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Inverted garbage can

  1. May 30, 2017 #1
    1. The problem statement, all variables and given/known data
    An inverted garbage can of weight ##W## is suspended in air by water from a geyser. The water shoots up the ground with speed ##v_0##, at a constant rate ##dm/dt##. The problem is to find the maximum height at which garbage can rides.

    2. Relevant equations



    3. The attempt at a solution

    Suppose the garbage can is at its highest position ##h##.

    Then the velocity of elementary mass ##\Delta m## just before collision with the can would be ##\sqrt{v_0^2 - 2gh}## and velocity after collision would be zero.

    So change in momentum would be ##\Delta P = - \Delta m \sqrt{v_0^2 - 2gh}##

    Or the force on the can would be ##F = \sqrt{v_0^2 - 2gh} \dfrac{dm}{dt}##

    Since the forces on the can is balanced, therefore ##W = \sqrt{v_0^2 - 2gh} \dfrac{dm}{dt}##,

    Solving for ##h## I got ##\displaystyle h = \dfrac{1}{2g}\left(v_0^2 - \left(W \over \dfrac{dm}{dt}\right)^2 \right)##

    This is incorrect.

    Where am I incorrect ?
     
  2. jcsd
  3. May 31, 2017 #2

    haruspex

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    It all looks right to me. Do you know what the answer is supposed to be?
     
  4. May 31, 2017 #3

    Nidum

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    So where does the water go ?
     
  5. May 31, 2017 #4

    haruspex

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    It falls. I do not see any basis for claiming that the water bounces off the bucket, or in any other way makes a greater contribution to supporting it. The bucket could be significantly wider than the jet.
     
  6. May 31, 2017 #5
    A clue was given; if ##v_0 = 20m/s##, ##W = 10 kg##, ##dm/dt = 0.5 kg/s## then ##h = 17 m##.

    These values does not match when I put it into my formula :(.

    @Nidum It falls down under gravity.
     
  7. May 31, 2017 #6

    haruspex

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    I assume you mean W=10N.
    That answer is clearly wrong. 20m/s is the speed at which the water needs to be moving when it reaches the bucket.
    If you use g=9.8m/s2 and v0=27m/s it works out quite accurately to 17m.
     
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