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Section 2-4 : Hydrostatic Pressure and Force

In this section we are going to submerge a vertical plate in water and we want to know the force that is exerted on the plate due to the pressure of the water. This force is often called the hydrostatic force.

There are two basic formulas that we’ll be using here. First, if we are \(d\) meters below the surface then the hydrostatic pressure is given by,

\[P = \rho gd\]

where, \(\rho \) is the density of the fluid and \(g\) is the gravitational acceleration. We are going to assume that the fluid in question is water and since we are going to be using the metric system these quantities become,

\[\rho = 1000{\mbox{ kg/}}{{\mbox{m}}^{\mbox{3}}}\hspace{0.25in}\hspace{0.25in}g = 9.81{\mbox{ m/}}{{\mbox{s}}^{\mbox{2}}}\]

The second formula that we need is the following. Assume that a constant pressure \(P\) is acting on a surface with area \(A\). Then the hydrostatic force that acts on the area is,

\[F = PA\]

Note that we won’t be able to find the hydrostatic force on a vertical plate using this formula since the pressure will vary with depth and hence will not be constant as required by this formula. We will however need this for our work.

The best way to see how these problems work is to do an example or two.

Example 1 Determine the hydrostatic force on the following triangular plate that is submerged in water as shown.
The plate is a triangle whose base is horizontal and at the water’s surface.  The point of the triangle is pointed directly down into the water and is directly under the middle of the base.  The length of the base is given as 6 meters and the height of the triangle is given as 4 meters.
Show Solution

The first thing to do here is set up an axis system. So, let’s redo the sketch above with the following axis system added in.

This sketch is almost identical to the sketch in the problem statement with a couple of exceptions.  First, the lengths have been removed and, more importantly, an x-axis has been added that starts in the middle of the base and goes straight down through the lower point of the triangle.  It is also acknowledged that the point will occur at a value of x=4 on the axis.

So, we are going to orient the \(x\)-axis so that positive \(x\) is downward, \(x = 0\) corresponds to the water surface and \(x = 4\) corresponds to the depth of the tip of the triangle.

Next we break up the triangle into \(n\) horizontal strips each of equal width \(\Delta x\) and in each interval \(\left[ {{x_{i - 1}},{x_i}} \right]\) choose any point \(x_i^*\). In order to make the computations easier we are going to make two assumptions about these strips. First, we will ignore the fact that the ends are actually going to be slanted and assume the strips are rectangular. If \(\Delta x\) is sufficiently small this will not affect our computations much. Second, we will assume that \(\Delta x\) is small enough that the hydrostatic pressure on each strip is essentially constant.

Below is a representative strip.

This is a sketch of the triangular plate and there is a line bisecting the triangle drawn from the middle of the base straight down to the point of the triangle below it.  It is noted that the distance from this bisecting line to the edge of the base is 3 meters.  There is also a representative horizontal strip shown at about the midpoint of the triangle.  On the strip, the distance from the bisecting line and the edge of the strip is given as “a”.  The distance from the base of the triangle down to the strip is given as $x_{i}^{*}$ and the distance from the strip to the bottom point of the triangle is given as $4-x_{i}^{*}$.

The height of this strip is \(\Delta x\) and the width is 2\(a\). We can use similar triangles to determine \(a\) as follows,

\[\frac{3}{4} = \frac{a}{{4 - x_i^*}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}a = 3 - \frac{3}{4}x_i^*\]

Now, since we are assuming the pressure on this strip is constant, the pressure is given by,

\[{P_i} = \rho gd = 1000\left( {9.81} \right)x_i^* = 9810x_i^*\]

and the hydrostatic force on each strip is,

\[{F_i} = {P_i}\,A = {P_i}\left( {2a\Delta x} \right) = 9810x_i^*\left( 2 \right)\left( {3 - \frac{3}{4}x_i^*} \right)\Delta x = 19620x_i^*\left( {3 - \frac{3}{4}x_i^*} \right)\,\Delta x\]

The approximate hydrostatic force on the plate is then the sum of the forces on all the strips or,

\[F \approx \sum\limits_{i = 1}^n {19620x_i^*\left( {3 - \frac{3}{4}x_i^*} \right)} \,\Delta x\]

Taking the limit will get the exact hydrostatic force,

\[F = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {19620x_i^*\left( {3 - \frac{3}{4}x_i^*} \right)} \,\Delta x\]

Using the definition of the definite integral this is nothing more than,

\[F = \int_{{\,0}}^{{\,4}}{{19620\left( {3x - \frac{3}{4}{x^2}} \right)\,dx}}\]

The hydrostatic force is then,

\[\begin{align*}F &= \int_{{\,0}}^{{\,4}}{{19620\left( {3x - \frac{3}{4}{x^2}} \right)\,dx}}\\ & = \left. {19620\left( {\frac{3}{2}{x^2} - \frac{1}{4}{x^3}} \right)} \right|_0^4\\ & = 156960\,N\end{align*}\]

Let’s take a look at another example.

Example 2 Find the hydrostatic force on a circular plate of radius 2 that is submerged 6 meters in the water.
Show Solution

First, we’re going to assume that the top of the circular plate is 6 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate. Setting the axis system up in this way will greatly simplify our work.

Finally, we will again split up the plate into \(n\) horizontal strips each of width \(\Delta y\) and we’ll choose a point \(y_i^*\) from each strip. We’ll also assume that the strips are rectangular again to help with the computations. Here is a sketch of the setup.

The water’s surface is shown at the top of the sketch.  Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system.  It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.  Also shown in the sketch is a representative strip that is located a distance of $d_{i}$ down from the surface of the water and is a distance of $y_{i}^{*}$ from the x-axis.  Also, the right half of the strip (i.e. from the y-axis to the edge of the strip on the edge of the circular plate) is shown to have a distance of $x=\sqrt{4-{{\left( y_{i}^{*} \right)}^{2}}}$.

The depth below the water surface of each strip is,

\[{d_i} = 8 - y_i^*\]

and that in turn gives us the pressure on the strip,

\[{P_i} = \rho g{d_i} = 9810\left( {8 - y_i^*} \right)\]

The area of each strip is,

\[{A_i} = 2\sqrt {4 - {{\left( {y_i^*} \right)}^2}} \,\,\Delta y\]

The hydrostatic force on each strip is,

\[{F_i} = {P_i}{A_i} = 9810\left( {8 - y_i^*} \right)\left( 2 \right)\sqrt {4 - {{\left( {y_i^*} \right)}^2}} \,\,\Delta y\]

The total force on the plate is,

\[\begin{align*}F & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {19620\left( {8 - y_i^*} \right)\sqrt {4 - {{\left( {y_i^*} \right)}^2}} \,\,\Delta y} \\ & = 19620\int_{{\, - 2}}^{{\,2}}{{\left( {8 - y} \right)\sqrt {4 - {y^2}} \,dy}}\end{align*}\]

To do this integral we’ll need to split it up into two integrals.

\[F = 19620\int_{{\, - 2}}^{{\,2}}{{8\sqrt {4 - {y^2}} \,dy}} - 19620\int_{{\, - 2}}^{{\,2}}{{y\sqrt {4 - {y^2}} \,dy}}\]

The first integral requires the trig substitution \(y = 2\sin \theta \) and the second integral needs the substitution \(v = 4 - {y^2}\). After using these substitutions we get,

\[\begin{align*}F & = 627840\int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{{{\cos }^2}\theta \,d\theta }} + 9810\int_{{\,0}}^{{\,0}}{{\sqrt v \,dv}}\\ & = 313920\int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{1 + \cos \left( {2\theta } \right)\,d\theta }} + 0\\ & = 313920\left. {\left( {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}\\ & = 313920\pi \end{align*}\]

Note that after the substitution we know the second integral will be zero because the upper and lower limit is the same.