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Yesterday my friends and I were solving this easy-looking question:

A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube:

(A) Will Increase

(B) Will Decrease

(C) Will remain the same

(D) Will become zero

At first we thought the answer to be (B) considering the buoyant force acting opposite to the weight of the cube. But then, a question popped up in my mind: "What causes this buoyant force even if no fluid was actually displaced?"

So I referred to Archimedes' Statement:

Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. — Archimedes of Syracuse

But according to the question, no such fluid was displaced by the block!

So, here are my questions:

  1. What is the basic reason behind buoyancy? Is it the pressure difference in fluids or the 'liquid-displaced' theory?

  2. What must be the answer to the question (That I was solving) ?

  3. Considering the case that you may say- "The reason behind buoyancy is the pressure difference in the fluids"; Will there be no buoyant force applied to the iron cube as there is no fluid layer present below the cube i.e. no pressure difference?

(Considering the iron cube is VERY dense and thus not allowing any fluid layer to push its way below it)

  1. Or you might add another statement to the above question that-"Yes, buoyant force is applied as there is always a fluid layer present below it..." .Please do explain yourself.

  2. If your answer is according to the Principle given by Archimedes of Syracuse , will there be no buoyant force on the cube as the cube didn't actually displace any fluid? Just like the case where you may consider it to be a part of the vessel itself.

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  • $\begingroup$ Your second question is clearly one of those "please do my homework for me !" types which are not welcome here. Also consider making your question more short and crisp. Please re-read How do I ask a good question? $\endgroup$ – Gaurav Dec 7 '14 at 9:18
  • $\begingroup$ Well actually, the second question is a reference for the answerer to prove his/her argument and not just "please do my homework for me!" as you state it. PS: I already do have an answer key. And the answer given is (C). I just raised this topic as it seemed to be a good topic to discuss over. $\endgroup$ – Priyansh Sangule Dec 7 '14 at 13:02
  • $\begingroup$ C? Well, that's an interesting answer. Does the answer key, by any chance, have some wordy explanation? $\endgroup$ – neverneve Dec 7 '14 at 14:05
  • $\begingroup$ Unfortunately it doesn't. But I think it perhaps considers that the block is JUST immersed, i.e. we don't consider any liquid above the cube. $\endgroup$ – Priyansh Sangule Dec 7 '14 at 15:12
  • $\begingroup$ It may, in that case the answer would in fact be C. $\endgroup$ – neverneve Dec 7 '14 at 17:19
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  1. The reason behind buoyancy is the pressure difference in fluids. More specifically, the difference in hydrostatic pressure on different levels, since the hydrostatic pressure of water increases with depth ($p=\rho gh$ wher $h$ is the distance below the surface). The part of the body which is subject to higher hydrostatic pressure will be pushed more upwards than the part of the body subject to lower hydrostatic pressure will be pushed downwards. Also, in space there would be no buoyancy because there would be no gravity, hence no difference in hydrostatic pressure.
  2. The answer is actually A. Since, as you said, no water gets below the cube, there will be no force directed upwards due to pressure. The liquid will actually push the cube downwards, to the floor, so the force on the bottom of the vessel in contact with the cube will be the cube's weight + the weight of the liquid above.
  3. Yes, since the reason behind buoyancy is the pressure difference, if there's no water below the cube, there will actually be no buoyant force. The liquid will push the cube downwards.
  4. I would argue that the cube does actually displace water (if you removed it, the water level would decline), but in this case you're right when you treat it as a part of the vessel, since there's no water below it. Every body immersed in a fluid will displace some part of it (because the fluid was there in the first place, when you immersed the body it had to go somewhere), but the Archimedes principle saying the force on the body will be equal to the weight of the displaced fluid is only right when the body is completely immersed in it. Actually, the derivation of this law is fairly simple:

Let's consider a volume of water $V$, let it be in the shape of, for instance, a potato. The water is subject to two forces: its weight and the buoyant force. Since it doesn't move, we conclude that the buoyant force is equal the the weight of $V$ of water. Now, let's say we displace the water with a potato of the same shape and volume $V$ - it occupies exactly the same space as the water discussed. Since the buoyant force is the result of the difference in pressure and does not depend on any characteristics of the body, it will be exactly the same as it was with the water volume. That's because the pressure around the body doesn't change. That's why we conclude the bouyant force is equal to the weight of the displaced fluid.

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  • $\begingroup$ Body does NOT need to be fully immersed for Archimedes to apply - it is indeed the weight of the displaced liquid that equals the buoyancy and that is still true with a partially immersed object (think a boat). $\endgroup$ – Floris Jan 2 '15 at 0:46
  • $\begingroup$ Yes, of course, I'm sorry. What I rather meant was that the body in this particular situation has to be "touching" water on all sides for the buoyant force to equal the weight of displaced fluid. $\endgroup$ – neverneve Jan 2 '15 at 17:48
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Your answer key is wrong. The force will not be unchanged. When the cube is fully submerged there will be a weight of the liquid on top of the cube which is pressing down.

Imagine for a moment the the cube has exactly the same density as the water around it. In that case it is clear that as the water level rises the pressure on the bottom of the vessel will increase. It will be no different when the cube is made of iron.

Now when the water level has not yet reached the top of the cube the question is more interesting. If the walls of the cube are completely vertical and no liquid can penetrate the gap under the cube, then there will be no net force on the cube due to the water.

This gets to the heart of your question. Buoyancy is caused by a pressure difference between surface of the object facing (partially) up or down. When you divide an object into infinitesimally narrow columns, each column having area $dA$ and length $\ell$ then the force on that element due to liquid pressure (given that pressure increases with depth as $\rho g h$) is $\ell dA \rho g$. And when you integrate that over the entire object the answer is the volume of the object ($\int \ell dA$) multiplied by the weight of unit volume of liquid $\rho g$ - giving the familiar "Weight of displaced liquid" relationship.

Note that when the bottom is sealed, non-vertical side walls will give rise to some (reduced) force - whether this is up or down would depend on the angle.

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  1. B is correct. It does NOT matter where the object in the fluid is, only how much volume lies under the surface of the liquid which = the "Volume" of fluid that is displaced (unless density changes at varying depths, then the force will increase in higher density areas and vice versa.)

  2. The buoyant force is not a result of pressure differences, but of density. For example, helium floats in air due to a buoyant force between the denser air and the less dense helium. A better way the look at this buoyant force is to not think of it as necessarily a force that pushes up, but more like a force that pushes heavier things down because the gravitational pull per molecule is greater on a more dense substance.

  3. The lead block does displace water. Since the entire block submerges, the volume that is displaced is equal to the volume of the block. The resultant buoyant force is equal to the mass*gravitational acceleration, which is probably pretty negligible compared to the weight of lead.

    Put a scale underwater or a fish scale or just a spring and see what happens. Do rocks or anything get heavier as they sink lower to the bottom?

Do Not use amount people, amount does not specify any unit and it turns out they make a difference.

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    $\begingroup$ How can a force be a result of density? Clearly objects with smaller density float in liquids of higher density (as helium floats in air), but that's because the buoyant force is equal to/greater than the weight of the object. From another perspective, if you put in the water a stone and a piece of styrofoam, both of the same shape and volume, the stone would sink and the styrofoam would float. The buyoant forces acting on them would be the same (what we can see from the formula for the buoyant force $F=\rho_{liquid} Vg$). $\endgroup$ – neverneve Dec 7 '14 at 14:17
  • $\begingroup$ This is very wrong. Buoyancy is absolutely the result of the difference between forces pushing down and forces pushing up. Pressure is a function of depth because of density of liquid and gravity. So density plays a role but not in the way you state. $\endgroup$ – Floris Dec 8 '14 at 4:00

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