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I Buoyancy question -- Pressure

  1. Feb 7, 2017 #1
    Hi there,

    I was just wondering if I'm thinking about buoyancy in the correct way. I understand that it is related to the water displaced and that water displaced would usually have pressure being pushed up on it by the water underneath. A boat that floats would be taking up the space of that water and if the weight of the ship is less than the water it's displaces, then the boat will float. It is also about the pressure differential and I believe Newton's third law comes into play but I am not exactly sure about that last part. Please let me know what I missed.
     
  2. jcsd
  3. Feb 7, 2017 #2

    boneh3ad

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    Absolutely it is about the pressure differential, and the familiar Archimedes' principle relating the force to the displaced weight of water can be derived directly from integrating the pressure distribution over a submerged object.
     
  4. Feb 7, 2017 #3
    I would add that the pressure distribution is integrated Vectorially.
     
  5. Feb 7, 2017 #4
    So this pressure from below from the water below the object or what would be water otherwise has to be equal in weight? Isn't the weight of the water a downward force?
     
  6. Feb 7, 2017 #5

    boneh3ad

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    Sure but pressure isn't. Pressure acts normal to whatever surface it touches.
     
  7. Feb 7, 2017 #6
    This means that anything filling up the space where the water use to be would experience the same amount of pressure that the water previously taking up that space had but if the weight of a ship let's say isnt greater then the fluid pressure underneath. I've read people talking about a race to the bottom or the object having less downward force which is obvious but in my opinion, a downward force means it will still end up going down. It's the fluid pressure underneath that applies the pressure to keep that smaller weight up. I guess it's dependent on what makes sense practically and maybe my understanding of pressure needs to be expanded. I'm trying visualize things bc I think that's how you come up with solutions easier. Let me know if I'm still on a good track.
     
  8. Feb 7, 2017 #7

    boneh3ad

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    It's a force balance. Any object has a weight that is acting downward. When submerged underwater, it experiences an upward buoyant force that is equal to the weight of the water it displaced. If the buoyant force is greater than the weight, the object with float. If the weight is greater than the buoyant force, it will sink.

    When it comes to floating near the surface, an object will sink into the water exactly as far as required until the weight of the displaced water is equal to the weight of the object. The rest will remain above the surface.
     
  9. Feb 7, 2017 #8
    Oh yes, I know all objects have a downward force. Although, why is it equal to the weight of the water displaced? How is that displaced water acting on the object? Is it pushing it the object upwards because this water wants to move back down due to gravity? I know it's pressure differential related too.
     
  10. Feb 8, 2017 #9

    boneh3ad

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    It's not really any sort of force as a result of water that was displaced wanting to "move back in". It's all about pressure. Pressure varies with depth in a fluid (gets higher as you go deeper) so if you submerged any object, it will have a higher pressure on the bottom than on the top. That's the fundamental source of the buoyant force.
     
  11. Feb 8, 2017 #10

    sophiecentaur

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    Do you mean their Weight?
    Or do you mean 'upwards' i.e. upthrust when submerged?
    The expression "weight of the water displaced" is a bit terse and needs a bit more explanation. For simplicity, choose a vertical submerged cylindrical object. It experiences upwards pressure on the bottom face and downwards pressure on the top face. The resulting force is the difference times the top area (the bottom is the same). The pressure difference is the height of the cylinder X density X g and the total upwards force is hρgA. The Weight of the water displaced is the volume of the cylinder times ρg. This is the same (not surprisingly) as the buoyant force. Things don't change if you tilt the cylinder at any angle but calculating the force would be a bit more complicated.
    If an object happens to float, the total upwards force will still be the weight of displaced water and this will equal the weight of the object (at equilibrium.
    OR is there something else you are having a problem with?
     
  12. Feb 9, 2017 #11
    I think I'm kind of strung up on newton's third law with all this. I think I'm overly concerned with acceleration due to gravity being a downward towards earth so you force should be in downward direction. Won't the object be having an equal and opposite reaction force from it's own weight being placed on the water? However, I know there is another component of all this that I'm currently missing.
     
  13. Feb 9, 2017 #12
    No. From Neeton's 3rd law, there is an equal and opposite force on the earth.
     
  14. Feb 9, 2017 #13
    Is the acceleration due to gravity involved because those water particles are all experience the downward force of gravity so the pressure will compounded once you get to lower depth? I guess this is related to all the particles trying to move downward and then experiencing interference from the other molecules around them also being forced downward?
     
  15. Feb 9, 2017 #14
    You are familiar with the development of the hydrostatic equation, correct?
     
  16. Feb 9, 2017 #15
    I looked it up just now. I would appear that the downward pressure due to the fluid above, weight of the object, and the upward pressure from the fluid below must equal if it is all at constant velocity. I guess that's what I know. I don't know much about the development itself.
     
  17. Feb 9, 2017 #16
    That's OK. In our situation, the upward pressure from the fluid below must be equal if it is all stationary, at zero velocity.
     
  18. Feb 9, 2017 #17
    I think I get it now from looking at this the post in this thread: https://www.physicsforums.com/threa...-density-of-the-liquid-vobj-submerg-g.765872/

    It would appear from the calculation done by vanhees71 that the force of gravity of, in this case, a parallel sheet is -Ahpgez and the pressure is the negative normal component which is given in the opposite direction (at least I think so). I'm hoping that the negative normal force is in the opposite direction. Therefore, the acceleration due to gravity needs to be there for the purposes of producing that pressure in the other direction.
     
  19. Feb 9, 2017 #18

    Dale

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    There are two relevant third law pairs here.

    There is a gravitational force from the earth pulling down on the boat. The third law pair for that is a gravitational force from the boat pulling up on the earth.

    There is a pressure force from the water pushing up on the boat. The third law pair for that is a pressure force from the boat pushing down on the water.

    The gravitational force on the boat and the pressure force on the boat are not a 3rd law pair.
     
  20. Feb 10, 2017 #19

    A.T.

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    The two equal but opposite forces in newton's third law are never acting on the same object, but on two different objects.
     
  21. Feb 10, 2017 #20

    sophiecentaur

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    N3 just doesn't figure in Archimedes' Principle. You need to read what N3 actually tells us. It does not deal with all the forces on a submerged object. N3 is not relevant here.
     
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