2
$\begingroup$

This question already has an answer here:

The situation I am thinking about is that, for example, lets say that there is a block that weighs 10N outside of water, when it is put into water its apparent weight is 2N. From this it can be inferred that the buoyant force is 8N. So if the buoyant force is less than the weight, I would think that the object is completely submerged, resting on the floor of the container. Is this correct? and if it is how can there be a buoyant force, when there is no water below the block to be pushing it up if it is resting on the bottom of the container

$\endgroup$

marked as duplicate by Jon Custer, John Rennie, stafusa, Kyle Kanos, Qmechanic Jan 23 '18 at 23:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ you have to know the volume, so you could get the density plus the density of the fluid en.wikipedia.org/wiki/Buoyancy $\endgroup$ – anna v Jan 21 '18 at 8:18
  • 4
    $\begingroup$ Possible duplicate of Does an object need fluid under it to float? $\endgroup$ – Solomon Slow Jan 22 '18 at 2:22
  • $\begingroup$ There is the minor technicality that a "buoyant" object (weighs less than the equivalent volume of water) can "stick" to the bottom of a water-filled container because no water can get under it. Until the water gets under it there is no buoyant force operating on it. But the standard assumption when buoyancy is being discussed is that some small amount of water can seep in below the object, and it only takes a tiny amount to get things moving. $\endgroup$ – Hot Licks Jan 22 '18 at 3:02
0
$\begingroup$

It's a paradox: if the buoyancy--an upward force--results from the difference in water pressure on the bottom and top faces (the contribution from the side forces being horizontal), how can there still be buoyancy when there's apparently no water underneath the bottom face?

Consider what happens as the block approaches the bottom of the container: as the gap between the two shrinks, a slight excess in pressure develops under the block, forcing the water in the gap out the sides. This process continues until whatever high points that microscopically exist on the block and the container bottom begin to make contact with each other. As the contact force gradually increases, there is less and less weight left over to maintain the slight horizontal pressure gradient underneath the block, so the the rate at which water is squeezed out the gap decreases. The pressure under the block decreases, approaches that elsewhere on the bottom of the container, and when everything reaches equilibrium the pressure underneath the block, pushing it up, is the same as it would be if the bottom of the container weren't there.

If you do something to relieve this pressure, you can indeed get rid of the buoyancy force. If, for example, after the block has settled into place, you cut a hole in the bottom of the container just inside the perimeter of the bottom of the block, the block will no longer experience a buoyancy force. Instead, the net force exerted on the block by the water is downward: it's the weight of the column of water above the block.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.