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.