Is Force the flux of pressure ?

Main Question or Discussion Point

According to general information I can find in introductory books of physics, area is a pseudo-vector while force is a vector. Pressure appears defined as the ratio between then. Put differently, force will come as the product of pressure and area. Does this product is in the same sense of a flux ?

I would also like to ask about the vectorial character of pressure.
Consider the following statement:
The balance of a small plate depends on the pressure up-down being equal to the pressure down-up.[\b]

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not exactly, flux is a scalar which is a product of force and area. The problem with pressure that it may be uniaxial (like vector) or hydrostatic (like scalar), but the force is always a vector.

so, if in a small cube, very small, if the pressure on its side does not cancel out, then some directional pressure appears. Otherwise, it is hydrostatic and is only scalar.
Is it the rigorous manner to deal with this concept ??

Best Regards,

DaTario

There's no such thing as directional pressure. Let's take an example of what I think you're talking about. Let's take an ideal fluid, rotating inside a vessel at an angular speed $$\omega$$.
Consider an element, at a distance x from the centre of the vessel. Let its width be dx, and its area be dA. Since the element is rotating about the centre, and the fluid is ideal, it is acted upon by the net force
$$\\ dF = \rho\omega^2 xdAdx \\$$
towards the centre.
The only force that can provide the net force is the net force due to pressure. The only way pressure can provide the required force is if it increases along the radial direction. Hence,
$$\\ dAdP = dF$$
Now, the direction comes in ONLY because of the area. The area direction can obviously be both radially inward and outward, but we needn't worry about that as long as it's in the radial direction. The point is, force is not the flux of pressure because it's exerted in ALL directions. No matter how the element is oriented, the pressure is the same scalar function of distance from the centre.
If, for instance, the element had a square shaped area A, and it were oriented so that one side of the square was closer to the centre than the other, you CANNOT take the component of area along the radial direction, and multiply it by the pressure gradient, as you probably would do if force were the flux of pressure. You would have to consider the pressure gradient at each point on the area, and then sum it up by integration! I hope this was clear...it was a little difficult for me to explain it without a diagram.

a complicated circuit of capacitor?how to solve that

neera said:
a complicated circuit of capacitor?how to solve that

To Arun:

shyboy said:
flux is a scalar which is a product of force and area. The problem with pressure that it may be uniaxial (like vector) or hydrostatic (like scalar), but the force is always a vector.
??

arun_mid said:
There's no such thing as directional pressure. Let's take an example of what I think you're talking about. Let's take an ideal fluid, rotating inside a vessel at an angular speed .
Consider an element, at a distance x from the centre of the vessel. Let its width be dx, and its area be dA. Since the element is rotating about the centre, and the fluid is ideal, it is acted upon by the net force

towards the centre.
The only force that can provide the net force is the net force due to pressure. The only way pressure can provide the required force is if it increases along the radial direction. Hence,

Now, the direction comes in ONLY because of the area. The area direction can obviously be both radially inward and outward, but we needn't worry about that as long as it's in the radial direction. The point is, force is not the flux of pressure because it's exerted in ALL directions. No matter how the element is oriented, the pressure is the same scalar function of distance from the centre.
If, for instance, the element had a square shaped area A, and it were oriented so that one side of the square was closer to the centre than the other, you CANNOT take the component of area along the radial direction, and multiply it by the pressure gradient, as you probably would do if force were the flux of pressure. You would have to consider the pressure gradient at each point on the area, and then sum it up by integration! I hope this was clear...it was a little difficult for me to explain it without a diagram.
Let's consider what happens when a straw, which is partially inside a cup of tea, with its upper terminal open, has this same terminal closed and the internal pressure is made to fall to zero instantaneously. Consider what happens in the tea surface. How do you describe the pressure(s) there ?

Comment: obviously that, after an instant, the tea level will rise inside the straw.