Finding the force on a camera window underwater

AI Thread Summary
The pressure on the camera window underwater is calculated to be 8.0442x10^7 Pa. The area of the window, a rectangle measuring 0.1 m by 0.15 m, is crucial for calculating the force using the formula F=PA. It is assumed that one side of the window is exposed to seawater while the other side is at 1 atm of air pressure. The discussion emphasizes that the net force calculation should consider the pressure difference, specifically using the formula ρgd for seawater. Overall, the approach involves understanding the pressures on both sides of the window to determine the correct force acting on it.
yourheartandsoul
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Homework Statement
Deep-sea photographs have been made at depths of 8000 m. (a) What is the pressure at this depth? (b) What is the force on the camera win dow if it measures 0.1 by 0.15 m?
Relevant Equations
P = dgh
F= PA
I figured out that the pressure is 8.0442x10^7 Pa and I’m not sure how to approach the second half of the question.

I’m thinking I use the formula F=PA but I’m realizing it’s not saying that .15m is a diameter or a radius.
 
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It's a 0.1 m × 0.15 m rectangle.
 
And what's on the other side of the window ?

:welcome:
 
kuruman said:
It's a 0.1 m × 0.15 m rectangle.
oh wow I'm on another planet today duh. So i would use the area of a rectangle with the F=PA?
 
yourheartandsoul said:
oh wow I'm on another planet today duh. So i would use the area of a rectangle with the F=PA?
Not so fast. To paraphrase @BvU, what's on each side of the window?
 
It is an assumption, but a good one, I think. The inside of the housing will be one atmo of air.
 
kuruman said:
Not so fast. To paraphrase @BvU, what's on each side of the window?
Water one side and a camera on the other.. lol not too sure what's on the inside of a camera lens. it was for a test so if it was wrong to approach it that way then oh well. I live and learn
 
DaveC426913 said:
It is an assumption, but a good one, I think. The inside of the housing will be one atmo of air.
Assuming, of course, that the person who assembled the camera didn't work under pressure. (Sorry, I had to say that.)
 
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BvU said:
And what's on the other side of the window ?
It doesn't specify net force, so my guess is they just want the force from the water side.
If it's 1atm inside then the net force will be just ρgd.
Note also it is seawater, not fresh water.
 
  • #10
haruspex said:
It doesn't specify net force, so my guess is they just want the force from the water side.
If it's 1atm inside then the net force will be just ρgd
Fully agree with ##\rho\, g\,\Delta h## -- it's more an awareness thingy

haruspex said:
Note also it is seawater, not fresh water.
With 8.0442x107/9.81/8000 = 1025 that seems to be in perfect order ..
 
  • #11
BvU said:
Fully agree with ##\rho\, g\,\Delta h## -- it's more an awareness thingy
Let me be clearer - I think the way it is worded implies 1atm +##\rho\, g\,\Delta h##, not that it will make much difference.
 
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