a DC-10 aircraft is flying at an altitude where the atmospheric pressure is 40% of sea level pressure. if the aircraft interior is maintained at sea level pressure, what is the size and direction of the net force acting on a cabin door that is 1.0 meters wide and 2.0 meters high?
Pressure at sea level is 101325 Pascals. One pascal exerts a force of one Newton per square meter.
1. Find 40% of 101325.
2. That number will also equal the force in N on each square meter.
3. Find the area of the door.
4. Multiply the answer of #1 above by the answer of #3.
THANK YOU VERY MUCH...
40530 = 40% of 101325
This number equals the net force (N per Square meter).
Area of door: Width x Height= 2x40530
= 81060
To determine the size and direction of the net force acting on the cabin door, we need to consider the pressure difference between the inside and outside of the aircraft.
First, let's calculate the pressure difference between the inside and outside of the aircraft. We're given that the atmospheric pressure outside the aircraft is 40% of sea level pressure. Let's assume the sea level pressure is P0.
Pressure outside the aircraft, P_outside = 0.40 * P0
Inside the aircraft, the pressure is maintained at sea level pressure, P_inside = P0
The pressure difference, ΔP, is given by:
ΔP = P_inside - P_outside
= P0 - 0.40 * P0
= 0.60 * P0
Now we can calculate the size of the net force acting on the cabin door using the pressure difference.
Force = Pressure × Area
The pressure acting on the door is ΔP, and the area of the door is given as 1.0 meters wide and 2.0 meters high, so the area of the door is:
Area = Width × Height
= 1.0 meter × 2.0 meters
= 2.0 square meters
Substituting the values into the formula:
Force = ΔP × Area
= 0.60 * P0 × 2.0 square meters
= 1.20 * P0 square meters
The size of the net force acting on the cabin door is 1.20 times the sea level pressure.
As for the direction of the net force, it acts inward since the pressure on the inside of the aircraft is higher than the pressure outside.