Confusion over Pascal's Experiment

AI Thread Summary
Pascal's barrel experiment demonstrates how hydrostatic pressure works, where the height of a water column generates pressure regardless of the column's width. The pressure at the bottom of the pipe is calculated using the formula P = ρgh, where ρ is the water density, g is gravity, and h is the height of the water column. Even with a narrow pipe, the pressure exerted on the barrel's sides can be substantial due to the large area of the barrel. The force applied by the water in the pipe is equal to the weight of the water, which can lead to the barrel bursting under pressure. Understanding this principle clarifies how a small volume of liquid can exert significant pressure.
Jimmy87
Messages
692
Reaction score
19
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!
 
Physics news on Phys.org
"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 g\rho Ah where "g" is the acceleration due to gravity (about 9.81 m/s^2) and \rho is the density of water. We get the pressure by dividing that by the area, A: g\rho h.

Now, the point is that the pressure at each point in the barrel is g\rho h (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) g\rho hB which can be very large. You do the calculations.
 
  • Like
Likes 1 person
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!
 
  • Like
Likes 1 person
Thanks for your time guys, those answers really helped me to understand!
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top