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Re: [Phys-l] buoyancy on a submerged pole



In case I wasn't clear in my post, I was referring to the force acting on the bottom of the aquarium, not on the box.

Bill


William C. Robertson, Ph.D.


On Nov 3, 2010, at 8:49 PM, John Mallinckrodt wrote:

On Nov 3, 2010, at 6:42 PM, Chuck Britton wrote:

Great - now I'm beginning to understand!!!
If there is no water under the box we can't call the upward force
Buoyancy even tho it's still equal to the weight of the displaced
water.

Right. After all, why would you want to call a force that is exerted on the box by the bottom of the aquarium a "buoyant force"? It's simply a force of support, a "normal force" if you want.

But that force is NOT equal to the weight of the fluid displaced by the box. It's equal to the sum of a) the weight of the box and b) the magnitude of the downward force exerted on the box by the fluid.

Again, the box is at rest and there are only three forces acting on it: Two downward (due to contact with the fluid and gravity) and one upward (due to contact with the bottom of the aquarium.)

If we let a monolayer of water seep under the box (but keep it tied
down with a thread) we can then call the upward force Buoyancy. Same
amount of force - different name. OK.

(I don't know that a monolayer would be sufficient, but I'm willing to call that a nit.) In this case there are still only three forces: Two downward (due to tension in the string and gravity) and one upward (due to contact with the fluid.)
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