1
$\begingroup$

enter image description here

I have such a weird misunderstanding about liquid pressure. On the diagram, I have two weightless and frictionless piston on the two openings of the container. There's atmospheric pressure on both of them. So one piston has area A1 (the smaller one) and other has the area A2 (the bigger one). So my misunderstanding is since the atmospheric pressure is equal for both the pistons and one has a bigger area than the other, so the force on larger piston would be greater than the smaller piston. Then larger piston should move down and smaller piston should move up. But that doesn't happen. So where did I misunderstand? Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ Well, the force is greater, but it's spread out over a larger area. $\endgroup$ – Javier Jun 21 '18 at 13:41
  • $\begingroup$ Do you mean that force felt by individual molecules on both the openings will be the same? $\endgroup$ – Sandeep Jun 21 '18 at 13:52
2
$\begingroup$

The upward pressure of the water on the bottom face of each piston is also atmospheric, so the net force on each piston is zero. Therefore, each piston is in equilibrium, and it won't move.

$\endgroup$
  • $\begingroup$ Ok! wouldn't that be equal even with different levels ? $\endgroup$ – Sandeep Jun 21 '18 at 14:41
  • $\begingroup$ I don't quite follow. Are you saying that you have different levels without having any external loads on the pistons other than atmospheric pressure? $\endgroup$ – Chet Miller Jun 21 '18 at 14:44
  • $\begingroup$ Yes. And I feel like due to greater force on the large piston , it should have gone downward and the smaller piston would go up. $\endgroup$ – Sandeep Jun 21 '18 at 14:50
  • $\begingroup$ If the columns aren't equal in height, then the system will not be in hydrostatic equilibrium, and the fluid in one column will be accelerating downward while the fluid in the other column will be accelerating upward. Have you actually drawn a free body diagram on an arbitrary layer of fluid in each column showing the pressure forces acting above and below, the gravitational force on the layer, and, if they don't sum to zero, the mass times acceleration? $\endgroup$ – Chet Miller Jun 21 '18 at 15:10
  • $\begingroup$ Yes, same thing would happen. I think there will be oscillation in the fluid column with that. The liquid level on both the opening remains same without having any external loads on piston . But why is that ? Force on the two open faces are different but still the system is in equilibrium . Why does that happen ? $\endgroup$ – Sandeep Jun 21 '18 at 15:27
1
$\begingroup$

First, let's consider a weightless scenario to demonstrate that, for the fluid to be in balance, forces $F_1$ and $F_2$ do not have to be equal.

In the weightless case, according to the Pascal's law, the pressure everywhere in the fluid should be the same.

enter image description here

If we apply force $F_1=P_0A_1$ to the left piston, while the right piston is fixed, it is easy to see that the same pressure $P_0$ will be at points $A$, $B$ or $C$. The pressure force of the fluid acting on the right piston, $F_2=P_0A_2$ will be determined by the pressure under the piston, $P_0$, and the area of the right piston. So, it won't depend on the area of the left piston and will be different than $F_1$, while the fluid will be in balance.

If we add gravity, the weight of the fluid will cause the pressure increase with depth, but, at any given depth, according to the Pascal's law, the pressure will still be the same. We can also see, that, if the horizontal pressure was not the same, the fluid would move from higher pressure areas to lower pressure areas. It should be rather obvious then, that to maintain the same pressure at any given depth everywhere, the heights of the left and right columns, which determine the pressure under them, would also have to be the same. Hence, the pistons have to be at the same level.

enter image description here

My last point should answer your implied question: how come we have two unequal forces, $F_1$ and $F_2$, acting on the fluid "from two different directions" and the fluid does not move? Is it a violation of the Newton's second law? It is not, if we consider all forces acting on the fluid, including the reaction forces from the walls. This should be clear from the diagram below and should be easy to prove, since we know the pressure at all points.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.