Hydrostatic Pressure and Force
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.5: “Physical Applications” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-5-physical-applications |
| Textbook used in class | Stewart, Calculus, Section 8.3: “Applications to Physics and Engineering” |
Opening Scenario
A dam holds back water. The deeper you go, the greater the pressure: the weight of water above pushes harder on surfaces below. If the dam face were horizontal, total force = pressure times area, a simple product. But the dam face is vertical, so every horizontal strip is at a different depth and feels a different pressure. The total force on the dam face is the integral of (pressure at each depth) times (strip area).
Quick Reference
Pressure at depth $d$: $$P = \rho g d,$$ where $\rho$ is the fluid density (water: $\rho = 1000$ kg/m$^3$ or $62.5$ lb/ft$^3$) and $g \approx 9.8$ m/s$^2$ (or $32$ ft/s$^2$).
Hydrostatic force on a vertical submerged plate with width function $w(d)$ at depth $d$, from $d = a$ to $d = b$: $$F = \rho g \int_a^b d\,w(d)\,dd.$$
Key Concepts
1. Pressure Increases with Depth
At depth $d$, the pressure from the fluid above equals the weight of a unit column of fluid of height $d$: $$P = \rho g d.$$
Key facts:
- Pressure acts equally in all directions (Pascal’s law).
- Pressure depends only on depth, not on the shape of the container or plate.
- On a horizontal plate of area $A$ at depth $d$: $F = PA = \rho g d A$.
2. Vertical Plate: Why Integration Is Needed
On a vertical plate, different horizontal strips are at different depths. A thin horizontal strip at depth $d$ with width $w(d)$ and thickness $dd$ has area $w(d)\,dd$ and faces pressure $\rho g d$. The force on that strip is: $$dF = \rho g d \cdot w(d)\,dd.$$
Total force: $$F = \rho g \int_a^b d\,w(d)\,dd.$$
3. Setting Up the Coordinate System
Choose a coordinate $x$ measuring depth downward from the surface. Label the top of the plate at depth $a$ and the bottom at depth $b$. The width function $w(x)$ describes how wide the plate is at each depth $x$.
For plates that are not rectangles, $w(x)$ must be expressed in terms of $x$ using the geometry of the plate.
4. Pressure vs. Force
Students frequently confuse these:
- Pressure $P = \rho g d$ is force per unit area (Pa or lb/ft$^2$). It depends on depth only.
- Force $F = \int P\,dA$ is total push on a surface. It depends on both depth and the size of the plate.
Worked Example
A triangular plate with base 4 m at the surface (depth 0) and vertex at depth 3 m is submerged vertically. Find the total hydrostatic force on the plate (use water with $\rho g = 9800$ N/m$^3$).
Set up: let $x$ be depth from the surface, $0 \leq x \leq 3$. The plate narrows linearly from width 4 at $x=0$ to width 0 at $x=3$.
Width at depth $x$: $w(x) = 4\left(1-\dfrac{x}{3}\right) = \dfrac{4(3-x)}{3}$.
$$F = 9800\int_0^3 x \cdot \frac{4(3-x)}{3}\,dx = \frac{9800 \cdot 4}{3}\int_0^3 x(3-x)\,dx.$$
$$\int_0^3 (3x-x^2)\,dx = \left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3 = \frac{27}{2}-9 = \frac{9}{2}.$$
$$F = \frac{9800 \cdot 4}{3}\cdot\frac{9}{2} = \frac{9800 \cdot 36}{6} = 9800 \cdot 6 = 58800 \text{ N}.$$
Boxed answer: $F = 58{,}800$ N $= 58.8$ kN.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Computing pressure instead of force | Reporting $P = \rho g d$ as the answer | Force requires integrating $\rho g d \cdot w(d)$ over the plate; pressure is force per unit area |
| Using the depth of the plate’s center instead of integrating | $F = \rho g \bar{d} \cdot A$ (incorrect in general) | This formula gives $F$ only for a rectangle (or can be derived from the centroid; see Pappus); for variable-width plates, integrate |
| Forgetting that width is a function of depth | Using a constant $w$ for a non-rectangular plate | Express $w$ as a function of depth $x$ using the plate’s geometry |
Common Misconceptions
the total hydrostatic force on a vertical plate equals pressure at the plate’s center times the plate’s area.
This is the rate-as-fixed-number error. Pressure $P = \rho g d$ varies with depth $d$; it is not constant across a vertical plate. Using the pressure at the centroid depth $\bar{d}$ and multiplying by area, $F = \rho g \bar{d} A$, gives the correct answer only for a rectangular plate (or can be justified through the centroid formula). For a triangular plate with vertex at depth 3 m and base at depth 0, the average depth $\bar{d}$ is 1 m, but the correct integral gives $F = \rho g \cdot 6$ N (from the example), not $\rho g \cdot 1 \cdot (\text{area})$, because the pressure-weighted depth is not the same as the arithmetic average depth.
pressure and force are the same quantity, so reporting $P = \rho g d$ is the same as reporting the hydrostatic force.
This is the concept-image-conflicts-definition error. Pressure $P = \rho g d$ is force per unit area, measured in pascals or lb/ft$^2$, and it depends only on depth. Force is pressure integrated over area, $F = \int P\,dA$, measured in newtons or pounds. A small plate at great depth can have the same pressure as a large plate at the same depth, but vastly different forces. Reporting pressure instead of force omits the plate’s size entirely and answers the wrong question.
Leveled Practice
Level 1: Rectangular Plate
Problem 1. A rectangular gate 2 m wide and 3 m tall is submerged vertically with its top edge at depth 1 m. Find the total hydrostatic force (use $\rho g = 9800$ N/m$^3$).
Show answer
Depth ranges from $x = 1$ to $x = 4$ (bottom is at $1 + 3 = 4$ m). Width $w(x) = 2$ (constant).
$F = 9800\displaystyle\int_1^4 2x\,dx = 19600\left[\dfrac{x^2}{2}\right]_1^4 = 9800(16-1) = 9800 \cdot 15 = 147{,}000$ N.
Boxed answer: $F = 147{,}000$ N $= 147$ kN.
Level 2: Variable Width
Problem 2. A semicircular plate of radius 2 m is placed with the diameter at the water surface and the curved edge below. Find the force on the plate (use $\rho g = 9800$ N/m$^3$).
Show answer
The flat diameter lies at $x = 0$ (surface). At depth $x$, the width of the semicircle is $2\sqrt{4-x^2}$.
$F = 9800\displaystyle\int_0^2 x\cdot 2\sqrt{4-x^2}\,dx = 19600\int_0^2 x\sqrt{4-x^2}\,dx$.
Sub $u = 4-x^2$, $du = -2x\,dx$. Limits: $u = 4$ to $0$.
$F = 19600\displaystyle\int_4^0 \sqrt{u}\cdot\left(-\dfrac{du}{2}\right) = 9800\int_0^4 \sqrt{u}\,du = 9800\left[\dfrac{2}{3}u^{3/2}\right]_0^4 = 9800\cdot\dfrac{2}{3}\cdot 8 = \dfrac{156800}{3} \approx 52{,}267$ N.
Boxed answer: $F = \dfrac{156{,}800}{3} \approx 52{,}267$ N.
Mastery Checklist
Mental Model
Hydrostatic force on a vertical plate is an integral because depth varies. Slice the plate into thin horizontal strips. Each strip is essentially horizontal, so pressure is constant across it. Force on one strip = pressure at that depth times strip area. The total force is the sum (integral) over all strips. The same principle that turns “area under a curve” into an integral here turns “force on a submerged plate” into an integral.
Connections
Looking back
- Definite integral as accumulation (Section 4.2): Force is the sum of forces on infinitesimal strips, exactly as area is the sum of areas of infinitesimal rectangles.
Looking ahead
- Center of mass (Section 8.3): The force formula $F = \rho g \int d\,w(d)\,dd$ is closely related to the moment about the surface, which determines the center of pressure.