# calculating work from to pump water over the top of a trough

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• May 4th 2010, 07:26 AM
victorxz90044
calculating work from to pump water over the top of a trough
A water trough is 6 feet long, its vertical cross section is as isosceles trapezoid with lower base 2 feet, upper base 3 feet and altitude 2 feet. if the trough is full, how much work is done in pumping all of the water over the top of the trough? Use 62.5 lb/ft^3 as the weight density factor for water.
• May 4th 2010, 07:46 AM
Grandad
Hello victorxz90044
Quote:

Originally Posted by victorxz90044
A water trough is 6 feet long, its vertical cross section is as isosceles trapezoid with lower base 2 feet, upper base 3 feet and altitude 2 feet. if the trough is full, how much work is done in pumping all of the water over the top of the trough? Use 62.5 lb/ft^3 as the weight density factor for water.

Here are the steps you need to solve the problem:

• Consider a horizontal cross-section through the water at a height $\displaystyle h$ above the base, and with thickness $\displaystyle \delta h$.
.
• Work out the area of this cross-section in terms of $\displaystyle h$, and hence its mass in terms of $\displaystyle h$ and $\displaystyle \delta h$.
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• This mass of water is raised through a vertical height of $\displaystyle (2-h)$ feet. Find an expression for the work done, $\displaystyle \delta W$, when this happens.
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• Write down an expression for $\displaystyle \frac{dW}{dh}$, and hence, by integrating between $\displaystyle h = 0$ and $\displaystyle h = 2$, evaluate the total work done.

Can you do this now? Let us know where you got stuck, if not.

Grandad

• May 4th 2010, 07:52 AM
victorxz90044
as far as i have gone...
(b1+b2)/2 y = A

pump distance= 2-y

y =0 , b=2
y=2, b=3

-b2 = (b1+(1/2)y)/2

6*b2*dy
6(b1+(1/2)y)dy

dw = int 62.5 [6(2+2y)dy] (2-y)

now what
• May 4th 2010, 11:45 AM
Grandad
Hello victorxz90044
Quote:

Originally Posted by victorxz90044
(b1+b2)/2 y = A

pump distance= 2-y

y =0 , b=2
y=2, b=3

-b2 = (b1+(1/2)y)/2

6*b2*dy
6(b1+(1/2)y)dy

dw = int 62.5 [6(2+2y)dy] (2-y)

now what

Each cross-section is a rectangle whose length is 6 feet. If the breadth is $\displaystyle b$ at height $\displaystyle y$ above the base, then, as you say:
When $\displaystyle y = 0, b = 2$

When $\displaystyle y = 2, b = 3$
and the relation is clearly linear, so:
$\displaystyle b = 2+\tfrac12y$
So the area of cross-section is $\displaystyle 6b = 12 + 3y$ square feet, and the volume of this thin slice is $\displaystyle (12+3y)\delta y$ cubic feet. Its mass is therefore $\displaystyle 62.5(12+3y)\delta y$ pounds. It is raised through a distance $\displaystyle (2-y)$ feet. So the work done is:
$\displaystyle \delta W = 62.5(12+3y)\delta y(2-y)$ foot pounds. (I don't know whether that's the correct unit; I usually work in SI.)

$\displaystyle \Rightarrow \frac{dW}{dy}=62.5(12+3y)(2-y)$

$\displaystyle \Rightarrow W = \int_0^262.5(12+3y)(2-y)\;dy$
Grandad