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anonymous
 one year ago
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)
anonymous
 one year ago
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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triciaal
 one year ago
Best ResponseYou've already chosen the best response.0work? Is this all of the question? we know the original volume and the new volume E = m c^2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have W = ∫9800π (2y)dy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal yes, that is the full question. And I have no idea what you just said.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Can 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0now that you have all of the question it makes sense

Empty
 one year ago
Best ResponseYou've already chosen the best response.3The 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I say "might" because I haven't actually derived this myself yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the diagram is like thisdw:1434954369546:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, it should be (2,0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have a choice between from 0 to 1 and 1 to 2

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Can you pump the water out of the bottom of the tank before pumping the water out of the top half?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, realistically, you can although nobody would do that

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Well 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, from 1 to 2 makes more sense. But if you pump from the bottom first, wouldn't the equation change?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 I don't think it's going to be similar

dan815
 one year ago
Best ResponseYou've already chosen the best response.3it is, u have to think how is your force changing as u pull it up

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the math only works if we lift the entire cable to the top of the building. But here we only lift half of its length

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty do you see the problem i'm trying to convey?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yes I see the problem, but it is not fundamentally different.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It'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?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah I'll work out a nice example step by step. Would you prefer I do the water or cable problem for my example?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3looks like a guy peeing off a roof dan

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:o i've been working on the problem whole time?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, just post the answer when you're done. I'll work on my end to see if our answer match

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Well I'm helping someone with a double integral right now too so I'm juggling back and forth lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 960 lbft. reminder: 10ft cable with linear density = 1lb/ft. Work need to lift half of its length to the top of building

Empty
 one year ago
Best ResponseYou've already chosen the best response.3g=32 ft/s^2 is that right?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I'm getting 1200 lbft so I'm trying to see if I messed up or not

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 800 lbft for the lower half.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got ∫ 32y dy, from 0 to 5, for the upper half

Empty
 one year ago
Best ResponseYou've already chosen the best response.3That's what I have as well.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops! I messed up. It should be 400 for the upper half lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now do you see why the conventional cutting method wouldn't work in this case?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3So 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 $$W=1200 ftlb$$ as advertised! =P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3A 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 usubstitution 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 (1y) or whatever I sometimes see in these problems.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome! thank you for your time :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Yeah definitely glad I could help =)
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