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Path: utzoo!linus!decvax!harpo!eagle!mhuxi!houxm!ihnp4!ihuxr!lew
From: lew@ihuxr.UUCP
Newsgroups: net.physics
Subject: Re: Hourglass (spoiler)
Message-ID: <679@ihuxr.UUCP>
Date: Mon, 3-Oct-83 14:47:45 EDT
Article-I.D.: ihuxr.679
Posted: Mon Oct  3 14:47:45 1983
Date-Received: Tue, 4-Oct-83 00:59:04 EDT
Organization: BTL Naperville, Il.
Lines: 30

You have to really BELIEVE Newton's second law for this one. Take the
hourglass (including sand) as a system. The sum of the external forces
equals the acceleration of its center of mass.

While the hourglass is running its center of mass has a constant
velocity. There is a short downward acceleration as it starts
and a short upward acceleration as it stops.

This means that it will "weigh light" for a moment as it starts, and
conversely. When it weighs light, the vector sum of its weight (force
of earth on hourglass) and the force of the scale on the hourglass
will be in the downward direction.

Note that if you regard the top chamber as a "rocket", it has zero thrust
becuse of a zero exhaust velocity. The impact at the bottom is just
great enough to account for the "missing mass" of the sand in freefall:

	V*(dm/dt) = (dm/dt)*T * g since V=g*T

During startup, the net force on the hourglass increases linearly from
0 to (dm/dt)*T*g during the time T that the sand takes to fall, then
drops to zero abruptly as the sand begins to hit bottom. The same thing 
occurs as the sand runs out, except the force is in the opposite direction.
With dm/dt ~ 1gram/minute and g*T = V ~ sqrt(2*1000 cm/sec2 * 5cm) = 100 cm/sec,
I estimate a peak force of about 1 dyne. That's about .001 gram-force.
Pretty subtle, but you should be able to measure it with the right scale.
I tried once on a crude balance with no success.  Of course, you have to
try for the heavy spike as it stops, as a practical matter.

	Lew Mammel, Jr. ihuxr!lew