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Angle of Inclination of a Water Surface

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    So the full problem is the following:
    "On the top of a truck, there is a cubic container with a side-lenght of 1,00 m and it's filled with 500 liters of water. The truck accelerates with 2.0 m/s2 in the x-direction during the whole experiment. When the watersurface has reached an equilibrium state, a smal cork is released from the middle of the bottom of the container, let's call this point x=0 , y=0. Calculate the coordinates where the cork will reach the surface, ignore swirls in the water."

    2. Relevant equations
    F = m*a
    M = F * l
    Fl = V * ρ *g
    ρcork ≈ 300 kg/m3

    3. The attempt at a solution
    With some (maybe not so good) assumptions, such as ignoring the friction from the water and setting a constant acceleration, I think I have a basic understanding of how to calculate where the cork will reach the surface if the angle of inclination of the surface is known. The problem is that I don't really know how to calculate the angle. I have tried modelling the surface of the water as somekind of rod which must be in torque-equlibrium, but this gave me unrealistic answers like -55,7° and 155,1°.
    So I would be really thankful is you could give me a hint of how this should be done, which forces are acting where and so on. :)
     
  2. jcsd
  3. Mar 8, 2015 #2

    PeroK

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    The key to finding the shape of the water's surface is to note that the surface cannot sustain any tangential force.
     
  4. Mar 8, 2015 #3

    gneill

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    If the truck was not accelerating you'd expect the cork to rise straight upwards through the water: It would surface directly above the release point, its path along the line of the acceleration vector, which is due to gravity.

    What happens then when the truck is accelerating (constantly)? What is the "new" net acceleration vector for objects in this moving frame of reference?

    What can you deduce about the direction of this vector with respect to the surface plane of the water?
     
  5. Mar 9, 2015 #4
    So, does this then mean that the surface will be parallell with the vector repersenting the vectorial sum of the forces acting on the surface? As far as I can see only the accelerating force from the container (parallell with the ground) and the weight of the water (perpendicular to the ground) is acting on the water (if we neglect surface-tension). But can I make the same assumption regardning the water surface? Is then the angle of inclination α given by tanα = mg/F , where F = m*a?
     
  6. Mar 9, 2015 #5
    I think that the corks motion in the x-direction will be descirbed by sx = 2t2/2 = t2
    And in the y-direction by: sy = ayt2/2
    where ay =( ρwater*g*Vcork - mg)/m = gρwatercork - g

    By making the two equations into one (substituting t2) I can get an expression for the motion of the cork. If the inclination of the watersurface is known (which is what I have trouble with calculating), an equation for the watersurface can be found. The point where the equation for the motion of the cork and the equation for the watersurface intersect is where the cork will reach the surface. Isn't it?

    But this only holds if we neglect the resistance from the water...

    Do you have any advice for a better solution? :)
     
  7. Mar 9, 2015 #6

    PeroK

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    I might have said that the surface of the water is perpendicular to the nett accelerating force (g + a). I think you're saying the same thing.
     
  8. Mar 9, 2015 #7

    gneill

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    I think you don't need to worry about drag or other water effects. In the frame of reference of the tank there will be a local net acceleration equivalent to gravity (equivalence principle) that will set both the angle of the water surface with respect to the tank's floor and sides, and define the straight-line trajectory of the cork in that frame.

    If you and the tank were in a sealed lab which was (unknown to you) constantly accelerating, you would measure a local "gravity" that would be indistinguishable from a real uniform gravitational field of the same magnitude and direction. All the physics associated with gravity would be the same, including how corks and water surfaces behave.
     
  9. Mar 10, 2015 #8
    No, I had missunderstod, but now I got it! According to my calculations the angle is ca 5,8 degrees, which seems reasonable. It's an interesting fact that a watersurface cannot sustain any tangential force, this is a concept I have not encountered before. Do you have any explanation to it? Maybe on a microscopic level? :)
     
  10. Mar 10, 2015 #9

    gneill

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    How did you arrive at that angle?
    Any tangential force along the surface must make the water flow along the surface. This will redistribute the water until any additional flow would have to be "uphill". The result is a surface of minimal potential energy, which in the case of a uniform field, is a flat surface perpendicular to the field lines.
     
  11. Mar 10, 2015 #10
    Do you mean that the angle of the trajectory of the cork should be the same as the for the water surface? Isn't the first one dependent on the magnitudes of the corks weight and the lifting force from the water? Or wait, aren't these two dependent on the new "gravity"?
     
  12. Mar 10, 2015 #11

    gneill

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    Exactly. The new "gravity" is the only "gravity" that matters in the moving frame of reference.
     
  13. Mar 10, 2015 #12
    I made a free body diagram with the accelerating force F = 2m on the x-axis and the weight mg on the y. The slope of the resulting force was then -mg/F , if the water surface is perpendicular to this, then it's slope k is given by k * (-mg/F) = -1 , which gives k = F/mg . But k is also equal to tanv, where v is the angle of inclination.
    Thus: v= arctan(F/mg) = arctan (2/g) ≈ 11,5° (I must have misspressed some button when I got 5,8...).
     
  14. Mar 10, 2015 #13
    But can't I divide it up in two composants, one equal to g and the other to a (=2 m/s2) and count the old x and y directions separately?

    Is it wrong to set sx = 2*t2/2 → t2=x
    and sy = ayt2/2
    where ay =( ρwater*g*Vcork - mg)/m = gρwatercork - g ≈ 1,1633g ? (where g is the "usual" gravity)
    And from this obtain sy = 1,1633g *x/2 ?
     
  15. Mar 10, 2015 #14

    gneill

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    Yes, that's better :smile:
     
  16. Mar 10, 2015 #15
    Thanks for the correction! :)
     
  17. Mar 10, 2015 #16

    gneill

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    The buoyant force will depend upon the new gravity direction, as the pressure gradient in the fluid will be maximized along the same direction. It is a lot more work to try to break the forces into components and compute a trajectory dealing with buoyancy and drag rather than just the simple straight path obtained in the moving frame.

    Fig1.gif
     
  18. Mar 10, 2015 #17
    Thank you so much! This is a much better solution than mine.

    However, I'm now wondering about the direction of the tilting water surface, how comes it's in a direction away from the direction of acceleration? My intuition says it should be the other way around... :S
     
  19. Mar 10, 2015 #18
    Wait, I think I understand! Will try to digest it! :)
     
  20. Mar 10, 2015 #19

    gneill

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    For my figure I assumed that the truck is accelerating to the right with an acceleration magnitude at. The inertial force that results on objects in that frame of reference (Newton's 3rd law) is in the opposite direction, so the at vector in the acceleration triangle is shown pointing left.

    The at vector in the diagram represents the acceleration "felt" by objects in that frame due to the acceleration of the frame of reference. It's an inertial force. An analog would be the classic elevator in space that is accelerating "upwards" at g, so the passengers inside feel an acceleration "downwards" which they interpret as gravity.
     
  21. Mar 11, 2015 #20
    Thank you so much for all your help Sir! :)
     
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