Finding the force on a camera window underwater

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SUMMARY

The discussion centers on calculating the force exerted on a camera window underwater, specifically using the formula F=PA. The pressure is established at 8.0442x10^7 Pa, and the dimensions of the window are confirmed as a rectangle measuring 0.1 m by 0.15 m. Participants emphasize the importance of considering the pressure difference on either side of the window, noting that the inside of the camera housing is at 1 atm of air pressure while the external pressure is from seawater. The net force calculation involves using the density of seawater and the gravitational constant, leading to a refined understanding of the forces at play.

PREREQUISITES
  • Understanding of fluid mechanics, specifically pressure calculations.
  • Familiarity with the formula F=PA for force calculations.
  • Knowledge of seawater density and its implications in pressure scenarios.
  • Basic grasp of atmospheric pressure and its effects on submerged objects.
NEXT STEPS
  • Research the properties of seawater, including density variations with depth.
  • Study the implications of atmospheric pressure on submerged objects.
  • Explore advanced fluid dynamics concepts, particularly hydrostatic pressure.
  • Learn about the design considerations for underwater camera housings.
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Students and professionals in physics, engineering, and marine technology, particularly those involved in underwater equipment design and pressure analysis.

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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