压强和深度的关系

Suppose that we have a tank of static incompressible fluid. Most liquids are close to incompressible, so this is fairly realistic. This means that we may assume that the liquid has a uniform density $\rho$.

Suppose that we have a thin plate of area $A$ at a depth $h$. The thickness of the plate doesn't matter.

The pressure on all of the vertical walls of the plate cancels out; thus, we only need to consider the pressure on the top and the bottom. Also, the pressure on the plate on the top and bottom due to Pascal's Principle also cancel out. Thus, the only remaining force is the weight of the fluid on top of it. The pressure on the surface results only from the weight of this fluid.

The weight of the fluid above it is: $$W=mg=\rho Vg$$ where $V=Ah$ is the volume of the fluid.

Calculating the pressure, $$P=\frac{F}{A}$$ $$P=\frac{W}{A}$$ $$P=\frac{\rho Vg}{A}$$ $$P=\rho gh$$

Thus, we have that the pressure at a depth $h$ is the product of the depth, the density of the fluid, and the gravitational acceleration.

This also means that the pressure difference between two points is directly proportional to the height difference: $$\Delta P=\rho g\Delta h$$ Notice that this derivation is only so simple because the fluid is incompressible and gravity is roughly constant. In a gaseous medium, or in an experiment with more variability in gravitational strength, an accurate calculation of the pressure would require the following integration: $$P=\int_{0}^{h}\rho(z)g(z)dz$$ However, in virtually all cases, the approximations are good enough for us to work with and get accurate results.

In my first lecture on pressure I have talked about the defination of pressure. Now we discuss the pressure in fluids. Suppose I have a cube that was made by iron. Now I try to detract the height of my cube. Gradually it will turn into a surface that has no height actually! Finally I get a surface and I dipped it in water(one kind of fluid). Look at the picture-

Does the surface ABCD feel any force in downwards?

Yes, of course.

What is the ringleader of downwards force?

Actually the answer is the weight of the surface ABCD and the wight of the water(ABCDEFGH)

If we give up the weight of surface, we can measure how much force does the surface ABCD feel that is created by the weight of water.

Now we try to calculate it.

Assume that,

the mass of water(ABCDEFGH) is = M

gravitational acceleration = g = 9.8m/s^2

the volume of ABCDEFGH is = V

the area of the surface ABCD is= A

the deepth of the water where the surface is staying =h

and the density of water is p(roh) = 1000kg/m^3

so,

the downwards force that is created by the weight of water 'F' is

= Mg

=Vpg [M=Vp]

=Ahpg [V=Ah ]

In fine, F=Ahpg

We know that pressure (symbol: P) is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

So the pressure on the surface ABCD which is produced by the weight of water(ABCDEFGH) is

= P

= F/A

=Ahpg/A

=hpg

We can apply this equation in any case of pressure in fluids where h is the deepth of the surface or alluded point in fluid, p(roh) is the density of fluid and g is the gravitational acceleration= 9.8m/s^2.