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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)

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  1. triciaal
    • one year ago
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    work? Is this all of the question? we know the original volume and the new volume E = m c^2 ?

  2. anonymous
    • one year ago
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    I have W = ∫9800π (2-y)dy

  3. anonymous
    • one year ago
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    @triciaal yes, that is the full question. And I have no idea what you just said.

  4. Empty
    • one year ago
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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?

  5. anonymous
    • one year ago
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    @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

  6. triciaal
    • one year ago
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    now that you have all of the question it makes sense

  7. Empty
    • one year ago
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    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.

  8. Empty
    • one year ago
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    I say "might" because I haven't actually derived this myself yet

  9. anonymous
    • one year ago
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    the diagram is like this|dw:1434954369546:dw|

  10. anonymous
    • one year ago
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    sorry, it should be (2,0)

  11. anonymous
    • one year ago
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    I have a choice between from 0 to 1 and 1 to 2

  12. Empty
    • one year ago
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    Can you pump the water out of the bottom of the tank before pumping the water out of the top half?

  13. anonymous
    • one year ago
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    well, realistically, you can although nobody would do that

  14. Empty
    • one year ago
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    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|

  15. anonymous
    • one year ago
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    yeah, from 1 to 2 makes more sense. But if you pump from the bottom first, wouldn't the equation change?

  16. dan815
    • one year ago
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    yes gravity

  17. anonymous
    • one year ago
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    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

  18. dan815
    • one year ago
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    |dw:1434955098800:dw|

  19. dan815
    • one year ago
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    |dw:1434955245009:dw|

  20. anonymous
    • one year ago
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    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

  21. dan815
    • one year ago
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    very similiar

  22. anonymous
    • one year ago
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    @dan815 I don't think it's going to be similar

  23. dan815
    • one year ago
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    it is, u have to think how is your force changing as u pull it up

  24. Empty
    • one year ago
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    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.

  25. anonymous
    • one year ago
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    @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.

  26. anonymous
    • one year ago
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    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

  27. anonymous
    • one year ago
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    @Empty do you see the problem i'm trying to convey?

  28. Empty
    • one year ago
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    Yes I see the problem, but it is not fundamentally different.

  29. anonymous
    • one year ago
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    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?

  30. Empty
    • one year ago
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    Yeah I'll work out a nice example step by step. Would you prefer I do the water or cable problem for my example?

  31. anonymous
    • one year ago
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    cable

  32. dan815
    • one year ago
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    |dw:1434956362609:dw|

  33. Empty
    • one year ago
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    looks like a guy peeing off a roof dan

  34. dan815
    • one year ago
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    hAHAHahaha

  35. anonymous
    • one year ago
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    @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

  36. Empty
    • one year ago
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    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.

  37. anonymous
    • one year ago
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    :o i've been working on the problem whole time?

  38. anonymous
    • one year ago
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    ok, just post the answer when you're done. I'll work on my end to see if our answer match

  39. Empty
    • one year ago
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    Well I'm helping someone with a double integral right now too so I'm juggling back and forth lol.

  40. anonymous
    • one year ago
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    ok lol

  41. anonymous
    • one year ago
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    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

  42. Empty
    • one year ago
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    g=32 ft/s^2 is that right?

  43. anonymous
    • one year ago
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    right

  44. Empty
    • one year ago
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    I'm getting 1200 lb-ft so I'm trying to see if I messed up or not

  45. anonymous
    • one year ago
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    I got 800 lb-ft for the lower half.

  46. Empty
    • one year ago
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    Same

  47. anonymous
    • one year ago
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    I got ∫ 32y dy, from 0 to 5, for the upper half

  48. Empty
    • one year ago
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    That's what I have as well.

  49. anonymous
    • one year ago
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    oops! I messed up. It should be 400 for the upper half lol

  50. Empty
    • one year ago
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    Ahh ok 800+400=1200 :)

  51. anonymous
    • one year ago
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    Now do you see why the conventional cutting method wouldn't work in this case?

  52. Empty
    • one year ago
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    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

  53. anonymous
    • one year ago
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    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

  54. Empty
    • one year ago
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    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.

  55. Empty
    • one year ago
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    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.

  56. anonymous
    • one year ago
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    awesome! thank you for your time :)

  57. Empty
    • one year ago
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    Yeah definitely glad I could help =)

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