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

• Dec 19th 2008, 04:38 AM
Eric69
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?]

Cheers

Ship
Shiperton Henethe

P.S.
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.
• Dec 24th 2008, 09:11 PM
TKHunny
W1*D1 + W2*D2 + W3*D3 + ... + Wn*Dn

This is the TOTAL force on one side.

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(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.

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

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A wieght in the middle is of little consequence. With a distance of 0 from the balance point, the weight is discounted to zero.
• Dec 26th 2008, 04:57 PM
Mush
What has this to do with business maths?