---
title: "Force on end of half-filled horizontal tank"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:half-filled-tank-end-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/half-filled-tank-end-force
site: BCT Engineering
---

# Force on end of half-filled horizontal tank

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2076 Asoj, 8 marks
- Video: [Hydrostatic force at end of cylindrical tank   half full](https://www.youtube.com/watch?v=cNFieN0tokY) — Stephanie Tran
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/half-filled-tank-end-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

Cylindrical tank 2 m diameter and 4 m long, with its axis horizontal, is half filled with water and half filled with oil of density 880 kg/m³. Determine the magnitude and position of the net hydrostatic force on one end of the tank.

## Short answer

For the 2 m diameter circular end, upper half oil (γ=8.633 kN/m³) and lower half water (γ=9.81 kN/m³) give total hydrostatic force 27.9 kN, acting 1.26 m below the tank top on the vertical centre line.

## Answer

**Data:** circular end, $D=2$ m ($R=1$ m). Upper half oil ($\rho_o=880$, $\gamma_o=8.633$ kN/m³), lower half water ($\gamma_w=9.81$ kN/m³); the top of the tank is at atmospheric pressure. Depths are measured from the top.

```text
   _______ top, p = 0
  /  oil  \
 |---------|  interface at depth 1 m
  \ water /
   -------
```

For a semicircle of radius $R$: $A=\dfrac{\pi R^{2}}{2}=1.5708\ \text{m}^2$, centroid $\dfrac{4R}{3\pi}=0.4244$ m from the diameter, $I_G=\left(\dfrac{\pi}{8}-\dfrac{8}{9\pi}\right)R^{4}=0.1098\ \text{m}^4$.

#### Oil part (upper semicircle)
Centroid depth $\bar h_o=1-0.4244=0.5756$ m.

$$
F_o=8.633(0.5756)(1.5708)=7.805\ \text{kN}
$$

$$
h_{cp,o}=0.5756+\frac{0.1098}{1.5708(0.5756)}=0.5756+0.1214=0.697\ \text{m (from top)}
$$

#### Water part (lower semicircle)
Pressure at the interface: $p_i=8.633(1)=8.633$ kPa, equivalent to $0.880$ m of water. Centroid is 0.4244 m below the interface.

$$
F_w=(8.633+9.81\times0.4244)(1.5708)=12.797\times1.5708=20.10\ \text{kN}
$$

$$
h_{cp,w}=1+0.4244+\frac{0.1098}{1.5708(0.880+0.4244)}=1.4244+0.0536=1.478\ \text{m (from top)}
$$

#### Net force and position
$$
F=7.805+20.100=27.91\ \text{kN}
$$

$$
\bar h=\frac{7.805(0.697)+20.100(1.478)}{27.91}=1.260\ \text{m below the top}
$$

**Answer: net hydrostatic force $\approx27.9$ kN, acting 1.26 m below the top of the tank (0.74 m above the bottom) on the vertical centre line.**

## Check yourself

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

1. For the upper semicircle (oil), what is the centroid depth from the top?
   A. 4R/(3π) = 0.424 m
   B. 1 - 4R/(3π) = 0.576 m
   C. 1 m
   D. 1 + 0.424 = 1.424 m
   Answer: B. The semicircle's centroid is 0.4244 m from its diameter, which is at depth 1 m, so it lies 0.576 m below the top. (Key point: Oil part centroid)

2. How is the pressure on the lower (water) half found?
   A. Oil pressure at the interface (8.633 kPa) plus water pressure from the interface
   B. Water pressure from the top only
   C. Zero at the interface
   D. Oil pressure only
   Answer: A. The oil layer adds 8.633 kPa at the interface, which is equivalent to 0.880 m of water. (Key point: Water part pressure)

3. What is the net force on the end?
   A. Only the water part, 20.1 kN
   B. Average of the two, 13.95 kN
   C. Sum of the two parts, 7.81 + 20.10 = about 27.9 kN
   D. The oil part x 2
   Answer: C. Forces on the oil and water portions add as they act in the same direction. (Key point: Net force)

4. How is the position of the net force found?
   A. Take moments of the two forces about the top: h = 1.26 m below the top
   B. Average of the two centres of pressure
   C. At the centroid of the full circle, 1 m
   D. At the interface
   Answer: A. h = (7.805 x 0.697 + 20.100 x 1.478)/27.91 = 1.26 m. (Key point: Line of action)

## Related in Fluid Mechanics

- [Net hydrostatic force on conical surface](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Anet-force-conical-surface) — Fluid Mechanics, chapter 3
- [Oil-water interface in enlarged-end U-tube](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Au-tube-enlarged-ends-oil-water-interface) — Fluid Mechanics, chapter 2
- [Force and position on inclined door](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Adoor-force-and-position) — Fluid Mechanics, chapter 3
- [Accelerated oil tank, spilled volume](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aaccelerated-cube-tank-spilling) — Fluid Mechanics, chapter 3
- [Oil-water manometer with gas cylinder](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aoil-water-manometer-gas-cylinder) — Fluid Mechanics, chapter 2
- [Force on 120° pipe bend, 0.23 m³/s](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apipe-bend-120-force-0085) — Fluid Mechanics, chapter 7, asked 2×

## Related in other subjects

- [Hydrostatic force on sector gate (10 m long)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asector-gate-10m) — Hydro Power, chapter 1, asked 2×
- [Hydrostatic force on sector gate (4 m long)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asector-gate-4m) — Hydro Power, chapter 1, asked 2×
- [Net force on panel with water and glycerin](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apanel-water-glycerin) — Hydro Power, chapter 1
- [Position of 2t force for zero net elongation](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abar-force-2t-zero-elongation) — Strength of Materials, chapter 3