In hydrostatics, the fluid in question is at rest (with regards to macroscopic flows), and internal forces within a fluid are balanced. The most important force in hydrostatics is the pressure; pressure is the force exerted by the fluid on a surface, normal to a surface, divided by the area of the surface (and in the SI system is measured in pascal. 1 Pa=1 =1 ). Today I will demonstrate some important properties of fluid pressure that are usually stated without proof, or with only minimal proof.

**I. **

Suppose we consider a small volume of fluid (incompressible with uniform density ρ ) in the shape of a triangular prism as in figure 1, with z being the upward direction. The volume of this shape is . Now, let us consider the forces on the three surfaces parallel to y:

Note that with the angle θ as given in the diagram, .

We have the force on the base: upwards

the force on the vertical side: to the right

and the force on the angled side: down and to the left at angle θ up from straight down (see figure 2).

Here we have pressures , , and for the three surfaces. Now, breaking the angled force into it’s horizontal and vertical components:

Now, basic trigonometry tells us and , so:

Horizontally, the forces must balance:

Vertically, the total forces must cancel, including the weight of the volume, which for fluid density ρ is downward. Thus our vertical force balance is:

Now, let us shrink this volume towards a point, while maintaining the same ratio between ∆x and ∆z, so that θ remains the same. Then the weight term goes to zero, and at a single point in the fluid (infinitesimal volume);

Thus the pressures on the surfaces are the same. As this holds for whatever θ we choose, we see that the pressure at a point is the same no matter what direction the surface is oriented: Pressure is isotropic. Thus, at any point, with an outward-oriented surface element , the force due to pressure is , with P(x,y,z) a scalar field.

** II. **

Consider a rectangular volume of fluid, as in figure 3. Then we have volume . Let us consider the forces on the sides, and remember that the forces must cancel in each direction.

First, we consider the forces in the x direcion. We have the force on the surface at x, and the force on the x+∆x surface, which must cancel:

.

Thus we see the pressure cannot vary in x. An analogous arguement holds in the y direction, so we see the pressure does not vary in y either.

Now, we consider the vertical forces:

.

Now let us shrink the volume toward the point (x,y,z). As , we see the left hand side becomes the partial derivative of the pressure with respect to z:

, so, along with from our horizontal force cancellations, we see:

. Using gauge pressure (P=0 at surface), we thus have , where h is the depth below surface. Thus we see hydrostatic pressure is a function of depth alone, and increases linearly with depth.

**III.**

Suppose we have some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. Let be the outward surface normal. Then for any point on the surface, the pressure is , and the force on an area element is:

And so total force is:

Using a form of the divergence theorem for a scalar field (equation #8 at the link):

Now,

.

Thus

which is an upward force equal to the weight of a volume V of the fluid. This is Archimedes’ Principle: the buoyant force on an object is equal to the weight of fluid the object displaces.

Tags: Archimedes, Buoyancy, Friday Physics, Hydrostatics, physics, Pressure

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