---
title: "Pressure on inclined rectangular plate"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:inclined-rectangular-plate"
source: https://www.bctengineer.com/q/semester_6/hydro_power/inclined-rectangular-plate
site: BCT Engineering
---

# Pressure on inclined rectangular plate

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

- Asked in: 2071 Bhadra, 8 marks
- Video: [Problem on Inclined Rectangular Plane Submerged in Water - Fluid Mechanics 1](https://www.youtube.com/watch?v=JMU8GZ1jqpw) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/inclined-rectangular-plate
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A rectangular plate 0.6 m wide and 1.2 m deep lies within a water body such that its plane is inclined at 40° to the horizontal and the top edge is 0.70 m below the water surface. Determine the total pressure force on the plate and the location of the center of pressure.

## Short answer

Inclined plate (0.6 m × 1.2 m at 40°) with top edge 0.7 m below surface yields total force 7.67 kN from F = ρgA h̄ (h̄ = 1.086 m). Center of pressure at 1.131 m vertical depth (0.671 m down plate) from h* = h̄ + (IG sin²θ)/(A h̄).

## Answer

For an inclined plane surface: $F = \rho g A\bar h$ and $h^* = \bar h + \dfrac{I_G \sin^2\theta}{A\bar h}$, where $\theta$ is the angle with the horizontal (free surface).

**Given:** width $b = 0.6$ m, length along the slope $d = 1.2$ m, $\theta = 40^\circ$, top edge 0.70 m below the surface; water.

```text
 free surface ===========================
          : 0.70 m
          +.                 
            \  .   plate 1.2 m long,
             \    .  0.6 m wide
              \ G    .
               \    P   .  40 deg to
                +----------- horizontal
```

#### Centroid depth
$$
\bar h = 0.70 + \frac{1.2}{2}\sin 40^\circ = 0.70 + 0.6 \times 0.6428 = 1.0857\ \text{m}
$$

#### Area and second moment
$$
\begin{aligned}
A &= 0.6 \times 1.2 = 0.72\ \text{m}^2 \\
I_G &= \frac{b d^3}{12} = \frac{0.6 \times 1.2^3}{12} = 0.0864\ \text{m}^4
\end{aligned}
$$

#### Total pressure force
$$
F = 1000 \times 9.81 \times 0.72 \times 1.0857 = 7668\ \text{N} = 7.67\ \text{kN}
$$

#### Centre of pressure
$$
\begin{aligned}
h^* &= \bar h + \frac{I_G \sin^2\theta}{A\bar h} \\
&= 1.0857 + \frac{0.0864 \times 0.4132}{0.72 \times 1.0857} \\
&= 1.0857 + 0.0457 = 1.131\ \text{m}
\end{aligned}
$$
Measured along the plate from the top edge: $\dfrac{1.131 - 0.70}{\sin 40^\circ} = 0.671$ m (the centroid is at 0.600 m).

**Answer:** total pressure force $\approx 7.67$ kN; centre of pressure at a vertical depth of 1.131 m below the free surface, i.e. 0.671 m down the plate from its top edge, on the centre line.

## Check yourself

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

1. The plate is 1.2 m long along the slope at 40 deg, top edge 0.70 m deep. How is the centroid's vertical depth found?
   A. 0.70 + 1.2 cos40
   B. 0.70 + 0.6
   C. 0.70 + 0.6 sin40 = 1.0857 m
   D. 0.70 + 1.2 sin40
   Answer: C. The centroid is 0.6 m along the slope from the top edge, so its vertical drop is 0.6 sin40 deg = 0.3857 m. (Key point: Centroid vertical depth)

2. Which I_G is used, with b = 0.6 m and d = 1.2 m measured along the plate?
   A. b d^2/6 = 0.144 m^4
   B. b d^3/12 = 0.0864 m^4
   C. d b^3/12 = 0.0216 m^4
   D. b d^3/3 = 0.3456 m^4
   Answer: B. I_G = b d^3/12 with d the length along the plate: 0.6 x 1.728/12 = 0.0864 m^4. (Key point: Second moment of area)

3. What is the total pressure force on the plate?
   A. About 15.3 kN
   B. About 3.84 kN
   C. About 11.5 kN
   D. About 7.67 kN
   Answer: D. F = 1000 x 9.81 x 0.72 x 1.0857 = 7668 N = 7.67 kN. (Key point: Total force)

4. Why does the centre of pressure formula include sin^2(theta)?
   A. Because theta is measured from the vertical
   B. Because the density changes with angle
   C. Because the area shrinks by sin(theta)
   D. Because force is resolved along the plate and the vertical depth scale is sin(theta) times distance along the plate
   Answer: D. Distances along the slope convert to depths by sin(theta), so the offset I_G/(A y_bar) in depth becomes I_G sin^2(theta)/(A h_bar). (Key point: Role of sin^2 theta)

## Related in Hydro Power

- [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 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
- [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
- [Centre of pressure below centroid proof](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Acentre-pressure-below-centroid) — Hydro Power, chapter 1
- [Pressure on isosceles triangular plate](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-plate-pressure) — Hydro Power, chapter 1
- [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×

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