An airtight box having a lid with an area of 12 in.² is partially evacuated. If a force of 108 lb is required to pull the lid off the box, and the outside atmospheric pressure is 15 lb/in.², what was the pressure in the box?

We are given an airtight box with a lid of area 12 in², and it is partially evacuated. A force of 108 lb is needed to pull the lid off, and the external atmospheric pressure is 15 lb/in². We need to determine the pressure inside the box.

Step 1: Understanding the Forces Acting on the Lid
The force required to remove the lid is due to the pressure difference between the outside and inside of the box. The total force exerted on the lid is given by:

Force = (Outside Pressure - Inside Pressure) × Area

Let P_inside be the pressure inside the box. Using the given values:

108 lb = (15 lb/in² - P_inside) × 12 in²

Step 2: Solving for P_inside
Rearrange the equation to solve for P_inside:

P_inside = 15 lb/in² - (108 lb / 12 in²)

P_inside = 15 lb/in² - 9 lb/in²

P_inside = 6 lb/in²

Final Answer:
The pressure inside the box was 6 lb/in².