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At 14:29 -0500 11/06/2010, brian whatcott wrote:
Four concrete cases:
1) 1 cc of lead is suspended on a fine thread in air
lead scale shows 11.3 gm (using lead density = 11.3 gm/cc)
Scale under brimming full aquarium reads x gm
2) 1 cc of lead is suspended at mid water depth
lead scale reads 10.3 gm ( using water density of 1 gm/cc)
scale under aquarium reads x gm)
3) 1 cc of lead rests on sandy bottom of aquarium.
lead scale reads 0 gm
Aquarium scale reads x + 10.3 gm
4) edges of 1 cc cube of lead are sealed to the aquarium floor excluding
all water beneath.
Lead scale reads 0 gm
Aquarium scale reads x + 10.3 gm
Floor Plate stress distributions:
Case 1&2) uniform area loading.
Case 3&4) increased area loading under cube, area loading elsewhere
unchanged .
(How am I doing?)
I don't think so.
Since no mass has been removed from the picture, something has to
support it. If it's not the string, it has to be scale under the
tank. So in case 2 the aquarium scale reads x + 1 gm, and in cases 3
and 4, the aquarium scale must read x + 11.3 gm. The upshot is that
this isn't going to tell us anything about how the buoyancy is
working.
The difference is that in case 3, the water is still providing 1 gm
of offset through its bouyant effect but that additional force has to
be ultimately provided by the aquarium structure which is not
supporting a slightly larger volume of water (which shows up as the
water displaced by the lead), while in case 4, the lead has become an
integral part of the aquarium itself and so the aquarium structure is
providing 100% of the support for the lead, and there is no buoyant
force.
There is, of course, no additional water in the tank, so the force is
provided directly in case four, and indirectly in case 3.
The situation is that to have a buoyant force there *must* be a
surface of the object that is capable of being pushed upward by the
water pressure. If there is no such surface, then the tank must
provide the required force directly. In either case the reading on
the aquarium scale will not change
If you do this with a 1 cc piece of impervious cork (I have no clue
what its mass might be), you should get results that are consistent,
but reflect the fact that the cork is less dense than the water it
will displace.
Hugh
--
Hugh Haskell
mailto:hugh@ieer.org
mailto:haskellh@verizon.net
It isn't easy being green.
--Kermit Lagrenouille
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