Re: A twist on the hourglass problem

[brian whatcott <inet@INTELLISYS.NET>]

At 06:19 11/3/99 -0500, you wrote:
> ...
> He knew the mass of the system didn't change (yeah!), but he lacked a
> "mechanism" to explain why/how the "free falling sand" was not an
> influence. ... he removed all but one grain of sand from
> the hourglass.  His question is:  While the mass of the system doesn't
> change, W=mg and the weight shouldn't change.  BUT, why?!
> .... No matter what
> I did, he still couldn't grasp the "mechanism" which made the statement
> true that thed weight didn't change -- given his other objections....

> Peter Schoch

This topic was discussed recently.
If he defines weight as the force which resists the tendency of objects
to fall in a gravity field, he can see that when a grain drops in a vacuum
it is not supported pro temp, so its weight is removed from the container
 until it hits the base, at which time it returns the
 missing force (=weight) times time.

Respectfully
brian whatcott <inet@intellisys.net>
Altus OK