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I'm an engineering student (first year) studying Physics 1 (now an introduction to fluid mechanics).

Q1

In my physics textbook, the "medium pressure" is defined as: $$p_m = \frac{\Delta F_{\perp}}{\Delta S}$$ and, intuitively, the "pressure in a point" is defined as: $$p = \frac{dF_{\perp}}{dS}$$ Regardless if the form it's written in, I cannot understand the meaning of putting a $\Delta$ in front of $F_{\perp}$: usually, we used $\Delta X$ to mean $X_2 - X_1$ for some physical quantity $X$, but here instead it seems to me that we mean:

  • For $\Delta F_\perp$, really just a perpendicular force $F_\perp$
  • For $\Delta S$, we mean a portion of a surface $S$

Q2

In a similar way, what is the meaning of the $\Delta$ in the definition of "medium density"? $$\rho_m = \frac{\Delta m}{\Delta V}$$ It seems to me that, in a similar way, $\Delta m$ just means "the mass $m$ of a portion of the volume $V$".

NOTE

My question derives also from the fact that, apparently for me, there is no reason to add the "differential" $d$ in these definitions, but all the equations work with this "differential form" (and probably wouldn't have sense/work in other forms)

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  • $\begingroup$ You have two answers to your Q1. Regarding Q2, I've never met the equation you quote. The mean density of the material forming a body of mass $m$ and volume $V$ is just $\rho=m/V$. If we add to the body an extra piece of material we could reasonably denote the mass of the extra material by $\Delta m$ and its volume by $\Delta V$, so the density of the extra material would be $\Delta m/\Delta V$. If the material is homogeneous, then $m/V=\Delta m/\Delta V$. But I see no advantage in defining $\rho$ in terms of the incremental quantities or the derivative $dm/dt$. $\endgroup$
    – Philip Wood
    Commented May 18, 2023 at 13:32
  • $\begingroup$ @PhilipWood thanks, but I repeat to you the question I asked to mikestone: how is it possible that someone can get the same results (while, for example, proving theorems) with both versions the definitions, one with the $d$ of the differential, one without? I don’t know why, but this confuses me a lot $\endgroup$
    – selenio34
    Commented May 20, 2023 at 10:42
  • $\begingroup$ It's because, as I said, if the material is homogeneous then the density of an extra bit of material (mass $\Delta m$, volume $\Delta V$) added to the body is the same as the density of the material already in the body, so $\Delta m/ \Delta V=m/V$. $\endgroup$
    – Philip Wood
    Commented May 20, 2023 at 17:22

2 Answers 2

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The $\Delta$ is there to indicate that it is the "bit" of the total force on the boundary due to the pressure pushing on the small area $\Delta S$.

$P= \Delta F/\Delta S$ is better written as a vector equation $$ \Delta {\bf F} = P \,{\bf n} \,\Delta S $$ where $F$ is the force on a small area $\Delta S$ of the fluid boundary., ${\bf n}$ is the unit normal vector to the small area pointing from the fluid into the boundary.

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  • $\begingroup$ Thanks, but it’s still not very clear to me, then, why many other sources just define $P=F/A$ (I used $A$ in this case to indicate the surface). In general, force shouldn’t depend from the area it’s applied to, so why should we take a “small bit” of it? Where am I mistaken? And how is it possible that someone can get the same results (while, for example, proving theorems) with both versions the definitions, one with the $d$ of the differential, one without? $\endgroup$
    – selenio34
    Commented May 20, 2023 at 10:39
  • $\begingroup$ See the answer by @Farcher below. $\endgroup$
    – mike stone
    Commented May 20, 2023 at 11:27
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Pressure is the amount of force applied perpendicular to the surface of an object per unit area.

A problem arises when the surface is not flat ie curved.

So what is done is to take an infinitesimally small element of the surface $dA$ with a clearly definable normal $\hat n$ and then define pressure $P$ via the relationship $\vec F= P\,d\vec A = A \,dA\,\hat n$.
With your notation it is assumed that the incremental area $\Delta A$ is taken to be small enough for the area to be assumed to be flat.

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