilbrium equation provide us with enough equations for unknowns. But how do we know it solves?
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    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    It is a linear algebra question.

    Note the equations,
    \left\{ \begin{array}{cc}x+y=2 \\ 2x+2y=4 \end{array} \right\}
    You have 2 equations, 2 varaibles.
    But no unique solution?
    Why?
    Because they are not linearly independent:
    Meaning, linear dependent:

    Let, the vectors,
    \bold{x}_1=<1,1>
    \bold{x}_2=<2,2>
    The coefficients of the system.

    Then we want to know "can be find non-zero (non-trivial) solution" to:
    k_1\bold{x}_1+k_2\bold{x}_2=\bold{0}
    Of couse, k_1=k_2=0.
    But I ask different solutions (non-trivial).
    And we can,
    2<1,1>-<2,2>=<0,0>.
    That tells us that the system has no unique solution.

    Another way, is through the determinant,
    \left| \begin{array}{cc} 1&1\\2&2 \end{array} \right|=0
    Since it is zero there cannot be a unique solution.
    ----
    Thus, I am asking that the static equilbrium equation provide us with enough equations for unknowns. But how do we know it solves?
    Yes we know what what linear independence means, the obscurity is in
    what you mean by the static equilibrium equations.

    For static equilibrium we require that the sum for the foces on a body/system
    is zero, as is the sum of the torques on the body/system.

    RonL

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Can someone show that the static equilibrium equations in two and three dimensions on any body are linearly independent.
    The answer is no.

    I can't provide you with an example off the top of my head, but it is quite possible that the static equilibrium equations do NOT give enough equations to solve a problem in general.

    The trick is that, in a static equilibrium case, we have two equations at our disposal: Newton's Second Law and it's rotational analogue, the torque - angular acceleration equation. The nice thing about the last equation is that if the system is not rotating we can pick any point as an axis of rotation and technically get a new equation for our system. The problem is that in practice you often find that two different points for the axis of rotation can produce linearly dependent equations.

    There may be a theorem out there to say just how many linearly independent equations you can get from a system, but I've never run across one. I have, however, run into problems in Intro level Physics where the system is "overloaded" with the unknowns...there simply aren't enough equations to solve the system in terms of constants such as the masses of the objects in the system and the location of the CM of those objects. (This doesn't necessarily imply that the equilibrium doesn't exist, it simply means that more advanced techniques or special symmetries of the system need to be exploited to obtain a solution.)

    -Dan
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