Hydrostatic Force

This is our final physics application: hydrostatic force. As the name suggests (hydro- (water) + static (not moving)), it is the force exerted by an unmoving body of water. Think, for example, of a dam; when designing a dam, we need to be able to calculate the amount of force it needs to withstand.

Dam sideview

Notice that the dam is thicker at the base than it is at the top. This is because the force exerted by the water at the bottom is greater than at the top.

Pressure

There are two pieces that we need to put together to talk about hydrostatic force.

The first piece is that water pressure can be calculated by $P = \rho g d$ where $\rho$ (Greek rho) is the density of water, $g$ is the gravitational acceleration, and $d$ is the depth. Since $\rho$ and $g$ are constants, all the determines pressure is depth: as you go deeper, the pressure increases.

Note on units

In English units, $\rho g = 62.5$ , and in metric units, $\rho = 1000$ and $g = 9.81$ , so $\rho g = 9810$ . In either case, the units on $\rho g$ are force per unit volume.

The units lead into the second thing we need to know: pressure is equal to force divided by area. $P = \dfrac{F}{A}$

Putting these two pieces together: $\dfrac{F}{A} = \rho g d$

Finally, we can solve for $F$ , since force is the part that we're interested in at the moment: $\ans{F = (\rho g d)(\textrm{Area})}$

In this section, we'll calculate the force that is exerted on the side of a vertical plate submerged in water, similar to the case of a dam.

Each time, we'll need to slice the plate into horizontal slices, since the force depends on depth, so we need each slice to have a consistent depth (hence a thin slice, so that the difference in depth between the top and the bottom is approximately zero).

Let's see how this works with an example.

Force on a Triangular Plate

Find the force exerted by the water on the plate shown below.

Triangular plate

Solution

First, cut a horizontal slice out of this plate at a position $x$ :

Slice of triangular plate

Remember that the force on a slice will be $F_{slice} = \rho g\ \cdot d_{slice} \cdot A_{slice}$

Area

To find the area of the slice, we need to find an expression for $y$ , which varies with $x$ : $\begin{array}{c c} x & y\ \hline 0 & 4\ 3 & 0 \end{array}$

$y = -\dfrac{4}{3}x + 4$

Then $A_{slice} = \left(-\dfrac{4}{3}x + 4\right) \ \Delta x$

Depth

Since $x$ starts at the top of the plate ( $1$ ft below the surface of the water), the depth of a slice at position $x$ will be $d_{slice} = x+1$

Force

Since this problem uses English units, $\rho g = 62.5$ so $F_{slice} = 62.5(x+1)\left(-\dfrac{4}{3}x + 4\right) \ \Delta x$

The total force will be the integral of this function over the range of the plate: $x$ goes from $0$ to $3$ : $\begin{aligned} F &= 62.5 \int_0^3 (x+1)\left(-\dfrac{4}{3}x + 4\right) \ dx\ &= 62.5 \int_0^3 -\dfrac{4}{3}x^2 + \dfrac{8}{3}x + 4 \ dx\ &= 62.5 \left[-\dfrac{4}{9}x^3 + \dfrac{4}{3}x^2 + 4x\right]_0^3\ &= \ans{750\ \textrm{lb}} \end{aligned}$

Force on a Trapezoidal Plate

Find the force exerted by the water on the plate shown below.

Trapezoidal plate

Solution

This slice, as always in these problems, is a thin rectangle:

Slice of trapezoidal plate

Area

Find $y$ the same way as before: $\begin{array}{c c} x & y\ \hline 0 & 4\ 2 & 2 \end{array}$

$y = -x + 4$

Then $A_{slice} = (-x+4)\ \Delta x$

Depth

We set the origin at the top of the plate (which helped with the geometry part), and the water surface is $1$ m below that point, so the depth at any $x$ value will be $1$ m less than $x$ : $d_{slice} = x-1$

Force

The force on a slice will be $\begin{aligned} F_{slice} &= \rho g d (A_{slice})\ &= 1000(9.81)(x-1)(-x+4)\ \Delta x \end{aligned}$

Finally, the total force will be the integral of this function from $1$ to $2$ , since that is the range over which water covers the plate: $\begin{aligned} F &= 9810 \int_1^2 -x^2+5x-4\ dx\ &= 9810 \left[-\dfrac{1}{3}x^3 + \dfrac{5}{2}x^2 - 4x\right]_1^2\ &= \ans{11,445\ N} \end{aligned}$

Find the force exerted by the water on the plate shown below.

$\begin{aligned} y &= \dfrac{4}{5}x\ A_{slice} &= \left(\dfrac{4}{5}x\right)\ \Delta x\ d_{slice} &= x\ F_{slice} &= 9810x\left(\dfrac{4}{5}x\right)\ \Delta x\ F &= 9810 \int_0^5 \dfrac{4}{5}x^2\ dx\ &= \ans{326,667 \ N} \end{aligned}$