Hydrostatic Pressure and Force

Main idea

Water pushes harder the deeper you go, so you can't just multiply one pressure by the whole wall. Instead, slice the wall into thin horizontal strips, each at a single depth where the pressure is constant, and add up pressure × area.

F = \int_a^b {\htmlData{sym=0}{\rho}} {\htmlData{sym=1}{g}}\, {\htmlData{sym=2}{d(y)}}\, {\htmlData{sym=3}{w(y)}}\,{\htmlData{sym=4}{dy}}

What the symbols mean

$\rho$
Density of the fluid
1000 kg/m³ for water.
$g$
Gravity
9.8 m/s². Together ρg = 9800 N/m³ for water.
$d(y)$
Depth of the strip
How far below the surface the strip sits. Pressure there is ρg·d.
$w(y)$
Width of the strip
How wide the wall is at that depth.
$dy$
Strip thickness
A thin horizontal slice.

Explanation

Hydrostatic force is simply the total push of still water on something underwater, like a dam or an aquarium window. ("Hydro" means water and "static" means not moving, so it's the force from water that's just sitting there.) It's what decides how thick a dam or how strong a porthole has to be. The formula $\int \rho g\, d(y)\, w(y)\,dy$ has a lot of letters, but each one is simple. $\rho g$ (how heavy the water is, a fixed number), $d(y)$ (how deep this strip is, since deeper water pushes harder), $w(y)\,dy$ (the area of a thin horizontal strip: its width times its thickness), and the integral (add up the push on every strip, top to bottom). It's pressure × area, done one strip at a time.

If you've ever dived to the bottom of a pool, you've felt this in your ears. At depth $d$ the pressure is $P = \rho g d$ , which is just the weight of the column of water sitting above each square meter. Pressure pushes equally in every direction, so a vertical wall feels it too.

If the pressure were the same everywhere, force would be pressure × area and we'd be done. The trouble is that it changes with depth. So what can we do? The same thing we've done all unit: slice the problem until each piece is simple. A thin horizontal strip sits at basically one depth, so the pressure on it is constant, and its force is just pressure × (strip area). Add the strips with an integral.

The recipe:

Pick a vertical axis. Measuring $y$ downward from the surface makes the depth simply $d = y$ .
Slice the plate into horizontal strips of thickness $dy$ .
Write the strip's width $w(y)$ from the plate's shape (similar triangles for triangles, Pythagoras for circles).
Force on the strip: $\rho g\,d(y)\,w(y)\,dy$ . Integrate over the depths the plate occupies.

In US units, use the weight density of water, $\rho g = 62.5$ lb/ft³.

A useful shortcut. The total force equals $\rho g$ × (depth of the plate's centroid) × (area). That's no coincidence: $\int d(y)\,w(y)\,dy$ is a moment, exactly like the ones in the center of mass lesson, and dividing it by the area gives the centroid's depth. So the integral is secretly computing the pressure at the centroid times the area.

See it

Water rising behind a 10 m wide dam wall. As you can see, the push gets stronger toward the bottom, since pressure grows with depth. The total force works out to F = ρgwd²/2, so it grows with the square of the depth: at 5 m it's about 1.2 MN, and at 10 m it's about 4.9 MN, four times as much. That's why dams are much thicker at the base.

Worked examples

Example 1

A dam is a vertical rectangle 10 m wide holding back water 6 m deep. Find the force on it.

1
Measure $y$ down from the surface, so the depth is just $y$ . Every strip of a rectangle has the same width, 10.

Example 2

The end of a trough is an upside-down isosceles triangle, 4 m wide at the top (the water surface) and 2 m deep. Find the force on it.

1
The triangle narrows steadily from 4 m at the surface to 0 at the bottom, so at depth $y$ its width is $w(y) = 4 - 2y$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A vertical gate is a rectangle 2 m wide and 3 m tall, with its top at the water surface. Find the force on it in newtons.

Problem 2+5 XP

The same gate is lowered so its top is 1 m below the surface. Find the new force.

Problem 3+5 XP

A triangular window has its 2 m base along the water surface and its vertex 3 m below. Find the force on it.

Problem 4+5 XP

A circular porthole of radius 1 m has its center 5 m below the surface. Find the force on it.

Problem 5+5 XP

An aquarium's viewing window is a square $2$ m on a side, set vertically with its top edge $3$ m below the surface. Find the water's force on it ( $\rho g = 9800$ N/m³).

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A dam's face is a vertical trapezoid: $60$ m wide at the water surface, $40$ m wide at the bottom, and $20$ m deep. Find the total water force on it ( $\rho g = 9800$ N/m³).

Where it's used

This is why dams are thicker at the bottom, why deep-sea submersibles are spheres, and how aquarium glass and ship hatches are sized.

How Hard Water Pushes on a DamCalc II · Physics

Center of Mass Probability