Find work done draining water from a tank

[Nuke787]

Homework Statement

A tank of water is 3m long and has a triangular cross section of height 1m and width at the top of 1m. How much work is done pumping the water to the top of the tank? The density of water is 1000kg/m³ and the gravitational constant g has value g m/s²

Homework Equations

Many ways to solve this problem, see below.

The Attempt at a Solution

I solved the problem already by doing 1000g*3*integral from 0 to 1 of ((1/2)x²).
This yields 3000g*(1/6*x³ evaluated at 0 to 1). This will yield 1/6*3000g or 500g.
My problem is my teacher doesn't understand this on the test that I was given and refuses to give me credit until I can explain it in a way he understands. The units in the above equations check out and give you N-m, and the number is correct as well. I have applied this to numerous problems and showed him it works. His problem is he doesn't see "the distance traveled". When solving this problem he does it by slicing in which he is able to see a distance from the top of the tank. Can someone explain to me why this works and hopefully their explanation will make more sense to him than mine. I understand this only works for triangles where the base is changing at the same rate as the height.

 

[tiny-tim]

Hi Nuke787!

I solved the problem already by doing 1000g*3*integral from 0 to 1 of ((1/2)x²).
This yields 3000g*(1/6*x³ evaluated at 0 to 1). This will yield 1/6*3000g or 500g.
My problem is my teacher doesn't understand this on the test that I was given and refuses to give me credit until I can explain it in a way he understands.
…
I understand this only works for triangles where the base is changing at the same rate as the height.

(is that 3000g ∫ (1/2)x² dx ?)

ok then …

explain that integral to us. ​

 

[Nuke787]

Hi Nuke787!


(is that 3000g ∫ (1/2)x² dx ?)

ok then …

explain that integral to us. ​

The integral is l*1000*g*∫(1/2)*b*h
The way I did this problem is I followed the units. I know that 1000(the mass density)*g will give me N/m³. Next I thought ok, the integral of an area will give me a cubic function. So I integrated the area of a triangle, or (1/2)*b*h. I know that for this triangle the base and height are equal and both change at the same rate, so I went ahead and said (1/2)h*h, or (1/2)x². I knew that 3 (the length) would be constant as this doesn't vary. When I solved this I get the above answer. I don't know how to explain it to my teacher so it makes sense to him, he believes this doesn't follow the rules of physics when the method works for similar problems, and the units check out. He integrates slabs with respect to y so he can see the distance each slab is from the top and physically see the force times the distance. So far I have a doctorate in math stumped and a doctorate in engineering stumped on how to explain it.

 

[tiny-tim]

… Next I thought ok, the integral of an area will give me a cubic function. So I integrated the area of a triangle, or (1/2)*b*h. I know that for this triangle the base and height are equal and both change at the same rate, so I went ahead and said (1/2)h*h, or (1/2)x2.

Sorry, but I'm not following this.

You seem to have merely made a plausible guess, which in this case happened to give the right answer.

Do you have any justification for this method, other than that it works?

 

[Nuke787]

None at all. That was my thought process, I follow units in problems I don't have a full grasp on, which works in many cases as the unit leads you to the answer. It works in this problem and all the other work problems for the triangle in which the base and the height are equal and changing at the same rate. What part of what I have done do you not understand, I can try and clear it up to the best of my ability. Right now all I have is a method that works and I can follow the units along with arriving at the correct answer.

 

[tiny-tim]

My problem is my teacher doesn't understand this on the test that I was given and refuses to give me credit until I can explain it in a way he understands.

I agree with your teacher.

You have no idea why your method works, and indeed you have no idea that it works, except that it gives the same result as your teacher gets.

Your teacher doesn't justify his method by saying "It gives the same result as Nuke787 gets!" … he can prove quite independently that it must work.

You can't. ​

 

[Nuke787]

He is correct, he asked me to support what I did, and I can't in a way that he and I understand. I can prove it works for problems of the same type, just not visually enough I guess. If I can show him the distance part he will give me credit, as that is his main discrepancy. My weekend is going toward understanding what I have done so I can get my points. Do you have any ideas where I should start I'm only in calculus 2 so I don't know if this is going to involve some upper math I don't know yet. Numerically, they are the same. It's just how can I relate them conceptually.

 

[tiny-tim]

… If I can show him the distance part he will give me credit, as that is his main discrepancy. My weekend is going toward understanding what I have done so I can get my points. Do you have any ideas where I should start …

You should certainly start by writing it out your teacher's way, so that we can see that you know how to do it that way.

Having done that, you said earlier …

I follow units in problems I don't have a full grasp on, which works in many cases as the unit leads you to the answer.​

… which of course does work in cases where everything is obviously in proportion.

But (I'm not clear on this) didn't you have to insert a factor of 1/2 for no particular reason?

Even if you can show that the result is proportional to each input measurement, you still have to prove what the factor of proportionality is.

(perhaps you could try an easier case first, like a rectangle?)

 

[Nuke787]

Sure thing, my teacher does it like so.
Mass=density*volume
So 1000*3*y
Force=mg
so 1000*3*y*g
And Work is force times a distance
The distance is going to be 1-y
this gives us 1000*3*g*∫y*(1-y) with the integral being from 0 to 1
This yields us 3000g*((1/2)y²-(1/3)y³) evaluated at 0 to 1
When solved this yields 500g.
The only step that doesn't make 100% sense to me the more I look at it is where he finds the mass. I'm assuming he only uses y because the volumes he is integrating is so small? The 1000 and 3 are constants, I can definitely see where those are coming from, but just a single y has me a tad confused. To think I wouldn't be in the pickle I'm in if I had just done it this way , but what fun is that.

 

[tiny-tim]

The only step that doesn't make 100% sense to me the more I look at it is where he finds the mass. I'm assuming he only uses y because the volumes he is integrating is so small? The 1000 and 3 are constants, I can definitely see where those are coming from, but just a single y has me a tad confused. To think I wouldn't be in the pickle I'm in if I had just done it this way , but what fun is that.

hmm … you really haven't grasped the principle at all

(i don't see any slices, and i don't see any "dy")

the essential thing in these integrals is to divide the volume into slices …

the typical slice here is goes from height y to height y+dy, so its volume is (to first order) 300gydy …

carry on from there ​

 

[HallsofIvy]

Yes, that is the whole point of the integral. To find the area under a curve, or the volume of a figure, or the mass of an object, we imagine dividing it into strips or pieces so small we can approximate one or nmore of the parameters by a constant. Adding them all together approximates the entire area (or volume or mass) and taking the limit as the size of the pieces goes to 0 changes the "Riemann sum" into the integral and gives the exact value.

 

[Nuke787]

Why is the volume 3000gydy? I just don't see the units working. I see a F/m³ *m*m. (Sorry for not putting the dy's, I know they go there, I'm just used to people assuming that it was there. It was never stressed those were important to write, but I guess I can see where I should have.)

 

[tiny-tim]

Why is the volume 3000gydy?

If you think I made a mistake, what do you think the volume is?

 

[Nuke787]

I'm not saying you made a mistake, you have a method that you can support with calculus and justify what you have done, and it works for this problem and others. I feel like the volume should be 3 times the area of a trapezoid with height y, I'm not sure what your bases would be. That would give you your mass density*the area of a trapezoid*length which gives you units of kg. I'm guessing I'm making some fundamental error in my thought process.

 

[tiny-tim]

I'm not saying you made a mistake …

I did make a mistake!

So what do you think the volume is?

(You need to be able to do these things yourself)

 

[Nuke787]

I was assuming that we were overlooking that you typed the 300gydy lol the mass is 3000gydy. Assuming that the bottom of the triangle is y, the height would be y+dy, so that makes sense.Scratch the trapezoid thing I said earlier, needed to draw a picture which says that is obviously wrong...oops. The volume should be 3*(1/2)y²dy. This brings me back to what got me into the whole problem though. Volume is going to be your length of the tank*the area. That is what makes the units check out.

 

[tiny-tim]

3*(1/2)y²dy.

hmm …

let's take this slowly …

what are the length breadth and height of this slice?​

 

[Nuke787]

You have a length of 3 a height of y and a breadth of y. The slice I'm picturing is the shape of a triangle. I'm thinking of our tank and just drawing a line through it horizontally with everything underneath it being the slice.

 

[tiny-tim]

… a height of y …

No, the slice has a height of dy.

 

[Nuke787]

So the height and breadth are dy. Does that make sense? Since the height and base are equal and change at the same rate. Is this where I'm messing up attempting to integrate triangles.

 

[tiny-tim]

So the height and breadth are dy.

No.

This is a slice between heights y and y+dy.

 

[Nuke787]

Ahh, ok I see what we are doing I believe. I can clearly see the height. Why is the breadth y? The breadth isn't constant, the only thing constant as far as the shape is concerned is the length of 3. And thanks for all the help Tim, whether I get to a point where I understand or not I really appreciate all the help you are providing.

 

[tiny-tim]

A tank of water is 3m long and has a triangular cross section of height 1m and width at the top of 1m.

Ahh, ok I see what we are doing I believe. I can clearly see the height. Why is the breadth y? The breadth isn't constant, the only thing constant as far as the shape is concerned is the length of 3.

If the slice is from height y to y+dy, then it is rectangular …

its length is 3, its breadth is y, and its height is dy.

Yes, the breadth depends on which slice we are talking about, but for any particular slice it's fixed, at y.

 

[Nuke787]

Why do we take rectangular slices out of a triangular object? If we were to transpose a rectangular slice over the triangle wouldn't it protrude over the edges of the tank giving us a larger area than we were looking for?

Just adressed the rectangle issue with some friends since n approaches infinity the error that I mentioned above dissappears. That makes sense they also addressed the trapezoid thing that ran through my mind.

 

[tiny-tim]

(just got up :zzz: …)

Just adressed the rectangle issue with some friends since n approaches infinity the error that I mentioned above dissappears. That makes sense they also addressed the trapezoid thing that ran through my mind.

So is everything ok now?

But i need to comment on this …

Why do we take rectangular slices out of a triangular object?

… we don't take rectangular slices, we take horizontal slices …

we do so because we ultimately need to multiply by the distance to the top, and that will be the same over the whole of each horizontal slice …

if we needed the distance from a point (for example, to find the moment of inertia ), we would have to take spherical slices instead of horizontal ones …

for a triangular tank, that would be really awkward, but we would have to do it

 

[Nuke787]

I think so, and horizontal slices would be correct. On a side note, I just saw where someone did a problem pretty much exactly like this one and for their volume they calculated the width using similar triangles and didn't keep it fixed at some value. Is this an ok method to use? It makes more sense to me than keeping it fixed at a value.

 

[tiny-tim]

… I just saw where someone did a problem pretty much exactly like this one and for their volume they calculated the width using similar triangles and didn't keep it fixed at some value. Is this an ok method to use? It makes more sense to me than keeping it fixed at a value.

I don't understand …

can you write it out in maths? ​

 

[Nuke787]

Sure, I'll just use the same tank that we have been dealing with the entire time.
Slicing the triangle into horizontal pieces, and measuring distance from the top of the tank.
We know that volume is = l*w*h
The length is constant, it is 3. The height is going to be dx.
The width is what is changing as well, so it needs to be written in terms of x. If we take our triangle and draw a horizontal line through it, we are left with our initial triangle with a base of 1 and height of 1 and the triangle underneath the horizontal line. Using similar triangles we have $$\frac{1}{1}$$=$$\frac{1-x}{w}$$. If we solve for w we get w=1-x.
So for volume we have 3*(1-x)*dx.

 

[tiny-tim]

Slicing the triangle into horizontal pieces, and measuring distance from the top of the tank …

This is exactly the same, except that you're measuring x (or y) from the top of the tank instead of from the bottom …

can you see that the two integrals will be the same?​

 

[Nuke787]

Yea I can the problem actually makes sense now . Thanks for helping me understand it. Now to try and prove my original method, getting docked 20 points sucks. All I'm at is that method works for all shapes that are proportional and symmetrical. Thanks for all the help Tim.