# Calculating hydrostatic force & locating centre of pressure

1. May 19, 2014

### kachi

1. The problem statement, all variables and given/known data
A circular viewing window of diameter D = 0.4m is placed in the Holding tank. The top of the window is H = 1.2m below the water surface. Find the hydrostatic force acting on the window and locate the corresponding centre of pressure (CP).

There is a diagram attached to the question. I have attached a file with the diagram (hydrostaticq.png).

2. Relevant equations

Pressure (kPa):
p = ρ x g x h

ρ density of water at 20 degrees celsius (kg/m^3)
g acceleration due to gravity (m/s^2)
h height (m)

Force (KN):
F = p x A
where,
p = pressure (Pa)
A = area (m^2)

3. The attempt at a solution

Given:
- Window has a diameter of 0.4m
- Top of window height measures 1.2m

Find:
Hydrostatic force acting on window
Centre of pressure (CP)

Pressure (kPa):
p = ρ x g x h

ρ density of water at 20 degrees celsius = 998 kg/m^3
g acceleration due to gravity = 9.81 m/s^2
h height = 1.2m

Force (KN):
F = p x A
where,
p = pressure
A = area = pi(diameter)^2/4 = pi(0.4m)^2/4= 0.12566m^2

---

p = 998 kg/m^3 x 9.81 m/s^2 x 1.2m
= 11748.456 N/m^2
= 11.75 kPa

F = 11.75 kPa * 0.12566 m^2
= 1.476 KN

Hydrostatic force acting on window is 1.476 KN

----

Is this correct?

Can someone please explain to me in simple English what Hydrostatic force is? What are some real-life examples of where I can see Hydrostatic force? For example, according the problem above is hydrostatic force, the force from water inflicted on the viewing window?

Also with the center of pressure, can you also please clarify what the 'centroid' and the 'centre of pressure' are? I have a slight understanding of what they are but not that much...

Locate the centre of pressure:
Would the centroid be located at the middle of the viewing window? Would the center of pressure be located on the middle of one side of the centroid? I have an image attached where I've marked the points. (hydrostaticq2.png).

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• ###### hydrostaticq2.png
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2. May 19, 2014

### paisiello2

You did not do it correctly.

A hydrostatic force is the force caused only by the weight of a body of water (or any liquid) exerting pressure on its container. Without a container the body of water would collapse so the container then must resist the force that tries to collapse the body. The force is static because the body of water and the container are in equilibrium, hence hydrostatic as opposed to hydrodynamic.

The centroid of a body would be the location of the average location of all points of a body.

The center of pressure on a body would be the location of the average location of all the forces caused by the pressure acting on the body.

You located the centroid of the window correctly but not the center of pressure correctly.

Last edited: May 19, 2014
3. May 19, 2014

### SteamKing

Staff Emeritus
To be more precise, the center of pressure is calculated by determining the first moment of the hydrostatic force acting on the window w.r.t. some convenient reference. The c.o.p. = 1st moment of hydrostatic force / total hydrostatic force.

4. May 19, 2014

### BvU

You can even feel it: try to push an upright empty bucket straight down in a bath. CP is where you have to put your finger to keep it steady.

You use the 1.2 m in your calculation. Don't you think that the pressure at e.g. 1.6 m is a little higher ?

5. May 19, 2014

### Staff: Mentor

Let zc represent the depth of the center of the window, and let y represent the distance measured upwards from the center of the window. Then at location y above the center of the window, the hydrostatic pressure is p = ρg(zc-y). The differential area of window between locations y and y + dy is

$dA=2\sqrt{r^2-y^2}dy$,

where r is the radius of the window. What is the differential force of the water pressure acting on this differential area of window? To get the total force of the water pressure on the window, you need to integrate this differential force from y = -r to y = +r.

Chet

6. May 19, 2014

### kachi

Ok thank you for clear the explanation!

Ok thank you!

Ah I see thanks!! Yes, it looked like my calculated pressure was too small.

Ok thank you! I'll try it again.

Thank you very much paisiello2, SteamKing, BvU and Chestermiller for your responses!