How Do You Calculate the Force on Different Faces of an L-Shaped Water Tank?

[Puchinita5]

Homework Statement

The L-shaped tank shown in Fig. 14-33 is filled with water and is open at the top. If d = 3.24 m, what is the force due to the water (a) on face A and (b) on face B?

IMAGE: http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c14/fig14_33.gif

Homework Equations

i know that F=PA (where P is pressure and A is surface area)
P-Pi= ρgh (where ρ is desnity and Pi is initial atmospheric pressure)

The Attempt at a Solution

Other than those formulas, i don't know what else i need. But i can't see how i have enough information to use those formulas. I don't know how i would calculate initial atmospheric pressure or desnity without them being given.

 

[LowlyPion]

The problem says it's filled with water, so you should have the density of that.

As for atmospheric pressure ... look it up.

 

[Puchinita5]

okay so atmospheric pressure and water density i know now (didn't know that I was supposed to know those at the top of my head)...but i still don't know what to do with this problem...i know i somehow need to integrate something, but don't know what...really need guidance on this one

 

[djeitnstine]

Pressure varies by depth. And is constant along anyone level of water.

So P= pgh is already evaluated, and A= w*h. where 'w' = width. so if it varies along the vertical direction how would your equation for force look like?

 

[Puchinita5]

okay, first tihng I'm confused with...should pressure not be p=p(atmospheric) +pgh?

i used this to get p=(1.013e5)+(1000)(9.8)(6.48)=164804
and then since p=F/A...(164804)(10.4976)=F
so F=1730046.47 N

is that not correct for part A? i would not be surprised if its wrong lol

 

[djeitnstine]

If the tank is fully exposed to the atmosphere, then we can say atmospheric pressure acts on both sides of all the surfaces. So any effect of it is canceled out.

 

[Puchinita5]

hmmm...so you mean, since it is putting pressure on the walls from the outside as well? so that it doesn't matter if the top is open?

 

[djeitnstine]

No it wouldn't matter if the top is open, as long as the pressure above the water surface is atmospheric, and the pressure outside the tank is atmospheric also.

Since pressure always acts normal to any surface, there will be no effect of the atmospheric pressure force since it is acting on both the inside and outside of the tank.

Edit: from your diagram this would appear so, especially since no extra information about an additional pressure acting on any surface is given.

 

[Puchinita5]

ok i see! so then, p=(1000)(9.8)(6.48)=63504...yes?

and then how would i go about part B? I'm pretty sure i need to integrate, so would i integrate F=pA=pghd^2 from d=0 to 3.24?? does tha make sense because I'm kinda guessing?...

 

[djeitnstine]

Since pressure varies with depth, wouldn't force vary with depth also? Since at anyone level the pressure and thus the force remains constant. Look at post #4 again.

 

[Puchinita5]

okay...so i would integrate pressure? so integrate pgh from h=2d to 3d?? and then whatever my answer is multiply it by the area?

 

[djeitnstine]

No you already evaluated it. pgh is already an integral evaluated (Specifically it is \int_0^H (\rho g) dh Where H= final height).

So let P=pgh - which you already calculated. then all you're left with is \int_0^H (Pw) dh , (Where w = width and dh is your differential height)

 

[Puchinita5]

i'm very sorry, I'm still confused...as you said, the pressure would be different as you vary the hieght...but how did i already evaluate pgh? because for part a, the face was horizontal so it didn't vary, but for part B the face is vertical so I'm still not sure what do do for its pressure? and then in ur final formula, how did you get Pwdh??

sorry if you feel like ur spoon feeding here, I'm just so confused!

 

[rl.bhat]

Pressure at h = 2d is ρgh. Take a small strip length d and width dh on face B.
Force on this strip df = ρgh*d*dh. d remains constant throughout B.
So the total force = ρgd*Intg.h*dh from h = 2d to h = 3d.

 

[djeitnstine]

In part A pressure is constant since the surface is at one height, so your Force acting on that is \rho g (2d) d^2 In part b as we said the force is not constant along this surface! since it encompasses a depth 2d - 3d. So you have to evaluate pressure \rho g h|^{2d}_{3d}.

And since force again varies with depth because pressure is NOT the same when you change height! so let P = \rho g h|^{h=3d}_{h=2d} (i e. pressure evaluated from height 3d to height 2d!), you now need to evaluate force in a similar fashion.

Thus we get F_p = Pwh|^{h=3d}_{h=2d}

PS. I edited the integrals in post 12 to be clearer