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Fluid statics with Pascal's barrel

  1. Mar 5, 2012 #1
    I am confused about the some of the concepts behind fluid statics. Here is an example to illustrate this, using Pascal's barrel experiment.

    Consider a cylindrical barrel with cross-sectional area A and height H. Now insert a straw of height h and cross-sectional area a so that the total of height of the barrel and straw is h + H. Assuming a fluid of uniform density ρ, the total weight of the fluid is

    [tex] W = m_{tot}g = \rho V g = \rho g (ha + HA)[/tex]

    If the barrel is sitting on a plank, then the magnitude of the normal force N of the plank on the barrel is simply N = W.

    However, using a different method, I come up with a different result for N. By Pascal's equation, the gauge pressure at the bottom of the tank is

    [tex] P_{bottom} = \rho g (h + H) [/tex]

    Thus the force due to pressure acting on the bottom of the barrel is

    [tex] F_P = P_{bottom}A = \rho g A (h + H) [/tex]

    In order to balance this force, the normal force N from the plank on the barrel must thus be equal to FP, yet this is a different result then from before: N = W ≠ FP = N.

    There must be some error in my reasoning here, but I cannot find it. Thanks a bunch for the help!
     
  2. jcsd
  3. Mar 6, 2012 #2

    tiny-tim

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    Hi Undoubtedly0! :smile:

    Pascal's barrel was firmly closed at the top (apart from the hole for the straw).

    So the water just under the top of the barrel is at higher than atmospheric pressure, pushing upward on the top of the barrel.

    By good ol' Newton's third law, the top of the barrel is also pushing down on the water. :wink:
     
  4. Mar 14, 2012 #3
    Thanks tiny-tim. This accounts for the difference perfectly. The force due to pressure exerted on the top of the barrel is

    [tex] F_{P_{top}} = P_{top}(A-a) = \rho gh(A-a)[/tex]

    And so it follows that

    [tex] W = F_P - F_{P_{top}} [/tex]
     
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