Why doesn't atmospheric pressure move a piston with unequal exposed areas?

[dmetrd]

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

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?

 

[Dale]

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.

 

[Janus]

Consider this: What happen to the volume of the evacuated region as the piston slides left to right?

 

[SammyS]

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 ?

 

[dmetrd]

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.

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?

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?

 

[Dale]

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.

 

[DocZaius]

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.

 

[jbriggs444]

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.