Volumes are Awesome

• Feb 1st 2009, 08:09 AM
zhupolongjoe
Volumes are Awesome
http://www.math.rutgers.edu/~greenfi...dfstuff/w2.pdf

Problem 2:

Basically I know how to calculate the work for the cylinder. Since they have the same volume, we know the volume of the cones, but since they are the same depth underground, then how is the answer going to be different for any of these three. I know the work is the integral of the force, and the force is basically found by taking the volume * density * g...so why are they going to be different and how can I know?
• Feb 1st 2009, 08:21 AM
earboth
Quote:

Originally Posted by zhupolongjoe
http://www.math.rutgers.edu/~greenfi...dfstuff/w2.pdf

Problem 2:

Basically I know how to calculate the work for the cylinder. Since they have the same volume, we know the volume of the cones, but since they are the same depth underground, then how is the answer going to be different for any of these three. I know the work is the integral of the force, and the force is basically found by taking the volume * density * g...so why are they going to be different and how can I know?

1. That must be a funny liquid (Thinking)

2. Compare the distance of the centroids to the surface:

The distance of the centroid to the surface of the first cone is 2.5 m

The distance of the centroid to the surface of the cylinder is 5 m

The distance of the centroid to the surface of the second cone is 7.5 m
• Feb 1st 2009, 09:00 AM
zhupolongjoe
So, would the difference merely be that the limits of integration change in each instance? (i.e., we integrate the force and since the force is the same in each case since the volume is the same, the only thing that changes is the limits of integration, which relates to the distance of the centroid to the surface?)
• Feb 1st 2009, 10:24 PM
earboth
Quote:

Originally Posted by zhupolongjoe
So, would the difference merely be that the limits of integration change in each instance? (i.e., we integrate the force and since the force is the same in each case since the volume is the same, the only thing that changes is the limits of integration, which relates to the distance of the centroid to the surface?)

I'll demonstrate how to do the calculation with the first tank(apex down):

1. The volume of this tank is the same as the volume of the cylindical tank:

$V = \pi \cdot (3)^2 \cdot 10 = 90\pi\ m^3$

Let D denote the density of the fluid then the weight of the fluid is $F = V \cdot D$.

$F = 90\pi \cdot 300\cdot g\ \frac{N}{m^3}$

The work which has to be done to empty the tank is

$W = F \cdot (\text{distance of the centroid to the surface)}$

$W = 300\cdot g \cdot 90\pi \cdot 2.5 = 67500\pi\cdot g\ Nm$

-------------------------------------------------------------------------------------------

2. I calculate the radius of the base circle of the cone:

$V = 90\pi=\frac13\cdot \pi \cdot r^2 \cdot h$ ....... With your values:

$90\pi=\frac13\cdot \pi \cdot r^2 \cdot 10~\implies~r=3\sqrt{3}$

Now calculate the volume of a small cylinder which is at a distance of s from the surface. Use proportions:

$\dfrac r{10-s}=\dfrac{3\sqrt{3}}{10}~\implies~r=\dfrac{3\sqrt{ 3}(10-s)}{10}=3\sqrt{3}-\frac{3\sqrt{3}}{10}\cdot s$

If the height of one cylinder is $ds$ you'll get the complete work which is necessary to empty the tanke by:

$W=\int_{0}^{10}\left(300\cdot g \cdot \pi \left( 3\sqrt{3}-\frac{3\sqrt{3}}{10}\cdot s \right)^2 \cdot s \right)ds$ ....... Factor out the constants:

$W=300\cdot g \cdot \pi \cdot (3\sqrt{3})^2 \cdot \int_{0}^{10}\left(s - \frac15 s^2 + \frac1{100}s^3 \right)ds =$ $8100\pi \cdot g \cdot \left[\frac12s^2-\frac1{15}s^3+\frac1{400}s^4\right]_{0}^{10} = 8100\pi\cdot g\cdot \frac{25}3$

which is exactly the same value as I've calculated at #1.
• Feb 2nd 2009, 04:59 AM
zhupolongjoe
I think I got it, thanks. I actually used a different method (slightly), but got answers that are likely equibalent. Thanks.