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



Yeah - you're right - I just had to poke at that old hornet's nest :-)

However, the "buoyancy" thread has been very useful. The fact that so many people posted is a testimony to phys-lers passion for understanding and getting things right. It's better to argue these things out here among friends than to be unprepared when asked by our students.

Bob at PC

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From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bernard Cleyet [bernardcleyet@redshift.com]
Sent: Saturday, November 06, 2010 11:17 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Finishing up with that Dead Horse.

Ha ha!

Someone wrote to another list (I think in response to my suggesting he "join" phys-l.) that the phys-lers were just a bunch of nitpicking profs, and had soon quit.

Well, to quote Robert Frost when asked about the mechanics of verse (rhyme, meter, etc.) suggesting why bother, he replied NO! We glory in it.


However, since this (weight) has occupied us several times, I think we can define it by majority vote, and then not argue.


bc on second thought, thinks maybe Bob just jokes.


On 2010, Nov 06, , at 18:56, LaMontagne, Bob wrote:

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

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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.


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