Now, let's start on an interesting topic....fluid pressure.

If we submerge a flat piece of, say, metal horizontally in water the pressure on it is straightforward. The force on it is $\displaystyle F={\rho}hA$.

Where $\displaystyle {\rho}=\text{weight density of the fluid}$ and

h is the depth and A is the surface area of the object.

Pressure is defined as force per unit volume. Therefore, the pressure p exerted at each point of a flat surface of area A submerged horizontally at a depth h is given by $\displaystyle p=\frac{F}{A}={\rho}h$

In the British system weight density is measured in pounds per cubic foot and pressure is measured in pounds per square foot.

In metric, weight density is measured in newtons per cubic meter and pressure in newtons per square meter.

One Newton per square meter is called apascal

One Pa is approximately 0.02089 pounds per square foot.

The weight density of water is approximately 9810 newtons per cubic meter and 62.4 pounds per cubic foot in the British system.

Here is a very trivial case:

If a flat circular piece of metal of radius 2 meters is submerged horizontally in water so that the top surface is at a depth of 3 meters, then the force on top of the plate is:

Since we are using metric, we use Newtons.

$\displaystyle F={\rho}hA={\rho}h({\pi}r^{2})=(9810)(3)(4\pi)=117 ,720{\pi} \;\ N\approx{369,828} \;\ N$

And the pressure at each point on the plate surface is

$\displaystyle p={\rho}h=(9810)(3)=29,430 \;\ \frac{N}{m^{2}}=29,430 \;\ Pa$

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Suppose we want to find the force on a vertical submerged surface. That is a little more involved than a simple submerged horizontal surface.

Let's introduce a vertical x-axis whose positive direction is downward and whose origin is at any convenient point.

Let's say the submerged portion of the surface extends from x=a to x=b on the x-axis. Also, suppose that a point x on the axis lies h(x)

units below the surface and that the cross section of the plate at x has width w(x). Next, divide the interval [a,b] into n subintervals with length

$\displaystyle {\Delta}x_{1}, \;\ {\Delta}x_{2}, \;\ ......., \;\ {\Delta}x_{n}$

and that each subinterval choose an arbitrary point $\displaystyle x_{k}$. We approximate the section of the plate along the kth subinterval by a rectangle

of length $\displaystyle w(x_{k})$ and width $\displaystyle {\Delta}x_{k}$.

Because the upper and lower edges of the plate are at different depths we can't use the simple formula $\displaystyle F={\rho}hA$.

But, if $\displaystyle {\Delta}x_{k}$ is small enough the difference in depth between the upper and lower edges is small and we can reasonably

assume the entire rectangle is concentrated at a single depth $\displaystyle h(x_{k})$ below the surface.

Therefore, we get: $\displaystyle F_{k}={\rho}h(x_{k})w(x_{k}){\Delta}x_{k}$

Where $\displaystyle h(x_{k})$ is the depth and $\displaystyle w(x_{k}){\Delta}x_{k}$ is the area of the rectangle.

Looks like a Riemann sum, huh?. Well, it is and that's where the integration comes in. We add up all those rectangles to find the total pressure.

Thus, the total force on the plate is $\displaystyle F=\sum_{k=1}^{n}F_{k}=\sum_{k=1}^{n}{\rho}h(x_{k}) w(x_{k}){\Delta}x_{k}$

To find theexactvalue of the force we take the limit as $\displaystyle {\Delta}x_{k}\rightarrow{0}$.

This gives us $\displaystyle F=\int_{a}^{b}{\rho}h(x)w(x)dx$

So, we have the formula for fluid force:

Assume that a flat surface is submerged vertically in a liquid that has weight density $\displaystyle {\rho}$ and that the submerged portion

extends from x=a to x=b on a vertical x-axis. For $\displaystyle a\leq{x}\leq{b}$, let w(x) be the width of the surface at x and let h(x) be the

depth at point x. Then the total fluid force on the surface is $\displaystyle F=\int_{a}^{b}{\rho}h(x)w(x)dx$

In the next thread we will begin some actual problems.