The Basics of Buoyancy [HELP!]

Hello people!

Yesterday my friends and I were solving this easy-looking question:

A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube:

(A)Will Increase (B)Will Decrease (C) Will remain the same (D)Will become zero

At first we thought the answer to be (B) considering the buoyant force acting opposite to the weight of the cube. But then, a question popped up in my mind: "What causes this buoyant force even if no fluid was actually displaced?"

So I referred to Archimedes' Statement:

Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. — Archimedes of Syracuse

But according to the question, no such fluid was displaced by the block!

So, here are my questions:

What is the basic reason behind buoyancy? Is it the pressure difference in fluids or the 'liquid-displaced' theory?
What must be the answer to the question (That I was solving) ?
Considering the case that you may say- "The reason behind buoyancy is the pressure difference in the fluids"; Will there be no buoyant force applied to the iron cube as there is no fluid layer present below the cube i.e. no pressure difference? (Considering the iron cube is VERY dense and thus not allowing any fluid layer to push its way below it)
Or you might add another statement to the above question that-"Yes, buoyant force is applied as there is always a fluid layer present below it..." .Please do explain yourself.
If your answer is according to the Principle given by Archimedes of Syracuse , will there be no buoyant force on the cube as the cube didn't actually displace any fluid? Just like the case where you may consider it to be a part of the vessel itself.

Please help!

Thank you !

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Comments

@Priyansh Sangule, I just did the following experiment. I placed a dry wooden cube into a flat bottomed plastic food container. I then poured water very slowly into the container so as not to move the cube. I successfully got the water level to approximate twice the height of the cube, and the cube stayed on the bottom. After twenty seconds, the cube then floated to the top. If I then put the cube back on the bottom, it would stay for half a second, then float up. I believe this suggests two things:

There is no buoyant force until the water actually makes its way under the wooden cube.
Once the cube is wetted and squeezed to the container, there is some surface tension at the interface of the cube bottom and the flat container bottom that resists the buoyant force but is quickly overcome.

However, the force between the metal cube and the bottom should increase because the cube will now have water on top of it, which has a weight, and is pushing down on the cube.

Josh Silverman Staff - 6 years, 6 months ago

Hmmm...to my level of understanding,I tink the question just want to test student level of understanding on Archemedes principle...4rm Archemedes principle which state that when an object is partially or fully immersed in a fluid,it experience an upthrust which is equal to the weight of fluid displaced...Archemedes principle is just xplaing dat when an object is partially or fully immersed in a fluid,Weight=Upthrust.These is also observed wen u immerse a bucket inside a pool of water..U will notice that wen the bucket is inside the water,it will loose weight,Thus,the weight reduces.But if u want to bring out the bucket of water containg water from the pool,the bucket will appear heavy..So from the question u asked,i think the answer is B..The weight or Force or Upthrust will decrease...Also concerning the question u asked next..Upthrust is caused by the change in pressure difernce in fluids......That my view sir.

Ayanlaja Adebola - 6 years, 6 months ago

I appreciate your view buddy. But think over it- Did the block actually displace any fluid ?

Priyansh Sangule - 6 years, 6 months ago

here are my answers in the order of questions asked:

First of all, the basic reason behind buoyancy is pressure difference in fluids. The Archimedean principle is just the other way of saying that, or perhaps, more quantitatively.
Well, i think the answer depends on the level of water. If the water level is exactly the height of the cube, the answer should be c). You can think of this answer by taking the cube and the extremely thin layer of water beneath the cube(Yes its there) as a system. Only 2 forces in vertical direction acts on this system:

i) Weight acting downwards ii) Normal force acting upwards

Since the system is at rest, the 2 forces must cancel out each other. Moreover, the thin layer of water can be taken to have negligible mass. Thus, even on addition of water, the normal force will remain the same.

However, if the water level gets more than the cube's height, the normal force will increase due to the weight of the water column above the cube.

and 4. No matter what you do or how much force you apply on the cube from the top, there will always be a thin water layer present below the cube. You can easily see this by considering a cube kept on floor in air. There has to be a very thin(microscopic) layer of air beneath the cube, otherwise there would be a vacuum and you would never be able to lift the cube again. Actually, the normal force does not act directly on the cube. It actually acts on that thin layer of air and this layer in turn applies the force on the cube.
As i have already mentioned, the Archimedean principle and the pressure difference are basically the same thing. You are misinterpreting the phrase" displacing the fluid". Think carefully and you will get it.

Mayank Khetan - 6 years, 6 months ago

I think its A because buoyant force will not act on it and the water above it will increase the normal force

Aditya Khandelwal - 5 years, 2 months ago

If the underside of the cube is not wet by the surrounding water i.e the water does not get under the cube, the upward force on the bottom as per Pascal's law will be missing.So no buoyancy force,so no change in the force acting at the bottom of the vessel till the time the cube is just immersed.Buoyancy has its origin in the net upward force acting on a submerged body as per Pascal's law.

satyendra kumar - 6 years, 6 months ago

So you're saying that if the body is submerged with it base touching the bottom with no fluid in between , the net force is due to the weight of the body plus the weight of the liquid above it? And that we don't consider any buoyant force on it.

That is right.

@David Mattingly @josh silverman @jatin yadav @Ishan Tarunesh @Peter Taylor - I need your aspects as well people! :)

I think, as long as the water doesnt seep through, the only two forces on the cube are

1) its weight

2) the water pressure from top (or the weight of the fluid column about it)

3) and the normal reaction from below

As the normal reaction is a self adjusting force, so the cube wont float up untill water seeps in, so yes you are correct, no force should act upon the face till water seeps in,

I really appreciate your view point instead of straight ticking away B by using Vpg as buoyant force , you are reflecting over it in detail...

But practically seen, there will always be a layer, otherwise as has been already pointed out, it would be impossible to lift an object placed on ground, also we would have to consider PA as well along with mg in all physics problems.

Also, in optics problems regarding paperweight, we always assume theres a thin layer of air below the paper weight and above the paper, and thus it directly enters the glass at critical angle

Mvs Saketh - 6 years, 5 months ago

I think the answer is C. The force asked about here is Normal force acting on the metal cube (due to its contact with bottom) which will always be equal to its weight. Even the buoyant force doesn't change it. Since the weight is same , the normal force with respect to metal cube is same.

Thejna Ros - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$