# Pressure on a window of a submarine

1. Dec 2, 2009

### indietro

1. The problem statement, all variables and given/known data
a research submarine has a 20 cm diameter window 8 cm think. the manufacturer says the window van withstand forces up to 1x106N. What is the submarine's maximum safe depth?

2. Relevant equations
p = po + $$\rho$$gd

3. The attempt at a solution
p = F/A = 1X106 / 0.03 = 3.2x107 Pa

so now do i use p = po + $$\rho$$gd ??

but if i do what would be my $$\rho$$?
$$\rho$$ = m/v = m/$$\pi$$r2h
how would i find mass?

2. Dec 2, 2009

### indietro

ok so i solved it with the help of a different post but i had a question:
she does p' = p1 + $$\rho$$gd
p' - p1 = $$\rho$$gd

and then she says that p1 = 1x105 Pa > atmospheric pressure. I was just wondering why?

can someone explain how
$$\Delta$$p = p1 - po relates or is used in conjunction with p' = p1 + $$\rho$$gd? is $$\Delta$$p = p'

3. Dec 2, 2009

anyone??

4. Dec 2, 2009

### indietro

i really just cant seem to wrap my head around pressure concepts, if someone knows how to explain it clearly or knows a website where i can see examples and stuff that would be great!

5. Dec 3, 2009

### PhanthomJay

it gets confusing with so many different letter variables like $$p, p_1, p_o, p', \rho$$ (and Pa for Paschals to boot), so let's define the variables:

let

$$p_o$$ = atmospheric pressure = 1.01(10^5) Paschals
$$p$$ = total pressure at depth d
$$\rho$$ = density of water = 1000kg/m^3)

Now when the top of the water is not sealed off (that is, it is exposed to the atmosphere),
$$p = p_o + \rho gd$$, that is , the total pressure at depth d is the sum total of the atmospheric pressure and the water pressure at that depth.