anyone good with calc 2 work problem? A cylinder tank with radius 1 m and height 2 m is full of water. How much work is needed to pump half of the amount of water out of the tank? (density of water is 1000kg/m^3)

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anyone good with calc 2 work problem? A cylinder tank with radius 1 m and height 2 m is full of water. How much work is needed to pump half of the amount of water out of the tank? (density of water is 1000kg/m^3)

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work? Is this all of the question? we know the original volume and the new volume E = m c^2 ?
I have W = ∫9800π (2-y)dy
@triciaal yes, that is the full question. And I have no idea what you just said.

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Can you explain how you got your integral @sourwing ? That looks about right though, since as you pump water out the height you'll have to pump increases so you have to do more work on the remaining water than you did to begin with right?
@Empty yes, that's right. The work exerted on the second half is greater than that of the first half. I'm only stuck on the limit of integration though
now that you have all of the question it makes sense
The limits of integration might go from 2 to 1 to represent the fact that you're pumping water from the top to halfway down since these are the values of y in which you are pumping an infinitesimal disk of water from with height dy.
I say "might" because I haven't actually derived this myself yet
the diagram is like this|dw:1434954369546:dw|
sorry, it should be (2,0)
I have a choice between from 0 to 1 and 1 to 2
Can you pump the water out of the bottom of the tank before pumping the water out of the top half?
well, realistically, you can although nobody would do that
Well what I am saying is that water empties from the top of the bucket instead of from the bottom like this, which isn't realistic according to what we're doing: |dw:1434954659687:dw|
yeah, from 1 to 2 makes more sense. But if you pump from the bottom first, wouldn't the equation change?
yes gravity
As a matter of fact, I already knew the answer but this made me think how would you do if this was a cable/chain problem. Obviously, you can not just cut the cable into pieces (although the math tells you to do so) and move those pieces up
|dw:1434955098800:dw|
|dw:1434955245009:dw|
what if the problem now is: A 10ft chain is hanged from a building (the building is higher than the cable btw). How much work is needed to lift half of its length to the top. Let just say the its linear density is 1lf/ft
very similiar
@dan815 I don't think it's going to be similar
it is, u have to think how is your force changing as u pull it up
Yeah imagine pulling up only one individual infinitesimal segment of the cable up the building. This will be a work integral. Now add up all the different pieces of the cable with another integral to get the total work. The only difference is we called it cable instead of water.
@Empty well, in the water case, those segment of upper volumes can move up while those at the bottom can stay still. But in cable case, they all have to move at the same time.
the math only works if we lift the entire cable to the top of the building. But here we only lift half of its length
@Empty do you see the problem i'm trying to convey?
Yes I see the problem, but it is not fundamentally different.
It's indeed not completely different from the case of water, but I still have no idea how to approach this problem. Do you have any ideas?
Yeah I'll work out a nice example step by step. Would you prefer I do the water or cable problem for my example?
cable
|dw:1434956362609:dw|
looks like a guy peeing off a roof dan
hAHAHahaha
@Empty I believe I've found a away to approach this. See if this makes sense. Obviously, the lower half of the cable has to move 5 feet. As for the upper half, we can just use usual the "cutting" approaches
Yeah that's exactly what I'm working on doing, just sorta polishing off this example so it's clear and simple to follow before I post it.
:o i've been working on the problem whole time?
ok, just post the answer when you're done. I'll work on my end to see if our answer match
Well I'm helping someone with a double integral right now too so I'm juggling back and forth lol.
ok lol
I got 960 lb-ft. reminder: 10ft cable with linear density = 1lb/ft. Work need to lift half of its length to the top of building
g=32 ft/s^2 is that right?
right
I'm getting 1200 lb-ft so I'm trying to see if I messed up or not
I got 800 lb-ft for the lower half.
Same
I got ∫ 32y dy, from 0 to 5, for the upper half
That's what I have as well.
oops! I messed up. It should be 400 for the upper half lol
Ahh ok 800+400=1200 :)
Now do you see why the conventional cutting method wouldn't work in this case?
So here's my work, because this is very straightforward I believe (ask me to clarify any steps!): The work of an infinitesimal piece of cable is: $$dW=sdF=sgdm=sg\sigma dx$$ where $$\frac{dm}{dx}=\sigma$$ and s = distance rope was pulled So now we can ad up the contributions from all the pieces by integrating to get: $$W=\int_0^{10}sg\sigma dx$$ we know s(x)=x from x=0 to x=5 and s(x)=5 from x=5 to x=10 so we break the integral up: $$W=\int_0^{5}xg\sigma dx+\int_5^{10}5g\sigma dx$$ Plug in more stuff and we get $$W=1200 ft-lb$$ as advertised! =P
I think this method can be used in the water case but with water being pumped from the bottom in the way a syringe pumps liquid out of the needle
I can do the water method pretty quickly here too: So here's my work, because this is very straightforward I believe (ask me to clarify any steps!): The work of an infinitesimal layer of water is: \[dW=sdF=sgdm\] where the density \[\frac{dm}{dV}=\rho\] Infinitesimal volume is related to the infinitesimal circular area: \[dV=Adh=\pi r^2 dh\] So similar to the cable we can take the dm from the force and write it in terms of height: \[dm=\rho\pi r^2 dh\] \[dW=sg \rho\pi r^2 dh\] So now we can add up the contributions from all the layers of water by integrating to get: \[W=\int_0^{1}sg \rho\pi r^2 dh\] In this particular instance I'm taking the perspective that from the top of the water is y=0 and 1 meter deep is +1. I can choose any coordinate system I want, it's my right as a mathematician to choose what's convenient for me since coordinate systems aren't real! we know s(h)=h, so we don't have the problem like we did with the cable, so it plugs in nicely. \[W=\int_0^{1}g \rho\pi r^2 hdh\] \[W=g\rho\pi r^2 \frac{1}{2}h^2|_0^1 \] plugging in g=9.8, r=1, rho=1000 we get: W=4900 J So you see I worked this out by just replacing values in the old problem, not really doing much different at all.
A note should be made that I could have used such a coordinate system where the integration could have run from y=1 to y=2 or something like this. This is literally what you might call a u-substitution away and how I happened to come up with my problem in a way that made sense to me. So saying "Oh does this integral go from 0 to 1 or 1 to 2" is completely arbitrary, it depends entirely upon how you have placed your physical object in your mental coordinate system. I hope this serves to clear up the debate of (1-y) or whatever I sometimes see in these problems.
awesome! thank you for your time :)
Yeah definitely glad I could help =)

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