Expert: Paul Klarreich - 12/14/2011

Question
A triangular trough has a base of 3 feet height of 3 feet and a length of 10 feet if the trough is filled with water how much work is required to pump out the water through a spout that is 2 feet above the top of the trough the weight of the water is 64.4lbs/ft(cubed)


our teacher dumped this on us for a test on Friday and we never went over this on the review sheet i was hoping you can please help.

Answer
Questioner: christopher deblasio
Country: New York, United States
Category: Calculus
Private: No
Subject: Calculus work
Question: A triangular trough has a base of 3 feet height of 3 feet and a length of 10 feet if the trough is filled with water how much work is required to pump out the water through a spout that is 2 feet above the top of the trough the weight of the water is 64.4lbs/ft(cubed)


our teacher dumped this on us for a test on Friday

>>>>>> I know who you mean.


and we never went over this on the review sheet i was hoping you can please help.
....................................................
You want two things:

1. The weight of the water.  You won't have any trouble with that part.  Area of a triangle times length of the trough times the density.

2. The centroid of the triangle.  You can do this in several ways, but since this is calculus, you will want to integrate:

Each cross-section of a triangular slice has width equal to its height.  Why?  Because the value of  h  goes from 0 to 3, and so does the width.  (That old similar triangle black magic)  The thickness is dh.

Cross-section area:

{h=3
|     width * thickness =
}h=0

{h=3
|     h * dh =
}h=0

h^2/2[0..3] = 9/2

but you knew that from part 1.

Moment =
{h=3
|   h * width * thickness =
}h=0

And when you have computed that, do:

h-bar = Moment/area

Now just pretend that ALL THE WATER is concentrated at that point, and must be lifted to a height of 5 feet.  (3 + 2, of course)

That should do it.