# B Direction of buoyant force on sunken object?

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1. Dec 8, 2018

### Logic hunter

We say that buyont force act upwards (in usual cases) and that the normal force exerted by the base of a container (of liquid) on a object is less than its true weight, so a weghing machine will give smaller reading (in terms of value) than expected. But suppose a cube sinks in water. Now water cannot exert force upwards on it since its lower surface which is the only one surface that can experience upward force from liquid is no longer in contact with water. Its vertical surfaces will experience force from pressure which will cancel out but the upper horizontal surface will experience pressure force downwards hence buyoant force will be downwards and more normal force (than weight) will be required to balance weight and buyoant force so an object should weigh more. What am I missing ?

2. Dec 8, 2018

### davenn

No .... forget about the bottom of the container ... it's irrelevent

No .... if the object sinks to the bottom, then the buoyant force = zero

buoyant force never points downwards

tis time for you to have a bit of a reading session

https://en.wikipedia.org/wiki/Buoyancy

Dave

3. Dec 8, 2018

### Staff: Mentor

Not clear how the weighing machine is hooked up or what it's measuring here.

You are correct that if the cube somehow has a liquid-tight seal with the bottom, then the force of the liquid on the cube will be downward instead of the usual upward force. And thus the normal force required to support the cube will be greater than its weight. (You have to support both the cube and the liquid that pushes down on it.)

4. Dec 8, 2018

### Buzz Bloom

HI Dave:

I am a bit confused by this. I have the thought that the buoyant force is what reduces the effective weight, which is what determines the rate of falling, which also takes into account the friction of the falling object with the water. If the effective weight is negative, then the object does not sink. Is this wrong?

Regards,
Buzz

5. Dec 8, 2018

### davenn

and the buoyant force then has a value >0

6. Dec 8, 2018

### Staff: Mentor

Not sure why you think that. In the "usual case" where there is a bit of fluid under the object as it rests on the bottom, the buoyant force should be the same as usual (and upward). Things are slightly different if a tight seal is made with the bottom.

7. Dec 8, 2018

### davenn

well yeah ... ok maybe non-zero, but the upward force would be very small and since the object sank to the bottom
it cant be very relevant ?

8. Dec 8, 2018

### Staff: Mentor

Do you think the buoyant force depends on its distance from the bottom? The fact that it sinks just means that the weight of the object is greater than the buoyant force, not that the buoyant force is near zero or negligible.

9. Dec 8, 2018

### Staff: Mentor

I think that you need to make a tight seal around the bottom and then release pressure underneath. For example, suppose the object is a plug and the bottom is a drain hole to a pipe. If the pipe is completely clogged then putting the plug in place doesn’t alter the net force from the fluid. It is only if the plug is put in place and then the pressure underneath is relieved e.g. by clearing the clog and allowing the pipe to empty.

10. Dec 8, 2018

### Logic hunter

Measuring device is the one which divides normal force downwards by g. But i have seen people describing Buyoant force as change in the weight of an object(which sinks. Weight is constant but the normal force on its base, which would be measured by the machine is the new weight), where the new weight (normal force) is less than original weight. (For eg: some questions say a sphere weighs 2.5 gram in air and 0.7 gram in H2SO4, 1.5 grams in water, find density of h2so4. Clearly, this assumes that buyont force acts upwards, hence nomral force(new weight/g) is reduced. So does this mean that all the liquid will not leave contact will bottom surface unless a tight seal is made externally.

11. Dec 8, 2018

### Droidriven

That was my thought exactly while I was reading through all the posts here....the buoyancy force doesn't "disappear" if the object sinks completely to the bottom, it just isn't greater than the downward force. Wouldn't the difference in forces have more to do with how quickly or slowly the object "can sink" instead of whether or not its buoyancy is unchanged/changed or non-existent?

Is the OP trying to ask whether or not that buoyant force is still trying to displace the object to occupy its space with water the same as it would if the object were floating on the surface or if partially submerged or if partially submerged but still touching the bottom?

Wouldn't it still be something like the weight of object - buoyancy, whether some distance just below the surface or on the bottom? I'm not understanding how sitting on the bottom would change anything, doesn't it still have the same amount of upward force because it is surrounded by material of a different density instead of it being an effect of material "under" it to "push" on the bottom(assuming the object is less dense than what it is surrounded by). Or is there some counter-intuitive factor involved that I'm not aware of?

Last edited: Dec 8, 2018
12. Dec 8, 2018

### Staff: Mentor

Sure. The example given here was a "cube" of something. By "tight seal" I just meant that there is no liquid between bottom surface and base, which eliminates any upward pressure from the liquid.

13. Dec 8, 2018

### Buzz Bloom

Hi Dave:

My point regarding this case is that the buoyant force is not only greater than zero, it is also greater than the weight force trying and failing to sink the object. You seem to be defining the buoyant force as the force difference:
weight minus (what I am thinking of as) buoyancy.​

The Wikipedia article says:
In physics, buoyancy or upthrust, is an upward force exerted by a fluid that opposes the weight of an immersed object. In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus the pressure at the bottom of a column of fluid is greater than at the top of the column.​

Regards,
Buzz

14. Dec 8, 2018

### Staff: Mentor

Yes, it's just that the "tight seal" being described cannot be achieved by just dropping an object into a container of water.

15. Dec 8, 2018

### Staff: Mentor

Right!

16. Dec 9, 2018

### sophiecentaur

If you 'squeegee' the object into the floor of the container, just expelling most of the trapped water and not re-admitting it, you can end up with less pressure underneath than on the rest of the floor. Then the buoyant force will be less. A Rubber Sucker works on this principle and the excess air pressure outside will support a heavy load.
There are many demonstrations of the 'breaking ruler' experiment where excess pressure on top of the newspaper keeps it down against the table (briefly, because the air takes time to flow in). Same thing at work in air or in water.
Horrible story about a guy who fell overboard into water with a muddy bottom (the water- not the guy!) His legs went into the mud and struggling to lift one leg would only increase the force on the other and push it down further. He was stuck because his net buoyant force was near zero (less than his weight. That's an extreme version of the experience of getting your wellies stuck in mud and stepping out of them.