Finding the force on a camera window underwater

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Homework Help Overview

The discussion centers around calculating the force exerted on a camera window submerged underwater, involving concepts of pressure and area. The problem involves understanding the pressure acting on the window and how to apply it to find the force, particularly considering the dimensions of the window and the conditions on either side.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the formula F=PA, questioning the dimensions of the window and the conditions on both sides. There is a focus on clarifying the assumptions regarding the pressure inside the camera housing and the nature of the water (seawater vs. freshwater).

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem. Some have offered insights into the assumptions about the pressure on either side of the window, while others are questioning the implications of these assumptions on the calculations. There is no explicit consensus, but several productive lines of reasoning are being developed.

Contextual Notes

Participants note that the problem does not specify whether to consider net force and highlight the importance of distinguishing between seawater and freshwater in the calculations. There are also references to the potential for different pressures based on the assembly of the camera.

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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