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I came across this figure that I'm currently unable to accept. pressure independent of mass

I've come up with a thought experiment. Please point out my error:

enter image description here

Assume zero atmospheric pressure and no boundary effects of the containers. We are measuring the pressure on the green ball. The red line represents a movable barrier.

In fig A and B, the force applied by the top columns of water are unequal. Top column B has more weight, and so the pressure felt by ball B is greater than the pressure felt by ball A.

Fig C and D are as in the first diagram, seemingly ball C and D feel the same pressure.

Since the barrier is movable, systems A and C are identical, as well as B and D.

But B > A = C = D = B.

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  • $\begingroup$ The pressure in liquid has to be such that it balances the weight of the column above the movable barrier. In B the total weight of the column above the barrier is more than in A but since it is distributed over larger area in B it is the same pressure at the barrier that is needed to maintain the balance. $\endgroup$ – Maxim Umansky Apr 5 '16 at 6:10
  • $\begingroup$ I think someone said it is because the container is rigid it is counterintuitive. Don't remember why though... $\endgroup$ – Emil Apr 5 '16 at 6:14
  • $\begingroup$ @MaximUmansky The movable barrier is rigid horizontally and constrained to allow only vertical movement. Only the net force matters. The systems are in equilibrium, so the greater weight of top column B implies the pressure exerted on the barrier is larger for system B. $\endgroup$ – roro Apr 5 '16 at 6:30
  • $\begingroup$ @Emil you are correct, system A and system C are not the same. System C has rigid edges, so pressure exerted by the liquid onto the barrier is not counteracting the pressure exerted by the top column. The pressure at the bottom of A is less than the pressure at the bottom of C. $\endgroup$ – roro Apr 5 '16 at 6:43
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A and C are not identical; that's where the thought experiment breaks down.

Consider the pressure in the fluids around the opening between the long thin neck section and the wide base section. In C, you have one continuous fluid, and the pressure is the same both above and below the neck (and equal to $\rho g h$ where $h$ is the height of the next and $\rho$ the density of the liquid).

This is not the case in A. The pressure just above the plate is $\rho g h$, as in C. But the pressure below the plate is very different. The plate is not moving, so the net force on it is zero.

Forces acting on the plate: its weight, the weight of the column of liquid above it, and the pressure of the liquid beneath it. Lets assume that the plate is weightless, the neck has cross sectional area $A_n$ and the base has cross sectional area $A_b$. The the balance between the weight of the column and the pressure $p$ at the top of the liquid below gives:

$\rho g h A_n = p A_b$

So $p=\frac{A_n}{A_b}\rho g h$ in A, while $p=\rho g h$ in C.

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