Does the shape of an object affect the pressure?

  • #1
abrek
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TL;DR Summary
Pressure
Suppose the speaker has a container into which pressure is pumped, a rod passes through the container with a disk strung on it with one flat side and on the other with a pointed side, will the disk move in any direction under the influence of pressure?
IMG_20240901_005110.png
 
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  • #2
No, why do you think it will?
 
  • #3
Arjan82 said:
No, why do you think it will?
I think that on the left side of the disk has a flat surface and the pressure will push it to the right, and the right side of the disk with angles and the pressure will also be at an angle, and will not be stronger than the pressure directed straight to the right, so the disk will move, please tell me what am I wrong?
 
  • #4
Thread closed briefly for Moderation...
 
  • #5
After fixing some Google Translate issues, the thread is reopened,
 
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  • #6
Why not calculate how much the force difference is? That will show you the right answer immediately.
 
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  • #7
abrek said:
please tell me what am I wrong?
Get a cone. Or even just a triangular prism. Put it on a table. Does it move due to air pressure?

You should always do maths rather than guessing. Guesses will mislead you. If the pressure is ##P##, what's the force due to pressure on a small area, ##\delta A##, of the curved surface of the cone in the horizontal direction? What's the horizontal component of that force and does it vary with position? What's the surface area of the cone (excluding the flat base) in terms of the radius of the base and the angle at the tip? So what's the total force? How does this compare to the force on the base?
 
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  • #8
Your previous thread was based on the same misunderstanding, applied to hydrostatic pressure. You've been told the answer several times now. Do you not believe/accept it?
 
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  • #9
abrek said:
I think that on the left side of the disk has a flat surface and the pressure will push it to the right, and the right side of the disk with angles and the pressure will also be at an angle, and will not be stronger than the pressure directed straight to the right, so the disk will move, please tell me what am I wrong?

The pressure on the right side will indeed be the same as on the left side. But if the pressure is not stronger as you say, why then would it move? And what direction do you think it moves to? Do you know how to compute a force from a pressure applied to an area?
 
  • #10
russ_watters said:
Your previous thread was based on the same misunderstanding
Rut-roh. And not just that one.

OP, you really have two options: do the calculation yourself so you can see the answer yourself, or choose not to do the calculation and accept the results from those who can. Guessing is not going to convince anyone of anything (other than that some people are bad guessers).
 
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  • #11
abrek said:
and the right side of the disk with angles and the pressure will also be at an angle, and will not be stronger than the pressure directed straight to the right, so the disk will move, please tell me what am I wrong?
The object is a disk with a cone attached. Call the object "X".

To tell if X is pushed left or right, you need to compare the total force pushing it left with the total force pushing it right.

The pressure on the right side of X acts on an angled surface exerting a force on it. You need to think about the component of this force which acts to the left.
 
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  • #12
This is one of those cases where solving the integral in general is easier than solving it for any particular case: all it relies on is that a 2-d picture taken from the left has the same shape as a 2-d picture taken from the right.
 
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  • #13
Arjan82 said:
The pressure on the right side will indeed be the same as on the left side. But if the pressure is not stronger as you say, why then would it move? And what direction do you think it moves to? Do you know how to compute a force from a pressure applied to an area?
it seems that yes, I cannot correctly calculate the water pressure per area, could you tell me the formula for this?
 
  • #14
abrek said:
it seems that yes, I cannot correctly calculate the water pressure per area, could you tell me the formula for this?
Pressure acts perpendicular to a surface. You can do some simple trigonometry to find the vector component of force in the horizontal direction. Does that help?
 
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  • #15
abrek said:
it seems that yes, I cannot correctly calculate the water pressure per area, could you tell me the formula for this?
Pressure per area makes no sense.

What you want is the force on a small area. Do you know how to get a force from a pressure?

Then you want the horizontal component of that force. Do you know how to calculate that from tge force?

Then you want to sum that over the whole curved surface of ths cone. Do you know the area of the cone?

If you don't know the answer to these, at least go and see what Wikipedia says.
 
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  • #16
The pressure is at an angle now, but the area of the surface is larger. The two effects cancel out.
 
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  • #17
We assume the OP thinks the object will move to the left because of the larger area (that's the usual mistake) but the reasoning of the OP seams to indicate he thinks the object will move to the right, that is why I asked. Unfortunately I didn't get an anwer.

abrek said:
it seems that yes, I cannot correctly calculate the water pressure per area, could you tell me the formula for this?

Citing formula's without background is pointless. It seems you don't know about the basics of (fluid) mechanics. We are not going to give a mechanics 101 course here. You really first need to figure some things out yourself. Forces and areas in a fluid (i.e. fluid statics) is the first chapter of an introductionary book/course on fluid dynamics. Please read that first and then come with some specific questions.
 
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