---
title: "Vertical force on cylinder with hemispherical ends"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_3/fluid_mechanics:hemispherical-cylinder-vertical-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/hemispherical-cylinder-vertical-force
site: BCT Engineering
---

# Vertical force on cylinder with hemispherical ends

Fluid Mechanics · Chapter 3: Hydrostatics

- Asked in: 2078 Bhadra, 8 marks
- Video: [HYDROSTATIC FORCE ON SUBMERGED HEMISPHERE WITHOUT INTEGRATION](https://www.youtube.com/watch?v=B6c5zbCxu0M) — PHYSICS GURUKOOL
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/hemispherical-cylinder-vertical-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3

## Question

A tank is hermetically sealed into two compartments by plate AB. A cylinder of diameter 0.3 m with two hemispherical ends protrudes above and below the seal AB and is welded to the seal AB. What is the vertical force in the cylinder? [Figure: top compartment with air at $p_1$ = 500 kPa gage above water (20 m deep); plate AB with oil (S.G. = 0.8) layer 3 m; lower compartment with water, $p_2$ = -360 kPa gage; cylinder D = 0.3 m; marked dimensions 0.2 m, 2 m, 25 m]

## Short answer

Net vertical force on the cylinder is downward, computed as (p_u - p_l)A minus the weight of the two hemispheres, resulting in 63.4 kN carried by the weld.

## Answer

The figure values are only partly legible, so the reading used is stated. Pressures are gauge, $g=9.81$.

**Assumed reading:** above the plate AB the fluids are, from the top, air at $p_1=500$ kPa, oil ($S=0.8$) 3 m thick, then water 2 m thick down to AB. Below AB is water with $p_2=-360$ kPa at the plate level. The cylinder ($D=0.3$ m) extends 0.2 m above and 0.2 m below the plate and ends in hemispheres of radius $r=0.15$ m.

```text
      air  p1 = 500 kPa
      oil  S=0.8   3 m
      water        2 m
   ---+===[ cap ]===+--- plate AB
         | cyl 0.2m|
         [ lower  ]
      water, p2 = -360 kPa
```

**Pressures at the bases of the two hemispheres** (flat plane at the end of each cylinder portion):

$$
p_{plate,top}=500+0.8(9.81)(3)+9.81(2)=543.16\ \text{kPa}
$$

$$
p_u=543.16-9.81(0.2)=541.20\ \text{kPa},\qquad p_l=-360+9.81(0.2)=-358.04\ \text{kPa}
$$

**Vertical forces** (the cylinder side has no vertical component). Projected area $A=\pi(0.15)^{2}=0.07069\ \text{m}^2$, hemisphere volume $V_h=\tfrac23\pi(0.15)^{3}=0.007069\ \text{m}^3$.

- Downward force on the upper hemisphere: $F_u=p_uA-\gamma_wV_h$
- Upward force on the lower hemisphere: the pressure there is lower, giving a net downward effect

$$
F_{net}=(p_u-p_l)A-(\gamma_w+\gamma_w)V_h
$$

$$
F_{net}=(541.20+358.04)(0.07069)-2(9.81)(0.007069)
$$

$$
F_{net}=63.57-0.14=63.4\ \text{kN}
$$

**Answer: the fluid exerts a net vertical force of about 63.4 kN downward on the cylinder, and this is the load carried by the weld to the plate AB.** The result is dominated by $(p_u-p_l)A$; use the same method with the exact figure heights.

## Check yourself

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

1. Which parts of the cylinder carry vertical pressure force?
   A. The two hemispherical ends; the cylindrical side has no vertical component
   B. The cylindrical side only
   C. The plate AB only
   D. The side and the ends equally
   Answer: A. Pressure on the vertical cylindrical side is horizontal, so only the hemispheres contribute. (Key point: Where the vertical force comes from)

2. What area multiplies the pressure difference?
   A. π (0.15)^2 = 0.0707 m^2, based on radius 0.15 m
   B. π (0.3)^2 = 0.283 m^2
   C. π (0.15)^3
   D. 2π (0.15) x 0.2
   Answer: A. D = 0.3 m so r = 0.15 m; the projected circular area of each hemisphere is π r^2. (Key point: Projected area)

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