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Moving water up an incline

  1. Apr 2, 2010 #1
    I would like to know if its possible to move water from a filled rectangular tank to another tank sliding down an incline using force generated due to the weight of the empty tank rolling down the slope.

    Attached is a figure depicting the scenario.

    I would like to know what should be the weight of the upper tank which is sliding down the incline and what should be the angle of the incline to displace 100 Lires of water from the stagnant tank below into the tank sliding down?
     

    Attached Files:

    • fig.JPG
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  2. jcsd
  3. Apr 2, 2010 #2

    mgb_phys

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    Presumably >100kg (ignoring friction and the mass of water in the pipe etc)
     
  4. Apr 2, 2010 #3
    Hi spalose,

    Welcome to PF...

    In answer to your first question; yes, it is possible to move water from a filled rectangular tank to another tank sliding (rolling?) down an incline using force generated due to the weight of the empty tank rolling down the slope. However, I'm not sure that the arrangement in your sketch would accomplish such a transfer...

    The answer to your second and third questions (weight of the upper tank and the angle of the incline) can not be answered with the data you provide...

    Before going further with this thread, please acknowledge that you are aware that the energy expended in lifting the wheeled tank to its start position on the ramp will be exactly equal to the total energy delivered to the system when the wheeled tank rolls down the incline. The "total energy delivered" includes the energy consumed by the flowing friction of the water and the rolling friction of the wheeled tank. This energy will be unrecoverable. Only the slight increase in the water's gravitational potential energy would remain when the wheeled tank ends its roll.

    .
     
  5. Apr 2, 2010 #4

    Dale

    Staff: Mentor

    Energy must be conserved so, let's say that you wanted to pump 100 kg water up a distance of 10 m when the moving tank drops a distance of 1 m. The increase in your water's GPE is mgh = 100 kg 10 m/s² 10 m = 10000 J, so the decrease in your tank's GPE must be at least as great 10000 J < mgh = m 10 m/s² 1 m, solve for m gives m > 1000 kg.
     
  6. Apr 2, 2010 #5

    Yes it will be rolling. The moving plates which are attached to each other acts like a floating piston. Actually the plates will always be stationary, the tank/container will move to displace water. I agree, the fig is wrong, the plates should be on the opposite ends from where they are right now.

    Yes there will be energy spent to take the empty tank up the slant. But having said that, if the weight of the tank is 1 ton and the water to be displaced is 100 KG then i think we will have enough kinetic energy to displace the entire water, cancel the friction components too. Am I right in my thinking?

    I have another thought, as the tank is rolling down and is pulling up the water, its weight is also increasing with the distance it rolls down. Shouldn't it have a compounding effect?

    If the incline is 60 degrees, the weight of the container/tank is 1 ton, weight of water is 100kg and diameter of the pipe is 10 inch, then what would be the distance traveled by the rolling container down the slant?
     
    Last edited: Apr 2, 2010
  7. Apr 2, 2010 #6

    LURCH

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    What you have here is a possitive displacement system. Therefore, the distance travelled will be the same regardless of the weight (so long as all the water gets moved). Assuming 1tn is indeed sufficient, then 2tns would not move the tank any farther. Given that information, do you know what the extra weight would change (not being patronizing; just seeing if I've communicated clearly)?
     
  8. Apr 2, 2010 #7


    Completely agree. It wont move more than the length of the container itself. Thanks for this insight.
     
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