In Archimedes' principle, is the buoyant force ever NOT equal to the weight of the floating object?

Question

I know that in Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced by the object. Since the object will sink until it has displaced fluid equal to its weight, why isn't the principle that the bouyant force is equal to the weight of the object? Isn't this true:

weight of object = weight of displaced fluid = buoyant force?

Answer 1

i dont believe so, the weight of the object is reduced, that is for sure, but it is more like

B + W(object in fluid) = W(object out of fluid)

in that respect, the water you displace isn't always going to equal your weight in water, because your density and volume really determine how much water you displace

say for instance we drop a cubic centimeter of gold into water. it should have some sort of boyancy force acting on it eqaul to the weight of 1 cubic cm of water

if we drop a cubic cm of bronze int the same tub, while it weighs less, it still displaces the same amount of water as the gold

Answer 2

An object can displace more or less fluid than its weight. This can occur if the object is in vertical motion with respect to the fluid. If you drop a closed container in water, it momentarily displaces first more, then less water.

It could occur also if there were centrifugal force applied to the object and the fluid. Sling a bucket of water in a circle, and an object in it will have a heavier apparent weight.

Answer 3

The key word here is FLOATING. Assume you had an object whose DENSITY you could adjust. Your interpretion would be true ONLY when the density of the object = that of the liquid. For a heavier object, Archimedes principle still holds true; the bouyant force is the volume of the object* the density of the liquid.

Answer 4

Yes, if the object sinks then the buoyant force + weight feels on the bottom of the tank = weight

Answer 5

Archimedes is out for lunch. I'll answer when he comes