Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A = 0.0145 m² . The air pressure in Denver is PD = 80500 Pa. and in New Orleans is PNO = 100750 Pa. Assume the lid is weightless.
Required :
Write an expression for the force FM, required to remove the container lid in New Orleans

Answer 1

Answer:

$Fm=\frac{A(Pno)}{2}$

Fm=730.7375N

Explanation:

Hello!

To solve this problem we must follow the following steps

1. Remember that force is the product of pressure and the perperndicular area on which it acts.

2. Draw a free body diagram of the container lid in New Orleans (see attached image).

3. Now we add the forces with their respective sign and find Fm equalizing it to zero, in this way we find the minimum force Fn that causes the lid to rise upwards

if you follow the previous steps and review the attached image you will find the following equation

Fm+(A)(Pc)=(Pno)(A)

Where

Fm=force required to remove the container lid in New Orleans

A=area=0.0145m^2

Pno=pressure in New Orleans= 100750 Pa

Pc=pressure in the tank(Remember that the pressure inside is half the pressure at sea level)=100750/2Pa=50375Pa=Pno/2

solving for Fm

Fm=A(Pno-Pc)

Pc=Pno/2

$Fm=A(Pno-\frac{Pno}{2})} \\Fm=\frac{A(Pno)}{2}$

this is the final expresion

if we solve

$Fm=\frac{0.0145m^2(100750Pa)}{2}=730.4375N$