3.0 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.
A pressure cooker is shaped like a can
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.
3 answers
OH! well Thank you! My teacher seems to leave like .. most of that out.. geez!
F=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
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
1 answer