Net force on a Submarine Window?

[colcol500]

Homework Statement

A submarine is 100 m below the surface of the water, which has a density of 1000 kg/m^3. Compare the force of the water outside the submarine and the air inside (1.0 atm) on a .1 m x .2 m window.

Apply the relationship P= P0+rho*g*h and any other principles needed to determine the net force that the air inside and the water outside exert on the window.

Relevant Equations

Apply the relationship P= P0+rho*g*h and any other principles needed to determine the net force that the air inside and the water outside exert on the window.

I have no idea, I honestly thought that the net force would be zero since the submarine window isn't moving...yikes

 

[mfb]

The window will be held in place by its support at the edge, that is not part of this problem.

You are given a formula where you know all parameters or can find them - just plug them into find the water pressure outside.

 

[collinsmark]

I honestly thought that the net force would be zero since the submarine window isn't moving...yikes

You are correct. There is also the force from the hull which counteracts the forces from the air and water.

I think the problem is asking you to compare these forces [individually].

 

[HallsofIvy]

There's a lot of information missing here! For example, the submarine is 100 m below the surface but the submarine is not a single horizontal line! What is the depth of the top of the window? I am going to assume that the top of the window is 100 m below the surface.

Also you say the window is "0.1m by 0.2 m" but don't say which measurement is horizontal and with is vertical. That's important to the answer! I am going to assume that the 0.1 m measurement is horizontal and the 0.2 m measurement is vertical.

Imagine dividing the window into n horizontal strips, each of thickness 0.2/n meters. Taking n sufficiently large, each strip will be narrow enough that we can treat it as a single depth. Taking x= 0 at the top of the window, x= 0.2i/n for the ith strip and it is 100+ 0.2i/n m below the surface. We can think of that as a narrow "stack" of water of length 0.1 m, width 0.2/n m, and height 100+ 0.2i/n m. That is a volume of 0.02(100+ 0.2i/n)= 2+ 0.002i/n cubic meters. Since the density of water is 1000 kg/m^3, the weight of that water is 2000+ 2i/n kg. That is, of course, downward but water, or any liquid, has the property that it exerts the same force in all directions. So that strip has a force of 2000+ 2i/n kg on it. To get the total force on the window, add all of those strips: $$\sum_{i=0}^n (2000+ 2i/n)$$. That is a "Riemann sum". Taking the limit as n goes to infinity gives the integral $$\int_0^{0.2} (2000+ 2x) dx$$.

 

[jbriggs444]

Also you say the window is "0.1m by 0.2 m" but don't say which measurement is horizontal and with is vertical. That's important to the answer!

It is unimportant at the level of precision suggested by the "100 m" depth, the "1.0 atm" inside pressure or the "0.1 by 0.2m" window dimensions. For all we know, both window dimensions might be horizontal.

 

[DEvens]

It is unimportant at the level of precision suggested by the "100 m" depth, the "1.0 atm" inside pressure or the "0.1 by 0.2m" window dimensions. For all we know, both window dimensions might be horizontal.

Urm... Umm... Trying to picture a window that is 0.1 m wide, and 0.2 m wide. LOL.

 

[jbriggs444]

Urm... Umm... Trying to picture a window that is 0.1 m wide, and 0.2 m wide. LOL.

Think glass bottom boat.

 

[mfb]

@HallsofIvy: I'm certain OP is expected to neglect pressure variations within the window area.