# Determine Force on 6ft Door Due to Water & Air Pressure

• MiMiCiCi22
In summary, the door has a total force of 10296 lbf acting to the left due to the water pressure and a total force of 73829 lbf acting to the right due to the air pressure.
MiMiCiCi22

## Homework Statement

Determine the total force and its direction (left or right) due to water and air pressure in the surface shown in Fig. P3.62 (lb f). The door is hinged at the top and is 6 feet wide. See attached image.

## Homework Equations

Fp (force of pressure) = specific weight of fluid * depth to center of gravity * Area
P=F/A
Pg (gauge pressure) = specific weight of fluid * depth
Specific weight of water = 62.4 lb/ft^2
lbf = pounds of force

## The Attempt at a Solution

I first tried to find the force of pressure from the water acting on the side of the door...
Fp = (62.4 lbf/ft^2)*(3ft+2.5ft)*(5ft)*(6ft)
Fp = 10296 lbf
I then thought that if I added the force of pressure from the water acting on the side of the door to the force of pressure from the air acting on the side of the door, I would be able to find the total force. I first converted 2 atm to lbf/ft^2...
2 atm * (0.0209 lb/ft^3)/(9.869x10^-3 atm) = 4235.48 lbf/ft^2
I then tried to find force using this pressure...
F = PA = (4235.48 lbf/ft^2)*(5ft)*(6ft)
At this point, my professor told me that I was solving it wrong.

#### Attachments

• IMG_9466.jpg
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• IMG_9466.jpg
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Note that at the top of the water there is atmospheric pressure of 1 atm.

(Welcome to PF!)

So how do I take the atmospheric pressure into account? What would this do to my calculations?

How do you find the pressure at a point in the water at a depth h?

So I would need to use the Equation P=Pa+(specific gravity of water)*(h) ?

MiMiCiCi22 said:
So I would need to use the Equation P=Pa+(specific gravity of water)*(h) ?
Yes [with "specific gravity" replaced by "specific weight"]

MiMiCiCi22
So I found the pressure for the water side of the door using the equation mentioned earlier. I then converted it to force using the equation F=PA. For the side with air, I converted 2 atm to lb f. I then used F=PA to convert this pressure to force. I'm currently getting a net force of 73829 acting to the left, however the correct answer is that there should be -53.184 lb f acting to the right. I'm confused as to where I may have made a mistake.

Please show the details of your calculation for the force due to the water.

## 1. What is the formula for calculating the force on a 6ft door due to water and air pressure?

The formula for determining the force on an object due to water and air pressure is Force = Pressure x Area. In this case, the area would be the surface area of the door.

## 2. How do the water and air pressure affect the force on the door?

Water and air pressure exert a force on objects due to their weight and density. When there is a difference in pressure on either side of the door, it can create a net force that pushes or pulls on the door.

## 3. Are there any other factors that can affect the force on the 6ft door?

Other factors that can affect the force on the door include the shape and size of the door, the material it is made of, and the direction and strength of the pressure.

## 4. How can we measure the water and air pressure on the door?

To measure the water and air pressure on the door, you would need to use instruments such as pressure gauges or manometers. These devices can measure the amount of force exerted by the water and air on the door.

## 5. What are some real-life examples of objects being affected by water and air pressure?

Some real-life examples of objects being affected by water and air pressure include doors and windows in a building, a submarine diving or surfacing in water, or a balloon rising or falling in the air.

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