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

[ship]

"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?]


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.

[Barb Knox]

In article

Post by ship
"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?).
Would this resulting torque mathematically be the
SAME THING as if you took a "weighted average"
of all the torques?

No. As you just discussed, it's the SUM of the torques (which is
admittedly a trivial weighted average, with all the weights the same,
but that isn't what you want).

The total torque is the weighted average of the WEIGHTS, where
(confusingly) the averaging weight for a given weight is its distance,
not its weight. :-)

Post by ship
[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?]
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.

No, because the averaging weight for that large weight is its distance
(not its weight), which is 0.