Understanding Liquid Pressure in a Cone: How Does It Compare to a Cylinder?

[Alexander350]

I am confused about the pressure depending only on the depth of the liquid, particularly for a cone.
Up until
$$P=mg/A=ρVg/A$$
I understand it, but the problem I have is cancelling the area with the volume to give the height. How does this work when the volume of the liquid in the cone is not given by $$V=A*h$$ but is instead given by $$V=1/3*A*h$$
Would this not mean that the pressure at the bottom of a cone is a third of the pressure at the bottom of a cylinder of the same height?

 

[russ_watters]

The V is what is directly above the A. You can't do the whole cone at once because it is sloped, so the pressure is different at different depths (just like the hydrostatic pressure equation tells you).

 

[Chestermiller]

Part of the weight of the liquid is being supported by the upward component of pressure exerted by the sloped sides of the cone.

 

[Alexander350]

How can you show that the pressure at the base is still equal to ρgh having to factor in the forces due to the sloped sides?

 

[russ_watters]

How can you show that the pressure at the base is still equal to ρgh having to factor in the forces due to the sloped sides?

My point was that you DON'T factor in the sloped sides. The pressure is the total height of the column times the weight density and that's it.

[Note: I thought at first you were describing it point down, but it really doesn't matter]

 

[Alexander350]

However, if the cone is point up, different points on the base have different heights of water above them. Wouldn't this mean that these different points have a different pressure?

 

[russ_watters]

However, if the cone is point up, different points on the base have different heights of water above them. Wouldn't this mean that these different points have a different pressure?

As @Chestermiller said, the water pressure acts on the sloped sides of the cone, so out toward the edges, it isn't just the water directly above that matters, it is still the total column height.

 

[Chestermiller]

It's supposed to be a cone with the point pointing upward, right? And the pressure at the top is assumed to be zero, right? If that is the case, I will show you how it is calculated in my next post.

 

[Chestermiller]

@Alexander350 I need you to answer the questions in my previous post before I can proceed.

 

[Alexander350]

It's supposed to be a cone with the point pointing upward, right? And the pressure at the top is assumed to be zero, right? If that is the case, I will show you how it is calculated in my next post.

Yes, it is a cone pointing upwards with zero pressure at the top.

 

[Chestermiller]

Let x be the depth below the upper point of the cone, and let $\theta$ be half the included angle of the cone. The pressure at depth x is $p=\rho g x$. Since the pressure is acting perpendicular to the sloped sides of the cone, the downward component of the pressure force acting on the side is $p\sin{\theta}=\rho g x \sin{\theta}$. At depth x, the radius of the cone is $x\tan{\theta}$, so between depth x and depth x + dx, the area over which this pressure acts is $2\pi x\tan{\theta} \frac{dx}{\cos {\theta}}$. So the downward component of pressure force acting on this area is $$dF=2\pi \rho g x^2 \tan^2{\theta}dx$$ If we integrate this between depth zero and depth h, we obtain:
$$F=\frac{2\pi \rho g }{3}h^3 \tan^2{\theta}=2\rho g \left(\frac{1}{3}\pi r_b^2h\right)$$where $r_b$ is the radius at the base of the cone. This is the downward force exerted by the sloping sides on the fluid. It is twice the weight of the fluid in the cone itself. If we add the weight of the fluid in the cone, we get the total downward force acting on the base of the cone:$$2\rho g \left(\frac{1}{3}\pi r_b^2h\right)+\rho g \left(\frac{1}{3}\pi r_b^2h\right)=\rho g h(\pi r_b^2)$$So the pressure exerted by the fluid on the base of the cone is $\rho g h$.

 

[Alexander350]

Is there a way to prove it's ρgh for every single shape? I don't see how going from $P=ρVg/A$ to $P=ρgh$ can account for the forces by the container in every case.

 

[Nidum]

 

[Chestermiller]

Is there a way to prove it's ρgh for every single shape? I don't see how going from $P=ρVg/A$ to $P=ρgh$ can account for the forces by the container in every case.

This can readily be done. Have you had Vector Calculus in you math courses yet (in which you learn about the so-called divergence theorem)?

 

[Alexander350]

This can readily be done. Have you had Vector Calculus in you math courses yet (in which you learn about the so-called divergence theorem)?

The furthest I have got to is what you showed in the cone calculation and I have never heard of that theorem. I have done basic vectors but only got as far as things such as the vector product.

 

[Chestermiller]

The furthest I have got to is what you showed in the cone calculation and I have never heard of that theorem. I have done basic vectors but only got as far as things such as the vector product.

Well, I don't want to have to go about deriving the divergence theorem for you before I can even get to actually solving this problem. If you want to learn about the divergence theorem, see Kreyszig, Advanced Engineering Mathematics.

 

[jbriggs444]

Assume that we have some fluid in a container. It has been allowed to sit quietly so that everything has settled into equilibrium. No movement. Assume that the fluid has a uniform density. The top is exposed to the air. We will take this to be at zero pressure. There is a point at the bottom of the container and we wish to determine the pressure at that point.

Let $\rho$ be the density of the fluid and $g$ be the acceleration of gravity (which we assume is uniform over the region in question).

Imagine a small cube-shaped volume at a fixed position somewhere in the midst of this fluid. This is just an imaginary cube. The boundaries are simply drawn in, they are not physical. There is pressure on all six faces of this cube. Top, bottom, left, right, front, back. Since everything is in equilibrium, we can conclude that the pressure on each of the faces is approximately equal. If the pressures were not equal, fluid would be flowing through this cube -- in through the face with the higher pressure and out through the face with lower pressure. The [average] pressure on the front, back, left and right faces must be exactly equal. The pressure on the top and bottom faces however must be different by just enough to support the fluid in the volume. (Otherwise fluid would flow in the top and out the bottom. Or vice versa)

Let $d h$ be the height of such a cube. Its volume is $(d h)^3$. Its mass is $\rho (d h)^3$. The force of gravity on the cube is $\rho g (d h)^3$. The area of the top and bottom faces is each $(d h)^2$. The pressure difference ($dp$) between top and bottom must then satisfy:
$$dp\ (d h)^2 = \rho g (d h)^3$$.
Simplifying, that becomes:
$$dp = \rho g\ dh$$

Now, back to our fluid. Starting at any point on the top of the fluid, draw a continuous line from there to the point where we want to determine the pressure. Keep this line a little bit away from the container walls. Now pick a cube size that is small enough that you can build a continuous chain of cubes following the path you just drew so that the path is entirely contained within these cubes. Line up the cubes so that they are dead centered face to face.

Follow this chain of cubes from beginning to end.

As you move from one cube to the next horizontally, the pressure stays the same.
As you move from one cube to the next downward, the pressure increases by $\rho g\ dh$
As you move from one cube to the next upward (for instance if the path wiggles up and down), the pressure decreases by $\rho g\ dh$.

By the time you get to the target point at depth $h$, it should be clear that you will have seen the pressure increase by a total of $\rho g\ h$.

 

[Phyics hooman]

This can readily be done. Have you had Vector Calculus in you math courses yet (in which you learn about the so-called divergence theorem)?

I am also wondering how this can be cone for any shape (or at least this upright pointing cone) and I am familiar with the divergence theorem (and probably 95% of vector calc). Do you mind demonstrating this approach?

 

[Chestermiller]

I am also wondering how this can be cone for any shape (or at least this upright pointing cone) and I am familiar with the divergence theorem (and probably 95% of vector calc). Do you mind demonstrating this approach?

The surface contact force per unit area exerted by the surroundings on the fluid is $-p\mathbf{n}$ where $\mathbf{n}$ is an outwardly directed normal from the fluid. So the contact force that the surroundings exert on the fluid is $$\mathbf{F}=-\int{p\mathbf{n}dA}$$ But, from the divergence theorem, $$\mathbf{F}=-\int{p\mathbf{n}dA}=-\int{\boldsymbol{\nabla} p dV}$$

But, from the hydrostatic equation, we know that $$\boldsymbol{\nabla} p=-\rho g \mathbf{i_z}$$So, $$\mathbf{F}=\rho g V\mathbf{i_z}$$

 

[Phyics hooman]

I follow the math but the conclusion F=(rho)gV(unit z vector) is just Mg directed along z which i believe this discussion has been saying isn't the case. Am i misreading the last line?

 

[Chestermiller]

The last line is correct, and this is what this discussion (at least my part of it) has been saying. The discussion has been saying that the net force of the container on the fluid (and of the fluid on the container) is equal to the weight of the fluid (as would be expected).