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[Phys-l] line of action == locus of constant lever arm projection



On 09/06/2010 09:04 AM, Edmiston, Mike wrote:
once a line of action (also called line of force) is defined, it can
act on an object by attaching to the object at any point along the
line of action and will end up producing the same moment about O no
matter what point of attachment (along the line of action) is chosen.
I don't seem to find this idea in books such as Tipler/Mosca or
Serway/Jewitt.

1) This is turning into quite an interesting discussion.

2) The line-of-action idea is familiar to me, although I
never knew it had a name.

It is based on a simple property of the wedge product, or
(gasp) the cross product (choke, choke).

That is, when before taking the product of A with B, you
can project off any component of A that is parallel to
B ... or vice versa ... but not both. You keep only the
perpendicular component.

Some books mention that it is possible to project the
force, keeping only the force-component that is perpendicular
to some pre-chosen lever arm. But it works the other way,
too: You can project the lever arm, keeping only the
component perpendicular to the force.

The aforementioned "line of action" is just the locus of
constant lever-arm projection ... unless I am misunderstanding
something, which wouldn't be the first time today.

This projection property is arguably the easiest way to
calculate wedge products. I suggest this method at
http://www.av8n.com/how/htm/motion.html#eq-torque-def