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I Something about Archimedes' Principle

  1. Jul 14, 2016 #1
    Hello,

    I was reading about Archimedes' Principle on the wiki ( https://en.wikipedia.org/wiki/Archimedes'_principle ) and a question popped up regarding the upthrusting force (also called buoyant force) that the liquid exerts on an object that is in the liquid.

    I understand that the upthrusting force is equal to the mass of the displaced liquid caused by the object in it. This force is exerted by the liquid right beneath the object. However, I was wondering what exactly causes this upthrusting force.
    I first thought that as soon as the object falls in the liquid, it compresses the liquid beneath it and thus increasing the pressure of that liquid beneath which results in a force upwards against the object. In reality however, the object actually pushes the liquid on the sides and the liquid rises in height.
    This means that the object isn't actually compressing the liquid beneath it (but just pushing the excess away), thus there shouldn't be an increase in pressure in the liquid beneath the object (the same amount of molecules is there).

    If that's so, then where does this upthrusting force come from other than pressure?
     
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  3. Jul 14, 2016 #2

    jambaugh

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    Hmmm.. how to explain. First you should avoid thinking in terms of "causing a force" when considering static systems. Rather think more in terms of balancing of forces necessary for the situation to be static. (Cause and effect is better applied to dynamic situations as in what causes a particle to change its direction of motion...)

    Now take the boat and ask why doesn't it sink further into the water. The answer is that in general it might but in so doing it will displace water. It also will be deeper into the water where the pressure is greater and the force on the bottom of the boat must increase.

    If you are not sure why the pressure deeper down is greater remember that water at a given depth must support the water higher up just like the boat so deeper water supports more weight over the same area and must have higher pressure.

    Once you have these facts you have the balance of weights since for the boat to go down the water it displaces must go up and vice versa.

    What you say about "compression" isn't quite right as the fluid need not be in any way compressible. I believe you're really thinking "pressible" as in being a fluid and thus able to accept and transfer pressure.

    To get into the "cause and effect" of that uou need only think about the atomic substructure of the fluid which is dynamic... a bunch of particles bouncing randomly off each other... the randomness causes the average forces (in a very small volume) to be distributed evenly in all directions in proportion to the areas across which the moving particles may transfer momentum.
     
  4. Jul 14, 2016 #3
    JohnnyGui,

    Are you comfortable with the idea that the water pressure increases with depth because it is supporting the weight of the water above it?
     
  5. Jul 14, 2016 #4

    jambaugh

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    In a static setting yes.
    In a static setting yes. Again this is not a "cause and effect" but rather a "solution to a constraint" type of "because".

    A fluid by definition cannot sustain a shear force, so the only sustainable force on a level cube of fluid is the normal pressure forces and gravitation. Stack em, add em up, and thar ya go.
     
  6. Jul 14, 2016 #5
    Thanks a lot for your reply. I see that you're partly associating the boat not sinking further with the pressure being higher at depth. Can I say that pressure is the cause lifting the boat up?

    I think I'm more focused on understanding the "cause and effect" of all this. Let's say we push a cube into the water. Can I say that the deeper I push the cube, the more particles bounce in a given time against the side of the cube that is pointed downwards? If that's the case, what exactly makes the number of particles that bounce against that side of the cube increase? Because if the cube pushes the excess water to the sides, then the number of particles bounces against the underside of the cube in a given time should stay constant. Unless it's the increasing pressure with depth that's causing this. But if we talk about pressure, doesn't this mean in a sense that water is somewhat compressed down there since its density is increased?
     
  7. Jul 14, 2016 #6

    jambaugh

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    Well... not necessarily "more particles" but an aggregate of that and the particles moving a bit faster and imparting more momentum when they bounce. And you are right in that there must actually be some level of compression i.e. volume change in practice. But not necessarily in principle one could have the pressure effect rather an increase in temperature keeping the number of particles per unit volume exactly the same, but they'd be moving faster causing more pressure.

    Whether it's faster particles or more particles causing the higher pressure doesn't matter in Archimedes' principle. What matters is that things balance out whatever the cause.
     
  8. Jul 14, 2016 #7
    I'm having this habit of trying to understand the exact cause of things, which is sometimes tiring :P.

    So if I understand correctly, the particles are moving faster beneath the cube the deeper down you push it because the pressure gets higher with depth? And even though there's enough room around the cube for the excess water to "escape" to, there's a net higher weight on the water beneath the cube (weight of cube + "weight of the water above" instead of just the water above) which makes the pressure down there even higher than before?
     
  9. Jul 14, 2016 #8

    Dale

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    I don't think that viewing a liquid as particles is a very useful mental model. It works well for gasses, where the interactions between particles are limited. For liquids the interactions are strong enough that they simply don't behave like the classical point particle approximation.

    If you really want particles for some reason, then you should limit yourself strictly to buoyancy in (ideal) gasses.
     
  10. Jul 14, 2016 #9

    jambaugh

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    faster or there's more of them or both... which is an unimportant detail to understanding the titular principle. Archimedes didn't need an atomic theory of fluids to develop it. (And Dale you're right... my hope was, in bringing it up was to show you don't want to/need to pay attention to this level of detail the "cause and effect" is a distraction) and JonnyGui, the cause and effect is a distraction. Your habit is a bad habit in this case.

    eeehhh... this isn't a great way of thinking about it. In an arbitrarily large "sea" the boat may raise the water level an arbitrarily small amount (though of course always by the fixed volume it displaces). Don't focus on the pressure at a given level changing because the surface rises with displacement. (and my apologies if this is not what you meant.)

    Focus on the static problem. The pressures on the surface of a volume of water (or boat) must balance the weight of that volume (or boat). At the heart of it is Newton's Law: F=ma. Since acceleration, a=0 in the static case, so too must net force.

    When I mentioned increase in pressure I was referring to the fact that bottom of the boat would sink deeper if the buoyancy was less than the weight. (Or I meant to imply that.) There is some equilibrium point (unless the boat as a whole is too heavy and sinks to the bottom). Where that equilibrium point is is the other part.

    Replace the volume of the floating boat below its water line with an identically shaped volume of water and the pressures at various points on boundary of these two volumes are identical. They also, given the static situation, must result in a net upward force equal to the weights of the two supported objects. Eureka! The volume of water and the boat weigh the same.
    (if the point by point pressures in the two cases were not equal then there would be a different net forces on the remaining body of water. Both could not be in equilibrium. That's a subtle argument and it's easier to reason that pressure of a liquid body is a function of depth only which gives you the same result.)
     
  11. Jul 14, 2016 #10

    jambaugh

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    I do see your point... the water could be supporting the weight of the boat above it.. :)
     
  12. Jul 15, 2016 #11
    This is exactly what I'm struggling with. If there's a displacement of water because of the boat, then the net pressure of water beneath the boat should be the same (it only carries the boat's weight), and yet as the boat sinks more in the water the buoyant force increases until there's an equilibrium that compensates for the weight of the boat. Where does this increase in buoyant force come from if the pressure of water beneath the boat doesn't change at all?

    I'm sorry if I've missed your explanation on this question.
     
  13. Jul 15, 2016 #12

    Dale

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    Be very clear here, please. Exactly which two scenarios are you comparing and claiming are the same?
     
  14. Jul 15, 2016 #13

    Nidum

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    The pressure in the water increases with depth in any case even if there is no boat present .

    Think about a very simple boat consisting of a short vertical axis cylinder with a flat bottom . As this boat sinks further into the water the pressure on the bottom surface increases thus increasing the upward force on the boat .
     
  15. Jul 15, 2016 #14
    I was struggling with why the water pressure beneath the bottom surface of the boat should increase with depth because the only thing that this body of water carries is the weight of the boat all the time. Even if the boat sinks further, any extra weight in the form of excess water is pushed to the sides by the boat so there's still a net weight of the boat that is being carried by the water beneath it.

    It is as if you're decreasing the volume of the water beneath it (= the boat sinking further), but at the same time you're actually removing water from that volume (the boat pushes water away as it sinks) and thus the pressure beneath the boat would stay the same. Unless ofcourse the increased water height on the sides causes an increase in pressure on the water beneath it and in turn exerts a force against the body of water that is beneath the boat.

    @Dale : I've made a simple drawing regarding this theory to explain what I mean (you need to zoom in to see the text): https://www.dropbox.com/s/wtx6um71apt7kbe/Archimedes.jpg?dl=0

    Could this theory be correct on why the buoyant force increases with depth?
     
  16. Jul 15, 2016 #15

    Dale

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    In your drawings h is irrelevant and can be made to go to 0 simply by making the water container very wide.

    Do you understand that, even without the boat, the water pressure increases with depth? The presence of the boat is not what drives this effect, just the weight of the water itself.

    Do you know the formula for calculating the water pressure as a function of depth. (in the fluid itself without any boat)
     
  17. Jul 15, 2016 #16
    Yes, I do understand that. Regarding the h being irrelevant; what I meant is that the increase in height of water in any volume is what eventually increases the pressure of the water beneath the boat as explained in my drawing. Without the boat, the increase in pressure with depth is because of the water above it.

    If you make the container very wide such that the water height is 0, this is just according to the previous container. In that wide container, the water would still decrease in height (albeit very little) if you take out the object.
     
  18. Jul 15, 2016 #17

    Nidum

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  19. Jul 15, 2016 #18

    Dale

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    I understand that is what you meant, but it is wrong. For example, suppose that the container itself expands or contracts in order to maintain an exactly constant height. According to your suggestion, objects would sink in such a situation. Again, h is completely irrelevant to this. You are preventing yourself from understanding by focusing on it.

    What is the formula for pressure as a function of depth, in the fluid itself without the boat?
     
  20. Jul 15, 2016 #19

    russ_watters

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    Oddly, it seems like you were closer yesterday than you are today. Somehow you now think the weight of the ship matters. But going back to yesterday's's start:
    The upthrusting force comes from the pressure only. There is no increase in pressure when the object is placed in the water: the pressure in the water (at a certain depth) is the same whether there is other water on top of it or an object on top of it. That pressure is where the buoyant force comes from.
     
  21. Jul 15, 2016 #20
    Ah ok, I think I get it now. So the weight of the object that the water beneath the object is carrying, is the same as the weight of the water that was above it before the object, correct? Thus the pressure of the water beneath the object is constant.

    Also, thanks @Nidum for drawing this, this helped me understand it better.


    This leads me to one last thing though; if the container (where the water is in) is fixed in volume, shouldn't the water displacement, caused by the object, lead to the fact that every other body of water in the container that doesn't carry the object above it, eventually carry that extra displaced volume of water? As a result, this would mean that the water beneath the object is just carrying the weight of the object, while the other bodies of water at the same water level (that is, at the level of water beneath the object) would carry the weight of the object in the form of water plus the weight of the displaced water. Shouldn't this lead to the object rising a bit until an equilibrium is achieved since the other bodies of water are trying to even out that extra displaced body weight?
     
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