Apollo 1 command Module Pressure/Force

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Homework Help Overview

The discussion revolves around calculating the force on the hatch of the Apollo 1 command module during the "plugs out test," considering the pressure differences inside and outside the module.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for the hatch area to calculate the force, with one suggesting an estimated area for rough calculations. Questions arise regarding how to handle pressure values and their differences.

Discussion Status

Participants are exploring the relationship between pressure and force, with some guidance provided on calculating pressure differences. There is an ongoing inquiry into the necessary parameters for the calculations, but no consensus has been reached yet.

Contextual Notes

There is a mention of specific pressure values (14.8 PSI outside and 17.5 PSI inside) and the need for the hatch area, which is currently unknown. Participants are working within the constraints of the problem without complete information.

mr.toronto
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Homework Statement



What would have been the force on the hatch of the Apollo 1 command module (Before the fire) during the "plugs out test"


Homework Equations


more info 14.8 PSI of pressure outside of the module and 17.5 PSI
1 PSI= 6 894.75729 pascals. 14.8 x 6894.74=102 042.152 Pascals
17.5 x 6894.74=120657.95 Pascals

so I'm kind of stuck here.



The Attempt at a Solution

 
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You need to know the area of the hatch. I had a quick look round but couldn't find it anywhere.

If you wanted to make a rough guess you could assume perhaps 0.25 m^2 and go from there. You need to use the formula:

Pressure = Force * Area
 
so what would I do for the pressure though
 
You just need to caclculate the pressure difference between the inside and the outside.

It's just like if you wanted to calculate the next force acting on a body if it had 4N acting on it one way, and 10N acting on it the opposite way. The net force is 6N, in the direction of the biggest force.

This is just the same, but instead of forces you're dealing with pressures, which are the force over a unit area.
 

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