# General Query on Archimedes' Principle

• TomK
In summary, when an object is floating, the volume of the partially submerged section of the object equals the volume of water displaced. If an object is held in-place underwater (with a hand), the forces on the object look like they act in a vertical direction and the object will not fall vertically. When the object has fully-sunk, which rules apply? The upthrust in Archimedes Principle is caused by the pressure acting on a surface of the body which has a horizontal component.
TomK
Homework Statement
Archimedes' Principle
Relevant Equations
upthrust = weight of fluid displaced
volume of object = volume of fluid displaced
I had some questions about Archimedes' Principle. This image shows an object being gradually submerged into fluid:

As far as I'm aware, when the object is floating, weight of fluid displaced by object = upthrust acting upwards on the object.
When the object is fully submerged, volume of the object = volume of fluid displaced.When an object is in the sinking process (i.e. upthrust cannot keep object afloat, so object gradually falls through fluid), does only the volume rule apply, or does the upthrust/weight rule apply too?

If an object is held in-place whilst fully submerged, which rules apply? If you hold an object in-place underwater (i.e. with your hand), what do the forces on the object look like (i.e. what direction does each force act in, and which stop the object falling vertically)?

When the object has fully-sunk (and has stopped moving vertically), which rules apply?

I'm guessing that upthrust is always acting on an object in fluid, no matter how much it has been submerged, but the magnitude of upthrust changes to equal the weight of fluid displaced (with upthrust being at its maximum when fully submerged). Is this right?

When an object is floating, does the volume of the partially-submerged section of the object equal the volume of water displaced?Please clarify these conceptions I have. Thank you.

Delta2
TomK said:
When the object has fully-sunk (and has stopped moving vertically), which rules apply?
Is it not obvious what happens if something is fully submerged?

TomK said:
When an object is floating, does the volume of the partially-submerged section of the object equal the volume of water displaced?
How could it be otherwise?

The upthrust in Archimedes Principle is caused by the fluid pressure acting on a surface of the body which has a horizontal component.

So, on a floating vertical cylinder, the upthrust is the pressure at the depth to which the cylinder has sunk into the fluid multiplied by the area of the circular base. The pressure on the curved side acts horizontally so does not contribute to the upthrust.

If you place the cylinder on the surface of the fluid there is no upthrust so the cylinder falls. As it descends, the pressure on the bottom face increases. If the pressure multiplied by the area equals the weight of the cylinder before the cylinder fully submerges the cylinder floats. If the cylinder fully submerges before the pressure multiplied by the area of the bottom face equals the weight, the cylinder sinks.

If the cylinder is fully submerged, the upthrust is the difference between the vertically up pressure on the bottom face multiplied by the area, and the vertically down pressure on the upper face multiplied by the area.

When the cylinder rests on the base of the vessel you can assume that fluid penetrates under the cylinder so the base of the cylinder still has a pressure applied upwards.

If there is a seal which prevents fluid getting under the cylinder, then a cylinder which normally floats will not rise to the surface as there is no upthrust force (pressure x area) on the base. The cylinder will remain at the base of the vessel.

Last edited:
Delta2
Frodo said:
The upthrust in Archimedes Principle is caused by the fluid pressure acting on a surface of the body which has a horizontal component.

So, on a floating vertical cylinder, the upthrust is the pressure at the depth to which the cylinder has sunk into the fluid multiplied by the area of the circular base. The pressure on the curved side acts horizontally so does not contribute to the upthrust.

If you place the cylinder on the surface of the fluid there is no upthrust so the cylinder falls. As it descends, the pressure on the bottom face increases. If the pressure multiplied by the area equals the weight of the cylinder before the cylinder fully submerges the cylinder floats. If the cylinder fully submerges before the pressure multiplied by the area of the bottom face equals the weight, the cylinder sinks.

If the cylinder is fully submerged, the upthrust is the difference between the vertically up pressure on the bottom face multiplied by the area, and the vertically down pressure on the upper face multiplied by the area.

When the cylinder rests on the base of the vessel you can assume that fluid penetrates under the cylinder so the base of the cylinder still has a pressure applied upwards.

If there is a seal which prevents fluid getting under the cylinder, then a cylinder which normally floats will not rise to the surface as there is no upthrust force (pressure x area) on the base. The cylinder will remain at the base of the vessel.

I think I understand now. As long as there is fluid in-contact with the base surface of the object, there will be an upthrust.

Does this upthrust always equal the weight of fluid which the object displaced, regardless of what percentage of the object is submerged?

I'm assuming that a base seal will still result in displacement of fluid equal to the volume of object that is submerged?

TomK said:
Does this upthrust always equal the weight of fluid which the object displaced, regardless of what percentage of the object is submerged?
Yes, provided the fluid can reach all of the object's surface that's below the fluid's surface.
TomK said:
I'm assuming that a base seal will still result in displacement of fluid equal to the volume of object that is submerged?
Yes, but Archimedes' principle will not apply.

As long as there is fluid in-contact with the base surface of the object, there will be an upthrust.
That is better written as "As long as there is fluid in-contact with surface of the object, where the resultant force from the pressure has a vertical component there will be an upthrust."

Pressure always acts normal to the surface it is pressing on. If a vertical cone is placed point downwards in a fluid the pressure acts normal to the face. You need to take the vertical component of the pressure for the upthrust and sum (or integrate) the vertical component of the pressure at all the different depths.

Archimedes Principle is a simplified way of expressing the calculation

Upthrust = pressure x horizontal component of area, or

Upthrust = vertical component of (pressure x normal area)

Archimedes Principle works but it can lead you astray - dealing with the pressure will always reveal what is happening.

Frodo said:
That is better written as "As long as there is fluid in-contact with surface of the object, where the resultant force from the pressure has a vertical component there will be an upthrust."

Pressure always acts normal to the surface it is pressing on. If a vertical cone is placed point downwards in a fluid the pressure acts normal to the face. You need to take the vertical component of the pressure for the upthrust and sum (or integrate) the vertical component of the pressure at all the different depths.

Archimedes Principle is a simplified way of expressing the calculation

Upthrust = pressure x horizontal component of area, or

Upthrust = vertical component of (pressure x normal area)

Archimedes Principle works but it can lead you astray - dealing with the pressure will always reveal what is happening.

Is the horizontal component of the area the same as taking a horizontal cross-section of the shape up to where the fluid line reaches?

For example, if you put in a cone with a circular base into fluid (tip-first - i.e. with the tip facing downwards), would the horizontal area be the same as the base of a truncated cone (based on how far up the cone the fluid reaches)?

Yes the total "horizontal area" is equal to the arae of the circle at the waterline ...

... but the bit of "horizontal area"close to the surface has only a very small pressure applied to it whereas the bit of "horizontal area" close to the tip has a large pressure applied to it as it is deeper.

To do the problem by pressure you need to integrate the vertical component of the pressure over the submerged area of the cone.

This case is more easily solved by thinking of Archimedes Principle - the volume submerged displaces fluid equal to its volume. As the cone is floating, the weight of the cone must be equal to weight of displaced fluid.

Choose the best method according to the problem.

Think about a sphere submerged to 3/4 of its diameter. You can integrate the pressure over the entire submerged surface, including the downward pressure on the bit just below the surface. This is fairly easy to do but not mentally. In this case it is easier just to calculate the volume of the sphere submerged.

Without studying the postings in great detail, the experiment here is something I have done previously in a "Gedanken" sense="thought experiment". It is the "effective" volume of fluid displaced, in the case of a beaker, rather than actual fluid displaced. e.g. It is possible to float a five pound boat shaped like the tub that it is placed in, using just one pound of water. The "effective" volume that is displaced=volume below the water line, will weigh 5 pounds, even if the actual amount of water that is used is much less.
Edit: Perhaps that is not the problem here, because it looks like the experiment does have the volume of displaced water will be the amount below the water line.

Last edited:
Frodo said:
Yes the total "horizontal area" is equal to the arae of the circle at the waterline ...

... but the bit of "horizontal area"close to the surface has only a very small pressure applied to it whereas the bit of "horizontal area" close to the tip has a large pressure applied to it as it is deeper.

To do the problem by pressure you need to integrate the vertical component of the pressure over the submerged area of the cone.

This case is more easily solved by thinking of Archimedes Principle - the volume submerged displaces fluid equal to its volume. As the cone is floating, the weight of the cone must be equal to weight of displaced fluid.

Choose the best method according to the problem.

Think about a sphere submerged to 3/4 of its diameter. You can integrate the pressure over the entire submerged surface, including the downward pressure on the bit just below the surface. This is fairly easy to do but not mentally. In this case it is easier just to calculate the volume of the sphere submerged.

Thank you for answering my questions.

I will look more into 'force due to pressure by integration'. I'm not at university yet, so I haven't had to use integration in real-world problems, but it could be useful to learn in-advance.

Without studying the postings in great detail, the experiment here is something I have done previously in a "Gedanken" sense="thought experiment". It is the "effective" volume of fluid displaced, in the case of a beaker, rather than actual fluid displaced. e.g. It is possible to float a five pound boat shaped like the tub that it is placed in, using just one pound of water. The "effective" volume that is displaced=volume below the water line, will weigh 5 pounds, even if the actual amount of water that is used is much less.
Edit: Perhaps that is not the problem here, because it looks like the experiment does have the volume of displaced water will be the amount below the water line.

Interesting. Is there a link to show/describe this experiment?

It is the "effective" volume of fluid displaced, in the case of a beaker, rather than actual fluid displaced
Which is why I don't like Archimedes Principle as taught in school - it is a special case.

I responded with pressure because that is what is really happening - the upthrust is caused by the fluid pressure.

## 1. What is Archimedes' Principle?

Archimedes' Principle is a scientific law that states that the buoyant force exerted on an object immersed in a fluid is equal to the weight of the fluid that the object displaces. In simpler terms, it explains why objects float or sink in a fluid.

## 2. Who discovered Archimedes' Principle?

Archimedes' Principle was discovered by the ancient Greek mathematician and scientist, Archimedes, in the 3rd century BC. He is also known for his contributions to mathematics, engineering, and astronomy.

## 3. What is the significance of Archimedes' Principle?

Archimedes' Principle is significant because it helps us understand the behavior of objects in fluids and allows us to make predictions about whether an object will float or sink. It is also the basis for many modern technologies, such as ships and submarines.

## 4. How is Archimedes' Principle applied in real life?

Archimedes' Principle is applied in various real-life situations, such as designing ships and submarines, determining the buoyancy of objects, and understanding the behavior of hot air balloons and blimps. It is also used in industries like oil and gas, where it helps to determine the volume of underground reservoirs.

## 5. Can Archimedes' Principle be applied to all fluids?

Yes, Archimedes' Principle can be applied to all fluids, including liquids and gases. However, it is most commonly used in liquids, as gases are less dense and have a lower buoyant force. The principle also applies to objects of any shape or size, as long as they are completely submerged in the fluid.

• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
824
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
25
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
8K
• Introductory Physics Homework Help
Replies
10
Views
10K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
3K