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



*Agreed*

particulary if you have in the back of your mind that your scale is properly calibrated, there are no buoyancy affects to worry about, and many of the other items that John Denker mentioned in his criticism, etc.

I think that defining as M times free-fall-acceleration is a cleaner way to encapsulate all of that in a definition. I'd have no problem with defining it *my* way then telling a class that this pretty much amounts to saying what the "spring scale reads", where I can mention the above kind of caveats in passing.

Joel Rauber


-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of LaMontagne, Bob
Sent: Monday, November 08, 2010 4:48 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] definitions ... purely operational, or not

But the comment

"... the weight of a mass M in a specified frame of reference is M
times the free-fall acceleration in that specified frame of
reference."

is the same as saying weight is what a scale reads.

In the frame of the space shuttle the free-fall acceleration is zero and
a scale reads zero.

Bob at PC

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of John Mallinckrodt
Sent: Monday, November 08, 2010 11:50 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] definitions ... purely operational, or not

On Nov 8, 2010, at 8:40 AM, Scott Orshan wrote:

I'm fine with that definition. Now can we get the astronauts to
stop
telling everybody that they are weightless, and in Zero G?

But, with that definition they *should* be saying that they are
weightless.

"... the weight of a mass M in a specified frame of
reference is M times the free-fall acceleration in that specified
frame of reference."

John Mallinckrodt
Cal Poly Pomona
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