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What's the answer?

  1. Dec 10, 2005 #1

    Alkatran

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    I spent all of yesrterday going over a set of puzzles. One of them was:

    Assume you have a container divided into two equal sections (one on the left and one on the right). One section is filled with water and the other with air. A log is placed a long the divider between these two sections using a frictionless axle and frictionless seals (to keep the water out of the air).

    Does the log spin? (Since one side is falling and the other is floating)
    (it would like something like this really bad drawing:)
    Code (Text):

        Water
    |Air|---|
    |   o   |
    |___|___|
     
    Basic physics (gravity, torque and bouyance) says that the log would spin. But even more basic physics (energy conservation) says it won't. What am I missing? Does it have something to do with the fact that the log won't actually float upwards? Will the log rotate for a short period of time?
     
  2. jcsd
  3. Dec 10, 2005 #2

    Danger

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    It'll just get waterlogged and stay put. :tongue:
    Common sense indicates that it would spin, but also points out that there are no such things as frictionless axles or seals. I don't know how much Brownian motion would affect it. I'm sure that it would spin if the water were boiling.
     
  4. Dec 10, 2005 #3
    Yes, but aren't boiling bubbles rather different from floating? I mean, the bubbles have a direct impact on the log, whereas the floating is actually an illusion; the log doesn't want to go up, the water merely tries to go down.

    This all sounds rather confusing. Ach, how I hate it when I can't even understand my own thoughts.

    EDIT: It won't spin, since the particles in the upper part of the log and the ones in the lower part all have the same forces on them.

    Yuch, I really need to get working on my physics-related english. My english is fine, but here... Would anyone know a list of commonly used terms and stuff for me?
     
    Last edited: Dec 10, 2005
  5. Dec 10, 2005 #4

    Danger

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    Hi Naz;
    The mention of boiling was actually just an extrapolation of my trying to figure out what effect Brownian motion would have. I've decided that it would have no effect (or rather, a zero net effect) because of its randomness. Boiling, on the other hand, would involve convection and bubbling, as you noted.
    I'm starting to seriously regret that 4th round of shooters last night. I don't usually get hangovers, but this puzzle is making my head hurt. :frown:
     
  6. Dec 10, 2005 #5
    If this were true, the log wouldn't want to spin, but sheer in two. It wouldn't have a torque on it, but a sheer force.
    On the water side of the tank the bulk of the force is not "up" but horizontal: the water wants to push the log sideways, hydraulically, into the less dense air.
     
  7. Dec 10, 2005 #6

    Danger

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    Regarding your first point, wouldn't a sheer force on one side of a freely suspended object automatically cause rotation?

    About part two... since the basic attempted displacement sideways should be equal along the full 'height' of the cylinder, wouldn't the slight bouyancy factor be the only net force?
     
  8. Dec 10, 2005 #7
    Ach, it's annoying we can't try this. But as a last resort: eventually the log will rot, and then the water will probably flow away, and your seals aren't perfectly frictionless anymore.
     
  9. Dec 10, 2005 #8
    The log is constrained by an axle so it has no sideways "give" to roll away from any sheer force, if you can see what I mean.
    Well, the sideways force is stronger the lower down you go, but even if it were equal what would make the sideways force go away leaving us left only with buoyancy, just because it was equal along the whole height? It would still be pushing sideways.
    The ratio of buoyancy to hydraulic pressure would depend on the density of the wood and the exact "head" of the water.
     
  10. Dec 10, 2005 #9

    Danger

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    Maybe I'm misunderstanding 'sheer'. I wish I was at my spare home so I could get on the Mac and make a diagram. None of the stupid PC's that I can access have Illustrator. (I suppose that I can degrade myself to the point of using Paint, but I'll save that as a last resort.)
     
  11. Dec 10, 2005 #10

    Alkatran

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    I'd like to point out that a 'sheer' force on the air side of the log would put a torque on it. (the furthest-from-the-center particle is being affected by gravity...)

    My intuition is that because the log is fixed it will 'float differently'. IE: because the log won't move (except rotation at most), the water won't continue to push on it (after some redistribution of the water, which does push).
     
  12. Dec 10, 2005 #11
    I think you probably understand what I mean, but don't see why the log is constrained such that the sheer force can't convert to torque in this situation. I'm not being very articulate, either. Maybe this will help: rolling wouldn't relieve any stress on the log-it wouldn't represent a more balanced situation. If the log rolls, it's right back where it started. Nothing's changed. It has no incentive to roll. Therefore, it doesn't roll.

    If you suppose the pressures will be relieved by pushing buoyant matter out into the air side, you have to remember that this requires rolling an equal amount of buoyant matter into the water side. All parties involved, the log, the water, the air, realize there's no point.

    Therefore, the upward force of buoyancy is relegated to simply wanting to cut the log in two lengthwise.
     
  13. Dec 10, 2005 #12

    Danger

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    Okay, Zoob... I've pretty much got it now. Intuitively, it seems weird, but it makes logical sense. Thanks for clearing it up.
     
  14. Dec 10, 2005 #13
    The fluid pressure everywhere of an incompressible fluid is the depth of fluid, times its density. On an immersed object, it acts perpendicularly to the surface everywhere. So for our immersed hemi-log, the total force is:
    [tex]\begin{align} \int_S P (-\vec{dA})&= \int_S \rho z (-\vec{dA}) \notag \\
    &= \rho L \int_0^\pi (-\cos \theta+D) \left(\hat{x} \sin \theta - \hat{z} \cos \theta\right) d \theta\notag \\
    &=\rho L \hat{z} \int_0^\pi (\cos^2 \theta + D \cos \theta) d \theta + \rho L \hat{x} \int_0^\pi \sin \theta (D-\cos \theta) d \theta \notag \\
    &=\frac{\rho L \pi}{2} \hat{z} + 2 \rho L D \hat{x} \notag
    \end{align}[/tex]
    (where D is the depth of the top of the log, and L is it's length (width); the radius is set to 1 arbitrarily).

    A net buoyant force, plus a net horizontal force into the side filled with air. But, there is no torque; due to the geometry of the log, the fluid only pushes at its surface in the direction into the axis of rotation, so there is no torque.
     
    Last edited by a moderator: Dec 10, 2005
  15. Dec 10, 2005 #14
    Another way to think of it; if you had half of a log (cut through the diameter), vertically oriented underwater, free to rotate about an axis located as before - then it would rotate to align itself upwards. The torque for this is applied on the side that is a straight line; the bottom half would experience a greater net force than the top half.

    Our situation is exactly this, minus the right side of the fluid (where the torque is applied). So no torque.
     
  16. Dec 10, 2005 #15
    A third way to think of it - we are accustomed to thinking of buoyancy as we think of gravity, as acting on the entire volume of an object, when in fact it only acts at the surface (Gauss' theorem allows us do this, I believe). But when the surface of the log exposed to the water is not closed, this no longer works.
     
  17. Dec 10, 2005 #16
    This makes perfect sense.
     
  18. Dec 11, 2005 #17

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    i thought that the bottom of the fluid will have more push(i remember an expariment where there is a cylinder and three holes one above next, all are opened at once and the water coming from the bottom hole goes the farthest and when the amount of water decreases the water goes less further) so there will be rotation in this bit, me thinks:
    air...........water
    |..........|...........|<low pressure to the log in this bit
    |..........0...........|
    |..........|...........|
    |______|_______|<high pressure to log in this bit

    this can be imagined this way, there is a container seperated by a log that can freely rotate on an axis, one side is empty (air) next side is water, the bottom bit of the log gets pushed the hardest, it will let go because there is no balancing force (there are forces acting on opposite direction 1. of air in the next side 2. water of the uper part or the part above the axle), but this will be lower than our water's force and the lg will move, and get a horizontal position and wont go any further because the water will stop pushing (it will pust the wall of the container instead).


    how about this one? please tell me if i am wrong, i know very less about physics.
     
  19. Dec 11, 2005 #18
    The lower you go the more push the water has, yes. But Rachmaninoff pointed out that the force gets directed toward the axle, the axis of rotation, so there's no torque (twisting force) on the log.

    Here's a drawing of what I think Rachmaninoff is saying:

    [​IMG]

    But this is subject to his confirmation.
     
    Last edited: Dec 11, 2005
  20. Dec 11, 2005 #19
    Yup, that's it.
     
  21. Dec 11, 2005 #20
    Another perpetual motion machine bites the dust.
     
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