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Re: [Phys-l] definitions ... purely operational, or not



Your rising hot air balloon is not in free-fall. Free-fall means, that there are no other influences acting on the object then gravitational. So to determine the weight of the hot-air balloon with the "Bartlett" definition we need to evacuate the atomosphere, or at least the local region where you are determining the weight force on the balloon, measure its free fall acceleration in the desired frame of reference and then compute m*g_free-fall.

So I conclude that it is not problematic but is well-defined.



-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of carmelo@pacific.net.sg
Sent: Wednesday, November 10, 2010 8:31 AM
To: phys-l@carnot.physics.buffalo.edu
Subject: Re: [Phys-l] definitions ... purely operational, or not

Quoting Hugh Haskell <haskellh@verizon.net>:

Buoyant forces are a bit tricky, and the recent thread here has
clearly shown. I think I would look at this from the POV that gravity
is still acting on the object, as mg, but the support force that
comes from the pressure differences is distributed over the surface
of the object. That won't be measured by a bathroom scale in contact
with the object, but it will be reflected by a lager reading on the
bathroom scale that is supporting the water tank in which the object
is immersed. So, I guess that I would not call the object in buoyant
equilibrium weightless (but I'm still thinking about it).

Suppose we want to know the weight of a hot-air balloon...
When we release this balloon, Oops, it has a "free-ascend" instead of
"free fall". Does this mean that the weight of a hot-air balloon is
negative?

Is not this definition of weight using "free fall acceleration" still
problematic? :-)


Best regards,
Alphonsus

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