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$\vec F$ = force vector,
$\vec A$ = area vector,
$P$ = pressure

Mathematically $\vec F = P \vec A$. By product rule we get,

$$ {\rm d}\vec F = P {\rm d}\vec A + \vec A {\rm d}P $$

Why do we often compute Force over a surface as $\vec F = \int P {\rm d}\vec A$ whilst ignoring the term $\int \vec A {\rm d}P$ ?

G. Smith
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Chaser
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    what would your second integral actually mean? – jacob1729 Sep 03 '20 at 13:19
  • Related question. – BioPhysicist Sep 03 '20 at 13:56
  • @jacob1729 The second integral could mean the change of force due to the change in applied pressure on a given surface. Eg/ consider a pressure of 2Pa on a 5m^2 surface. Pressure is changed to 3Pa. so change of force = ∫AdP = 5*(3-2) = 5N – Chaser Sep 03 '20 at 13:59
  • @Bob D Oh! so is F = PA wrong? I found this definition in many places. And if Pressure is the dot product of the force and normal vector, then where do the integral expressions come from? – Chaser Sep 03 '20 at 14:04

2 Answers2

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If the area under consideration is a more complicated surface than a flat plane, and, if the pressure varies with position spatially on the surface, then the force per unit area at a given location on the surface is $p\mathbf{n}$, where $\mathbf{n}$ is the unit normal to the surface, and the differential element of area over which the force is applied is dA, so the vectorial force on the surface is $\mathbf{F}=\int{p\mathbf{n}dA}$.

Chet Miller
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  • consider a pressure of 2Pa on a 5m^2 surface. Pressure is changed to 3Pa. so change of force = ∫AdP = 5*(3-2) = 5N. Can ∫pndA describe this phenomenon? I'm hoping for a relationship that describes both phenomena simultaneously. – Chaser Sep 03 '20 at 15:34
  • @Chaser Wouldn't that just be $\text dF/\text dt=\int(\text dP/\text dt),\text dA$? – BioPhysicist Sep 03 '20 at 16:27
  • @BioPhysicist -- (t) is time I suppose? Well, the issue is the final Force per se. Not the rate of change nah. And even if we integrate dF/dt=∫(dP/dt)dA we end up back again with F = ∫PdA right? – Chaser Sep 03 '20 at 17:54
  • @chaser No. that would be correct only if A is flat and p is uniform. – Chet Miller Sep 03 '20 at 17:56
  • @Chaser You shouldn't view $\int P,\text dA$ as a "change in force", just like how you don't view $M=\int\rho,\text dV$ as a "change in mass". It is just the total amount of force. But if you want to look at how that total force changes due to a change in pressure over time, you would look at the equation in my previous comment. Then it would tell you the change in the total force over time. – BioPhysicist Sep 03 '20 at 18:05
  • @Chet Miller Exactly. My example is a very simple one. But it cannot be explained using F=∫pndA right? So I thought we need a more general representation and that's how I came up with F⃗ =∫PdA⃗ + ∫A⃗ dP based on product rule. can u point out my mistake? – Chaser Sep 03 '20 at 18:07
  • @BioPhysicist well isn't it a change when u integrate usually with integral limits? like for example, we can talk about a total displacement only if the initial displacement is added to the integral of velocity wrt time? bdw i don't really get how your equation would fit in? can u please elaborate using my own example if u don't mind? – Chaser Sep 03 '20 at 18:19
  • @chaser I already said your method is imcorrrct unless the pressure is uniform and the area is flat. So stop wasting your time. – Chet Miller Sep 03 '20 at 18:24
  • @Chaser Comments are not places to have such discussions. If you have a specific scenario in mind then you should have put it into your question. – BioPhysicist Sep 03 '20 at 19:55
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The definition $\vec F = P \vec A$ is only valid if P is a constant, and in such case your second term is zero. The definition that makes physical sense when P varies with position is $d\vec F = P d\vec A$.

  • Actually ∫ AdP does make sense ryt? as in when Pressure changes when Area is fixed as in the below example. Eg/ consider a pressure of 2Pa on a 5m^2 surface. Pressure is changed to 3Pa. so change of force = ∫AdP = 5*(3-2) = 5N – Chaser Sep 03 '20 at 17:59
  • But in your interpretation that is the change in force across time, not the total force. As is, that term does not make sense in the original equation, which refers to the force at a given time. –  Sep 03 '20 at 19:03