Re: Apparent weight

[Leigh Palmer <palmer@sfu.ca>]

John Mallinckrodt wrote:

> The problem with the "what a scale reads"  definition of "weight" is that
> you have to decide *which* scale you're going to read. Why should it
> necessarily be the one you're standing on?  What about the one between
> your head and the ceiling when you stand up in a room that is too short to
> accommodate your height?  What about the one between your back and the
> hand of a friend as she pushes you out the door?  What about the one
> between your neck and the rope wrapped around it as ...  Well, you get the
> picture.

John, you are quibbling when you know the answer perfectly well.
Position the item you wish to weigh in the middle of your laboratory.
It will drift towrd some bounding surface. Interpose your scale
between the item and that surface. After equilibrium is established
the reading on the scale will be the weight. Of course you have to be
careful! You must lubricate the scale platform to make sure there is
no transverse component of the weight missed, and you must assure
that the item is uncharged, or else that electric fields have been
excluded from the laboratory, and all the other precautions that
might be added. Those are necessary to an experimental determination
of the weight; they are counterproductive to establishing a grasp in
a student being introduced to the concept for the first time. I just
had this experience this semester in regard to the establishment of
the ideal gas temperature scale using a constant volume gas
thermometer. The Devil is in those details!

> Since *all* contact forces can be turned into scale readings, the only
> sensible procedure is to add the forces they indicate as vectors and say
> that the vector sum of all contact forces that you exert on other objects
> is your weight.  This brings you perilously close, however, to concluding
> that your weight is equal to your mass times negative your acceleration.
> Hmm.  What a simplifying idea that might be.  I wonder if anyone else has
> ever thought of it?

Even though I now see where you were headed, I'm afraid this won't
do. The question here is one which relates the concept of weight to
the force of gravity, so I'm afraid that much simpler solution is
out. Somehow the concept of weight must be accommodated.

(At least I'm glad to see that we're both on the same side.)

....and then he said, in a later posting:

> Because I haven't heard anyone explicitly correct the misconception that
> "microgravity" is just a weak "gravitational force" or that it is somehow
> related to "apparent weight" allow me to do so:  "Microgravity" refers to
> the tidal effects which are gravity's only essential contribution to the
> observable world and which are exceedingly small (hence, "micro") near the
> earth. In a freely falling frame like the shuttle--which eliminates the
> overwhelming and misleading appearance of a nearly uniform "gravitational
> field"-- we find it much easier to observe these tidal effects.

Now I am confused. I explicitly stated that the principle of
equivalence applied *locally*. So far as I know there's no locally
observable effect of a tidal force. By "microgravity" I mean
"microgee", or gravitational fields being less than or of the order
of 10^-5 N/kg. As John correctly points out, the tidal effects are
observable in the space shuttle (see the cited article), but I don't
think they are the effects that are studied in what are usually
termed "microgravity" experiments.

Leigh