---
title: "Net force on panel with water and glycerin"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:panel-water-glycerin"
source: https://www.bctengineer.com/q/semester_6/hydro_power/panel-water-glycerin
site: BCT Engineering
---

# Net force on panel with water and glycerin

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

- Asked in: 2073 Bhadra, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/panel-water-glycerin
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Find the net hydrostatic force per unit width on rectangular panel AB in the figure given below and determine its line of action. [Figure: vertical wall with water on the left and glycerin (sp. wt. = 12.36 kN/m³) on the right; panel AB is 2 m high; on the water side A is 3 m (2 m + 1 m) below the water surface and B is 5 m below it, with the wall continuing 1 m below B; the glycerin surface is 2 m below the water surface, i.e. 1 m above A.]

## Short answer

Panel AB (2 m high) with water left (depths 3–5 m) and glycerin right (depths 1–3 m) yields net force 29.04 kN/m towards glycerin. Line of action 1.059 m above B from moment balance of individual fluid forces.

## Answer

Work per metre width. Find the force from each liquid and its line of action, then combine them by taking moments about B.

**Given:** panel AB is 2 m high. Water side: A at 3 m depth, B at 5 m depth. Glycerin side: A at 1 m depth, B at 3 m depth. $\gamma_w = 9.81$ kN/m³, $\gamma_g = 12.36$ kN/m³.

```text
 water surface ==
                 |  2 m
       glycerin  |  == glycerin surface
                 |  1 m
               A +
   water   ->    |   <-  glycerin
   (left)  ->    | 2 m <-
               B +
                 | 1 m
```

#### Force from water (left)
$$
\begin{aligned}
F_w &= \gamma_w A\bar h = 9.81 \times (2\times1) \times 4 = 78.48\ \text{kN} \\
h^*_w &= 4 + \frac{1\times 2^3/12}{2 \times 4} = 4.0833\ \text{m}
\end{aligned}
$$
So it acts $5 - 4.0833 = 0.9167$ m above B.

#### Force from glycerin (right)
$$
\begin{aligned}
F_g &= 12.36 \times 2 \times 2 = 49.44\ \text{kN} \\
h^*_g &= 2 + \frac{0.6667}{2 \times 2} = 2.1667\ \text{m}
\end{aligned}
$$
So it acts $3 - 2.1667 = 0.8333$ m above B.

#### Net force
$$
F = 78.48 - 49.44 = 29.04\ \text{kN per m width (towards the glycerin side)}
$$

#### Line of action (moments about B)
$$
\begin{aligned}
F\,y &= 78.48(0.9167) - 49.44(0.8333) \\
&= 71.94 - 41.20 = 30.74\ \text{kN·m} \\
y &= \frac{30.74}{29.04} = 1.059\ \text{m above B}
\end{aligned}
$$

**Answer:** net hydrostatic force $= 29.04$ kN per metre width, acting from the water side towards the glycerin, at 1.059 m above B (0.941 m below A, i.e. 3.941 m below the water surface).

## Check yourself

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

1. Water force is 78.48 kN and glycerin force is 49.44 kN, acting on opposite faces. What is the net force?
   A. 29.04 kN towards the water
   B. 29.04 kN towards the glycerin side
   C. 78.48 kN towards the glycerin
   D. 127.92 kN towards the water
   Answer: B. Forces on opposite faces subtract: 78.48 - 49.44 = 29.04 kN, in the direction of the larger force (water to glycerin). (Key point: Net force direction)

2. Why is the water force 78.48 kN per metre width?
   A. gamma x 2 x 5
   B. gamma x 2 x 4 = 9.81 x 2 x 4, with centroid depth 4 m
   C. gamma x 2 x 3
   D. gamma x 4 x 4
   Answer: B. Panel area per metre = 2 m^2, water centroid depth = (3 + 5)/2 = 4 m. (Key point: Water side force)

3. On the glycerin side, what is the centroid depth of panel AB?
   A. 4 m
   B. 2 m
   C. 3 m
   D. 1 m
   Answer: B. Glycerin surface is 1 m above A, so A is at 1 m and B at 3 m depth; the centroid is at 2 m. F = 12.36 x 2 x 2 = 49.44 kN. (Key point: Glycerin side depths)

4. How is the line of action of the net force found?
   A. Moments of both forces about B equal the moment of the net force
   B. Use the water centre of pressure only
   C. Average the two centres of pressure
   D. Take the midpoint of AB
   Answer: A. F y = 78.48(0.9167) - 49.44(0.8333) gives y = 1.059 m above B. (Key point: Line of action method)

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