---
title: "Pressure on rectangular plate 3 m × 5 m"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:rectangular-plate-3x5"
source: https://www.bctengineer.com/q/semester_6/hydro_power/rectangular-plate-3x5
site: BCT Engineering
---

# Pressure on rectangular plate 3 m × 5 m

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

- Asked in: 2077 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/rectangular-plate-3x5
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A rectangular plate 3 m × 5 m is immersed vertically in water such that the 3 m side is parallel to the water surface. Determine the hydrostatic force and the centre of pressure if the top edge of the surface is (i) flush with the water surface, and (ii) 2 m below the water surface.

## Short answer

For the 3 m × 5 m vertical plate: (i) top flush gives 367.88 kN at 3.333 m depth; (ii) top 2 m below gives 662.18 kN at 4.963 m depth. Forces from F = ρgA h̄, centers from h* = h̄ + IG/(A h̄).

## Answer

The plate is vertical with the 3 m side horizontal, so width $b = 3$ m and depth $d = 5$ m. Use $F = \rho g A\bar h$ and $h^* = \bar h + \dfrac{I_G}{A\bar h}$.

$$
A = 3 \times 5 = 15\ \text{m}^2, \qquad I_G = \frac{b d^3}{12} = \frac{3 \times 5^3}{12} = 31.25\ \text{m}^4
$$

#### (i) Top edge flush with the water surface
$$
\begin{aligned}
\bar h &= \frac{5}{2} = 2.5\ \text{m} \\
F &= 1000 \times 9.81 \times 15 \times 2.5 = 367\,875\ \text{N} = 367.88\ \text{kN} \\
h^* &= 2.5 + \frac{31.25}{15 \times 2.5} = 2.5 + 0.8333 = 3.333\ \text{m}
\end{aligned}
$$
(Check: $h^* = \frac{2}{3}d = \frac{2}{3}\times 5 = 3.333$ m.)

#### (ii) Top edge 2 m below the water surface
$$
\begin{aligned}
\bar h &= 2 + 2.5 = 4.5\ \text{m} \\
F &= 1000 \times 9.81 \times 15 \times 4.5 = 662\,175\ \text{N} = 662.18\ \text{kN} \\
h^* &= 4.5 + \frac{31.25}{15 \times 4.5} = 4.5 + 0.4630 = 4.963\ \text{m}
\end{aligned}
$$

| Case | $\bar h$ (m) | $F$ (kN) | $h^*$ (m) | $h^* - \bar h$ (m) |
|---|---|---|---|---|
| (i) Flush | 2.5 | 367.88 | 3.333 | 0.833 |
| (ii) 2 m below | 4.5 | 662.18 | 4.963 | 0.463 |

**Answer:** (i) $F = 367.88$ kN at 3.333 m depth; (ii) $F = 662.18$ kN at 4.963 m depth. The deeper the plate, the closer the centre of pressure comes to the centroid.

## Check yourself

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

1. The 3 m side is parallel to the water surface. Which are the width b and depth d in I_G = b d^3/12?
   A. b = 15 m, d = 1 m
   B. b = 3 m, d = 3 m
   C. b = 5 m, d = 3 m
   D. b = 3 m, d = 5 m
   Answer: D. The side parallel to the surface is the width (3 m); the vertical side is the depth (5 m), so I_G = 3 x 125/12 = 31.25 m^4. (Key point: Which side is the depth)

2. With the top edge flush with the surface, where is the centre of pressure?
   A. At 3.333 m depth, 2/3 of the depth
   B. At 5 m depth
   C. At 2.5 m depth, the centroid
   D. At 4.0 m depth
   Answer: A. h* = 2.5 + 31.25/(15 x 2.5) = 3.333 m = (2/3) d. (Key point: Flush top edge)

3. For case (ii), top edge 2 m below the surface, what is the centroid depth used in F = rho g A h_bar?
   A. 4.5 m
   B. 7.0 m
   C. 5.0 m
   D. 2.5 m
   Answer: A. h_bar = 2 + 5/2 = 4.5 m, giving F = 662.18 kN. (Key point: Force when 2 m below)

4. What happens to the gap between the centre of pressure and the centroid as the plate goes deeper?
   A. It grows
   B. It stays the same
   C. It shrinks
   D. It reverses sign
   Answer: C. The gap I_G/(A h_bar) falls as h_bar rises: 0.833 m flush, 0.463 m when 2 m below. (Key point: Trend with depth)

## Related in Hydro Power

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