Help Me Solve a Math Problem Involving Trapezoidal Cross Sectional Area

[alane1994]

I have asked for help from a math professor and it didn't really help. It would be absolutely fantastic if one of you could help me out.

Here is the problem verbatim.
A trough has a trapezoidal cross section with a height of 1 m and horizontal sides of width 0.5 m and 1 m. Assume the length of the trough is 11 m.

How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full? Use 1000kg/m³ for the density of water and 9.8 m/s²
for the acceleration due to gravity.

I have done some work on the problem.
The equation that I have been taught for this is.
W=\int^b_a{\rho g A(y)D(y)}dy
A(y) is the cross-sectional area of the horizontal slices.
D(y) is the distance the slices must be lifted.

I have figured out this much.
W=\int^1_0{(1000)(9.8)(A(y))(1-y)}dy

What I need help with is figuring out what A(y) is. Any and all help is GREATLY appreciated!

 

[MarkFL]

First, what shape is a horizontal slice of water?

 

[alane1994]

It will be a rectangle, but as the water is pumped out, the width will change.

I have something that looks like this...
A(y)=0.5+2(\frac{1}{0.25}...
this is where I get stuck...:(

 

[MarkFL]

Yes, the width will vary, but the height will not, it will remain constant at 11 m.

Can you say what type of variation the width has?

 

[alane1994]

I have the height as 1 metre... which goes down as the water is pumped out. The length is 11 metres.

 

[MarkFL]

So far we have:

$\displaystyle A(y)=11w(y) \text{ m}^2$

where $\displaystyle w(y)$ is the width of a rectangular slice as a function of y.

Now, what I am asking is what type of function do we expect w to be? We know the width decreases as the water level drops, but how does it decrease?

 

[alane1994]

Here is my work...

 

[alane1994]

I have to go to my next class soon, I will try to monitor this on my phone as best as I can. Thanks for all the help you have given so far!

 

[Jameson]

I have to go to my next class soon, I will try to monitor this on my phone as best as I can. Thanks for all the help you have given so far!

If you can spare the $1.99 and have an iPhone or Android device I would highly recommend buying Tapatalk. We're set up to be used with this software and it makes browsing forums very easy and convenient. The only downside is Latex doesn't render, but I still use it all the time on my phone.

 

[MarkFL]

Yes the width changes linearly, and we know:

$\displaystyle w(0)=1,\,w(1)=\frac{1}{2}$

So, knowing 2 points on the line, you may find the equation of the line.

 

[alane1994]

I got an answer of 17967. In the instructions for the answer it says round to nearest whole unit. Am I close or way off?

 

[MarkFL]

This is how I would set this up (bear in mind it has been a while since I took physics, so I will have to work from some basic principles):

$\displaystyle dW=F\cdot d$

$\displaystyle F$ = weight of rectangular slice

$\displaystyle d=1-y$ = distance rectangular slice must be lifted

$\displaystyle F=mg=\rho Vg=\rho ghw\,dy$

$\displaystyle \rho$ = mass density of fluid

$\displaystyle V$ = volume of rectangular slice

$\displaystyle g$ = acceleration due to gravity

$\displaystyle h$ = height of rectangular slice

$\displaystyle w$ = width of rectangular slice

$\displaystyle w=-\frac{1}{2}y+1$

$\displaystyle dy$ = thickness of rectangular slice

So, we now have:

$\displaystyle dW=\rho gh\left(\frac{1}{2}y-1 \right)(y-1)\,dy$

$\displaystyle dW=\rho gh\left(\frac{1}{2}y^2-\frac{3}{2}y+1 \right)\,dy$

Summing by integration, we have:

$\displaystyle W=\rho gh\int_0^1\left(\frac{1}{2}y^2-\frac{3}{2}y+1 \right)\,dy$

With:

$\displaystyle \rho=1000\,\frac{\text{kg}}{\text{m}^3}$

$\displaystyle g=9.8\,\frac{\text{m}}{\text{s}^2}$

$\displaystyle h=11\text{ m}$

I get a result (in Joules) substantially different than yours.

 

[alane1994]

What did you get for an answer? I exceeded my maximum number of guesses(3), and it gave me an answer. I can try a different but similar problem, but what did you get.
Here is what was "Correct"...

Spoiler

35,933 Joules

 

[alane1994]

I have gotten a new problem. All the lengths are the same except that the 11 m, is now 15 m.

 

[MarkFL]

I made a silly blunder...I should have:

$\displaystyle d=y$

and so:

$\displaystyle W=\rho gh\int_0^1y-\frac{1}{2}y^2\,dy$

Then, the result is:

$\displaystyle W=35933.\bar{3}\text{ J}$

 

[alane1994]

Can you explain to me where you get the A(y) function?

 

[MarkFL]

The width of the rectangular slices decreases linearly as we go from the top to the bottom. The width at the top is 1 and at the bottom it is 1/2. This effectively gives us 2 points on the line. We know the w-intercept is 1, and the slope is -1/2, so we have:

$\displaystyle w=-\frac{1}{2}y+1$

 

[alane1994]

So... for this latest problem... would the answer be 61,200 Joules?

 

[MarkFL]

Close.

Could you show your work in evaluating the integral?

 

[alane1994]

I give up on this problem. It is the only homework problem that I have not gotten correct. I thank you for all your help with this problem. I have to study chemistry and Engineering Computations. Thank you again!

 

[MarkFL]

Aw, don't give up now! You have the product:

$\displaystyle 1000\cdot9.8\cdot15=147000$

And this needs to be multiplied by:

$\displaystyle \int_0^1 y-\frac{1}{2}y^2\,dy$

In fact, given the result of the previous problem, you could in fact just multiply that by 15/11. But, I don't recommend this, it is important to be able to evaluate the given integral.

I'm not trying to make you do this, but just offering encouragement.:) I hate to see someone not make it to the end of a problem, particularly when I have initially made a mistake with my help. I would feel better if you triumph over this problem! ;)