$P=F/A$ and $P=\rhogh$ give different answers

Question

I have searched for this, and while I have found questions that brush on the same issue, my question wasn't fully adressed by them. This one, namely, is very similar, but I still don't totally get it.

So, let us have two containers, both of them have a volume of liquid $V$, bottom area $A$ and filled with the same fluid of density $\rho$. One of hem has a liquid column of height $h$ and the other one $H>h$.

According to hydrostatics, the pressure at the bottom of the containers is the following

$P1=\rho*g*h$

$P2=\rho*g*H$

$P2>P1$

On the other hand, we could look at the fluid as if all of its weight was being supported by the bottom wall (I realized that this might be my mistake while I was typing this. I'm gonna finish anyway to get some input from people who understand the issue better.)

That assumption would lead us to:

$P1=\rho*V*g/A$

$P2=\rho*V*g/A$

$P2=P1$

Which is inconsistent with reality.

I arrived at this problem by imagining what would happen if we put a very thin and long straw filled with fluid on top of a container, which would make the pressure $P2=\rho*g*H$ go really high by increasing $H$, which would mean that we could make any container fail by excessive stress if we put a long enough straw on top of it, which is absurd. In this case, the second analysis kinda makes more sense because if you don't increase the volume, you basically also don't increase the pressure, but I know this has to be wrong since pressure does depend on height of fluid column.

I guess this is kind of two questions in one, as I have one question that asks for correction on the concepts I used to calculate something and another one which is the question about how the absurd situation doesn't happen. If you prefer to focus on "one question", please focus on the second one, since it is the one that I feel less able to tackle myself.

FINAL ANSWER: I have been convinced. I thought the situation on the following video was absurd/impossible, and it isn't. My mind is settled. Thanks for your answers people!

https://www.youtube.com/watch?v=EJHrr21UvY8

Precisely what is the definition of your volume 'V'? The volume of the container itself? The volume of the liquid in the first container? The volume of the liquid in the second container? — , Mar 19 '18 at 19:22
How do the two scenarios have different heights and identical volumes then? — BowlOfRed, Mar 19 '18 at 19:25
You declared them to be cylinders, and to have identical bottom areas. If you expect them to have differing geometries, you need to put that information into the question. — BowlOfRed, Mar 19 '18 at 19:27
No I didn't declare them to be cylinders, only to have the same bottom area. — Walther Wennholz Johnson, Mar 19 '18 at 19:27
I see, so the straw isn't a continuation of the container. I think if you can include a diagram of two potential situations in your question, it would help. — BowlOfRed, Mar 19 '18 at 19:28
Can I disagree with this duplicate thing? That question doesn't come nearly as close to what I asked when compared to the one I linked on the first paragraph.(Realized the link was broken. My bad. Fixed it.) — Walther Wennholz Johnson, Mar 19 '18 at 21:10

score 3 · Accepted Answer · answered Mar 19 '18 at 19:42

"we could make any container fail by excessive stress if we put a long enough straw on top of it, which is absurd."

Absurd it might seem, but it's what happens. Barrels have been burst by putting tall thin pipes into their tops and filling with water.

You correctly diagnosed your mistake earlier on. When containers have sides that aren't vertical, these sides exert on the liquid forces that have a vertical component, either up or down according to whether the container is becoming broader or narrower as we go upwards. The thing to remember is that a liquid at rest exerts on a surface a force at right angles to the surface. The surface exerts an equal and opposite force on the liquid.

My favourite derivation of $p=h\ \rho\ g$ equates the work done pushing a piston at depth $h$ against the liquid to the gravitational PE acquired by the displaced fluid as it (effectively) rises to the top of the liquid. A bonus of this method is that it shows the pressure to be the same whatever the angle of the surface that the liquid is pushing against.

Well, if they do burst in that scenario, that kind of settles all of my questions ahaha. I'm gonna let the question run a little bit more to see if some people come up with other great explanations. — Walther Wennholz Johnson, Mar 19 '18 at 19:50
@WaltherWennholzJohnson I suggest reading the article on Pascal's Law to help reassure yourself about that. Specifically Pascal's barrel. It explains a lot of fluid behavior; such as why we can use hydraulics as a type of force multiplier. — JMac, Mar 19 '18 at 21:27
Thank you @JMac , I've added a video regardind the law to my question. — Walther Wennholz Johnson, Mar 19 '18 at 21:36

score 1 · Answer 2 · answered Mar 19 '18 at 19:29

Your problem is indeed assuming that all of the weight is supported by the bottom of the container. With walls sloped outwards, the walls themselves will support some of the weight of the fluid. With walls sloped inwards, the force on the bottom of the container will actually be greater than the weight of the fluid.

This is easy to see: just draw a container with sloped walls and draw the direction of pressure at each point. It is clear pretty quickly that the net force on the walls is not zero.

score 0 · Answer 3 · answered Mar 19 '18 at 19:30

0

Your second equation comes from the first. For a given prismatic volume of fluid, the volume can be given as $V = A h$.

Substituting that into $P = \frac {\rho g V}{A}$, you get $P = \rho g h $.

The volumes are different in both equations, so they give different pressure. If you filled a straw up incredibly high, it would take a lot of energy to pump it up there; and it would generate a pretty high pressure (you would have trouble setting this up without things bursting). If everything could contain it; it would either build up really high pressure, or there will be some way for the fluid to leave the container when the water comes back down the straw.

answered Mar 19 '18 at 19:30

JMac

15,298

"The volumes are different in both equations" what do you mean by this? – Walther Wennholz Johnson Mar 19 '18 at 20:53
@WaltherWennholzJohnson I assumed you were talking about prismatic containers; where the area doesn't change with height. That's where those two equations are the exact same; except for in the higher one, you have a greater volume. The volume equation can only be applied for a situation where it's a uniform cross-section. In the case where it isn't, you have to use the height to determine the pressure. In the case of a long tube, like in your picture, you have to use the height, because volume isn't constant. – JMac Mar 19 '18 at 20:59

$P=F/A$ and $P=\rhogh$ give different answers

3 Answers3

Linked

$P=F/A$ and $P=\rho*g*h$ give different answers

3 Answers3

Linked

$P=F/A$ and $P=\rhogh$ give different answers