Is a half submerged object submerged in its own displaced water?

[lost captain]

I feel like this is the dumbest question I've ever asked in my life, and honestly I'm sorry for taking anyone's time that's willing to answer.

So I have a rectagular prism container that has 4ml of water and i submerged a metallic plate that has volume 1ml, just like the picture below.
The plate fits snuggly in there and I'm holding the plate with a string so basically it is neutrally buoyant.

1) If the plate is fully submerged, no mater the depth, the displaced water will be 1ml. So we will see the water level rise from 4ml to 5ml. Right?

2) When I half submerge the plate, half of it's volume will be under water so 0.5 ml gets submerged and also 0,5 ml of water will get displaced on top...so the plate ends up fully submerged ? How can the plate be fully submerged with only 0,5ml of water displaced, it can't...,but then again that displaced water will go on top so...? What am i missing here?

I understand that if the submerged volume is 0.5ml then the water level should rise at 4,5ml, but then it will be like the submerged volume is submerged in the displaced water...

Thank you, and I'm sorry

 

[Baluncore]

Maybe you should be measuring the submergence relative to the bottom of the object, and the surface of the water, not the top of the object and the bottom of the container.

Your last diagram is wrong because you have a tight-fitting piston in a cylinder, the water level should be 5.0.

 

[lost captain]

Maybe you should be measuring the submergence relative to the bottom of the object, and the surface of the water, not the top of the object and the bottom of the container.

When the object is half submerged, i am measuring that from the bottom of the object. And if I'm measuring from the surface of the water, then that surface is at 4.5 ml?

 

[Baluncore]

As you lower the object, the bottom meets the water at 4.0 . An instant later, it has displaced one drop of water, that has half filled the narrow space around the piston. The bottom of the object is only just below 4.0, and the water has risen to 4.5 which makes it 50% submerged.

 

[lost captain]

Your last diagram is wrong because you have a tight-fitting piston in a cylinder, the water level should be 5.0.

I submerge half of the piston and 0.5ml of water gets displaced on top, so where am I making a mistake?

 

[lost captain]

The bottom of the object is only just below 4.0, and the water has risen to 4.5 which makes it 50% submerged.

Okay so the volume of the water that got displaced goes up to the 4.5ml mark even though only a small portion of the volume of the object got submerged

 

[Ibix]

I think the problem here is that water cannot be beside a "tight fitting" piston - it just teleports from below the piston to above it. In other words, it's an idealised model and the idealisation means that you cannot partially submerge it.

If you have a 9.999999 millimetre square piston then you can partially submerge it because the tiny volume between the wall and the piston can partially fill with water.

(Edit: I see a conversation along these lines has happened while I was typing.)

 

[lost captain]

I think the problem here is that water cannot be beside a "tight fitting" piston - it just teleports from below the piston to above it. In other words, it's an idealised model and the idealisation means that you cannot partially submerge it.

If you have a 9.999999 millimetre square piston then you can partially submerge it because the tiny volume between the wall and the piston can partially fill with water.

(Edit: I see a conversation along these lines has happened while I was typing.)

The plate has volume of 1ml, it's length is smaller than the container wall length so it can fit inside. In the picture i drew the length in this way, that why the height of the plate exceeds the 1ml mark a little

 

[lost captain]

If i understand correctly from the answers above, i can't figure out how much water was displaced from the volume markings unless the object is fully submerged below the original water level surface, in my case below 4ml

 

[Baluncore]

When the bottom of the object is at 4.0 , the object is 0% submerged and the top of the object is at 5.0 . As you lower the object into the water, the surface of the water covers the object with the water remaining at 5.0
The object goes from 0% submerged, through 50%, to 100% submerged very rapidly as it begins to be covered by the displaced water.

Part of the confusion comes from calibrating the vertical axis in millilitres, not in centimetres.

 

[A.T.]

I submerge half of the piston and 0.5ml of water gets displaced on top, so where am I making a mistake?

Most likely you are confusing yourself with your own misleading terminology. When you say: "I submerge half of the piston", you apparently actually mean: "I bring half of the piston below the initial water level". This are very different things, unless the area of the water surface is much greater than the cross section of the piston.

If i understand correctly from the answers above, i can't figure out how much water was displaced from the volume markings unless the object is fully submerged below the original water level surface, in my case below 4ml

If you know the cross sectional areas of container and piston, then you can figure that out, regardless of how much of the piston is submerged.

But for arbitrarily shaped and oriented objects, that you don't have the exact geometry of, you have to submerge them fully, in order to get the displaced volume only from the water levels and container geometry.

 

[Herman Trivilino]

I'm holding the plate with a string so basically it is neutrally buoyant

If it were in a state of neutral buoyancy you wouldn't need to hold it (with a string or anything else).

 

[Herman Trivilino]

The plate has volume of 1ml, it's length is smaller than the container wall length so it can fit inside.

The plate fits snuggly in there

Well, which is it? Slightly smaller or snug?

The issue is this: Can water move around the edges of the object or are you making it more complicated by arranging things so that water cannot leak around the edges?

 

[sophiecentaur]

The plate fits snuggly in there

There is an 'unknown' in this statement has to be resolved. You don't seem to have committed yourself either way yet and there are two possible answers.

If it is a totally snug fit then the object could not be where you have drawn it because the water is incompressible; it would be sitting on top of the water with a force acting upwards on it equal to its weight.

If there is a path for the water to flow around it then the situation is the same as if it's in the middle of a bucket of water. The upward force from the water will be equal to the weight of water displaced and the rest of the force would be supplied by the string. If there is a small but finite path around the object then this situation will be reached eventually, after a delay.

Tension in string (reducing) = Weight of object (constant)- weight of the volume of water it displaces (increasing)

 

[Baluncore]

Two reference heights are changing, the bottom of the object, and the water level. It is the difference between those heights that determines the displacement.

 

[haruspex]

Two reference heights are changing, the bottom of the object, and the water level. It is the difference between those heights that determines the displacement.

Yes, my mistake… I was using displacement to mean the volume that was occupied by water and now isn’t, which is not the definition. Will edit.

 

[haruspex]

I submerge half of the piston and 0.5ml of water gets displaced on top,

Revised post:
As @Baluncore reminded me, ‘displacement' in this context does not have its every day meaning. The displacement of an object immersed in a liquid is the volume of the object below the surface, i.e. the volume of liquid you would have to add on removing the object to avoid the level dropping.
Presumably you mean that 0.5ml of water is moved from its original position to be on top, but this is not so.

Say the horizontal area of the tank is A and that of the object is A'. If we start with the object sitting on the surface then push it down a small distance x, the moved volume is xA'. The volume of water below the object has reduced by xA, so that much water must now surround the object. Since that has a horizontal area A-A', it has a depth $\frac{xA}{A-A'}$. The surface has risen by $\frac{xA}{A-A'}-x=\frac{xA'}{A-A'}$.
Moved volume = $xA'$
Submerged volume = Displacement = $\frac{xAA'}{A-A'}$.
For x>0, moved volume / submerged volume = $\frac{A-A'}{A}$.
So the two are only equal for x=0 or A'=0.

 

[A.T.]

For x>0, moved volume / submerged volume = $\frac{A−A′}{A}$.
So the two are only equal for x=0 and A'=0.

Or approximately equal for $A \gg A'$, which is a common case in every day life, and thus likely the root of the confusion by the OP, as I noted here:

When you say: "I submerge half of the piston", you apparently actually mean: "I bring half of the piston below the initial water level". This are very different things, unless the area of the water surface is much greater than the cross section of the piston.

 

[sophiecentaur]

But for arbitrarily shaped and oriented objects, that you don't have the exact geometry of, you have to submerge them fully, in order to get the displaced volume only from the water levels and container geometry.

Yes; the details are very important in this sort of problem.

I remember, as a child, visiting the Lighthouse on Plymouth (UK) breakwater, The shutter rotated to sweep the beams around 360 degrees and the mechanism was (or had been) driven by clockwork and needed to be very efficient . We were told the the cylinder with the slots in was 'floating on mercury', which could have been taken to mean 'a tank full of mercury'. In fact there was only a deep circular groove which held a minimal amount of mercury; just enough to provide enough buoyancy for the lightweight shutter cylinder. We're talking 1950s health and safety legislation. No doubt the mechanism switched to electric and used a regular bearing system. Still, displacement means displacement.

 

[Baluncore]

Still, displacement means displacement.

You can float a battleship in one bucket of water, if you make a pool the same shape as the battleship below the waterline, but only very slightly larger.

 

[sophiecentaur]

You can float a battleship in one bucket of water, if you make a pool the same shape as the battleship below the waterline, but only very slightly larger.

Too right but this round shutter appears to have actually been floated in displaced mercury. Over the past decades, I guess the mechanism has changed so there's no way I could actually see it again.

That school visit also gave me a chance to see a basking shark really up-close. Both events were memorable for me with the Physics experience coming out just-ahead.

 

[Herman Trivilino]

You can float a battleship in one bucket of water, if you make a pool the same shape as the battleship below the waterline, but only very slightly larger.

Our brains do that.

 

[sophiecentaur]

Our brains do that.

Up to a point - depending on the brain. It actually happened in the lighthouse. What I saw was much easier to contemplate than a fictional battleship in a fictional dock.