# Buoyancy-a bit of explanation please

1. Nov 9, 2005

### beginner16

hello

Buoyancy is the upward force exerted on an object immersed in fluid, and upward force on an object in a fluid is equal to the weight of the fluid that was displaced.
If this bouyant force is less than the weight of the object itself, the object will be left with a net downward force and will sink. If the object floats, it floats enough that the bouyant force exactly balances its weight.

Object and water pushing on object from above together have force F[Now] (which is weight of an object + weight of a water above object ), so shouldn't the water below object also be pushing back with equal but opposite force, since that is what the 3. newton's law says?

Afterall, the pressure on water of area A at depth H is caused by the weight
( F[water_above] ) that the water (its volume is A*height) located directly above has. And that is why water at depth H is pushing upwards with force equal but opposite to F[water_above].
So if a part of fluid was displaced with an object it would suggest that now instead of F[water_above] force exerted on water at depth H is

F[Now]=F[water_above] - F[displaced_fluid] + F[object]

So why doesn't water at depth H push back with equal but opposite to the force F[Now]?
Instead it pushes with force equal but opposite to F[water_above], as if the displaced water was still in its place?

thank you

2. Nov 9, 2005

### Fermat

I'm afraid I find your post a bit confusing, so I'll make a few comments about buoyancy and see if that helps.

When the body floats, it is completely supported by the fluid and the fluid force upwards is equal to the weight of the body.

When the body sinks, the weight of the body exceeds the buoyant force. There is thus a net force acting on the body as it accelerates downwards through the fluid. As it moves through the fluid, it experiences a resistive force, or drag, which is usually of the form. R=kv or R = kv². Eventually, the body will reach a speed such that there in no longer a net force on the body and it will continue to sink at constant velocity.

When a body is completely immersed in a fluid, and held static say, there will be pressure force acting down on the top of the body and a pressure force acting up on the bottom of the body.
These forces are not equal, since the fluid pressures are different at different depths. If the pressure at the top of the object is P1 and the length of the body is h, then the pressure on the bottom of the body is P1 + rho.gh

Hope this helps.

3. Nov 9, 2005

### Staff: Mentor

I'll just add a few remarks to Fermat's answer. (I also found your description a bit confusing.)
It's certainly true, from Newton's 3rd law, that whatever force the bottom of the object exerts on the water must equal the force that the water exerts on the bottom of the object. But that force will only equal the "weight of the object + the weight of the water above the object" if the object is in equilibrium.

Also note that the pressure of the fluid at some depth below the surface is the same at any point (at the same depth), whether an object is there or not. Of course, when you put an object in the fluid, the fluid level rises, which increases the pressure.

Let's take a simple example. A perfect cube is put, rightside up, into the fluid. The force on the bottom of the cube from the fluid is due entirely to the pressure of the fluid times the area of the bottom surface. The deeper the cube is in the fluid, the higher the pressure. If the cube is submerged, then there is a net force on the cube due to the fluid above the cube pushing down and the fluid below the cube pushing up. Since the pressure is greater at the bottom of the cube, there's a net upward force on the cube due to the surrounding fluid. That's the buoyant force.

Here's another interesting thing to think about. Say I have a pot of water. I take that cube and put it in the water and it floats, with half of it submerged. Before I put the cube in the water, the force of the water on the bottom of the pot equalled the weight of the water. (And, of course, the force that the pot bottom exerted on the water was the same.) After I put the cube in the water, the water level rose just enough so that the water pressure at the bottom increased a bit. The increased force on the bottom of the pot due to the increased depth of water will exactly equal the weight of the "water + cube".

I hope this helps a bit also.

4. Nov 9, 2005

### beginner16

I really hope you can find the time and explain this to me

What confuses me the most is why water at the bottom of an object
always exerts up same force,no matter if there is an object above it or just water?

I know that since an object is not in equilibrium the net force is greater than zero and thus object will accelerate
But from my understanding of newton's laws : whatever force bottom of the object exerts on water,water must exert same force on bottom of an object -> the sum of forces is zero...end of story

5. Nov 9, 2005

### Staff: Mentor

For fluid in hydrostatic equilibrium (as opposed to a moving fluid), the pressure must be the same at any point along a horizontal line in the fluid. If this wasn't the case, the pressure difference would cause the fluid to flow from one side to the other until the pressure equilized.

I think you need to better understand Newton's 3rd law: For one thing, the two forces in Newton's 3rd law never (by themselves) produce equilibrium since they act on different bodies. What matters (for equilibrium) is the sum of the forces acting on the same body. The fact that the water and the surface exert equal and opposite forces on each other does not mean they add to zero: you can't (meaningfully) add forces that act on different bodies.

6. Nov 10, 2005

### beginner16

I know that. When I said sum of forces is zero I meant forces applied on an object. Two forces are exerted on an object from above:

*F[object]=Force of gravity + weight of the water

*So object exerts F[object] on water below it

*Due to Newton's 3. law water should exert same force on bottom of an object

To sum it up: We have F[object] pushing on object from above and -F[object] pushing on object from below . And since those two forces acting on object are of opposite direction, sum should be zero

But fact of the matter is that with accordance to newton's 3. law water below object should exert back same force as object exerted on it, and that force should be

F[object]=Force of gravity + weight of the water

I find it confusing that water below object doesn't "obey" newton's 3. law just because otherwise fluid would flow from one side to the other until the pressure is equilized

7. Nov 10, 2005

### Staff: Mentor

This is the force on the object due to the water pressure pushing down on it and its weight. OK.
Only if the object is in equilibrium.
Newton's law says that the object and water underneath will exert equal and opposite forces on each other. That's always true. And if the object is in equilibrium, that force will equal what you call F[object].
If the object is in equilibrium, that's certainly true.

Absolutely!
Only if the object is in equilibrium.
Why do you keep saying that the water below doesn't obey Newton's 3rd law? Of course it does!

Here's another example (with solids, not fluids) that may help. Consider two blocks of wood, one on top of the other, both sitting in the palm of my hand. Newton's 3rd law says that whatever force my palm exerts on the bottom block must equal the force that the bottom block exerts on my palm. Always true! But that does not mean that the force my palm exerts must equal the weight of both blocks. That's only true if the blocks are in equilibrium. But they don't have to be. I could be dropping my hand down (compare to an object sinking) or lifting it up (compare to the forces on a balloon forced underwater).

8. Nov 10, 2005

### beginner16

Please have patience with me. I would really need some thorough explanation on this since I'm totally lost. Below are just few questions I have but i'd appreciate if you could offer as much explanation you can think of with regards to this subject

Why exactly must object be in equilibrium for the water to push back with F[object] ? If object immersed in fluid is accelerating downwards is then the force it exerts on water greater than if it was in equilibrium?

But object in equilibrium has zero net force precisely because water at the bottom of an object exerted force equal to F[object]. Yet you are saying as if water exerted force equal but in opposite direction as F[object] BECAUSE object is in equilibrium. Isn't it just the other way around?! Object is in equilibrium BECAUSE water exerted force - F[object] on it

if a brick with a mass 1 kg drops on my head I won't be hit with force 10 N ?

By blocks in equilibrium you mean to in equilibrium to eachother(as in even if both blocks are moving with acceleration a in same direction they are in equilibrium to eachother ) ?

So unless in equilibrium object doesn't push down with force F[object]? Because it is already moving down and thus force of the water above it can't affect it that much? How much does weight of the water affect this object if object is sinking? Lordy am I confused :(

9. Nov 10, 2005

### BerryBoy

From Archimedes' principle:

The upthrust force on an object is equal to the weight of the fluid displaced. So if I submerge a large 2x2x2metre cube into water and it floats so that a metre of the height of the cube is submerged, then the force upwards will be the weight of the water displaced:

Density of water (at surface) = 1000kgm-3
Volume of water displaced = 2 x 2 x 1 metres

So the Weight of the water displaced = 1000 x 4 x 10 = 40 000 N

Now if I pushed the block futher down, more water would be displaced and so the force would increase.

Does this help at all?

Regards,
Sam

10. Nov 10, 2005

### Staff: Mentor

Let's keep it nice and simple. The force that the water exerts upward on the object merely depends on the pressure of the water at that depth. It has nothing to do with equilibrium. But if the object is in equilibrium, then you know that that upward force on the object must equal the downward force on the object.
Ignoring the motion of the water for simplicity, the force that the water exerts on the object (and vice versa) is always the same (if the object is fully immersed). The buoyant force depends only on the weight of the displaced water. If you take two identically shaped objects, one less dense than water (cork, say), the other more dense (iron), and submerge both under water: The water will exert the exact same force on both.

For the cork, that net upward force will be greater than the object's weight, so unless there's another force (like your hand) holding it under, it will move towards the surface. It's not in equilibrium. For the iron, the buoyant force is less than its weight, thus it sinks.

The object always exerts the same force on the water as the water exerts on the object. This is true whether the object sinks, floats, or whatever. (That follows from Newton's 3rd law.) But if you know that the object is in equilibrium, you can DEDUCE that the upward force (on the bottom of the object) must be equal to what you call F[object]. (That's a trivial statement following from Newton's 2nd law.)

Of course not! The force will depend on how far it dropped before landing on your head (and on how quickly your head stops it). Since your head must de-accelerate that brick, it ends up exerting a force much greater than the weight of the brick. But if you gently rest the brick on top of your head, then the force will equal the weight of the brick.

Equilibrium means: No net force, no acceleration. (Period!)
Right.
As I say above, the buoyant force (the net force that the water exerts on the object) is the same as long as the object is submerged. Submerge it 1 inch under water or 10 feet under water: the buoyant force is the same! (Of course the water pressure on the object is greater the deeper it goes. But the net force of the water on the object remains the same.)

11. Nov 10, 2005

### Staff: Mentor

Because if it isn't in equilibrium, some of the force goes toward accelerating the object. Using the example of a balloon forced underwater, there is an extra force (from your arm) making the situation balanced. Remove your hand and the unbalanced forces accelerate the balloon up.
No, the force on the water is the same, the force exerted by gravity on the object is where the imbalance is. I think you may be confusing the dynamic forces associated with movement (hydrodynamic drag) with the static forces of buoyancy. Consider the instantaneous case of a block that is accelerating, but not moving:

Consider a 2lb block of some material that displaces 1lb of water. It sinks, right? Now suspend it from a string in the water. Now 1lb of it's weight is held up by the water and 1lb of the weight is held by the string. The entire system is in static equilibrium.

Now cut the string. The instant you cut the string, what has changed? The block is still stationary (it hasn't moved yet, at time t=0) so the water around the block has not "felt" any change in anything. So the water is still at a static equilibrium. All that has changed is that now you have an imbalance of forces on the block, causing an acceleration.
I think you may be mixing the cases here: step back and think about what is in equilibrium with what.

For the case of the suspended weight, cutting the string does not change the static situation for the water at all. It still has a stationary object submerged in it at time t=0, so the water is still in static equilibrium with it's environment. But the weight is no longer in static equilibrium with it's environment. It feels a net downward force of 1lb.

12. Nov 11, 2005

### beginner16

Gosh, subject really got me confused so I hope you can help me at least one more time

Could part of the problem be that so far I only dealt with forces acting on objects in equilirium? As such, using my previus example, so far I haven't really learned how to calculate the force falling brick exerts on a head etc.

I'd like to google on the subject, but could you tell me what to write in search field ( I tried "dynamic forces" but nada )?

But that just doesn't make sense. If gravity exerts force on object, then object will exert same force on water.Example: I hold a rock with mass 1 kg in my hand. Gravity exerts force 10 N on the rock,and rock exerts that force(10N) on my hand. So at time t>0 when object feels weight pushing down on it it will exert that weight to the environment ( the way stone in my hand did )

13. Nov 11, 2005

### Staff: Mentor

Not necessarily. I'm going to guess that you think that this follows from Newton's 3rd law? If so, you are misinterpreting that law. The proper statement would be: If gravity (that is, the earth) exerts a force on an object, then that object will exert the same gravitational pull back on the earth. In the case of gravity, the two interacting bodies are: (1) the object; (2) the earth. They exert equal and opposite gravitational forces on each other. In the case of the water exerting a force on the object, the two interacting bodies are: (1) the object; (2) the water. These exert equal and opposite contact forces on each other.
Realize the weight of the rock is the earth pulling on the rock. This has nothing directly to do with force between the rock and your hand--that's a totally different interaction. The only reason that the rock and your hand would exert a 10 N force on each other is that you are purposely holding the rock still. It's in equilibrium. But it doesn't have to be. Lower your hand quickly--the force between hand and rock will decrease; accelerate your hand upward, the force increases. Even though the weight of the rock remains the same, it's easy to change the force it exerts on your hand.

Unlike the water, you can change the amount of force your hand exerts on the object. The water can only exert a force due to the water pressure at a given depth. If that force happens to equal the weight of the object, the object floats.

Last edited: Nov 11, 2005
14. Nov 11, 2005

### beginner16

Sorry to bring this up again ... I'd like to read on forces when object is in motion (force needed to stop a brick at certain speed etc)

What must I write in search engine to produce the correct results?

But you are contradicting yourself a little(probably not, but I see it as such) !

First you say whatever force object exerts to the water, due to 3. newton's law water exerts equal force on object. Now you said water can only exert a force due to water pressure, the size of which depends on depth of the water in question

15. Nov 11, 2005

### Staff: Mentor

Read up on Impulse which equals Force x Time. You might find this helpful: http://www.glenbrook.k12.il.us/gbssci/phys/Class/momentum/u4l1b.html [Broken]

Both statements are true (in the hydrostatic case that we are considering). Where's the contradiction?

Last edited by a moderator: May 2, 2017
16. Nov 15, 2005

### beginner16

hi, I really hope you again can spare some your time for this thread

Doesn't the water get more dense and thus more heavy the deeper you go, and so buoyant force gets bigger?

I understand that if an object is sinking it has momentum and since the point of contact with water on any level is only brief the water can't stop an object but it can't decrease its momentum. But the law of momentum conservation says that

I assume we can think of a water and a sinking object as isolated system. So I assume water that collided with an object now also gained momentum downwards?

I understand that on same depth water pressure is the same else there would be movement, but at same time water pressure is caused by water above it and so if there is an object of volume V above it, then water pressure below the object should be greater or smaller compared to if there was no sinked object. How does it all work out?

So if it would be possible to take a drop of water and made it collide with some object,water wouldn't be able to exert force on the other object because there is no pressure?

The thing of momentum is that for same impulse force in a collision can vary depending on the duration of contact. Since water at certain depth exerts certain force, it would seem reasonable that when colliding with moving object it will exert that force.
But on the other hand why can't it exert some other force?

17. Nov 15, 2005

### Staff: Mentor

Sure, but I'd like to stick to basic hydrostatics and ignore the complications of fluid dynamics. Your original issues were related to buoyancy, a hydrostatic phenomenon.

Yes. But that's a very small effect, especially at a depth of 10 feet. For practical purposes, consider water to be incompressible.

The water that contacts the object and the sinking object are not an isolated system (It's surrounded by other water). But as the object sinks, water does move around the object.

But recall a principle of hydrostatics: The pressure must be the same at any point along a horizontal line in the fluid. Imagine I suspend a block of iron underwater (hanging from a rope, but totally submerged). Directly under the iron the water pressure is exactly the same as it is at the same depth not under the iron. A submerged object at rest has no effect on the water pressure (except for the rise in the height of the water due to the object having displaced some water). Same thing with a floating object: If you swim under a boat (or other floating object) there's no change in pressure as long as you keep swimming in a horizontal line.

My statement "The water can only exert a force due to the water pressure at a given depth. " was meant for hydrostatics only. Of course you can take moving water and have it exert forces that have nothing to do with hydrostatic pressure.

18. Nov 15, 2005

### beginner16

Yes, if it is at rest then it means its density is same as that of displaced water. But say object is sinking due to having greater density than water. How does that work out ?

19. Nov 15, 2005

### Staff: Mentor

All that matters is that the submerged object be at rest (so we can ignore the dynamics of moving water and things like viscosity). The example I gave was a block of iron--its density is greater than water, but that doesn't matter. The point is that the buoyant force, which is the net force that the surrounding water exerts on the object and vice versa, doesn't depend on the density of the object. The effect of the object on the surrounding water is exactly the same as if it was replaced by water.