# Fluid Mechanics of a diving bell

[SOLVED] Fluid Mechanics

1. Homework Statement

You are designing a diving bell to withstand the pressure of seawater at a depth of 230.

a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.)

got 2.32x10^6 Pa - masteringphysics said this is correct

b) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 34.0cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You can ignore the small variation of pressure over the surface of the window.)

2. Homework Equations

rho = M/V

p = p0 + (rho)gh

p = F/A

F1/A1 = F2/A2

gauge pressure --> p-p0

3. The Attempt at a Solution

F = pA = (2.32x10^6).340m = 7.9x10^5 N - incorrect

F = pA = (2.32x10^6)(.340*230) = 1.81x10^8 N - incorrect

F = pA = (2.32x10^6)(.340^2) = 2.7x10^5 N - incorrect

I think I'm having a hard time visualizing what the question is asking, any help is greaty appreciated.

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alphysicist
Homework Helper
Hi clope023,

You have the equation F=pA. What does the variable A stand for?

"A" should stand for area.

clope023,

When considering the force of the water on an object (hydrostatic force), remember that the greater the area which the pressure acts upon, the greater the force. You would not expect a 34cm window to experience the same hydrostatic force as, say a 3m window. To take this into account for a given object (in this case a circular window), you must calculate the area A which the pressure acts upon or the surface area.

The surface area of the window is $$A = \pi\cdot r^2$$ - or in your case $$A=\pi\cdot(d/2)^2$$.

Thus, your final result should be $$F = pA = (p_0+\rho gh)\cdot(\pi(d/2)^2)$$

alphysicist
Homework Helper
Hi ok123jump,

Your equation would give the force of the water on the window; however, they want the net force on the window. So I believe the approach in the original post using F=pA, with p being the gauge pressure of the water, would give the correct answer (once the correct value of A that you gave is used).

thanks guys, got the correct answer using that area formula.

alphysicist,

Since the capsule is pressured to standard atmospheric pressure, this formula provides net force.

If the pressure inside the capsule was not the same pressure as $$p_0$$ (atmospheric pressure in this instance) we would need to account for that in the $$p$$ term - you're right. We would do that by defining $$\Delta p = p_{capsule} - p_0$$ and then $$p$$ would become $$p = p_0+\rho gh+\Delta p$$.

Summarily, the previous formula is only valid for the set of conditions which were stated by the OP.

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