Confused about why atmospheric pressure won't move piston

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SUMMARY

The discussion centers on the mechanics of a piston within a housing under atmospheric pressure conditions. Participants clarify that if the piston is in a vacuum, the forces acting on it are balanced due to equal surface areas on both sides, despite the atmospheric pressure acting on the larger area. The conversation emphasizes that as the piston moves, the volume of the evacuated region decreases, which does not create an unbalanced force. Ultimately, the forces remain equal when the internal pressure matches the external atmospheric pressure.

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dmetrd
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I'm feeling pretty silly not being able to understand this. Let's say that you have a piston (red) sitting inside of a housing (black) like this
4W67VIE.jpg

Let's say that something like this was lying on its side on a flat table. By my understanding, if the contact between the piston and the housing is frictionless and all of the air is sucked out of the central cavity, then the only forces acting on the red piston come from the atmospheric pressure acting on the two surfaces. Since atmospheric pressure acts on a greater surface area on the left compared to the right, it would seem that there would be a greater force acting on the left than on the right, and the piston would slide to the right. This result seems completely inane, but I can't for the life of me justify the expectation of nothing moving using a statics force balance. Could someone help me out with this?
 

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dmetrd said:
it would seem that there would be a greater force acting on the left than on the right, and the piston would slide to the right
That is correct.
 
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Consider this: What happen to the volume of the evacuated region as the piston slides left to right?
 
Yet another way to look at this ...

Suppose the two piston are not connected, and you are able to keep them form moving by applying an external force to each (in the outward direction of course). Which piston will require the greater force ?
 
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Dale said:
That is correct.
Hm, I guess that makes sense, but it doesn't seem intuitive to me. Is there anything about this that would make this impossible to recreate in the real world? Even without creating a vacuum in the center cavity, the area on the inside of the cavity is smaller than on the outside due to the connecting rod so it wouldn't balance the force.

Janus said:
Consider this: What happen to the volume of the evacuated region as the piston slides left to right?
The volume of the evacuated region would decrease, but would that affect anything if there was no air in the cavity?
SammyS said:
Yet another way to look at this ...

Suppose the two piston are not connected, and you are able to keep them form moving by applying an external force to each (in the outward direction of course). Which piston will require the greater force ?

With no rod between the pistons, I suppose the larger one would require more external force.

I'm not sure what to take from all of this. So it seems like I never actually described anything that isn't correct?
 
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dmetrd said:
Even without creating a vacuum in the center cavity, the area on the inside of the cavity is smaller than on the outside due to the connecting rod so it wouldn't balance the force.
If the pressure inside the cavity is the same as the external pressure then the forces are completely balanced. The area of all the "left-facing" surface is inherently always equal to the area of all the "right facing" surface.
 
Janus said:
Consider this: What happen to the volume of the evacuated region as the piston slides left to right?
It would decrease. I am curious what you are leading to, though.
 
DocZaius said:
It would decrease. I am curious what you are leading to, though.
If you put a vacuum in a cylinder with a piston, it sucks the piston in. More generally, if a mechanism can reduce the volume of a vacuum then, all other things being equal, it is energetically favorable for it to do so.
 

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