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Re: [Phys-l] Finishing up with that Dead Horse.


Ah, the next long thread! There is a 3rd mis-definition of weight found
in this month's TPT.
Currently, I'm in the "weight is the gravitational force" camp. The
scale under your feet simply measures the normal force that the scale
must exert to keep you in static equilibrium (no bouncing allowed,
please). If we don't agree on definitions, how can we possibly teach
anyone about this stuff.

I need more reference to understand your statement that there is a "3rd mis-definition" in the TPT article?? What were the first two?

I'm not sure that there is any such thing as a "mis-definition" of a definition, unless it is somehow ill-posed or not self-consistent. You may not agree with a definition and you may have your own reasons for preferring another definition, but that is different than something being a mis-definition. Is the TPT article definition ill-posed in some fashion or not self consistent?? If so, in what way?

Weight is gravitational force is a well-posed definition; but I will need some clarification on how to calculate it?

GMm/R^2 or mg

If mg, what's g?

The TPT article definition is well posed, and we calculate the weight via mg and g is the free-fall acceleration in the local frame of reference for which we desire the weight.

Joel Rauber


-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of Bill Nettles
Sent: Monday, November 08, 2010 1:14 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Finishing up with that Dead Horse.

Ah, the next long thread! There is a 3rd mis-definition of weight found
in this month's TPT.
Currently, I'm in the "weight is the gravitational force" camp. The
scale under your feet simply measures the normal force that the scale
must exert to keep you in static equilibrium (no bouncing allowed,
please). If we don't agree on definitions, how can we possibly teach
anyone about this stuff.

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of LaMontagne, Bob
Sent: Saturday, November 06, 2010 8:57 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Finishing up with that Dead Horse.

Whoa - weight is the force of a scale under your feet. The Gravitational
force exerted by the earth is the gravitational force exerted by the
earth. If you jump off a chair you are "weightless" - but gravity is
still operating.

Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of William Robertson
[wrobert9@ix.netcom.com]
Sent: Saturday, November 06, 2010 8:02 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Finishing up with that Dead Horse.

Two major points. I think we should never, ever use the concept of
"weight reduction" when talking about buoyant forces. No matter what
you do to this cube, as long as it remains where it is compared to the
center of the Earth, it will have the same weight because weight is
the gravitational force exerted by the Earth. When we are not careful
with our language, then all we do is confuse students.

Second, since you're still talking about a cube on the bottom but
don't want the "complications" of a "suction effect" (bad choice of
words for my taste), that means the cube is not really on the bottom.
There must be water underneath it if there is to be a buoyant force.
There will be no difference in the "flex" of the bottom of the
aquarium unless you tether the cube to the bottom or remove all fluid
between the cube and the bottom of the aquarium. I thought this was
cleared up, but no? If you want to know what's happening to the bottom
of the aquarium, you look at forces acting on the aquarium bottom. If
you want to know what's happening to the cube, you look at forces
acting on the cube. It seems you are confusing the two systems, and at
the heart of the second law is choosing a system and then assigning
forces accordingly.

Bill


William C. Robertson, Ph.D.


On Nov 6, 2010, at 9:36 AM, Chuck Britton wrote:

The flexing of the aquarium bottom can be used to quantify the
'weight' of an object placed on the bottom. This procedure can be
used whether the aquarium is filled to the brim with water or is
empty.

A lead cube will 'weigh' less if the aquarium is filled to the brim
with water than if the aquarium were empty.

It will weigh less by an amount equal to the weight of the displaced
fluid.

A spring scale could be (often is) used to test this weight reduction
also - but something called the 'suction cup effect' might complicate
things when the cube is on the bottom. I want to avoid the
possibility of any such 'suction cup effect'.

Is my weight reduction hypothesis correct?

(Full to the brim - so , yes, that weight of water WAS removed from
the system.
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
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_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l