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Buoyancy Force at the base

  1. Oct 11, 2006 #1
    I was doing a test and the there is a picture with the box at the bottom of the water.First they ask me to find the weight.I found it.Then later they ask me to calculate the buoyant force.My teacher said that there is no buoyant force at the bottom.But the Archimedes Principle states that the weight of object=buoyancy force.So which one is true?
  2. jcsd
  3. Oct 12, 2006 #2
    Firstly, that is not what Archimedes Principle about. It states the upthrust(buoyancy force) is equal to weight of fluid diplaced. Indeed, there is no such thing unless water can somehow get under the object (which it often does in most cases).

    You must know how upthrust works in the first place. You should have already covered how to calculate pressure in a fluid right? You see, there is an upthrust because there is DIFFERENCE in pressure because the pressure of the bottom is greater than the pressure at the top. Hence, we have a 'resultant pressure' which manifests itself as upthrust.

    You see, pressure in the first place results from the particles of the fluid bombarding (maybe hitting is a better word) the surface of the object. There is no fluid at the bottom, hence no particles, no pressure and therefore no upthrust.
  4. Oct 14, 2006 #3
    Yes. Archimedes principle states the above and not that upthrust is equals to the the weight of the object.

    The upthrust is equals to the weight of the object when it FLOATS. This is in accordance to the principle of floatation. However, at the base, there SHOULD be an upthrust.

    Think about this.. Imagine a light block that has a weight lesser that its upthrust. No matter how you push the block to the bottom of the water, it will still rise up to the surface. You dun expect the block to stick to the waterbed right? Why? This is because when u push the block down to the waterbed, upthrust is already there and hence it is regardless of whether there is difference in pressure or not, there will still be an upthrust.
  5. Oct 14, 2006 #4
    I just finished this chapter as well.

    I don't understand :

    "You see, pressure in the first place results from the particles of the fluid bombarding (maybe hitting is a better word) the surface of the object. There is no fluid at the bottom, hence no particles, no pressure and therefore no upthrust." ---QuantumCrash

    doesnt archimedes principle state that

    bouyancy force=the weight of fluid displaced

    well...if the object is right at the bottom of its container, didn't any fluid get displaced?!

    wouldn't the total of fluid displaced when the object is simply submerged half way through the fluid will equal to to the same amount of fluid displaced when the object reaches the bottom!?

    pls o pls correct me if i'm wrong!

  6. Oct 14, 2006 #5


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    Archimedes principle is wonderfully useful when properly applied, but buoyancy is not a fundamental force like gravity or electrical force. In order for an object to float, the force pushing from underneath the object must cancel the weight of the object plus any additional forces pushing it down from above. The beauty of Archimedes principle is that for any object of any weird shape the difference between the upward and downward forces on the object due to the fluid pressure surrounding it is just the weight of the fluid displaced.

    The pressure of a fluid acting on an object is the average force applied by molecules of the fluid bombarding the surface of the object. If there is no fluid below the object, there are no molecules to crash into the object.

    Fluids have no rigid structure, so any pressure is isotropic, meaning independent of direction. Solids have rigid structure, so pressure applied to one side is not spread uniformly throughout the material (anisotropic). As a result, an object on the hard bottom of a body of water does not feel the pressure of water at the bottom that it would feel if raised even slightly off the bottom.

    In reality, it is very hard to keep the water from leaking in between the object and the hard bottom, so even if the object momentarily stays on the bottom, it will soon be lifted. A porous bottom like sand will not be able to keep an object that normally floats on the bottom.

    You can do an experiment to demonstrate this with a small glass and a fairly large sheet of rubber material. Fill a reservoir of water (your sink will do). Invert the glass and let it fill with water until just enough air is left for the glass to float upside down. Press the sheet of rubber against the rim of the glass. You might have to put a bit of a dent in the rubber to squish a bit more water out of the glass. Now let the rubber sheet and the glass sink to the bottom. If the seal is good, the glass will remain in contact with the rubber sheet at the bottom of reservoir for a long time, in spite of the fact that the inverted glass parrtially filled with water is displacing more than its own weight of water. If the seal is not perfect, eventually water will seep into the glass increasing the pressure at the bottom of the air until the glass breaks free of the rubber and floats to the surface.
    Last edited: Oct 14, 2006
  7. Oct 14, 2006 #6
    so archimedes principle is just for any weird shape? how about regular shape objects?

    so wat about a using a non porous bottom like the bottom of the beaker of water?

    Correct me if i am askin stupid qns.
  8. Oct 15, 2006 #7
    Thank you to OlderDan for the detailed explanation.

    Sure, it might work, in the way using a plunger might work as well, if you think about it. How, chances are, it won't. Continuing from OlderDan's explanation, not only does the surface you are talking about would play a factor, but whether the object itself is porous plays a factor. Firstly, chances are if you do sink something to the bottom there will still be water molecules at the bottom and this may lead to leakage and a building of upthrust. In practice, the situation posted by kindaichi is more of a theoretical model with lots of If's and you won't necessaryly see it in common practice.

    Of course, here's something you can try, put a rubber block in a dry glass beaker and THEN fill it with water, chances are it will stay there. (of course you'll have to hold it down while you are filling the beaker.)
  9. Oct 15, 2006 #8


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    Archimedes principle works for any shape, including shapes for which it would be very difficult if not impossible to calculate the buoyant force from fluid pressure. The only shape for which this calculation is easy is a rectangular block of uniform density, where all the forces from pressure are either horizontal or vertical. Even for a block, the horizontal force varies with depth, so it is not trivial to calculate, but the horizontal forces do not contribute to buoyancy. Since the vertical force is exerted at surfaces of constant depth, the pressure over the surface is constant. Even for a simple shape like a sphere or cylinder (log), the pressure calculation is far more difficult than for a rectangular block.

    An object that would normally float placed on a non-porous bottom would have to make perfect contact to seal the water out of the space between the object and the bottom. If any water was able to leak into the space, more would surely follow and the object would eventually float away. In most cases, eventually would be a very short time.
  10. Oct 15, 2006 #9
    corect me if i'm wrong....

    this means that, there is no bouyancy force for a regular object when it reaches the bottom of a non-porous, eg, beaker's bottom.

    but there is bouyancy force for an irregular object or any object on a porous surface or when both of these factors are involved, when the object is in afluid.

    Am i right?

  11. Oct 15, 2006 #10


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    Yes, if by regular object you mean one that has only vertical walls. Imagine a wooden disk resting on a perfectly flat hard surface. The normal force pushing up on the disk is the weight of the disk. Suppose the hard surface is the bottom of a tank. On top of the disk you place a cylinder of water contained by a solid material that has the same density as water, but slowly disolves in water. The normal force is now the weight of the water above the disk, plus the weight of the disk. Now slowly fill the tank with water until it reaches the top of the cylinder. The pressure on the vertical walls of the disk increases, but the forces are all horizontal and cancel. The normal force on the bottom of the disk is still just the weight of the disk plus the water above it. It cannot lift the disk. The cylinder walls dissolve, and you are left with a wooden disk stuck at the bottom of the tank.

    Now repeat the process, but instead of using a disk use a frustum of a cone (shaped like a rubber stopper) made of wood, with the bottom end much smaller than the top end. As water fills the tank, the pressure on the side walls of the wood exerts a force that is somewhat upward instead of horzontal, and the net force will be upward. At some point the pressure will become great enough so that this net force exceeds the weight of the wood and the cylinder of water above, so the wood will be lifted off the bottom and float. The cylinder walls will disolve (or the cylinder will tip over because the whole thing is unstable) and the wood will float to the surface.

    The key is the presence or absence of non-vertical walls with sufficent area to create enough buoyant force to lift the wood off the bottom. The frustum of a cone is still quite regular, but the walls are not vertical. If the only irregularity in the shape of the object still leaves it with only vertical walls, it will not float off a sealed hard bottom, but even a regular shape with sufficiently non-vertical walls that water can get beneath is going to float.
  12. Oct 16, 2006 #11

    Oh, ok now i get it.
    That means that fo any shaped object that is at the bottom of a hard solid sealed container will not float up as long as all the walls are vertical.

    But now I don not ge the part net force will be upwards(bold in blue).

    Bouyant force of a floating object=Weight of object
    Resultant/Net Force = 0
  13. Oct 16, 2006 #12


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    For an object that is floating on the surface there is no acceleration, so the buoyant force equals the weight. But the buoyant force on an object (not resting on the bottom) is always the weight of the displaced water. If you push an object under water and then release it the buoyant force is greater than the weight of the object and the object accelerates upward. Motion under water finds resistance that increases with speed, so the acceleration may stop before the object reaches the surface. This is like the terminal velocity of a skydiver falling through air. The difference between the buoyant force and the weight is offset by the resistance at some terminal velocity.
    Last edited: Oct 16, 2006
  14. Oct 16, 2006 #13
    Indeef, remember that buoyant force is equal to weight of fluid displaced, and not the weight of the object. It is equal to the weight of the object when the weight of fluid displaced = weight of object, for static objects.

    The resistance that you get because of travelling is due to the viscosity of the fluid.

    Notice that I use the word fluid: this means that Archimedes's principle works in both gases and liquids. A prime example of this would be applying it to hot air baloons or airships. The air in the baloon displaces the air in the atmosphere, hence the upthrust is greater than the weight initially. Because the air inside the baloon is less dense, we have an acceleration upwards and flight!

    Here is a thought. Archimedes Principle practically doesn't work when their is no gravity since then; water does not have weight! Only viscosity will matter then.:rofl: :rofl:
  15. Oct 17, 2006 #14

    Pls explain .

    Does that mean that Archimedes Principle only works in places with gravity?
    ooo, never knew that part.

    oh yea, OlderDan, thank you!

    I get it at last

    Just to reaffirm, was my first statement:

    "That means that fo any shaped object that is at the bottom of a hard solid sealed container will not float up as long as all the walls are vertical"

  16. Oct 17, 2006 #15


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    Right. Water pushes down on its top. No water pushes it up from below.
  17. Oct 17, 2006 #16


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    And of course without gravity none of us or the fluids would be here. :smile:
  18. Oct 17, 2006 #17
    Ok now I'm gonna start of asking a million questions about gravity...


    Is there anywhere where there is 0 gravity!?
  19. Oct 17, 2006 #18
    Of course, but remember, there are places where the EFFECTS of gravity does not apply. Anyone for free-fall in a space station?
  20. Oct 17, 2006 #19
    Technically, there is no place in the universe where gravity does not exist. But you see, Archimedes principle works because of Newton's 3rd Law. The fluid, like the ocean around us, is pulled toward the earth by gravity. To simplify things, lets just think of it that the the ground or whatever it is that is stopping the fluid from falling to the centre, exerts and equal and opposite force on the fluid.

    In a space station which is constantly accelerating around the Earth, as far as objects in the station is concerned, there is NO gravity. Sure you could argue that gravity is accelerating anyone inside and the station itself, but as far as occupants are concerned, there is 0 gravitational force! Everyone is in free fall. Just think of any movies with space stations and the fluids seem to float around, again I say, if the Earth does not exert a gravitational force, you would not be able to tell the difference.
  21. Oct 18, 2006 #20
    Thank you.

    I've got the big picture already
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