An airtight box, having a lid of area 80 cm2, is partially evacuated.

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Homework Help Overview

The problem involves an airtight box with a lid that has a specified area, which is partially evacuated. The atmospheric pressure and the force required to remove the lid are given, leading to a question about the internal pressure of the box.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the internal and atmospheric pressures, with attempts to express the internal pressure in terms of the forces acting on the lid. There are questions about unit conversions and the calculations being performed.

Discussion Status

Several participants have shared their calculations and expressed uncertainty about the results. There is a recognition that the approaches taken may be valid, but a consensus on the correct internal pressure has not been reached. Some guidance has been offered regarding the relationships between forces and pressures.

Contextual Notes

Participants are working within the constraints of the problem as presented, including specific numerical values and the requirement to find the internal pressure based on the forces involved. There is an acknowledgment of potential unit conversion issues affecting the calculations.

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Homework Statement



An airtight box, having a lid of area 80 cm2, is partially evacuated. Atmospheric pressure is 1.01 105 Pa. A force of 108 lb is required to pull the lid off the box. The pressure in the box was:

Homework Equations





The Attempt at a Solution



I figure that the force that the pressure in the box exerted on the lid is equal to the force exerted on the lid by the atmosphere minus the applied force.

No matter how "creative" I get with the math, I can't come to any of their answers.

Am I on the right track?
 
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I think you are on the right track. Maybe units are causing you grief. Show what you are trying.
 


I've tried a boatload of things, most recently:

Box pressure = 1.01x10^5 - 479.52/.008

479.52 being 108lbs in Newtons.
 


I think you need to try, like you said, F(due to P internal) = F(due to P atm) - 479.52 N.

Once you find F(due to P internal), you can find the internal pressure.
 


Which would be:

0.008(B) = 0.008(1.01x10^5) - 479.52


This was one of my first processes. I just get the same answer of 41060.
 


Oh, right--which is what you were doing eariler... That still seems right. What did you get and what is the answer?
 


My answer is 41060.

The homework gives me the choices of:

7.50 e4 Pa
6.35 e4 Pa
2.60 e4 Pa
1.76 e5 Pa
1.38 e5 Pa
 


You got more or less what I got. I don't think your approach is wrong. Anyone else see something we are missing?
 


Anyone got any other ideas?
 

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