Possible title: How Deep Can a Tank be Filled Without the Window Failing?

[Wyoming Physics]

Given: A Large tank has a plastic window on one wall that is designed to withstand a force of 90,000N. The square window is 2m on a side, and its lower edge is 1m from the bottom of the tank.

Question: a) if the tank is filled to a depth of 4 m, will the window withstand the resulting force?
b) What is the max depth to which the tank can be filled without the window failing?

- I am very confused as to what to do for a) and b). After setting up a sketch I got the integral from 1to4 (9.8m/sec^2)(1000kg/m^3)(4m^2)(4-y). I am not very confident about the Depth which I put at 4-y.

 

[andrewkirk]

The formula is close, but contains two errors.

First, your integration limits should be 1 to 3, assuming y is measured as height above the bottom of the tank.

Next, look at your units. You have $ms^{-2}\times kg\,m^{-3}\times m^2\times m\times m$ where the last $m$ comes from integrating over height. That means your answer will have units $kg\,m^2s^{-2}$ whereas you want your answer to be Newtons, which is $kg\, ms^{-2}$.

Where in your formula do you think the superfluous $m$ came in? If you can find that, it will lead you towards a correct formula.

 

[Wyoming Physics]

The formula is close, but contains two errors.

First, your integration limits should be 1 to 3, assuming y is measured as height above the bottom of the tank.

Next, look at your units. You have $ms^{-2}\times kg\,m^{-3}\times m^2\times m\times m$ where the last $m$ comes from integrating over height. That means your answer will have units $kg\,m^2s^{-2}$ whereas you want your answer to be Newtons, which is $kg\, ms^{-2}$.

Where in your formula do you think the superfluous $m$ came in? If you can find that, it will lead you towards a correct formula.

I am assuming my error is with the height for which I put (4-y), since the window start 1m about the ground do I just subtract 4-1 and use 3m as my height?

 

[leright]

The hydrostatic pressure is rho*g*h. As andrew pointed out, you need to multiply the hydrostatic pressure rho*g*h by an area to get total force, not just height. Integrating over height is akin to multiplying by a height. You need to integrate over height in one direction, but multiply by the width of the window as well to get total force (you do not need to integrate in this direction, however, because the pressure is constant in this direction.

 

[leright]

I think the total force for a.) is the integral from 1 to 3 of w*rho*g*h*dh, where w is the width of the window.

for b.) you can simply generalize the itnegral and the sole for the unknown height after integrating and setting the integral equal to the total force allowed.

 

[leright]

I get 78,400 N for a.)

 

[andrewkirk]

I am assuming my error is with the height for which I put (4-y)

No, that part is correct. Even if it were wrong, it would not change the units, as the unit of height is metres and so are the units of 4-y.

The problem is as pointed out by leright, where you have multiplied by area of the window rather than width. You only want width because the vertical dimension is brought in when you integrate over height.

 

[haruspex]

To expand a little on what others have posted...
In general, the force is the pressure integrated over the area, and to be completely correct this should be as a scalar product of vectors: $\int \vec P.\vec{dA}$. In the present case, the pressure is collinear with the area element vector, so it simplifies to $\int \int P.dxdy$. Since the pressure is constant at a given depth, the integral wrt width is trivial: $\int \int P.dxdy=\int dx. \int P.dy$.

 

[Wyoming Physics]

I get 78,400 N for a.)

I get 78,400 N for a.)

So when I set ip the integral from 1to3 of w*rho*g*h*dh, I get the integral from 1to3 of (2m)(1000kg/m^3)(9.8m/sec^2)(4-y), although when I solve for this I get 58800N.

 

[leright]

So when I set ip the integral from 1to3 of w*rho*g*h*dh, I get the integral from 1to3 of (2m)(1000kg/m^3)(9.8m/sec^2)(4-y), although when I solve for this I get 58800N.

If you integrate from 1 to 3 you want to integrate w*rho*g*h*dh...not w*rho*g*(4-h)*dh.

 

[leright]

I see why you want to set up the integral that way though...in orde to determine part b.) Seems like they should both work out the same.

edit: ok, they won't work out the same actually. My approach measures depth from the top of the tank, not fro mthe bottom.

 

[Wyoming Physics]

Thank You!

 

[leright]

Thank You!

Not sure how helpful I was but I hope you figured it out!

 

[Wyoming Physics]

Not sure how helpful I was but I hope you figured it out!

Wouldn't (4-h) measure the height I need? resulting in the answer being 58800N? sorry for so many questions, I am just very confused about this specific problem because we have not gone over this material in class

 

[haruspex]

Wouldn't (4-h) measure the height I need? resulting in the answer being 58800N? sorry for so many questions, I am just very confused about this specific problem because we have not gone over this material in class

Since you are integrating from 1 to 3, it should make no difference whether the integrand is h or 4-h. It's symmetric. But I don't get 58800. Please post the details of your working.

 

[Wyoming Physics]

thank you I totally understand it now, although I am still confused about part b)

 

[andrewkirk]

I am still confused about part b)

Can you write a formula for the force on the window in terms of d, where d is the depth to which the tank is filled?

 

[Wyoming Physics]

The only thing I can really think of is the integral from 1to3 (9800)(2)(d+4)dd

 

[haruspex]

The only thing I can really think of is the integral from 1to3 (9800)(2)(d+4)dd

In andrewkirk's suggestion, d is the depth to which it is filled, not the depth at some arbitrary point on the window below that depth. The integration variable represents the depth at an arbitrary subaqueous point. So think again about the role that d plays in the integration.