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I'm having a difficulty deriving hydrostatic pressure. To be specific, I don't understand this image from Wikipedia:

enter image description here

How does the fluid below act on the below surface? Is it the normal force? Or something else?

And doesn't the buoyancy force act on the volume of fluid too?

another image:

enter image description here

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    $\begingroup$ In the figure you have shown, you actually are determining the buoyant force. It is the upward force due to pressure from below minus the downward force due to pressure from above. This net force supports the weight of the fluid in-between. $\endgroup$ – Chet Miller Feb 4 '17 at 13:28
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To explain the first figure we need to recall that a fluid exert pressure in all directions. This is because the fluid's particles moves in all direction, colliding with neighbors and transferring momentum. Therefore both layers (above and below) the particular "volume of fluid" are pushing it.This force is called contact force. The other force acting on that volume is of course its weight. In order to the fluid remain static we can see that the upward pressure must be greater than the downward one. That is the reason pressure increase with deepness.

The second figure just represents what was said above. As you can see (by the length of the arrows) there are more pressure upwards. This difference in pressures and therefore in forces is what we call buoyancy force.

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To answer your first question, fluids exert the same pressure in every direction. One can think of pressure as a consequence of reactionary forces between the molecules of the fluid itself.The fluid layer experiences an upward force for the same reason an object placed on a solid surface experiences reaction. The pressure gradient arises simply due to the confinement of the fluid in a container.

Thus to maintain equilibrium, one can say that the differential pressure, dp is going to be such that it balances the weight of the layer, thus Adp = gdm where 'dm' is the mass of the layer, on dividing by the volume of the cylindrical cross section, $ \frac{dp}{dh} = -g\rho $ which is probably the formula you were trying to derive.

And like Diracology said, the force of buoyancy is simply due to the differential pressure across the section, and for that reason need not be added. It is already accounted for.

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