Hydrostatic Force & Calculus
Step-by-Step Guide

Imagine you are underwater in a pool. When you are just below the surface of the water, not too deep, you can feel a small amount of force on your head and face. As you go further down in the water, you can feel a lot more pressure.

The sensation that you feel is hydrostatic force!

Hydrostatic force is a branch of fluid mechanics where we want to find the fluid force against an object, like:

• a river against a dam,
• an ocean against a ship’s hull,
• or even blood against your blood vessels.

Cool, right?

Types of Hydrostatic Force

Okay, so we will focus on three types of hydrostatic forces.

1. Submerged Below Water Level
2. Submerged at Water Level
3. Partially Submerged

Understanding Pressure and Hydrostatic Pressure Equation

But before we can find fluid pressure and fluid force, we have to have a better understanding of pressure on an object.

Pressure is the force per unit of area on the surface of an object or body. And the deeper an object is submerged, the greater the pressure upon that object.

So, how do we define pressure?

The pressure on an object at a depth, \(\mathrm{d}\), in a liquid is \(P=\rho g d\), where \(\rho\) is the density of the liquid per unit volume and \(g\) is the gravitational acceleration.

More specifically, the hydrostatic pressure equation describes the pressure at a certain depth within a fluid.

\(P = \color{red}{\rho}\color{black}{g}\color{blue}{d}\)

Where:

• \(P\) = hydrostatic pressure
• \(\color{red}{\rho}\) (rho) = fluid density
• \(\color{black}{g}\) = gravitational acceleration (approximately \(9.81\,m/s^2\))
• \(\color{blue}{d}\) = depth of the fluid column above the point of interest

The equation is based on the assumption that the fluid is both incompressible and stationary, and it also takes into account that the pressure remains constant in the horizontal direction. In line with Pascal’s Principle, the pressure generated by a fluid at a particular depth is experienced uniformly in all directions, and can be calculated using the appropriate formula.

\begin{align*}
\begin{gathered}
\text { Fluid Force }=(\text { Pressure })(\text { Area }) \\
F=P A
\end{gathered}
\end{align*}

Weight-Density Constant – Don’t Make This Mistake

Now this formula seems super easy to use, right?

It is! But there is something sneaky going on that most students miss.

Pressure is found using a variable depth, \(\mathrm{d}\), and a constant weight-density \(\rho g\).

And it’s this weight-density constant that seems to cause unnecessary confusion. So, let’s make sense of it, so you won’t be fooled. The weight-density of a liquid is a constant (number) that behaves like a coefficient. We multiply this constant by the variable depth to find the fluid pressure, \(P\).

Here is a list of some commonly used weight densities, \(\rho g\), of fluids in pounds per cubic foot.

\begin{equation}
\begin{array}{|l|l|}
\hline \text { Ethyl Alcohol } & 49.4 \\
\hline \text { Glycerin } & 78.6 \\
\hline \text { Mercury } & 849 \\
\hline \text { Sea Water } & 64.0 \\
\hline
\end{array}
\end{equation}

But sadly, the weight-density of water is still a heated debate, as scientists bicker over rounding. With that being said, depending on your textbook and the units of measure you are working in, you will see a slight variation in the weight-density constant that is used.

\begin{equation}
\begin{array}{|c|c|c|c|}
\hline \text { English Units } & 62.4 & \text { or } & 62.5 \\
\hline \text { Metric Units } & 9800 & \text { or } & 9810 \\
\hline
\end{array}
\end{equation}

And if you are not told the liquid or units you are working with, then you will simply use a generic letter or symbol, like \(\delta\), to represent the weight-density constant.

For example, \(\delta=9800\) if we are dealing with metric units for water, but \(\delta=78.6\) if we are working with English units for glycerin.

Again, all this means is that we don’t need to multiply \(\rho\) times \(g\) because it’s already been done for us!

All we have to do is grab the right weight-density constant from the tables above.

Calculating Force Exerted by a Fluid

Okay, so now we’re ready to use our knowledge to find the force exerted by a fluid on an object or body!

Integral Formula for Hydrostatic Force

Then the hydrostatic force acting on a vertical plate with a constant weight density is.

\begin{equation}
\delta \int_c^d(\text { strip depth })(\text { length of strip )( width of strip ) }
\end{equation}

\begin{equation}
\delta \int_c^d h(y) L(y) d y
\end{equation}

Now, this integral formula probably looks a bit overwhelming at first, but it’s not so bad once you know how to find all the pieces.

Let’s take a closer look by walking through an example together.

Example: Rectangular Plate

A rectangular plate that has a width of 3 meters and a height of 4 meters is submerged vertically in water so that the top of the plate is 5 meters below the water’s surface. Find the hydrostatic force against one side of the plate.

Step 1

First, we will redraw the vertical, rectangular plate so it is positioned on the \(x-y-a x e s\) at the origin and indicate the water level on the \(y\)-axis.

Step 2

Second, we will place a horizontal strip in the rectangular plate with a width \(\Delta y\)

Step 3

Third, we will determine our strip depth. Notice that the total water depth is 9 meters, so the depth from the water level to our horizontal strip is \(h(y)=9-y\)

Step 4

Next, we will calculate the strip length, length \(=b-a\), or simply \(L(y)=3-0=3\).

Now we are ready to find our limits of integration, or our “c” and “d” values, and our weight density constant. What is important to note is that these values measure the height of the rectangular plate only. Thus, our submerged plane is from \(y=0\) to \(y=4\). And since our rectangular plane is given to us in meters, we will use our metric units of \(9800 \mathrm{~kg} / \mathrm{m}^{3}\).

Step 5

Lastly, we drop all our expressions into our handy-dandy formula, simplify, and integrate!

\begin{align*}
\begin{aligned}
& \delta \int_{c}^{d} h(y) L(y) d y \\\\
& 9800 \int_{0}^{4}(9-y)(3) d y \\\\
& 29400 \int_{0}^{4}(9-y) d y
\end{aligned}
\end{align*}

\begin{equation}
29400\left[\left(9 y-\frac{y^2}{2}\right]_0^4\right]
\end{equation}

\begin{equation}
=29400\left[\left(9(4)-\frac{(4)^2}{2}\right)-\left(9(0)-\frac{(0)^2}{2}\right)\right]
\end{equation}

\begin{equation}
=823200
\end{equation}

This means the force acting on this rectangular plate is \(823,200 \mathrm{~N}\).

Next Steps

While tackling hydrostatic force problems, it’s essential to:

• Draw and label diagrams: This will assist you in visualizing the problem and organizing the necessary information.
• Recognize symmetry: Identifying symmetry in your problems can simplify calculations and help you find solutions more efficiently.

Together, we’ll explore numerous examples with various shapes, such as:

• Squares, rectangles, triangles, trapezoids, and semicircles

During the process, I’ll help you recall essential algebra, geometry, and trigonometry rules while we calculate fluid force.

Though it may seem overwhelming initially, I assure you that this video will keep you afloat and guide you through hydrostatic force problems with ease. Join me as we investigate hydrostatic force together!

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