Calculating Air Pressure to Produce 200N of Force

AI Thread Summary
To produce 200N of force with a piston that has a radius of 25mm, a pressure of approximately 101,859 Pa (or 1 bar) is required. This pressure calculation assumes the force is applied against the area of the piston. Atmospheric pressure is already present on both sides of the piston, resulting in no net force when the piston is left open. Therefore, to achieve the desired force, the compressor must generate pressure above atmospheric levels, effectively doubling the pressure to create a net force of 200N. Understanding that gauge pressure is referenced to atmospheric pressure clarifies the need for additional pressure to achieve the desired force.
Kalus
Messages
36
Reaction score
0
Ok, this is a very basic question, but for some reason my head can't work out the last bit and its making me doubt the workings...

As the attached diagram shows, there is an air compressor attached to a piece of metal pipe that then has a piston on the end. The goal is to make this piston produce 200N of force. It has a radius of 25mm or 0.025 m, meaning an area of 1.96 x10^-3 m^2.

This means that if you use pressure= force/ area

pressure = 200/1.96 x10^-3
=101,859 Pa

This is almost equivilant to 1bar, which is around 14.5 PSI.

Does this mean that to produce that force i need to have a compressor that will pump air to just over one atmosphere?

This is the bit that confuses me, because surely air is at atmospheric pressure already, so it should be exerting that pressure if i just didnt have a compressor at all and left the tube pointing up open topped?

So, is that calculation correct, and if so, what am i not understanding about the way atmospheric pressure works?

Many Thanks, Kalus.
 

Attachments

  • pressure example.gif
    pressure example.gif
    7.9 KB · Views: 558
Physics news on Phys.org
I didn't check your math, but what you're missing is that the pressure you read on a gage is already referenced to atmospheric. 1 bar gage pressure is 2 bar absolute pressure. And atmospheric pressure doesn't do anything to the piston because it is equal everywhere, pushing it in and out at the same time, ending up doing nothing.
 
The piston has pressure = 1 atm on both sides, so the net force is near zero (as per above reply). If you push the pistion to compress the air to 2 atm, then the net force on the piston is 1 atm, or 200N.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top