Physics Friday 15: Hydrostatics

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.

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2 Responses to “Physics Friday 15: Hydrostatics”

  1. Physics Friday 96 « Twisted One 151’s Weblog Says:

    […] Friday 96 By twistedone151 Suppose, as in this post, we have a fluid of density ρ, with some arbitrary three-dimensional object immersed in the […]

  2. Physics Friday 97 « Twisted One 151’s Weblog Says:

    […] the ratio of the densities of the cube and the fluid as xi;, so that . Then 0<xi;<1. Then via Archimedes’ Principle, the weight of fluid displaced equals the weight of the cube, and thus the volume Vd of fluid […]

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