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Thread: Do "torques" (on a bar) work the same way as "weighted averages"?

  1. #1
    Dec 2008

    Do "torques" (on a bar) work the same way as "weighted averages"?

    "Torques" (on a bar) vs. "weighted averages"

    For example, suppose that I have a solid, horizontal bar
    that is 10 units long. And suppose that the bar is pivoting
    about a point half way along the length of the bar.

    Now suppose there is a mixed/'random' collection
    of weights sitting on the bar at different points along
    the bar.

    Now if some of the weights are on opposite sides of the
    pivot, it is clear that to some extent they would cancel
    each other out and, if I am correct, whatever is left over
    would form a rotational "torque".

    From my O-level physics I recall "torque" is basically
    "force x distance".

    So I presume that if you just added up all of
    the individual torques on one side of the bar, and
    subtracted it from all of the torques on the other
    side of the bar, that whatever would be left over would
    be would be the resulting torque that the bar would
    exert on any objectthat tried to stop it from rotating.
    (Am I right so far?).

    But my question is this:
    Would this resulting torque mathematically be the
    SAME THING as if you took a "weighted average"
    of all the torques?
    [And if not, is there anything simple one could
    do with mechanical levers etc to visualise what
    the heck a "weighted average" actually means
    in mechanical terms?]


    Shiperton Henethe

    One issue that concerns me is that if there is
    a huge weight placed at the very middle of the bar
    (i.e. immediately above the pivot point) it obviously
    doesnt affect the bar's torque
    (because force x zero is stil zero!)

    But would a "weighted average" be affected by
    a large weight in the middle of the bar.
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  2. #2
    MHF Contributor
    Aug 2007
    W1*D1 + W2*D2 + W3*D3 + ... + Wn*Dn

    This is the TOTAL force on one side.


    (W1*D1 + W2*D2 + W3*D3 + ... + Wn*Dn)/(W1 + W2 + W3 + ... + Wn)

    This is the "Weighted Average" distance on one side.

    (W1*D1 + W2*D2 + W3*D3 + ... + Wn*Dn)/(D1 + D2 + D3 + ... + Dn)

    This is the "Weighted Average" force on one side.


    You can see that they are very closely realated, but somewhat different animals.


    A wieght in the middle is of little consequence. With a distance of 0 from the balance point, the weight is discounted to zero.
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  3. #3
    Super Member
    Dec 2008
    What has this to do with business maths?
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