Liquid Pressure: Understanding p=RO*g*h

[fawk3s]

http://img18.imageshack.us/img18/3276/lalballs.png

This started to bother me today. In the first picture, there's a cup wheres a certain amount of water in it. A ball made of aluminium hasnt been dropped in the cup yet.
In the second picture, the ball has been dropped in.

The ball has not sinked, a string is holding it.

Now please tell me, how did the pressure on the botton of the cup created BY THE WATER change?
I mean, the water level just rose. So the pressure should also.
But there seems to be another solution to the problem, and I just can't figure it out.

The equation for liquid pressure is p= RO*g*h.

Thanks in advance,
fawk3s

 

[kanato]

The weight of the ball has increased the pressure. Once it's in the water, there is a buoyant force on the ball, which means that the string is not holding the entire weight of the ball anymore, the water takes on some of the weight. That's what increases the pressure.

 

[fawk3s]

The buoyant force is caused due to the pressure differences of the upper and lower points of a body.
I don't understand how that's supposed to increase the liquid pressure. Could you explain in abit more detail?

Thanks in advance,
fawk3s

 

[russ_watters]

I don't think you should insert buoyancy into this at all. It is completely irrelevant to the question. Anyway, you gave the correct answer to the problem, so why are we here? What do you mean by this:

But there seems to be another solution to the problem, and I just can't figure it out.

We're not mind readers: what is this "other solution" that you are referring to that you are looking for help with?

 

[kanato]

The buoyant force is caused due to the pressure differences of the upper and lower points of a body.
I don't understand how that's supposed to increase the liquid pressure. Could you explain in abit more detail?

Thanks in advance,
fawk3s

Well I was thinking your objection was something like "why would the pressure increase if the weight of the ball is supported by the string and not the cup?" The pressure at the bottom of the cup before the ball is put in is $$m_w g/A$$ if the mass of the water is $$m_w$$ (this is just force/area definition of pressure). Then it might be counter intuitive that the pressure should increase if you believe the ball is supported by the string, because surely the mass of the water doesn't increase. The answer is that the water provides some of the upward force on the ball, reducing the tension in the string, so there is a 3rd law force of the ball pushing the water down, which in turn increases the downward force on the bottom of the cup, and hence the pressure.

You can also do this analysis by looking at the change in height of the water and the pressure formula. Both approaches yield the same answer, as they should.

 

[fawk3s]

Actually I am pretty embarrased at the moment because I didnt explain the problem fully. There was actually abit different problem, but I posted my own problem here. And to my own problem, there is no second answer. To the original problem, the answer was tricky and simple, but I am not going to post that problem here.

As for konato, the ball is only "pushing the water down" when you drop it in there. If its in there, not moving, its not pushing anything. The cubage of the water which was supposed to be in the balls place, is now somewhere else, which increased the height, which increased the pressure. And this was the original answer.
The upward force is just the water's pressure doing it's job. Nothing else. It doesn't increase the pressure.
At least I understand it that way.

 

[Cleonis]

As for konato, the ball is only "pushing the water down" when you drop it in there. If its in there, not moving, its not pushing anything.

Let me give another case, that has has an essential feature parallel with what you are asserting here.

A dumbbell weight is on the floor. You pick it up, and you lift it up to, say, elbow height, with you in upright position. Obviously, during the lift you were exerting a force upon the weight, to make it gain height.

You do not place the dumbbell on a table or so, you physically keep it at that elbow height.
You seem to claim that when the dumbbell isn't moving anymore, if it's height is sustained, you no longer have to exert a force upon it.

As for konato, the ball is only "pushing the water down" when you drop it in there. If its in there, not moving, its not pushing anything.

As I understand your personal theory of physics, you believe that pushing is only done during displacement, but that no push is necessary to sustain any situation.

Cleonis

 

[fawk3s]

My English is not always the best, especially when explaining anything related to physics. But ok, I am going to change my sentence:
"If its in there, not moving, its not pushing anything AROUND."
But the result is the same. Its not compressing the water under it. The water escapes upward and that increases the pressure.

Btw, I sometimes forget to add in the OP that I am not a genius. So I would like to add it now: I don't find people who mock others with phrases like "your personal theory of physics" or something else very much smarter. If you want to help and explain, go ahead. If you want to mock, don't even bother because it just has no point.