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Buoyant force concept

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data

    If you had a rectangular prism and split it in half, half iron, half wood, and let if float on water, would the buoyant force be different on the different sides?

    3. The attempt at a solution

    I originally thought since the iron was more dense than the wood, it would submerge more in the water and displace more fluid, thus the buoyant force would be greater then the wood.

    Then I thought, no wait, the density of the entire thing would be a certain value so they would both be the same regardless of what side you put them on.

    So which is it?
  2. jcsd
  3. Sep 9, 2008 #2
    one more time...

  4. Sep 9, 2008 #3
    Ok, see I originally thought that the side with that was more dense would displace more water, but then I realized its a whole system, and if you add the masses and get the volume that that block has 1 overall density. And I was fine with that for about an hour but then I got this picture in my head of someone floating along in an inner tube. That person displaces a lot less water when his body (more density) is on top of the tube rather then the tube (less dense) is on top of him. So now I am most certain the denser side would displace more water, I just need someones approval.
  5. Sep 9, 2008 #4


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    If an object is floating, the buoyant force is in equilibrium with the object's weight (i.e. the net force on the object is zero). Since the weight of the object doesn't change in either position, the buoyant force will be the same.

  6. Sep 9, 2008 #5
    So you would consider the density of the block as 1 system even though 1 side is more dense than the other?

    Also if the force of the buoyancy is equal to the weight of the fluid displaced where does this thought process fail?:

    A person clinging to an inner-tube, a floatable, is displacing nearly his whole bodies' volume of water, since only his head is above water. When he sits on that inner-tube it displaces less water, and the inner-tube is very high on the water. So if the force is equal to the displaced water then there would be more force of bouyancy when the denser object is on the bottom.
  7. Sep 9, 2008 #6


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    Yes, treat it as one system.

    You are making too many assumptions about the position of the tube and the person.

    For this type of conceptual problem, I would not use complex bodies to picture this since there are more variables to consider. Stick to a simple object like a box to see the concept.

    Remember that the buoyant force is the result of the hydrostatic pressure field acting on the object (box in this example). Pressure acting over the area of the bottom of the box will result in an upward force. This is the buoyant force. Since the box has some total weight (iron + wood in your question) the total force due to gravity (i.e. the weight of the box) is acting at the center of mass of the object. Since the box is floating, the buoyant force is in equilibrium with the weight of the box. That is to say that the box is not sinking or rising, just floating, or as I said, in equilibrium. Therefore the forces must balance (i.e. the buoyant force equals the weight of the object).

    Try drawing a FBD and then sum the forces using Newton's Second Law, remembering that the object is in equilibrium.

    Does that help?

  8. Sep 10, 2008 #7
    Yeah, it helps, thanks. I still can't get that picture out of my head though
  9. Sep 12, 2008 #8
    What if the box had one side that was taller and thinner and another side that was shorter and fatter. They would both have the same volume just in different shapes. Would both sides still have the same buoyant force sine they are the same weight? Or does it involve area?
    You said it was a result of hydrostatic pressure. And pressure is dependent on area. So since the thinner side would have less area would it have more buoyant force?
  10. Sep 12, 2008 #9


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    The net buoyant force would be the same regardless of orientation of the object. It does involve area, however it is the resultant force when considering the entire pressure field acting on the surface of the object. This resultant force is also equal to the weight of the displaced fluid for a closed pressure field per Archimedes' Principle.

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