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February 27, 2015

February 27, 2015

Posted by **mNoellb** on Saturday, February 19, 2011 at 10:41pm.

with a lid 25 cm in diameter. If the pressure

in the cooker can reach 3.0 atm, how much

force must the latches holding the lid onto the pot be designed to withstand?

I have F=PA

I have my P: 3.03 * 10^3 N/m^2

A=pi(.25 * 10^2m^2)^2

I'm not sure If I am doing it right.

- Physics -
**drwls**, Sunday, February 20, 2011 at 12:49am3.0 atm is 3.04*10^5 N/m^2. There is also a pressure of 1 atm on the outside, so the latches only have to withstand a net pressure force of 2 atm = 2.03*10^5 N/m^2.

Your lid area is incorrect also. It is (pi/4)*D^2 = 4.91*10^-2 m^2.

- Physics -
**moi**, Sunday, February 20, 2011 at 12:47pmOH! well Thank you! My teacher seems to leave like .. most of that out.. geez!

- Physics -
**mohanad**, Thursday, April 3, 2014 at 4:22pmF=P*A

A= Pi*r^2

A=Pi*(25/2)^2

A=0.04909 cm^2

Pgauge=Pabs- 1 atm

p abs =P gauge+1

P abs= 4 atm

F= (0.04909 cm^2 * (1m^2)/(100cm)^2)*(4 atm*1.013*10^5 N/m^2)

F=1.99 N

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