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# GS-5 Section III Question 44

### GS-5 Section III Question 44

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### Re: GS-5 Section III Question 44

Hi, I dont understand why you use the difference in pressure on the door to calculate the answer to question 45. Why dont you add the pressures pushing on the door to find the total pressure exerted on the door?
SarahMcKenna

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### Re: GS-5 Section III Question 44

SarahMcKenna wrote:Hi, I dont understand why you use the difference in pressure on the door to calculate the answer to question 45. Why dont you add the pressures pushing on the door to find the total pressure exerted on the door?

Imagine that you need to move a couch in your living room. So you are trying to push it into position. If someone is besides you pushing in the same direction, then it is easier for you. However, if someone is on the other side of the couch, pushing in the opposite direction, then it becomes much more difficult.

There is pressure mostly on the inside of the plane ("a pressurized cabin"). You can see by the shape of the door, which side has more pressure to wedge the door in place. Nonetheless, we are told: 20 kPa outside (which pushes the door to the inside), and the cabin is pressurized to about 75 kPa (which pushes the door in the opposite direction, to the outside). We want to know how to pull the door in, which is aided by the pressure from outside to in (20 kPa) but opposed by the pressure in to out (75 kPa). So, we can simplify these differences by saying that the net pressure is 55 kPa inside to out, so that must be overcome to open the door while in flight.
goldstanda3269

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