When to include hydrostatic pressure term in force problems?

[elchoi]

equation help -- water pressure at bottom of a pond

I'm trying to solve this problem "A car is flipped upside down in a pond, .50m below the surface, how much force is needed for the driver to push open the door." I got this wrong because the formula that was chose was F = (P - (P+rho x g x y)) So my question is when do include rho x g x y when solving problems?

 

[Spinnor]

Similar problem here,

https://www.physicsforums.com/showthread.php?t=363351

Found by Googling "physics force of water on swimming pool wall physics examples"

You have not given enough information to solve the problem, important details were left out. Also as the door is hinged there is a subtle reduction in force required to move the door, if one simply used the total force on the door the answer would be off?

Good luck!

 

[ehild]

The pressure is the force on unit area. To get the force exerted on the door, you need to know the area of of the door.
The difference of the outside and inside pressures at depth y is rho g y. As the door is on the side the car, the pressure varies along the height. You can use the average height, but it is not clear what is at 0.5 m depth from the surface. Taking 0.5 m the average depth, calculating the average pressure, and multiplying it by the area (what is the typical area of a car door?) you can get an estimate of the force and you will see that
it is very big .ehild

 

[elchoi]

Thank you both

 

[Jobrag]

If the driver can hold his breath till the inside and outside pressures equalise then the answer is close to zero.