# Lifting Problem

#### alane1994

##### Active member
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

Staff member
First, what shape is a horizontal slice of water?

#### alane1994

##### Active member
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

Staff member
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

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

#### MarkFL

Staff member
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

##### Active member
Here is my work...

#### alane1994

##### Active member
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!