Calculating the pressure on a surface exerted by opposing forces?

In summary: A on the other was something unusual because I have mostly calculated only one pressure value for a point. In summary, the pressure on any point of the surface would be 100/A on one side and 30/A on the other side, due to the 100N and 30N forces being applied on opposite faces of the circle. This different pressure on each side would lead to acceleration, but the exact acceleration will depend on the specific scenario and constraints of the system. In most practical applications, we refer to pressure in terms of gauge pressure, which is the difference between pressure on one side of a surface and that on the other side. However, in scenarios involving massless objects like a 2D circle, the
  • #1
Logic hunter
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Suppose there is a circle with area 'A' and on one of the faces of the circle a force of 100N is applied and on the other face a 30N force is applied (since all 2D surfaces have 2 faces) such that these forces are opposite to each other, perpendicular to the faces, and forces are equally distributed on the surface, then what will be the pressure on any point of the surface? Will it be 100/A or (100-30)/A=70/A or (100+30)/A=130/A and why?
 
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  • #2
The pressure will be 100/A on one side and 30/A on the other side. The different pressure on each side will lead to acceleration.
 
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  • #3
+1

You didn't say anything about the disc being constrained so it will accelerate as Dale says.

Is the disc constrained? Perhaps its a rigid window in the wall between two tanks of water at different pressures?
 
  • #4
Dale said:
The pressure will be 100/A on one side and 30/A on the other side. The different pressure on each side will lead to acceleration.
thank you for replying, but are you saying that each point on the surface will have 2 values of pressure ? In hydrostatics I was told that each point of the liquid will have only one pressure value by Pascal's law(P=p+xdg, where p is pressure at free surface, g is g effective(resultant of pseudo acceleration and g) and x is the distance of the point from free surface along the direction of g effective), if the container accelerates
 
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  • #5
When working with pressurized systems in most applications, we nearly always refer to pressure in the term “gauge pressure “. Gauge pressure is the difference between pressure on one side of a surface and that on the other side. For most applications, this is the pressure that matters. In fact, if you have an air compressor nearby, you can look at the pressure gauge and it probably says “”PSIG”, which stands for the pressure in units “per square inch; gauge”.

It is this pressure differential that makes fluid-powered systems go, or tells us how close we are getting to the maximum pressure a vessel can handle. For example; if a pressure vessel is rated to hold 20 lbs psi, what that really means is that it can hold 20 psi more than the pressure on the outside. If the vessel has 30 lbs of absolute pressure inside, and 14 lbs on the outside, it will be fine. Similarly, if the vessel is under water, with 50 lbs of absolute pressure on the outside and 60 lbs absolute on the inside, we’re still ok. However, if you fill it to 60 when it’s down deep, and then bring it up to the surface, it’s going to burst.

An instrument that only shows you the absolute pressure would give no warning, because the absolute pressure doesn’t change. But a gauge that shows pressure differential , or Gauge Pressure, will start to climb as soon as external pressure begins to drop.

There are some practical applications for which absolute pressure is important, like chemistry and biology, but most of the time, gauge pressure is the important value.
 
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  • #6
Logic hunter said:
are you saying that each point on the surface will have 2 values of pressure ?
Perhaps you should provide more detail about your scenario. I was envisioning a metal disk with pressure applied on both sides of the disk.
 
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  • #7
If your circle is truly 2D and flat, it has no mass, and you can't sustain a pressure difference across it. If you try, you will have an infinite rate of acceleration. In systems involving a massless frictionless piston, the pressures on both faces of the piston must always match irrespective of how rapidly the gas is expanding or compressing.
 
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  • #8
Chestermiller said:
If your circle is truly 2D and flat, it has no mass, and you can't sustain a pressure difference across it. If you try, you will have an infinite rate of acceleration. In systems involving a massless frictionless piston, the pressures on both faces of the piston must always match irrespective of how rapidly the gas is expanding or compressing.
I was talking about theoretical case like those involving calculations on moment of inertia of circles, rectangles about a axis etc, assuming they have weight and dealing with 1 dimensional objects (points) in force and laws of motion.
 
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  • #9
Dale said:
Perhaps you should provide more detail about your scenario. I was envisioning a metal disk with pressure applied on both sides of the disk.
Yes the disk is a metal one. Its just that a single point on disk having two pressure 100/A on one side and 30/A on the other was something unusal because i have mostly calculated only one pressure value for a point.
 
  • #10
Logic hunter said:
Yes the disk is a metal one. Its just that a single point on disk having two pressure 100/A on one side and 30/A on the other was something unusal because i have mostly calculated only one pressure value for a point.
Think about the hull of a submarine. It has a huge pressure outside and a low pressure inside.
 
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  • #11
Dale said:
Think about the hull of a submarine. It has a huge pressure outside and a low pressure inside.
And if you wonder why the walls of a sub don't accelerate due to the huge pressure difference its because there are other forces acting on the walls so the net force on any bit of wall is zero (assuming the sub is stationary).
 
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Related to Calculating the pressure on a surface exerted by opposing forces?

1. How do you calculate the pressure on a surface?

The pressure on a surface can be calculated by dividing the force exerted on the surface by the surface area. This can be represented by the equation: Pressure = Force / Surface Area.

2. What units are used to measure pressure?

The SI unit for pressure is Pascal (Pa), which is equivalent to 1 Newton per square meter (N/m²). Other commonly used units for pressure include pounds per square inch (psi) and atmospheres (atm).

3. What are the two types of forces that can exert pressure on a surface?

The two types of forces that can exert pressure on a surface are normal force and shear force. Normal force is perpendicular to the surface, while shear force is parallel to the surface.

4. How do you calculate the pressure on a surface when there are multiple forces acting on it?

In this case, the total pressure on the surface can be calculated by adding all the individual pressures exerted by each force. This can be represented by the equation: Total Pressure = Pressure1 + Pressure2 + ... + PressureN.

5. Can you use pressure to determine the direction of the net force on a surface?

Yes, the pressure on a surface can indicate the direction of the net force. If the pressure is greater on one side of the surface compared to the other, then the net force is acting in the direction of the higher pressure.

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