How much force will be required to remove the lid?

[iseidthat]

Homework Statement

An airtight box has a removable (massless) lid of area 4·10-2 m2. A lid is placed on it while it is on top of a mountain (where Patm=0.81·105 Pa). It is then taken to sea level, where Patm = 1.013·105 Pa. How much force will be required to remove the lid?

Homework Equations

F=PA

The Attempt at a Solution

For the first part:
F=PA=.04 x 81000=3240 N
For the second part:
F=PA= .04 x 101200=4052 N

is there a change in pressure type of equation?

 

[mgb_phys]

You are only interested in the change of pressure. - this gives you the pressure differnce between the inside and outside of the box and so the force to open it

 

[thomate1]

I would like to clarify a few things before providing a response. Firstly, it is important to note that the question does not specify whether the lid is being removed slowly or quickly. This can affect the amount of force required as it can create a pressure difference between the inside and outside of the box.

Assuming that the lid is being removed slowly, the force required to remove the lid can be calculated using the equation F=PA, where F is the force, P is the pressure, and A is the area of the lid.

For the first part, the pressure inside the box is equal to the atmospheric pressure at the top of the mountain, which is 0.81 x 10^5 Pa. Therefore, the force required to remove the lid would be F=0.81 x 10^5 x 4 x 10^-2 = 3240 N.

For the second part, the pressure inside the box is equal to the atmospheric pressure at sea level, which is 1.013 x 10^5 Pa. Therefore, the force required to remove the lid would be F=1.013 x 10^5 x 4 x 10^-2 = 4052 N.

In conclusion, the amount of force required to remove the lid depends on the pressure difference between the inside and outside of the box. As the atmospheric pressure increases, the force required also increases. This can be seen in the calculations above.