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Hydrostatics: Immersing a metal cube in water.....

  • #1

Homework Statement


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 same (D) will become zero

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The Attempt at a Solution


Well, I think it should stay the same, because initially, the force of cube and force due to atm pressure is acting on the bottom, so when water is added the block would displace some water, due to which the increase in force(due to increase in water column by the cube) will be balanced by the upthrust on block, so the overall resultant will stay the same.(atm pressure+pressure of block). Am i correct?
You can ask me for clarifications regarding my attempt, if you find it unclear.
 

Answers and Replies

  • #2
.Scott
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Will the vertical force applied to the top of the metal cube increase, decrease, or stay the same?
 
  • #3
Archimedes says 'EUREKA' for principle
 
  • #4
.Scott
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Archimedes says 'EUREKA' for principle
That was when he discovered a way to measure the volume of an irregular object (a crown).
Although both have "metal object submerged in water" in common, it is unrelated to the solution of this problem.
 
  • #5
Is not the Archimedes principle here? I'm then confused, 'when a body is inmerssed in a fluid....'
 
  • #6
.Scott
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Is not the Archimedes principle here? I'm then confused, 'when a body is immersed in a fluid....'
Yes, it is related to the Archimedes principle. But not to his supposed "Eureka" revelation.
However, once you can answer the question I posed in post #2, the answer should be very apparent - even if you are not briefed on the Archimedes principle.
 
  • #7
Charles Link
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@subhradeep mahata They don't specify in this problem how tight the seal is between the bottom surface of the cube and the bottom of the vessel, but they are asking for the force on the bottom of the vessel. The case of a tight seal is easy to see, and is actually an exception to Archimedes principle in the buoyant force that is experienced. In that case, there is no buoyant force and the force from pressure on the top side will get transferred to the bottom side. See also post 54 of https://www.physicsforums.com/threads/ball-floating-on-the-surface-of-water.944587/page-3#post-5977539 ## \\ ## Suppose though that there are 4 skinny legs that elevate the cube just slightly of the bottom. Without any water, the force from the legs on the bottom of the vessel will be ## F=mg ##. With the water filled just to the top surface of the cube, there will be a buoyant force on the bottom of the cube from water pressure, so that the legs will have less force on the bottom of the vessel, (this difference is the buoyant force as given by Archimedes principle), but the total force on the bottom of the vessel in the region directly below the cube is the same because the upward pressure force on the cube (buoyant force) is equal to the downward force on the bottom of the vessel from the water pressure. Adding more water to totally immerse the cube brings you to @.Scott 's question above.
 
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  • #8
.Scott
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If we try to think of the problem using Archimedes principle, this is what it might look like:
Let's say the cube is 10cm x 10cm x 10cm and weighs 5Kg.
We will place that cube of four tiny beads - just so that water can pass under the cube. The beads will have zero mass and near-zero volume.
We will have two scales: Scale "B" for weighing the beads, one scale "C" for weighing the 10cm x 10cm area under the cube.

Without water, scale "B" will show 5Kg (full weight of the cube) and scale "C" will show 0Kg.

As the water level rises from 0cm to 10cm, scale "B" will show a decreasing weight (because of boyancy) while scale "C" will show an increasing weight because of the deepening water column. But at all times, the total weight measured by "B" and "C" will be 5Kg.
When the water level reaches 10cm, scale "B" will show 4Kg and scale C will show 1Kg. However, at that point, the cube has not been completely immersed as specified in the OP.

Since the total weight below the cube does not change because of buoyancy, Archimedes principle is not very useful here.
 
  • #9
Charles Link
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@.Scott 's 4 beads and 2 scales of post 8 is very much the same idea that I have with the 4 legs in post 7. The addition of the scales is helpful in illustrating the concept.
 
  • #10
@.Scott the vertical force on the top of the cube should stay the same since it is only the atmospheric force
 
  • #11
  • #12
Charles Link
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  • #13
Ok, so the answer is that the force stays the same, as there is no buoyant force in this case and the pressure on the top face will get transferred to the bottom face which is actually an exception to archimedes principle ( in case of a tight seal).
Am i right?
 
  • #14
Charles Link
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Ok, so the answer is that the force stays the same, as there is no buoyant force in this case and the pressure on the top face will get transferred to the bottom face which is actually an exception to archimedes principle ( in case of a tight seal).
Am i right?
It stays the same until the water starts to cover the cube. ## \\ ## What happens if we make the water 10 feet deep, covering the cube by 9+ feet? Do you think the force on the bottom of the vessel would stay the same?
 
  • #15
No, it wont. Now i get it, as long as the water does not cover the cube, the force stays same.
 
  • #16
But the question is saying that it is completely immersed in water, dosen't that mean the cube is completely covered?
 
  • #17
Charles Link
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But the question is saying that it is completely immersed in water, dosen't that mean the cube is completely covered?
Yes, there is some water above the cube. They don't say how much, but they only want a qualitative answer, which I think you should be able to figure out.
 
  • #18
Yes.
I get it.
Thanks again Charles sir :)
 
  • #19
haruspex
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the total force on the bottom of the vessel in the region directly below the cube
But that is not what the question asks for. It says the region "in contact with the cube".
Your four skinny legs (ok, not your four skinny legs - the four you invented) can be thought of as representing that region. In practice, it would consist of lots of very small areas, but the result is the same; we are back to Archimedes.
 

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