# Basic question - flotation

1. Nov 27, 2009

### elvinc

Hi,

Can someone explain to me what happens when an object sinking in a liquid, hits the bottom of the container holding the liquid. I accept this is an "ideal" situation but it has turned up something I don't understand.

So, say we have a box with rectangular faces (a cuboid) that is immersed in a liquid in a larger container. The liquid is less dense than the material the box is made of, say a lead box placed in water. The box experiences upthrust equal to the weight of the volume of water displaced. This upthrust happens because the pressure on the bottom face of the box is greater than the pressure on the top face (P = hrg) . As the top and bottom faces of the box have equal area we can replace pressure by force and see there is an upthrust (force) on the box. However as the box is made of material more dense than water the weight of the box will exceed the upthrust and the box will sink in the liquid. So far, so good. Now assume the surface of both the immersed box and the floor of the container of liquid are "perfectly flat". When the box settles on the bottom of the container there is no longer any water underneath the box so there is no upward pressure and no upward force. Any forces on the sides of the cuboid will be symmetrical and opposite and will cancel - therefore not contributing to upthrust. Therefore as the box settles on the bottom of the container it will experience a sudden increase in apparent weight as the upthrust has disappeared. However, this just "feels" wrong to me - but I can't explain why. I would expect the measured weight (actual weight - upthrust) to be almost identical just before the box settles on the bottom of the container and as it comes in contact with the bottom of the container.

Any ideas?

Thanks

Clive

2. Nov 27, 2009

### wywong

When the bottom of the box no longer experiences upthrust exerted by the liquid, it experiences normal reaction from the bottom of the large container. That force is equal to the total downward force (weight+water pressure) experienced by the box so that the box is in equilibrium.

Wai Wong

3. Nov 29, 2009

### elvinc

That makes sense. Thanks.

A couple of follow on questions.

1. Am I correct in thinking that the upthrust, T, that occurs when an object is immersed in a fluid is a result of the difference between: the upward force (U) (resulting from the pressure exerted on the underside of the object Pu) and the downward force (D) (resulting from the pressure exerted on the "topside" of the object Pt). i.e. T = U - D ? The sideways forces (resulting from the sideways pressure on the object) cancel. This is a bit easier to visualise if one thinks of a cuboid lying horizontally - but the argument would still be valid however irregular the immersed object.

2. Archimedes principle is "just" a method for calculating the upthrust on a body immersed in a fluid (equal to weight of liquid displaced). A.P. doesn't explain where the upthrust comes from? For that you require 1. above?

3. The flotation/buoyancy explanation in 1. above works whether one is considering an object immersed in a (relatively dense) liquid such as water or a fluid with a (relatively) low density such as a gas? The problem I have understanding is my simple model of liquids is the pressure due to the liquid varies with depth calcuated by hrg but the pressure of a stationary (at the bulk level) gas is equal everywhere. So in a gas there is no pressure differential between the upper and lower surface of the immersed object to explain the net upthrust. So I fall back to Archimides principle to calculate a value for the upward force due to weight of gas displaced but without an explanation.

Thanks

Clive

4. Dec 1, 2009

### cortiver

The pressure of a stationary gas is not equal everywhere. It increases as you go down just as in a liquid, and for the same reason (the gravitational force on a piece of liquid/gas must be balanced by a pressure differential). Of course, since a gas is less dense, the pressure varies a lot more slowly, but that's why the buoyant force from a gas is much smaller.

Edit: And your other two points are basically right.

Last edited: Dec 1, 2009
5. Dec 1, 2009

### Naty1

" Am I correct in thinking that the upthrust, T, that occurs when an object is immersed in a fluid is a result of the difference between: the upward force (U) (resulting from the pressure exerted on the underside of the object Pu) ...

Instead of a cube, think of a dense ball shape for example...what happens when it touches bottom? Or a pyramid shape if it happens to land and sit on an apex??

An advantage to the Archimedes displacement idea is that it makes it clear the shape of the object has no bearing on its buoyancy.

6. Dec 1, 2009

### stewartcs

The apparent weight of the object is dependent upon the pressure field surrounding it. If there is no pressure on the bottom of the perfect cube you are referring to, then there will be no upward force from the fluid.

Archimedes Principle is just a convenient way to introduce buoyancy (do to it's ease of understanding). It is true for most situations but not all if taken at face value. That is, if you neglect the pressure filed developed by the fluid surrounding the object.

For example: Take steel pipe and support it vertically in the ocean by a wire such that it does not touch the bottom of the ocean and the top is just above sea level. How much will it weigh? Well, it will be its in-air weight (i.e. weight on a scale on the surface) less about 13%. The apparent weight of the pipe is smaller due to the pressure field created by the fluid. This just so happens to be the same value as found by using Archimedes Principle.

Now, what happens if we take this pipe and drive it into the ground (neglecting the small force from the top which is just beneath the sea level now)? Well, the pressure field has now changed, it no longer acts on the cross-sectional area of the bottom of the pipe, it only acts perpendicular to the pipe. So how much does it weigh? Well, it weighs the same as it does in-air since there is no more pressure acting on the bottom of the pipe (it's in the ground now)!

But, if you would have used Archimedes Principle, you would have calculated the wrong apparent weight. With Archimedes you would find essentially the same apparent weight as you did in the first part of the example...which would be wrong for the conditions given in the second part of the example.

Does this help?

CS

7. Dec 1, 2009

### sophiecentaur

I'm not sure what you mean by 'apparent weight'. Do you mean what it will weigh when hanging from a spring balance? When it is in the fluid its apparent weight will be 'normal weight' minus upthrust. When it is resting on the ground its apparent weight will be zero.

8. Dec 1, 2009

### stewartcs

I mean just what I wrote. It's apparent weight, that is, how much the object appears to weigh in a fluid.

The pipe's apparent weight is not zero while resting on the ground. Whether the ground is supporting the pipe completely or if the wire is supporting it completely (for the case where it is driven into the ground), the apparent weight is the same, and none-zero.

For example, if the ground was supporting the pipe completely then I attempted to pick the pipe up, the force in the wire would have to be the same as the force provided by the ground in order to move it (actually slightly higher to get it to move). The result being that the weight in either situation is the same, and equal to the in-air weight. It's still called the apparent weight for consistency since it is submerged in a fluid.

This apparent weight (the second case) when compared to the first case is different, which was my point (and the OP's question).

If the pipe were floating it's apparent weight would be zero, but not otherwise.

So again, the apparent weight of a submerged object is dependent on the pressure field, and not just the displaced fluid in all cases. It just happens to work out for a floating object that the buoyant force is equal to the weight of the displaced fluid due to the resultant of the forces acting on the body from the pressure field. That is, Archimedes Principle.

CS

9. Dec 1, 2009

### sophiecentaur

All you can really say is that its total weight turns up, one way or another, as a force on the ground. Displacing some water will just mean that the container will weigh a bit more.
I'm not sure where all this is going - except that Archie was right all those years ago.

There is another issue, however, and that is why the pipe will, left to its own devices, float horizontally an not vertically. That's all a matter of potential energy being at a minimum when it's horizontal.

10. Dec 1, 2009

### stewartcs

What does a container have to do with this? In my example I'm referring to a pipe in the ocean and only talking about the weight of the pipe.

The OP's question is about how much an object would weigh if it were sitting on the bottom of a tank without any fluid beneath it. Hence my example to show how it is not a simple matter of the weight of the displaced fluid as Archimedes Principle states. Archimedes was correct to a certain extent. However, his principle does not always hold as indicated in the example I posted, that is, if the pressure field from the fluid does not exist at the bottom of the object.

That's not an issue in this post...re-read the OP.

CS

11. Dec 1, 2009

### sophiecentaur

Doesn't the ocean press on the ground, then? Your pipe will raise the level of the ocean just enough to increase it's total weight. Are you looking for something more in this? There isn't anything.

12. Dec 1, 2009

### stewartcs

Yes the ocean presses on the ground...so what??

Who cares what the total weight of the ocean or the sea floor is...it has nothing to do with the weight of the pipe.

Your comments have nothing to do with the original post or mine so I don't know what YOU are looking for in this.

If you have a different question, start a new thread. If you are refuting a point, then make it clear, specific, and relevant.

CS

13. Dec 1, 2009

### sophiecentaur

Here is another scenario. You push a cork up through a hole in the bottom of the same tank. Neglecting friction, the force on the cork is downwards and equal to its weight plus the pressure (edit - at the top), times the area of the hole.
You are pushing up with a perfectly fitting piston so no water flows into the hole. If any water can flow in underneath the cork, it will bob upwards because of the pressure difference between top and bottom (or displacement, if you like).
That is the inverse of your scenario and, again, there is 'fudge' at the point when water enters or leaves the gap but there is a distinct difference between before and after.

Last edited: Dec 2, 2009
14. Dec 2, 2009

### stewartcs

That's the point I was making all along...if the pressure field doesn't exist on the bottom then the cork (in your example) will weigh the same as it does in air (plus the added force due to the pressure since it's submerged).

And, if you use Archimedes Principle to determine the apparent weight, you will get the wrong answer (i.e. the cork is submerged and displaces the fluid but it's apparent weight is the same as the in air weight plus the force due to pressure, not the in air weight minus the displaced fluid).

CS

15. Dec 2, 2009

### sophiecentaur

I think the problem is in defining exactly what is meant by 'apparent weight'. If it's apparent then it is being measured. The actual measurement situation can affect it. If, instead, we use the term 'tension in a string suspending the object' then it will be different with and without the water on its bottom but not using the actual word 'weight' may not offend the senses so much.
You could imagine a cylinder with a 'squeegee' inside its lower face which could be used to push all the water out of the gap between it and the floor of the tank. The force on the floor of the tank would gradually get more and more.
When people get 'trapped' by mud under water there could be something like this in operation - they get less and less buoyant as more and more of them goes into the mud.
Then there's the trick of breaking a ruler, laid flat on a table with one end covered by a light newspaper (you don't need the Sunday Heavies) and the other end overhanging. The force due to atmospheric pressure is great enough (briefly) to hold the end of the ruler down on the table. You strike the overhanging end and . . . crack!