Pressure: Force x Area - Q&A

  • Thread starter christym
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In summary: I get for trying to do too many things at once. Thanks. I'll try to be more careful next time!In summary, pressure is typically measured in mm or inches of mercury or kilopascals, using the traditional mercury barometer method. The force and area are inversely proportional, meaning that as area increases, pressure decreases. This relationship is also seen in observations such as water pressure decreasing as the area of the hose increases, and the impact of an object having less deviation on a surface as the surface area of impact increases. The formula for pressure is actually pressure = force/area, not force x area.
  • #1
christym
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If pressure = force x area

1) why do we measure it in mm of Hg (distance)
2) why do I observe pressure to decrease as area increases

?
 
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  • #2
1) Because historically an easy way to measure the variations (due to weather) in atmospheric pressure is by inverting a column of mercury (into an open-topped reservoir). The liquid continues to fall (leaving near-vacuum above) until the weight of the column equals the force applied at the bottom, i.e., the atmospheric pressure determines the height above a reservoir that mercury can be sucked by a vacuum.
2) You should explain fully the circumstances surrounding your observation.
 
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  • #3
Welcome to PF, Christym.
1) It's normally fluid, rather than mechanical, pressure that's measured as mm or inches of mercury or kilopascals. That's based upon the mercury barometer; the higher the air pressure, the higher it forces the column of mercury up the tube.
2) If the force remains the same, the pressure has to decrease as the area increases. That's dictated by the formula that you quoted.

edit: Hi, Cesiumfrog. Once again, you sneaked in while I was composing.
 
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  • #4
cesiumfrog said:
1) Because historically an easy way to measure the variations (due to weather) in atmospheric pressure is by inverting a column of mercury (into an open-topped reservoir). The liquid continues to fall (leaving near-vacuum above) until the weight of the column equals the force applied at the bottom, i.e., the atmospheric pressure determines the height above a reservoir that mercury can be sucked by a vacuum.

Thanks cesiumfrog,
Referring to that formula (pressure=force x area) then shouldn't it matter what the the area is, over which atmosphere is applied to the column of mercury? However I can't find any standards and these mercury columns seem to come in all sizes. It should matter shouldn't it?


2)
cesiumfrog said:
You should explain fully the circumstances surrounding your observation.

A couple of observation examples are, if I widen the aperture of a hose (increase the area) pressure of the water falls when measured at the end of the hose even if pressure at the source remains constant.

The impact of an object striking a surface causes less deviation in the surface as the surface area of the impact increases.
 
  • #5
Thanks for your welcome Danger. This looks like an interesting place ..

Danger said:
2) If the force remains the same, the pressure has to decrease as the area increases. That's dictated by the formula that you quoted.

I'm missing something...
Just using simple values for an example:
If Force =5 and Area =3 then Pressure = 3 x 5 (=15).
If I increase area to 5 then the new formula would be Pressure = 5 x 5 (=25)

According to the formula pressure increases if area increases.

No?
 
  • #6
No. To maintain the same pressure over a larger area, you would have to increase the force. It might be helpful for you to look into hydraulics. That is entirely based upon these principles. I'm certainly no expert, though, so you should listen to those such as FredGarvin, Brewnog, Astronuc, etc..
 
  • #7
That should be pressure = force / area, for example in english units it's pounds / in^2.
 
  • #8
christym said:
Thanks cesiumfrog,
Referring to that formula (pressure=force x area) then shouldn't it matter what the the area is, over which atmosphere is applied to the column of mercury? However I can't find any standards and these mercury columns seem to come in all sizes. It should matter shouldn't it?
Larger cross-sectional area implies greater atmospheric force but proportionally greater quantity and weight of mercury. So no, the only important variables would be ones like temperature, not shape.

christym said:
A couple of observation examples are, if I widen the aperture of a hose (increase the area) pressure of the water falls when measured at the end of the hose even if pressure at the source remains constant.
You are deceived: if you put your thumb over the end of the hose, the flow slows and the pressure becomes the same at both ends. If you release your thumb the pressure does not remain constant but decreases along the length of the hose (due to friction of the flow).

christym said:
The impact of an object striking a surface causes less deviation in the surface as the surface area of the impact increases.
Change in momentum = force = pressure x area. This observation confirms the relationship.

Danger said:
edit: Hi, Cesiumfrog. Once again, you sneaked in while I was composing.
Mwahahaha..
 
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  • #9
christym said:
Thanks cesiumfrog,
Referring to that formula (pressure=force x area) then shouldn't it matter what the the area is, over which atmosphere is applied to the column of mercury? However I can't find any standards and these mercury columns seem to come in all sizes. It should matter shouldn't it?
Just to make sure you caught it, your equation is wrong: p=f/a, not p=fa

So, the force depends on the weight of the column and the weight of the column is density times volume. Volume varies with area just as force does. So no, it doesn't matter what the area is. Try doing the math and you'll see that area cancels out of the equations.
A couple of observation examples are, if I widen the aperture of a hose (increase the area) pressure of the water falls when measured at the end of the hose even if pressure at the source remains constant.
That's a dynamic situation. Pressure varies with motion. That's called dynamic pressure. It isn't the same as static pressure and it is somewhat complicated to calculate if you've never been exposed to the idea. Let's get you understanding static pressure situations first.
The impact of an object striking a surface causes less deviation in the surface as the surface area of the impact increases.
That isn't true unless you are holding the force constant and changing the area. So that's not constant pressure. Constant pressure requires changing the force and the area together.
 
  • #10
Jeff Reid said:
That should be pressure = force / area, for example in english units it's pounds / in^2.

Dang! I should have caught that, but I still keep getting terms like 'work', 'force', 'energy', 'power' etc. mixed up. Maybe I should just type out all of the definitions and make it a 'stickie' on my screen. :redface:
 

1. What is pressure?

Pressure is defined as the amount of force applied per unit area. It is a measure of the intensity of force acting on a surface.

2. How is pressure calculated?

Pressure is calculated by dividing the force applied on an object by the area over which the force is applied. This can be represented by the formula P = F/A, where P is pressure, F is force, and A is area.

3. What is the unit of measurement for pressure?

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

4. How does pressure affect objects?

Pressure can have various effects on objects, depending on their material and structure. It can cause objects to compress, deform, or break. Pressure can also be used to measure the strength and durability of objects.

5. What are some real-life applications of pressure?

Pressure is an important concept in many fields, such as engineering, physics, and meteorology. Examples of its applications include hydraulic systems, scuba diving, weather forecasting, and blood pressure measurements in medicine.

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