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Cylinder in water

The formula for fluid pressure of any point at depth $h$ in a fluid is $p = hρg$. Here, the formula is derived by calculating the mass of the fluid column above the point (which is equal to $Ahρ$, and multiplying it by $g$ (thus, getting the weight of the fluid column) and dividing by the area.

In the above picture, there's a cylinder inside the water. My physics textbook says that the fluid pressure of the bottom of the cylinder which is at depth $h_2$ is $h_2ρg$. However, in this case, the weight of the column above the bottom of the cylinder isn't equal to $Ah_2ρg$ like before, since the column isn't entirely made of water, the cylinder is there as well, but it's equal to the weight of the cylinder + the weight of the water above the cylinder (so it is $mg + Ah_1ρg$). Why then is the correct formula for fluid pressure at the bottom of the cylinder still $h_2ρg \ $?

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    $\begingroup$ Does the water around the solid know if it surrounds a solid or a volume of water? $\endgroup$
    – basics
    Sep 22, 2022 at 9:42
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    $\begingroup$ @basics if the cylinder is not neutrally buoyant then it should, since the cylinder cannot displace a weight of water equal to its own weight. If it is sinking, then technically it displaced less, so the water below it should feel more force than on either side...? $\endgroup$
    – Basican
    Sep 22, 2022 at 9:51
  • $\begingroup$ Pressure at any point within the water depends only on the depth at that point. You can calculate the pressure by imagining an unobstructed, vertical column of water that extends from the point in question all the way to the surface, but that unobstructed column does not actually have to exist. So long as there is any path through the water, no matter how convoluted, that connects the point to the open surface, then the formula for pressure at that depth will work. $\endgroup$ Sep 22, 2022 at 14:37
  • $\begingroup$ The pressure at a given depth is totally independent of what object exists in the water column. $\endgroup$ Mar 12, 2023 at 16:30

6 Answers 6

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Buoyancy

Let's write the resultant force due to pressure acting of the surface of the body. For a body of general shape, we would integrate the elementary action on the surface

$\mathbf{F}^{buoy} = - \displaystyle \oint_{\partial V} P \mathbf{\hat{n}}$.

For a cylindrical volume, this computation is quite easy, since we can identify 3 contributions,

$\mathbf{F}^{buoy} = \mathbf{F}^l + \mathbf{F}^u + \mathbf{F}^{lat}$

  • from the upper surface, $\mathbf{F}^{u} = -P^u A \mathbf{\hat{z}}$, where $P^u$ is the constant (because each point on the surface has the same depth w.r.t. the free surface) pressure acting on the upper surface;
  • from the lower surface, $\mathbf{F}^{l} = P^l A \mathbf{\hat{z}}$, where $P^l$ is the constant pressure acting on the lower surface;
  • from the lateral surface, $\mathbf{F}^{lat} = \mathbf{0}$ by symmetry.

Now we need to evaluate $P^u$ and $P^l$, using Stevino's law, or in general the pressure in the water at depth $z$

$P(z) = P_0 + \rho g z$
$ \qquad \rightarrow \qquad P^u = P(z_u) = P_0 +\rho g z_u $
$ \qquad \rightarrow \qquad P^l = P(z_l) = P_0 +\rho g z_l = P_0 +\rho g (z_u + h)$

Thus,

$\mathbf{F}^{buoy} = \mathbf{F}^l + \mathbf{F}^u + \mathbf{F}^{lat} =$
$\qquad \ = \left[ P_0 +\rho g (z_u + h) A \right] \mathbf{\hat{z}} - \left[ P_0 +\rho g z_u A \right] \mathbf{\hat{z}} + \mathbf{0} = $
$\qquad \ = \rho g A h \mathbf{\hat{z}} = \rho g V\mathbf{\hat{z}} = M^{fl} g \mathbf{\hat{z}}$

where I introduced the definition $M^{fl} = \rho V$ as the mass that the volume would have if it was made of water (volume of the solid $\times$ density of water).

This is the bouyancy exerted by the water on the solid.

Equilibrium

If you need to write the equilibrium equation of the cylinder, you need to consider all the force acting on the cylinder:

  • the bouyancy pushing the solid upwards, $\mathbf{F}^{buoy} = \rho V^* g \mathbf{\hat{z}}$, being $V^*$ the immersed volume of the solid;
  • the weight of the cylinder pushing the solid downwards, $\mathbf{F}^{weight} = - \rho^{solid} V g \mathbf{\hat{z}}$

Some Remarks

Immersed volume. Pay attention that the volume $V^*$ appearing in bouyancy is the immersed volume of the solid in the water. This goes from $V^* = 0$ (i.e. for an infinitely light solid, with $\rho^{solid} = 0$, and no other external force, think at a table-tennis ball), to $V^* = V$ when the solid is fully immersed.

Steady conditions. When:

  • we write the resultant of the force exerted by the fluid on the solid considering only pressure stress
  • we use Stevino's law to evaluate pressure in a fluid in presence of gravity,

we are implicitly assuming that the system is in steady conditions.

External reaction for the equilibrium. In general, you may need to add some external force $N_z$ to guarantee the equilibrium, so that the equilibrium equation in the vertical direction becomes

$0 = \rho g V^* - \rho^{solid} g V + N_z$

Examples:

  • solid lighter than water $\rightarrow$ floating solid, e.g. $\rho^{solid} = \dfrac{\rho}{2}$: we need no external force to reach equilibrium, $N_z = 0$:

    $\rho g V^* = \rho^{solid} g V \qquad \rightarrow \qquad V^* = \dfrac{\rho^{solid}}{\rho} V = \dfrac{1}{2}V$

  • if we want to push the light solid into the water, $V^* = V$, we need to push it down with a force:

    $N_z = (\rho^{solid} - \rho) g V = -\dfrac{1}{2} \rho g V$

  • solid heavier than water, e.g. $\rho^{solid} = 2 \rho$: we need an external force to prevent the heavy solid to sink.

    $0 = \rho g V^* - \rho^{solid} g V + N_z \qquad \rightarrow$
    $N_z = g \rho ( 2 V - V^* )$

    • if the solid if fully immersed, $V^* = V$, and thus $N_z = \rho g V$,
    • if the solid is hold outside the fluid we get no buoyancy, and thus the reaction force must equilibrate its weight, $N_z = 2 \rho g V$
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    $\begingroup$ this is way too much to answer a simple question. "Help me understand this diagram" should not be answered starting with "let's integrate the elementary action on the surface" $\endgroup$
    – Señor O
    Mar 11, 2023 at 19:17
  • $\begingroup$ who are you to tell. Read other answers and don't bother me $\endgroup$
    – basics
    Mar 11, 2023 at 19:28
  • $\begingroup$ its actually pretty normal thing for people to leave comments about other answers here. you should probably take my advice, but you do you $\endgroup$
    – Señor O
    Mar 12, 2023 at 0:22
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    $\begingroup$ You'd probably know that not all the people learn with, like, find useful the same type of answer. So, what's wrong in having 2-3 answers at different levels of detail? Who are you to tell which answer is too much and which answer is not enough (while I assume that you always perfectly tune your answer to the taste and the need of the OP)? As an example, for me your answer is not clear enough, but it could be there in a bunch of answers of the original post without deserving any comments or downvoting $\endgroup$
    – basics
    Mar 12, 2023 at 10:58
  • $\begingroup$ fair enough, nothing wrong with having multiple answers $\endgroup$
    – Señor O
    Mar 12, 2023 at 13:52
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All these answers are so convoluted.

Clearly this principle at least works just outside the cylinder, right?

So if the pressure just outside the bottom of the cylinder is $P_b$, then the water pressure just under the cylinder has to also be $P_b$ otherwise there would be a net flow of water to whichever pressure is lower. That's all there is to it.

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You may be describing a non-equilibrium situation.

Consider a cylinder of water, density $\rho_{\rm w}$, of the same diameter as the solid cylinder, density $\rho_{\rm c}$, which is immersed in the water.

I have redrawn your diagram so that the depths are measured from the top surface of the water.

enter image description here

The weight of the cylinder of water above the cylinder is $h_1 A \rho_{\rm w}g$ and the weight of the cylinder is $h A \rho_{\rm c}g$.

The pressure at a depth $h_2$ below the water surface is $h_2\rho_{\rm w}g$ and so the upward force exerted by the water on the bottom of the cylinder is $h_2 A \rho_{\rm w}g$.

So one can look at this situation in one of two ways.

The downward force exerted by the water above the cylinder and the cylinder is $h_1 A \rho_{\rm w}g + h A \rho_{\rm c}g$ at the bottom of the cylinder and the upward force exerted by the water on the bottom of the cylinder is $h_2 A \rho_{\rm w}g$.

Noting that $h_2=h+h_1$, in general $h_1 A \rho_{\rm w}g + h A \rho_{\rm c}g\ne h_2 A \rho_{\rm w}g$, ie the forces on the bottom of the cylinder are not equal, which means that it is a non-equilibrium situation unless $\rho _{\rm w} = \rho_{\rm c}$.

The other way of looking at the situation is to note that the weight of water displaced by the cylinder is $h A\rho_{\rm w}g$ which is the upthrust (Archimedes) and the weight of the cylinder is $h A \rho_{\rm c}g$ and these two forces are not necessarily equal unless $\rho _{\rm w} = \rho_{\rm c}$.

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  • $\begingroup$ So does that mean that the formula for fluid pressure at the bottom of the cylinder $h_2ρg$ only work when the cylinder is at rest inside the fluid? $\endgroup$
    – Arxd
    Sep 23, 2022 at 11:23
  • $\begingroup$ @Arxd Equal pressures at the bottom of the cylinder would mean that there was no net force on the cylinder which in turn means that it is at rest (or moving at constant velocity). $\endgroup$
    – Farcher
    Sep 23, 2022 at 11:30
  • $\begingroup$ I just noticed that when you wrote "The downward force exerted by the water above the cylinder and the cylinder is $h_1Aρ_wg+hAρ_cg$ at the bottom of the cylinder and the upward force exerted by the water on the bottom of the cylinder is $h_2Aρ_wg$" you're taking the upward force on the bottom to be $h_2Aρ_wg$, which is exactly what I was confused about in my original question. Since the cylinder and the water column above it are exerting a force $h_1Aρ_wg+hAρ_cg$ downwards, shouldn't the water beneath the cylinder's bottom exert an equal force upwards due to Newton's third law? $\endgroup$
    – Arxd
    Sep 23, 2022 at 18:08
  • $\begingroup$ @Arxd All I was trying to show was that on the bottom of the cylinder the upward and downward forces were not equal unless the densities were the same. If there is a net force on the cylinder then it will accelerate. $\endgroup$
    – Farcher
    Sep 23, 2022 at 22:19
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    $\begingroup$ You are not annoying me and I have obviously not explained the reason well enough. The water dies not move horizontally at depth $h_2$ so the pressure due to the water at this level everywhere must be the same. $\endgroup$
    – Farcher
    Sep 26, 2022 at 14:43
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As you said, the downward pressure at depth h is p=hρg because this is the weight of a column of water above this point divided by the area of the base of the column.

As the molecules of water at this point are not moving anywhere, the upward pressure from the liquid underneath (transmitted via the liquid all around the sides of the column) must be equal to the downward pressure.

What Archimedes noted was that if you replace part of the column of water with something else, it will change the downward pressure, but it will not change the upward pressure as the liquid underneath and around the sides has not changed.

Therefore, the upwards pressure on the yellow object will still be hρg.

And as you noted, the upwards and downwards forces will not be equal if the yellow object's density is different to that of water. The resulting net force will cause the yellow object to move up or down.

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What you think (that the weight of the cylinder adds to the pressure just below the cylinder bottom) is not correct. The water is in an equilibrium situation. That means in this case that the pressure at a depth $y$ is the same everywhere. If it would not be the same, the pressure differential would lead to fluid flow in the horizontal plane which is not the case. There is no water flowing from just under the cylinder sideways. In the non-equilibrium situation (when you push down the cylinder into the water) the water is pushed sideways from the bottom of the cylinder to the sides of it, just to add to the water column aside the cylinder. So the answer to your question is: you're looking at a static equilibrium situation, not a dynamical situation.

(Same answer thus as Señor O).

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Despite the many answers given to the actual question "how does it work" (in a real physical sense) some similar questions ask for a more general derivation of this 'law' in terms of mathematical oriented first principles.

One of such principles is the principle of minimal energy. A closed system with no external forces at work will maintain an equilibrium such that the total energy of the system is an extremum (minimum).

This principle can easily be applied to derive Archimedes's law. The derivation only consists of basic algebra.

In short the derivation is as follows: consider an object floating in a fluid contained in some container like depicted (in 2D, it doesn't matter for the result). enter image description here The fluid as a whole has a potential energy $V_o(h_3)$ while the object has a potential energy $V_1(h_3)$. Both these energies depend on how deep the object is immersed in the fluid (this depth being $h_3$). We have \begin{equation} V = V_0 + V_1, \end{equation} where \begin{align} &V_0=A_0h_1\rho_0\frac{1}{2}h_1+(A_0-A_1)h_2\rho_0(h_1+\frac{1}{2}h_2)\\ &V_1=A_1l\rho_1h_3. \end{align} Rewriting in terms of $h_3$ gives \begin{equation} \begin{split} V=A_1l\rho_1h_3-\frac{1}{2}A_0\rho_0h_3^2\\ -\frac{1}{8}A_0\rho_0l^2+\rho_0h_3v_0-\frac{1}{2}l\rho_0v_0+\frac{1}{2}l\rho_0A_0h_3+\\ \frac{1}{2}\frac{\rho_0v_0^2}{A_0-A_1}-\frac{\rho_0}{A_0-A_1}v_0A_0(h_3-\frac{1}{2}l)+\frac{\rho_0A_0^2}{2(A_0-A_1)}(h_3-\frac{1}{2}l)^2. \end{split} \end{equation} Solving for $\frac{\partial V}{\partial h_3}=0$ we get \begin{equation} h_3=\frac{A_1l\rho_1+\rho_0v_0-A_0l\rho_1+\frac{1}{2}A_0l\rho_0}{A_0\rho_0}. \end{equation} or equivalently \begin{equation} A_{1} gl\rho_{1}=A_{1} gh_{2}\rho_{0}. \end{equation} The left side of this equation equals the weight of the object, the right side equals the weight of the amount of fluid displaced by the object. Thus we have found Archimedes's principle.

Note again it is derived from a higher physical principle instead of from fluid mechanics. The principle of minimal energy is very basic yet very powerful. The reason this works is that fluid mechanics itself can also be derived using this or a similar principle.

I thought this derivation lacked from all answers given to these sort of questions but should for completeness be included.

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