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May 27th, 2014, 02:55 AM   #1
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Big Tank full of Water

Here is the link for the problem Gyazo - 1c8e3a691f1093592dfde92a094a9d77.gif


Now, Using a right triangle I have my radius as 6ft, and using an xy coordinate I used 0 to 10

I have radius labeled as ri / 10 - xi = 6/10 so ri = 3/5[10-xi]

Volume = pi r^2 h delta x = pi(9/25)(10-xi)^2 delta x

Mass = Density * Volume Thus, Mi = 62.5 pound/ft^3 * pi(9/25)(10-xi)^2dx

and now Force = mi * gravity so I just added the product of (9. to above equation to get Force F.

Now, I simplified and cleaned up some ready to Integrate and have the following:

Work = lim as n goes to INF ... and thus the Riemann sum takes us to the definite integral and I have: The Integrand from 0 to 10 ... wait here I need to stop because of confusion. This problem (as you can see via the link pasted above) tells us Use the fact that water weighs 62.5 lb/ft3.

In another similar problem where I have been to attempt to gain insight, it says Mass = Density * Volume and gives Density as 1,000 kg/ meter ^3

Now I am new into this chapter and remember they were saying when you have a weight you can use as the force but at other times we have to multiply times gravity , etc to get force. I am thinking for this problem that it clearly says at start of problem to use water weight at 62.5 and if so than I am thinking that Work accomplished is W = Force * Displacement (or distance I forget which one) but here is where I am starting to get lost in the weeds and where it may be simple for many of your to fix all this in your minds I am a new calc 2 student who needs lots of help...lol

Anyways, I thought I was on a role for a minute because I got to integrating and did so and then plugged and chugged and got incorrect answers. I feel I am doing better with spending hours with math and getting better acclamated with it. I had a teacher a couple yrs ago in trig that told me mathematics is not something you can lay down and listen to YouTube videos and get it.

I am getting more comfortable with the looks and feel of equations so that must be a good thing.

This problem seems detailed and I want to ask the forum to help/assist me with the problem in the link I posted at top. I know we are told to show what work we can and I feel I have done so.

Thank you.
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May 27th, 2014, 08:43 PM   #2
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Hi mr. jenkins,

By saying that water weighs 62.5 $\displaystyle \frac{lb}{ft^3} \ $, they mean $\displaystyle \frac{lbf}{ft^3}$ so the force is kind of 'built in'. To show you that this is the case, I will work the problem using the given units and then metric units.

First, instead of thinking in terms of a radius, think in terms of a rectangular sheet of water. I will use the set of coordinates with x along c in the figure, y along b in the figure, and z along a in the figure with z=0 where the spout is and the positive direction of z going down away from the spout. The direction is unconventional, but I like the way the equations set up this way.

First, we need x as a function of z. The relationship is:

$\displaystyle \large x=\left(10-\frac{5}{3}z\right)\text{ft}$

This may be checked by setting z=0 with a result of x=10 and by setting z=6 with a result of x=0, which matches the figure.

So the area of each sheet is:

$\displaystyle \large A=9 \text{ft} \cdot \left(10-\frac{5}{3}z\right)\text{ft}=(90-15z)\text{ft}^2$

The volume of each sheet is:

$\displaystyle \large V=(90-15z)\text{ft}^2 \ dz \ \text{ft}=(90-15z)dz \ \text{ft}^3$

The work required for each sheet is:

$\displaystyle \large W=(90-15z)dz \ \text{ft}^3 \cdot 62.5 \frac{\text{lbf}}{\text{ft}^3} \cdot z \ \text{ft}=62.5(90z-15z^2)dz \ \text{ft} \cdot \text{lbf}$

The total work is:

$\displaystyle \large W_t=62.5 \int_0^6 (90z-15z^2)dz \ \text{ft} \cdot \text{lbf}$

$\displaystyle \large W_t=\left. 62.5\left(45z^2-5z^3\right)\right|_0^6=62.5(1620-1080)=33,750 \ \text{ft} \cdot \text{lbf}$

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

Using the metric system with the density of water as $\displaystyle 1000 \frac{kg}{m^3}$ then:

$\displaystyle \large x=\left(3.048-\frac{5}{3}z\right)\text{m}$

The area of each sheet is:

$\displaystyle \large A=2.7432 \text{m} \cdot \left(3.048-\frac{5}{3}z\right)\text{m}=(8.3613-4.572z)\text{m}^2$

The volume of each sheet is:

$\displaystyle \large V=(8.3613-4.572z)\text{m}^2 \ dz \ \text{m}=(8.3613-4.572z)dz \ \text{m}^3$

The mass of each sheet is:

$\displaystyle \large M=1000 \frac{\text{kg}}{\text{m}^3}\left(8.3613-4.572z \right)dz \ \text{m}^3=1000\left(8.3613-4.572z \right)dz \ \text{kg}$

The force (weight) of each sheet is:

$\displaystyle \large F=9.8 \frac{\text{m}}{\text{s}^2}\cdot 1000\left(8.3613-4.572z \right)dz \ \text{kg}=9800\left(8.3613-4.572z \right)dz \ \text{N}$

The work required for each sheet is:

$\displaystyle \large W=9800\left(8.3613-4.572z \right)dz \cdot z \ \text{N-m}=9800\left(8.3613z-4.572z^2 \right)dz \ \text{J}$

The total work is:

$\displaystyle \large W_t=9800 \int_0^{1.8288}\left(8.3613z-4.572z^2 \right)dz \ \text{J}$

$\displaystyle \large W_t=\left. 9800\left(4.18065z^2-1.524z^3\right)\right|_0^{1.8288}=9800(13.9822-9.32145)=45,675 \ \text{J}$

Per a google search:

$\displaystyle 1 \ \text{ft} \cdot \text{lbf}=1.3558 \ \text{J}$ so:

$\displaystyle \large W_t=45,675 \text{J} \cdot \frac{1 \ \text{ft} \cdot \text{lbf}}{1.3558 \ \text{J}}=33,689 \ \text{ft} \cdot \text{lbf}$

which compares well with the result given above.

I hope that this helps you with similar problems.
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May 29th, 2014, 08:02 AM   #3
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@jks I am working through this one and have been since last night. I have not done one of these before. You mentioned the spout and I have seen others with spout but I just now actually realized that there is a tiny spout in the illustration (which is I guess one would say at the 1/2 way point on line a (which you have re-assigned as line z) right?

Also, jks I could not seem to understand the figure as I thought I was supposed to straighten it up so that is what I did (which may be wrong or simply unnecessary or irrelovent ) but in any case I re drew the figure on graph paper as best I could using 2 triangles on sides (or for the bases)... a rectangle on top and the points of the triangles (or the vertices is correct way to say I beleive)... although on the few problems I have found that have the figure like this ... they all contain the tank only 1/3 full of water or 3/5 , etc etc (for my problem it states that the tank is FILLED with water.) and all these others are 1/2 filled or 1/3 filled etc etc. (I am also thinking here that this to0 may be kind of irrelovent and that the only difference will come in when inegrating and instead of going top to say 3 like a = 3 and b =10 we will have limits when tank is full
of like a =0 and b =10. Do you follow my thinking here? I assume you do and are probally laughing at my childlike way of thinking...lol I have so far to go and am so green but anyway... I wanted to get some thoughts down in text as to demonstrate my thanks to you for taking time out of your day to assist me. I didn't want you to think your time was unwarrented.

Now my next issue at hand is where you have stated we need x as a function of z

and I may be overthinking here but we have this figure and have re-labeled it instead of a,b & c we are using x,y & z. so again (and keep in mind I made like a C- in calc 1) and am just now in calc 2 ] I do not see equations, just constants ( naming x = 10, y =9 and z = 6) so are we either somehow setting all these constant values equal to themselves and solving for a particular variable OR are we using the concept of y = mx + b (the equation for a line) to find these equations.

So, again, when you soved for x as a function of z and I start getting confused or overthinking and I really just need some help or assistance so I can see what I am not seeing because I really feel like the dumb kid at the back of the class.....A LOT!

Ok...enough jibber jabber from me for now. I shall patiently await your reply.
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May 29th, 2014, 08:12 AM   #4
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the vertices pointing down I meant to say. so nothing changed with the figure except that I lterally revolved it (problem is : Now that I see there is a little spout on the original illustration..it was prob right in the first place and I shoujld have left it alone
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May 29th, 2014, 04:43 PM   #5
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Hi mr. jenkins,

Quote:
I could not seem to understand the figure ...
I interpreted the figure as in the first attachment, Fig. 1.

Quote:
Now my next issue at hand is where you have stated we need x as a function of z

and I may be overthinking here but we have this figure and have re-labeled it instead of a,b & c we are using x,y & z. ... I do not see equations, just constants ( naming x = 10, y =9 and z = 6) so are we either somehow setting all these constant values equal to themselves and solving for a particular variable OR are we using the concept of y = mx + b (the equation for a line) to find these equations.
Since a, b, and c were given as constants, I wanted to keep them as such so I assigned a coordinate system of x, y, and z. We do not really need y because as the water is pumped out, this dimension stays constant at the given value of b=9.

However, by looking at the second attachment, Fig. 2, you can see that as the water is pumped out, z is going to change, and so will x, which is why I made them variables. Furthermore, since we are going to integrate, it would be most convenient to have x in terms of z.

For x in terms of z, you are correct in that we will use the concept of x = mz + d. We know that at z=0, x=10 and at z=6 (going down is positive z), x=0. So the equation of the line is:

$\displaystyle \large x=-\frac{5}{3}z+10$

As the water is pumped out the surface of the water starts at z=0 and eventually reaches z=6 when the tank is empty. So it is most convenient to integrate from z=0 to z=6.

So in my previous post I divided the volume of water up into 'sheets' and I gave the formula for the area A, which is b*x = 9x. But since we are going to integrate over z, instead of x, we will use the expression for x in terms of z, resulting in:

$\displaystyle \large A=9 \text{ft} \cdot \left(10-\frac{5}{3}z\right)\text{ft}=(90-15z)\text{ft}^2$

Since the height of the sheet is dz, the volume of the sheet is:

$\displaystyle \large V=(90-15z)\text{ft}^2 \ dz \ \text{ft}=(90-15z)dz \ \text{ft}^3$

In my previous post, I first defined the amount of work required for each sheet and then I integrated to get the total work. Please refer to my previous post for those expressions, noting that z is the distance that each sheet needs to be lifted and that the weight of the water is 62.5 lbf/ft^3.

Anyway, I hope that I interpreted the problem correctly, and if so, I hope that I have explained it correctly. Feel free to reply if you have further questions.
Attached Images
File Type: jpg water_tank_2014_05_29_1.jpg (45.7 KB, 3 views)
File Type: jpg water_tank_2014_05_29_2.jpg (32.1 KB, 3 views)

Last edited by jks; May 29th, 2014 at 04:52 PM.
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May 30th, 2014, 08:56 PM   #6
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I just finished working through this problem in its entirety.
I made sure to give THANKS too, hoping that that means something or accumulates and is beneficial, etc.

I do like your method and did want to ask you if you could rather quickly explain another method (preferably.. the one that would most likely be submitted by other students and'/or perhaps a method that you would anticipate the textbook or software to give for an/the answer.

Many thanks again jks as I have all your text copied, and slowly read and worked through it and am going to now have permanently to use as a template, etc.
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May 30th, 2014, 09:04 PM   #7
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also, 33,750 is approx 33,689 so is that sufficient for submissions or are exact value required normally?
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May 30th, 2014, 09:34 PM   #8
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I would expect that the exact value is required unless it specifically says that you should round is off.
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May 31st, 2014, 05:02 AM   #9
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It took 33,750 but most of the problems do say to round to 2 places
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