# Submerging an unknown cube

1. Jan 2, 2008

### nonequilibrium

No question that it's simple, but for some reason I can't fully understand this.

I stumbled upon this problem asking "If we put a glass of water on a digital weighing scale and we reset the scale to zero, after which we submerge a cube of an unknown material in the water attached to a string (like on the picture attached), the number on the scale will be:"

a) equal to zero
b) different from zero and give information about its volume
c) different from zero and give information about its mass
d) different from zero and give information about its weight

I presumed the answer was a, as the cube is still suspended on a string, meaning the force of gravity should be canceled out by the upward muscle force (or whatever is holding the string), so I thought there shouldn't be any force acting down, onto the scale. But it turned out to be answer b. It probably has to do with water being shifted up, but I can't figure it out exactly. Quite frustrating, as it should be a simple question.

Any help?

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2. Jan 2, 2008

### Shooting Star

The water is exerting a buoyant force upward on the cube, which is numerically equal to the weight of the displaced water. The cube in turn is exerting an equal an opp reaction on the water and thus to the scale. The reading on the scale shows the weight of the volume of water displaced, and so gives information about the volume of the cube.

Last edited: Jan 2, 2008
3. Jan 2, 2008

### nonequilibrium

I still can't agree completely. I get the formula of the buoyant force (I'm used to calling it the Law of Archimedes) is independent of the object's mass, so I get how you can calculate the displaced water etc. But how does the submerged object exert a force if it's suspended, just like a suspended object in the air doesn't exert a force on the ground. I do know the answer b is right -- I've just tested this, and indeed, the scale went up higher than zero -- but I'm not yet sure why. I felt the object getting lighter as it went into the water, due to the buoyant force, I get that. So is it Newton's law (action reaction) that an opposite force is pointed downwards?

EDIT: The total amount of water doesn't change in the experiment and the suspended object can't weigh anything, that's where I'm stuck

EDIT2: also this wasn't a homework question, just a general physics question

Last edited: Jan 2, 2008
4. Jan 2, 2008

### Shooting Star

If you do agree that the submerged object does experience a force upward, then you have to agree that it exerts a reaction downward. Yes, it's the good old third law. It has got nothing to do with the total amount of water.

I didn't quite get what you meant by "the suspended object can't weigh anything, that's where I'm stuck".

5. Jan 2, 2008

### nonequilibrium

Well, if you suspend an object (in air) over a scale, it obviously won't weigh anything, seeing as the object isn't exerting a downward force, so I found it - at first sight - illogical that it would influence the scale if the medium was water instead of air. I get that the buoyant force exerts an upward force on the submerged cube and that due to Newton's 3rd law, there is also a force downwards. So I get that there must be a force acting onto the scale, but I don't know how/why, as the object is suspended and doesn't press anything downward -- there basically was no mass added to result in the scale going higher than zero (but it did).

I asked somebody else, and that person told me the buoyant force actually is the cohesion force of water pushing the object at all sides, so it's a rise in pressure, which is higher lower down (due to the formula of pressure in a fluid), resulting in a higher pressure under the object which pusses it up. And as a pressure (here: due to the cohesion force) spreads itself in a fluid, it also exerts a force on the bottom, pressing onto the scale, making it weigh something.

So is that a correct explanation?

6. Jan 2, 2008

### Shooting Star

First of all, when you say that if you hang an object over a scale, it won't weight anything, I think you mean to say that it does not exert any force anything below it, like a scale. That is true, but the weight of the object is always there. In this case, the weight is being exerted on which it is hanging.

Secondly, when you have submerged the body in water which is on the scale, you have allowed the body to interact with the scale through the water, and the first condition that the body isn't interact with the scale has been removed. Now, part of the weight is being supported by the water and thus the scale, and the rest of the weight is being supported from wherever it's hanging.

The buoyant force is due to pressure difference between the bottom and top of the object. The pressure is more at greater depths. Pressure exerted on a body submerged in a fluid arises as a result of the motions of the molecules of the fluid bombarding the body from all sides.

7. Jan 2, 2008

### HallsofIvy

mr. Vodka, you seem to be assuming that the string is holding all of the weight of the cube. That isn't necessairily so. The string is supporting some of the weight, the water is supporting some[//b] of the weight. If the water is supporting any of the weight of the cube at all, what happens to the scale?

8. Jan 2, 2008

### turbo

And can the scale's readout vary based on the density of the cube?

9. Jan 3, 2008

### Shooting Star

Since the scale is showing the weight of the volume of water displaced, then, no, it will not depend on the density of the cube, if you keep the volume the same, and it is fully immersed as shown in the picture.

If you take the mass to be the same, then the reading will vary with density.

10. Jan 3, 2008

### nonequilibrium

Thank you both for answering. I feel kind of stupid that there is still one aspect I'm not sure about. I get how the buoyant force works and that it results into an upward force on the object. This must lead to a force downward also (it's a necessary result of the force upward), making the scale's reading go higher than zero, but which consequence is doing that?

The submerged object pushes water up, so you could think "the new layer of water on top exerts a higher total pressure on the bottom" (making it weigh more), but the water under the object has a smaller height than when there is no object above it, meaning the pressure there is lower, canceling the higher pressure on the sides, so this can't be the reason for the increased weight.

A wild guess of mine is this: without any submerged objects, on the bottom (very simplified) half of the water particles move up and half down. The 50% going down adds weight for the scale to read, but of course we've zeroed the scale at that. But with a submerged object, the 50% going up below the object bounce onto the object and continue their path in the other direction, away from the object, downward, toward the scale. This results in the 50% going up eventually also contributing to the weight, increasing the scale's number.

Is this a crazy thought?

11. Jan 3, 2008

### Shooting Star

So you want the justification of buoyancy at the atomic level? You want to know the exact mechanism at the microscopic level, if I understand correctly?

12. Jan 4, 2008

### nonequilibrium

Well, that would be the only way I could really understand it. And most importantly, it would explain the consequence of the extra weight (with more than just saying it's necessary as the water lifts the object partially up).

So, yes. If possible.

13. Jan 4, 2008

### Shooting Star

According to this logic, the flatter the area presented horizontally by the body in its lower part, the more will be the number of molecules bouncing down after hitting the object, thus contributing more to the weight on the scale, which is not true.

Perhaps we can discuss this after you read up a bit on the “kinetic theory” of liquids; and why the pressure of liquids increases with depth.