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Pressure is defined as: $$P=\frac{F}{A}$$ My question is what if we want to find the total force in area of $\frac{A}{2}$. Would be equal to $\frac{F}{2}$? I think that is not true unless we know that the force is equally distributed along the surface contact. It is like trying to find mass from the formula: $\rho=\frac{m}{V}$ when the density is not constant. Why pressure is not defined as: $$P=\frac{dF}{dA}$$ Consider a cube made of two equal smaller cubes. The one half has a mass $m_1=500$ $g$ and the other half has mass $m_2=1$ $g$. The cube is sitting on the table. According to the first formula the pressure would be the same in each side of the area of contact irrespective of the mass which is above that area. Is this true? I came up with this this question when I thought about the pressure from gas particles inside a container. We know that if we measure the pressure with a manometer we would find the same value in every point of the container. But knowing the total force that is exerted on the wall of the container and dividing by the area of the wall does not implies that the pressure must be equal in every point of the wall. Does a more formal definition of pressure exist which takes into account the fact that force maybe is not equally distributed?

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It seems like you're basically flip-flipping between "average pressure" and "actual pressure" with your $P$.

If the $P$ you use in the equation just represents some average pressure across the measured area, not the actual pressure at each point; of course you might run into issues when you try to use a smaller area than the one the average was calculated on.

That said, you typically just wouldn't do that, because it would be confusing. If someone says the pressure at the surface is just $P$, you would typically assume that it's $P$ uniform across the entire surface, unless told otherwise. If you $P$ isn't uniform (it could be a function of time and/or position), then you need to be careful to calculate the forces based on the actual pressure.

This can be seen from your $P = \frac {dF}{dA}$. You can rearrange to find the net force: $$F = \int_{x_1}^{x_2}\int_{y_1}^{y_2} P(x,y) dx dy$$ The integral basically just adds up all the infinitesimally small "areas" that the different pressures act on, and sum them into the total force. (there's a bit more with vectors typically, but I omitted all that here just to illustrate my main point)

As far as your cube example goes, I don't see how you conclude that they would have the same pressure. Pressure is force over area, and the heavier cube would apply more force to the same area, and thus have a higher pressure acting on it.

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