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Confusion over Pascal's Experiment

  1. Jul 20, 2013 #1
    I have been trying to understand pascal's barrel experiment in a more qualitative sense but I can't get my head around it at all. Pascal supposedly had a strong wooden barrel which he attached onto it a very thin pipe which was 20m tall. He then poured water into the top of the pipe until both the barrel and pipe were filled. The outcome of the experiment was that the barrel burst due to the pressure generated by the pipe. If the pipe was 20m high then this would generate 20m of hydrostatic pressure apparently?! How does this work? I know quantitatively that it needs to work as there is no volume term in the pressure equation (P = ρgh). How can such a tiny mass of liquid exert so much pressure? Apparently the pipe was only 0.4cm wide! It almost seems like your getting a huge amount of pressure from almost nothing. If anybody can help explain how this experiment works that would be great!
  2. jcsd
  3. Jul 20, 2013 #2


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    "pressure" is the force applied to a surface divided by the area of the surface. Now, water, or any "fluid", applies the same force in any direction (that is, essentially, the definition of "fluid".). We can calculate the force the water in the pipe applies to an imaginary surface, of area A, the cross section area of the pipe, at the bottom of the pipe by calculating the weight of the water in the pipe: density times volume. Because the volume is "cross section area times length" that volume is Ah where h is the height of the pipe, the weight of water pressing on that surface is [itex]g\rho Ah[/itex] where "g" is the acceleration due to gravity (about 9.81 m/s^2) and [itex]\rho[/itex] is the density of water. We get the pressure by dividing that by the area, A: [itex]g\rho h[/itex].

    Now, the point is that the pressure at each point in the barrel is [itex]g\rho h[/itex] (actually slightly more as we go down into the barrel because we are not taking the weight of the additional water in the barrel into account- but certainly "at least" that value. If the side area of the barrel is B, then the total force on the sides of the barrel is (at least) [itex]g\rho hB[/itex] which can be very large. You do the calculations.
  4. Jul 23, 2013 #3

    Philip Wood

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    A hand-waving approach. Suppose the barrel is straight-sided: a cylinder. I imagine you'd find it less hard to accept that the barrel would burst if, instead of the narrow tube, the pressure were provided simply by extending the barrel upwards, i.e by making the head of water wide as well as just tall? Well, with the narrow 'header' tube full of water, the pressure just under the lid of the barrel will be the same all the way across the lid, and the same as the pressure at the bottom of the 'header tube'. So the water pushes on the lid of the barrel, and the lid of the barrel pushes back on the water with the same force that a column of water as wide as the barrel would push!
  5. Jul 23, 2013 #4
    Thanks for your time guys, those answers really helped me to understand!
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