2.6: Hydrostatic Force and Pressure

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Note to the Instructor (click to expand): This topic flows naturally from Work. It's a short section, but can be entertaining and strangely useful.

To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics.

Prerequisite Skills and Support Topics (click to expand)

Linear Functions
- Equations of Lines: In problems involving non-rectangular submerged surfaces, such as the trapezoidal dam in Example 1.7.2, it is necessary to find a linear function for the width of the plate, $w(x)$. This often involves finding the equation of the line that forms the slanted edge of the object.

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

Find the hydrostatic force against a submerged vertical plate.

Hydrostatic Force and Pressure

Let's look at the force and pressure exerted on an object submerged in a liquid. Pressure, $ P $, is force per unit area. That is,\[ P = \dfrac{F}{A}. \label{PFA} \]Therefore, in the English system, we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system, we have newtons per square meter, also called pascals.

We begin with the simple case of a plate of area $A$ submerged horizontally in water at a depth $s$ (Figure $\PageIndex{1}$). The force exerted on the plate is simply the weight of the water above it, which gives\[F = \rho A s, \label{FrAs} \]where $\rho$ is the volumetric density of water. To find the hydrostatic pressure - that is, the pressure exerted by water on a submerged object - we substitute $ F $ from Equation \ref{FrAs} into Equation \ref{PFA} so that\[P = \dfrac{F}{A} = \dfrac{\rho A s}{A} = \rho s. \nonumber \]

$This image has a circular plate submerged in water. The plate is labeled A and the depth of the water is labeled s.$
Figure $\PageIndex{1}$: A plate submerged horizontally in water.

By Pascal's principle, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal's principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot, however, apply the Equation \ref{FrAs} directly, because the depth varies from point to point on a vertically-oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.

Suppose a thin plate is submerged in water. We choose our frame of reference such that the $x$-axis is oriented vertically, with the downward direction being positive, and point $x=0$ corresponding to a logical reference point.¹ Let $s(x)$ denote the depth at point $x$. Note we often let $x=0$ correspond to the surface of the water. In this case, depth at any point is given by $s(x)=x$. However, in some cases we may want to select a different reference point for $x=0$, so we proceed with the development in the more general case. Finally, let $w(x)$ denote the width of the plate at the point $x$.

Assume the top edge of the plate is at point $x=a$ and the bottom edge of the plate is at point $x=b$. Then, for $i=0,1,2, \ldots ,n$, let $P=\{x_i\}$ be a regular partition of the interval $[a,b]$, and for $i=1,2, \ldots ,n$, choose an arbitrary point $x^∗_i \in [x_{i−1},x_i]$. The partition divides the plate into several thin, rectangular strips (Figure $\PageIndex{2}$).

$This image is the overhead view of a submerged circular plate. The x-axis is to the side of the plate. The plate's diameter goes from x=a to x=b. There is a strip in the middle of the plate with a thickness of delta x. On the axis, this thickness begins at x = x sub (i-1) and ends at x = x sub i. The length of the strip in the plate is labeled w(c sub i).$
Figure $\PageIndex{2}$: A thin plate submerged vertically in water.

Let's now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth, $s(x^∗_i)$. We then have\[F_i =\rho A s = \rho [w(x^∗_i) \Delta x] s(x^∗_i). \nonumber \]Adding the forces, we get an estimate for the force on the plate:\[F \approx \sum_{i=1}^n F_i = \sum_{i=1}^n \rho [w(x^∗_i) \Delta x] s(x^∗_i). \nonumber \]This is a Riemann sum, so taking the limit gives us the exact force. We obtain\[F = \lim_{n \to \infty }\sum_{i=1}^n \rho [w(x^∗_i) \Delta x] s(x^∗_i) = \int ^b_a \rho w(x) s(x) \, dx. \label{eqHydrostatic} \]Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.

Problem-Solving Strategy: Finding Hydrostatic Force

Sketch a picture and select an appropriate frame of reference. (Note that if we select a frame of reference other than the one used earlier, we may have to adjust Equation \ref{eqHydrostatic} accordingly.)
Determine the depth and width functions, $s(x)$ and $w(x).$
Determine the weight density of whatever liquid with which you are working. The weight density of water is $62.4 \,\text{lb/ft}^3$, or $9800 \,\text{N/m}^3$.
Use the equation to calculate the total force.

Example $\PageIndex{1}$: Finding Hydrostatic Force

A water trough 15 ft long has ends shaped like inverted isosceles triangles, with a base of 8 ft and a height of 3 ft. Find the force on one end of the trough if the trough is full of water.

Solution

Figure $\PageIndex{3}$ shows the trough and a more detailed view of one end.

$This figure has two images. The first is a water trough with rectangular sides. The length of the trough is 15 feet, the depth is 3 feet, and the width is 8 feet. The second image is a cross-section of the trough. It is a triangle. The top has a length of 8 feet, and the sides have a length of 5 feet. The altitude is labeled with 3 feet.$
Figure $\PageIndex{3a}$: A water trough with a triangular cross-section.
Figure $ \PageIndex{ 3b } $: Dimensions of one end of the water trough.

Select a frame of reference with the $x$-axis oriented vertically and the downward direction being positive. Select the top of the trough as the point corresponding to $x=0$. The depth function, then, is $s(x)=x$. Using similar triangles, we see that $w(x) = 8 − \frac{8}{3} x$. Now, the weight density of water is approximately $62.4 \,\text{lb/ft}^3$, so applying Equation \ref{eqHydrostatic}, we obtain\[ \begin{array}{rcl}
F & = & \displaystyle \int ^b_a \rho w(x) s(x) \, dx \\[16pt]
& = & \displaystyle \int ^3_0 62.4 \left(8−\dfrac{8}{3}x\right) x \,dx \\[16pt]
& = & 62.4 \displaystyle \int ^3_0 \left(8x−\dfrac{8}{3}x^2 \right) \, dx \\[16pt]
& = & 62.4 \left[4x^2−\dfrac{8}{9}x^3\right]\bigg|^3_0 \\[16pt]
& \approx & 748.8. \\
\end{array} \nonumber \]The water exerts a force of approximately 748.8 lb on the end of the trough.

Checkpoint $\PageIndex{1}$

A water trough 12 m long has ends shaped like inverted isosceles triangles, with a base of 6 m and height of 4 m. Find the force on one end of the trough if the trough is full of water.

Solution: $156,800$ N

Example $\PageIndex{2}$: Finding Hydrostatic Force

We now turn our attention to the Hoover Dam. The actual dam is arched rather than flat, but we are going to make some simplifying assumptions to help us with the calculations. Assume the face of the Hoover Dam is shaped like an isosceles trapezoid with a lower base of 750 ft, upper base of 1250 ft, and a height of 750 ft (see the following figure).

$This figure has two images. The first is a picture of a dam. The second image beside the dam is a trapezoidal figure representing the dimensions of the dam. The top is 1250 feet; the bottom is 750 feet. The height is 750 feet.$

When the reservoir is full, Lake Mead's maximum depth is about 530 ft, and the lake's surface is about 10 ft below the top of the dam (see the following figure).

$This figure is a trapezoid with the longer side on top. There is a smaller trapezoid inside the first with a height labeled 530 feet. It is also 10 feet below the top of the larger trapezoid.$
Figure $\PageIndex{4}$: A simplified model of the Hoover Dam with assumed dimensions.

Find the force on the face of the dam when the reservoir is full.
The southwest United States has been experiencing a drought, and the surface of Lake Mead is about 125 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

Solution

We begin by establishing a frame of reference. As usual, we choose to orient the $x$-axis vertically, with the downward direction being positive. This time, however, we will let $x=0$ represent the top of the dam, rather than the water's surface. When the reservoir is full, the surface of the water is $10$ ft below the top of the dam, so $s(x)=x−10$ (see the following figure).
$This figure is a trapezoid with the longer side on top. A smaller trapezoid is inside the first with a height labeled s(x)=x-10. It represents the depth of the water. It is also 10 feet below the top of the larger trapezoid. The top of the larger trapezoid is at x=0.$
Figure $\PageIndex{5}$: We first choose a frame of reference.
To find the width function, we again turn to similar triangles, as shown in the figure below.
$This figure has two images. The first is a trapezoid with a larger side on the top. The length of the top is divided into three measures. The first measure is 250 feet, the second is 750 feet, and the third is 250 feet. The height of the trapezoid is 750 feet. The length of the bottom is 750 feet. Inside the trapezoid, the width is labeled w(x). The second image is the same trapezoid. It has a height labeled as 750 feet. Inside the trapezoid, the height is divided into two segments. The first is labeled x, and the second is labeled 750-x. On the side of the trapezoid, a triangle is formed by a vertical line from the bottom to the top. Inside of the triangle is a horizontal line segment labeled r.$
Figure $\PageIndex{6}$: We use similar triangles to determine a function for the width of the dam. (a) Assumed dimensions of the dam; (b) highlighting the similar triangles.
The figure shows that $w(x)=750+2r$. Using properties of similar triangles, we get $r = 250 - \frac{1}{3} x$. Thus,\[w(x) = 1250 - \dfrac{2}{3}x \nonumber \]Using a weight density of $62.4 \, \text{lb/ft}^3 $ and applying Equation \ref{eqHydrostatic}, we get\[\begin{array}{rcl}
F & = & \displaystyle \int^b_a \rho \, w(x) s(x) \, dx \\[16pt]
& = & \displaystyle \int ^{540}_{10} 62.4 \left(1250−\dfrac{2}{3}x\right)(x−10) \, dx \\[16pt]
& = & 62.4 \displaystyle \int ^{540}_{10}−\dfrac{2}{3}(x^2−1885x+18750)\,dx \\[16pt]
& = & −62.4\left(\dfrac{2}{3}\right)\left(\dfrac{x^3}{3}−\dfrac{1885x^2}{2}+18750x\right) \bigg|^{540}_{10} \\[16pt]
& \approx & 8,832,245,000 \,\text{lb} \\[16pt]
& = & 4,416,122.5\,\text{tons}. \\
\end{array} \nonumber \]Note the change from pounds to tons ($2000$lb = $1$ ton).
This changes our depth function, $s(x)$, and our limits of integration. We have $s(x)=x−135$. The lower limit of integration is 135. The upper limit remains $540$. Evaluating the integral, we get\[\begin{array}{rcl}
F & = & \displaystyle \int^b_a \rho \, w(x) s(x) \, dx \\[16pt]
& = & \displaystyle \int ^{540}_{135}62.4 \left(1250−\dfrac{2}{3}x\right)(x−135)\,dx \\[16pt]
& = & −62.4(\dfrac{2}{3}) \displaystyle \int ^{540}_{135}(x−1875)(x−135)\,dx \\[16pt]
& = & −62.4\left(\dfrac{2}{3}\right) \displaystyle \int ^{540}_{135}(x^2−2010x+253125)\,dx \\[16pt]
& = & −62.4\left(\dfrac{2}{3}\right)\left(\dfrac{x^3}{3}−1005x^2+253125x\right)\bigg|^{540}_{135} \\[16pt]
& \approx & 5,015,230,000\,\text{lb} \\[16pt]
& = & 2,507,615\,\text{tons}. \\
\end{array} \nonumber \]

Checkpoint $\PageIndex{2}$

When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

Solution: Approximately 7,164,520,000 lb or 3,582,260 tons

Footnotes

¹ As before, you could treat this entire subsection using the standard coordinate system. The reader is trusted to be able to derive the relevant mechanics.

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Hydrostatic Force and Pressure

Problem-Solving Strategy: Finding Hydrostatic Force

Example \(\PageIndex{1}\): Finding Hydrostatic Force

Checkpoint \(\PageIndex{1}\)

Example \(\PageIndex{2}\): Finding Hydrostatic Force

Checkpoint \(\PageIndex{2}\)

Footnotes