Re: Apparent weight

["Rauber, Joel Phys" <RAUBERJ@mg.sdstate.edu>]

I agree with what Roger says below,  I usually say something more than 
defining weight as what the bathroom scale says; by saying that the bathroom 
scale is a device that measures the normal force that pushes up on the 
object that is standing on it.  This naturally leads into applying Newton's 
laws as Roger describes below and avoids some of the "weightless" and no 
gravitational paradoxes (confusion) with the space shuttle, i.e. the 
necessary centipetal force for the orbiting is present provided for by 
gravity and is not confused with weightlessness.

BTW I don't introduce gravitational field as Roger does.  I simply say 
gravitational is calculate by "mg" or "GMm/r^2" and weight is what your 
bathroom scales say.  I then proceed to compute the bathroom scale reading 
for the person on the scale in the accelerating and non-accelerating 
elevator.  (No explanation given, for why you are standing on a bathroom 
scale in a public elevator; I leave that for the psychology classes).

> I think you are making it too hard, Rick. Leigh asked for an operational
> definition. What I offered was such a way of defining weight. I don't
> care to know how the bathroom scale works, nor do I think that is
> important. The number I get when I stand on it is my weight.

> I take the scale with me in the shuttle and "stand" on it and get zero.
> Therefore, I'm weightless.

> The force of gravity is acting on me in both situations, but I choose to
> not talk about the force of gravity on me as my weight in my beginning
> classes. Thus I avoid many problems. I have written before that in the
> equation F=mg that I refer to the g as being the gravitational field and
> introduce this concept early. F=qE then follows naturatlly.