# Pressure in airplane

1. Nov 18, 2007

### gills

1. The problem statement, all variables and given/known data
I know this is probably a very easy question, but i'm new to this and never worked with it before.

3. The attempt at a solution

so you have 0.25atm pressure going into the window from outside, and 0.75atm from inside pushing out.

Am i assuming that with 0.50atm more pressure coming inward from a person would equal the force of pressure pushing out and therefore any more than that would theoretically pull the window inward?

P = F/A

2. Nov 18, 2007

### Staff: Mentor

You're on the right track. Figure out the force required to pull the window open. Is it likely that a passenger can exert such a force?

3. Nov 18, 2007

### gills

Well, if i'm doing this right, it seems pathetically small, but I guess this is assuming there are no latches of some sort. Maybe i need to convert my dimensions?

so P = F/A ---> F = PA

so for F to be able to pull the window in, it must be $$\geq$$ (0.75 atm/(0.50 * 0.90).

If the outside pressure is 0.25atm, then only 0.50 more is required..(?)

so one must apply a force of (0.50atm) * (0.50 *0.90) = 0.225 N

I thinking the dimensions need to be converted?

4. Nov 18, 2007

### Staff: Mentor

What is atmospheric pressure?

5. Nov 18, 2007

### gills

at sea level 1 atm = 14.7 psi = 760mm mercury = 101.3 kPa......a ha, i need to use Pascals...?

6. Nov 18, 2007

### gills

0.50 atm x 101.3kPa = 50.65 kPa

50.65 kPa * (0.50 * 0.90) = 22.8 kN which is a lot for a human being. Looks good?

7. Nov 18, 2007

### Staff: Mentor

Looks good.

8. Nov 18, 2007

### gills

to me, that's a dumb questions.

Thanks again Al