---
title: "Centre of pressure below centroid proof"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:centre-pressure-below-centroid"
source: https://www.bctengineer.com/q/semester_6/hydro_power/centre-pressure-below-centroid
site: BCT Engineering
---

# Centre of pressure below centroid proof

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2075 Bhadra, 2+6 marks
- Video: [Fluid Mechanics: Topic 4.2 - Center of pressure on a plane surface](https://www.youtube.com/watch?v=J1-4X1gKjFQ) — CPPMechEngTutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/centre-pressure-below-centroid
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Define total pressure and center of pressure in static fluid. Prove that the center of pressure for a vertically immersed plane surface is always below its centroid.

## Short answer

Total pressure is the resultant force, centre of pressure its point of action. For a vertically immersed plane, h* = h_bar + I_G/(A h_bar). Since I_G, A, h_bar are positive, h* > h_bar, so the centre of pressure always lies below the centroid.

## Answer

**Total pressure** is the resultant force exerted by a static fluid on a surface, acting normal to it. **Centre of pressure** is the point on the surface where this resultant force acts.

#### Derivation for a vertical plane surface
Let a plane surface of area $A$ be immersed vertically in a liquid of density $\rho$. Let $\bar h$ = depth of centroid G and $h^*$ = depth of centre of pressure P below the free surface.

```text
 free surface ===========================
        |                |
        | h-bar          | h*
  +-----G------+         |
  | ---strip-- | dh at h |
  |        P   |  <------'
  +------------+
```

**Total pressure:** for a horizontal strip of area $dA$ at depth $h$, $dF = \rho g h\, dA$.
$$
F = \rho g \int h\, dA = \rho g A \bar h
$$

**Centre of pressure:** moment of all strip forces about the free surface equals the moment of $F$:
$$
F h^* = \int \rho g h^2\, dA = \rho g I_o
$$
By the parallel axis theorem, $I_o = I_G + A\bar h^2$, where $I_G$ is the second moment of area about the horizontal axis through G. Therefore
$$
\begin{aligned}
h^* &= \frac{\rho g (I_G + A\bar h^2)}{\rho g A\bar h} \\
&= \bar h + \frac{I_G}{A\bar h}
\end{aligned}
$$

#### Proof that P is below G
- $I_G$ is a second moment of area, always positive.
- $A$ and $\bar h$ are positive for a submerged surface.

So $\dfrac{I_G}{A\bar h} > 0$ and hence $h^* > \bar h$: the centre of pressure always lies **below** the centroid.

**Physical reason:** pressure increases linearly with depth, so the lower part of the surface carries more force than the upper part, pulling the resultant below the centroid.

**Note:** as the surface goes deeper ($\bar h$ large), $I_G/(A\bar h) \to 0$ and P approaches G. Example: rectangle $b \times d$ with top edge at the surface: $h^* = d/2 + \frac{bd^3/12}{bd \cdot d/2} = \frac{2d}{3}$, which is below $\bar h = d/2$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a vertical plane surface, the depth of the centre of pressure is:
   A. h-bar - I_G/(A h-bar)
   B. h-bar + I_G/(A h-bar)
   C. h-bar
   D. I_G / A
   Answer: B. h* = h-bar + I_G/(A h-bar) from the parallel-axis theorem. (Key point: Centre of pressure formula)

2. Why is h* always greater than h-bar?
   A. I_G, A and h-bar are all positive, so I_G/(A h-bar) > 0
   B. I_G is negative
   C. The surface is heavy
   D. Pressure is constant with depth
   Answer: A. The added term is a ratio of positive quantities, so h* > h-bar. (Key point: Proof of P below G)

3. Physically, the centre of pressure lies below the centroid because:
   A. Pressure increases linearly with depth, so lower parts carry more force
   B. Pressure decreases with depth
   C. The free surface pulls the force upward
   D. The surface weight acts downward
   Answer: A. Greater force on the lower part pulls the resultant below the centroid. (Key point: Physical reason)

4. As the surface is immersed deeper (h-bar large), the centre of pressure:
   A. Moves far below the centroid
   B. Approaches the centroid
   C. Moves above the centroid
   D. Moves to the free surface
   Answer: B. I_G/(A h-bar) tends to 0 as h-bar increases, so P approaches G. (Key point: Deep immersion)

## Related in Hydro Power

- [Total pressure and centre of pressure derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atotal-pressure-centre-pressure) — Hydro Power, chapter 1, asked 2×
- [Pressure on rectangular plate 3 m × 5 m](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-plate-3x5) — Hydro Power, chapter 1
- [Pressure on circular plate (2.5 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-5m) — Hydro Power, chapter 1, asked 2×
- [Pressure on circular plate (2.6 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-6m) — Hydro Power, chapter 1, asked 2×
- [Pressure on circular plate (2.2 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acircular-plate-2-2m) — Hydro Power, chapter 1
- [Pressure on vertical rectangular plate 2 m × 3 m](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-plate-2x3) — Hydro Power, chapter 1

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