Calculating hydrostatic force & locating centre of pressure

• kachi
In summary, the problem involves finding the hydrostatic force acting on a circular viewing window of diameter 0.4m, which is placed in a holding tank with its top 1.2m below the water surface. The hydrostatic force is calculated using the formula F = p x A, where p is the pressure and A is the area. The pressure is determined using the formula p = ρ x g x h, where ρ is the density of water, g is the acceleration due to gravity, and h is the height. The center of pressure is the location of the average force caused by the pressure acting on the window, and is calculated by taking the first moment of the hydrostatic force acting on the window and dividing
kachi

Homework Statement

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).

Homework 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)

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).Thank you in advance for your help. I appreciate it!

Attachments

• hydrostaticq.png
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• hydrostaticq2.png
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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:
1 person
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.

1 person
What are some real-life examples of where I can see Hydrostatic force?
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 ?

1 person
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

1 person
paisiello2 said:
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.

Ok thank you for clear the explanation!

SteamKing said:
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.

Ok thank you!

BvU said:
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 ?

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

Chestermiller said:
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

Ok thank you! I'll try it again.

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

1. What is hydrostatic force and how is it calculated?

Hydrostatic force is the force exerted by a fluid on a surface due to the pressure of the fluid. It is calculated by multiplying the density of the fluid by the gravitational acceleration and the depth of the fluid above the surface.

2. How do you determine the location of the center of pressure?

The center of pressure is the point at which the total hydrostatic force can be considered to act. It can be determined by dividing the moment of the hydrostatic force by the total hydrostatic force.

3. What factors can affect the calculation of hydrostatic force?

The calculation of hydrostatic force can be affected by the density of the fluid, the gravitational acceleration, and the depth of the fluid above the surface. Other factors such as the shape and orientation of the surface, as well as any obstructions in the fluid, can also affect the calculation.

4. How is hydrostatic force used in engineering and construction?

Hydrostatic force is an important concept in engineering and construction, particularly in the design of structures such as dams, ships, and underwater pipelines. It is also used in the calculation of buoyancy and stability in shipbuilding and the design of offshore structures.

5. Can hydrostatic force be negative?

No, hydrostatic force is always positive as it is a result of the pressure of the fluid on a surface. However, the direction of the force can be negative if the force is acting downwards, for example in the case of an upside-down object submerged in a fluid.

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