Calculating Force on a Diving Bell Port at 147m Depth | Simple Pressure Problem

[BuBbLeS01]

Homework Statement

A spherical diving bell containing a camera is in the ocean at a depth of 147 m. It has a flat, transparent, circular port with a diameter of 12.7 cm. Find the total force on the port (use ρsea water = 1025 kg/m3).

Homework Equations

p = po + pgd
P = F/A
F = pA

The Attempt at a Solution

So do I find the total pressure and then multiply it be the area? I get a really big number (2.00 x 10 ^ 8) which doesn't seem right?

 

[mgb_phys]

Are you in the same class as https://www.physicsforums.com/showthread.php?t=198333&highlight=spherical+diving+bell ?

Yes, pressure is force / area so force = pressure * area.
Make sure your units are correct.
Yes the force is large - that's why diving bells have quite thick windows!

 

[BuBbLeS01]

No I don't think so cause it is worded differently and in our class they are all worded the same just different numbers...but that's weird its almost identical!

 

[BuBbLeS01]

yea that answer isn't right...I just tried it :(

 

[BuBbLeS01]

does anyone know how to do this?

 

[mgb_phys]

Sorry - I thought your comment meant you had solved it!
The area of the window is A = pi r^2
There is a column of water on top of it 147m high
So the volume of water sitting on the window is V = pi r^2 * depth
Add the density of water and you get the mass of water on the window.

V = pi * (0.127/2)^2 * 147 = 1.89m^3
Density is 1025 kg/m^3 so mass of water is 1900kg
Total force is F = mass * g = 18.7KN

Note there is also an extra 1 atmosphere of pressure pushing down on the surface of the water, but there is also an atmosphere of pressure inside the bell pushing out which cancel out.

 

[BuBbLeS01]

thats ok...but its telling me that's incorrect?

 

[BuBbLeS01]

this is the msg it gave me...
Calculate the total pressure at the given depth by adding up the atmospheric pressure (1.01e5 N/m2) plus the pressure due to the sea water. This assumes that there is no air in the bell, (probably a bad assumption).

 

[mgb_phys]

Alternate method - the weight of the column of water = m g h = 1.47MN
Area A= pi r^2 = 0.0127m^2
Force = pressure * area = 1.47MN/m^2 * 0.0127m^2 = 18.67N

EDIT - then just add atmospheric pressure to the water pressure, answer should be 20KN.

 

[Bill Foster]

I'm having deja vu and amnesia at the same time...I think I've forgotten this once before.

 

[BuBbLeS01]

LoL...

 

[BuBbLeS01]

I don't understand I enter 0.02 N (it doesn't recognize units like KN or MN) and it says incorrect??

 

[mgb_phys]

20 KiloNewtons ie 20,000N or 2E4 N