A vector quantity has magnitude and direction. Pressure is force applied per area. Wouldn't that be the force having a direction towards the area? And 'as Newton came to realize, within the body of a liquid, pressure acts equally in all directions so that there is no resultant pressing in any direction.' Wouldn't that pressure applied on water and 'pressure acts equally in all directions so that there is no resultant pressing in any direction' just be on a plane that is bent, thus the direction are towards each area calculated towards all direction? And in terms of base quantities it would be kg/ms^2, containing meter in it. But other vector quantities contains meter at the top, does this explain why it is not a vector quantity? And if so how does it work?
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3Does this answer your question? Define Pressure at A point. Why is it a Scalar? – Chemomechanics Jun 13 '23 at 15:27
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Pressure is magnitude of force per unit area. Thus, it is a scalar quantity. – Jun 13 '23 at 17:23
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The force from pressure is always directed perpendicular to the surface that it contacts. – David White Jun 13 '23 at 18:11
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For an incompressible fluid, pressure really represents the isotropic part of the stress tensor, which is a 2nd order tensor (multidirectional). – Chet Miller Jun 14 '23 at 09:50
2 Answers
The formula you are using is:
$P = |\vec{F}_\perp/A|$
You have to note that the force has a direction but so does the surface. When you define the pressure a force exert on a surface, you mean the intensity of the component perpendicular to the surface. If you want to take in account the other components you need an object with more components, called Stress Tensor.
I'll give you a visual reference for new people into the subject:
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Just to clarify, if there is a plane [1][2][3] above [4][5][6] at the middle [7][8][9] below like a 3x3 plane and pressure is exerted on box 5, will the pressure spread among the plane to the other boxes? – Sturburn Jun 14 '23 at 10:33
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Im not sure what you mean by plane above/middle/below, but usually we deal with microscopic quantities, so imagine the boxes like minuscule points. In that case the pressure is exerted only on that point. On a macroscopic object, you need to sum (integrate) over all your object volume to find out how your object reacts – LolloBoldo Jun 14 '23 at 10:59
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If the pressure does exert on that point, wouldn't it be a force towards the direction on that point/area? – Sturburn Jun 14 '23 at 11:44
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Yes, the force has a direction. Also the area orientation has a direction. The pressure is the scalar product of the force and the area direction, so it has no direction. Pressure is the component of the force along the direction perpendicular to the surface. The components not perpendicular are called the stresses. All of them are included in the stress tensor given above – LolloBoldo Jun 15 '23 at 15:46
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A vector quantity has magnitude and direction. Pressure is force applied per area.
Pressure has units of force (mass times length divided by time divided by time) divided by units of area (length times length), but that is not exactly what it "is."
Generally, there are body forces (e.g., due to gravity) and surface forces. The surface force $\vec F$ on a small area $\delta A$, oriented in the $\hat n$ direction, can be written as: $$ \vec F = \vec \Sigma(\hat n)\delta A\;. $$
Following Batchelor "An Introduction to Fluid Dynamics" chapter 1, we see that the Sigma vector components must take the form: $$ \Sigma_i = \sigma_{ij}n_j\;, $$ where $\sigma$ is called the "stress tensor."
Following Batchelor, it can be shown that the stress tensor is symmetric. Further, it can be shown that, for a fluid at rest, the stress tensor can be taken as diagonal: $$ \sigma_{ij} = -p\delta_{ij}\;,\tag{fluid at rest} $$ where $p$ is the pressure (or "static-fluid pressure").
When a fluid is not at rest the stress tensor can have off-diagonal components, due for example to the viscosity and change in fluid velocity with respect to space. But the diagonal part of the stress tensor is still used to define the pressure as: $$ p = -\frac{1}{3}\sum_i \sigma_{ii}\;. $$
So, for example, if you are interested in a fluid at rest and want to determine the force on a small area $\delta A \hat n$ you use the equation: $$ \delta F_{i} = \sigma_{ij}\hat n_j \delta A = -p\delta_{ij}\hat n_j \delta A = -p\hat n_i\delta A\tag{fluid at rest}\;. $$ Or: $$ \frac{\delta F_i}{\delta A} = -p \hat n_i $$
In other words, the force indeed has a magnitude and a direction. The magnitude of the force is $p\delta A$ and the direction is opposite the unit normal that defines the small area of interest.
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