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Pressure $(P_o)$ outside the container is $10^5 \ \ \mathrm{Pa}$. A force of $480 \ \ \mathrm N$ is required to remove the lid. The area of the lid is $50 \ \ \mathrm m^2$. Find the pressure $(P_i)$ inside the container.

enter image description here

I drew the diagram. It was not given with the question.

From the diagram I equated force,

$F + P_i \times A = A\times P_o$

$\implies F + P_i \times 50 = 50\times P_o$

$\implies 480 + P_i \times 50 = 50\times 10^5$

$\implies P_i \times 5 = 5\times 10^5 - 48 = 499952$

$\implies P_i = 99990.4 \ \ \mathrm {Pa}$

The answer is $4 \ \ \mathrm {kPa}$.

I don't know why I got the pressure this much off, I equated forces acting on the lid of the container properly.

This is a easy question still after trying so much I can't get the answer :((.

Please explain me the problem in my answer.

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closed as off-topic by John Rennie, Jon Custer, heather, Bill N, Kyle Kanos Jan 17 '17 at 10:58

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  • $\begingroup$ Are you certain about the givens and units? I see an error in your math. $\endgroup$ – Inquisitive Jan 15 '17 at 18:53
  • $\begingroup$ @Inquisitive I think yes because I just kept everything in SI units. $\endgroup$ – A---B Jan 15 '17 at 19:09
  • $\begingroup$ @Inquisitive The question is from Principle of physics by Halliday. Chapter is fluid and question number 15. $\endgroup$ – A---B Jan 15 '17 at 19:13
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    $\begingroup$ What edition of the book did you use? I just checked the 9th edition (of which a complete pdf is online) and question 2 in Chapter 14 talks about an area of 77 m$^2$ (but otherwise it's obviously the same problem). Their "negligible mass" is even more outrageous with those dimensions... $\endgroup$ – Floris Jan 16 '17 at 23:43
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    $\begingroup$ I googled him. He is at facultyprofile.csuohio.edu/csufacultyprofile/… $\endgroup$ – Floris Jan 17 '17 at 13:05
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Probably a typo in the textbook. If you go backwards from the provided answer of $P_i=4\,\mbox{kPa}$, you find that the force should have been $4{,}800\,\mbox{kN}$, instead of $480\,\mbox{N}$ which is a suspiciously small force on such a huge lid. $50\,\mbox{m}^2$ is the size of a small apartment, mind you. In New York city, it would be the size of quite a stately apartment ;-)

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  • $\begingroup$ Thanks for the answer and the analogy, That made be giggle. $\endgroup$ – A---B Jan 15 '17 at 19:38
  • $\begingroup$ Are you sure about that 480 kN working out to the correct answer? $\endgroup$ – Inquisitive Jan 15 '17 at 19:40
  • $\begingroup$ Sorry, typo on my part now. Fixed. $\endgroup$ – Pirx Jan 15 '17 at 19:46
  • $\begingroup$ @Pirx You corrected it to work. Alas, we'll never know the true intent of the textbook. $\endgroup$ – Inquisitive Jan 15 '17 at 19:46
  • $\begingroup$ Yeah, it's not a simple typo now anymore. Who knows, stuff happens. $\endgroup$ – Pirx Jan 15 '17 at 19:47
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It's obvious the book has a typo.

50 square meters is not a "container" - it's a handful of buses...

Let's assume they meant 50 cm$^2$ and see where that gets us.

Then the force on the lid due to the external atmosphere is 500 N, and we need an internal force of 20 N to give a net force of 480 N. That makes the internal pressure 20/500 of the external pressure, or 4 kPa.

Seems a lot more reasonable.

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