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Acceleration of the container

  1. Feb 6, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A light container is kept on a horizontal rough surface of coefficient of friction μ=Sh/V. A very small hole of area S is made at depth 'h'. Water of volume 'V' is filled in the container. The friction is not sufficient to keep the container at rest. The acceleration of the container initially is?


    3. The attempt at a solution
    Velocity of efflux = √2gh

    Force on container = rate of change of momentum.
    Let density of water be ρ. Total mass = ρV.
    After time Δt, change in mass of container=ρS(√2gh)Δt.
    Maximum frictional force = μmg, where m is mass of container at any time.

    Edit:
    [itex]a = \mu m(t) g - \dfrac{\delta m}{m_i - \delta m}[/itex]
    Substituting the values does not give correct answer.
     
    Last edited: Feb 6, 2014
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  3. Feb 6, 2014 #2

    BvU

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    Work things out using symbols for what you don't know. It may well be that they cancel by the time you have an expression for the inital a. A "light" container is probably a container that weighs a lot less than its contents, so if that mass does not cancel out, ignore it.
     
  4. Feb 6, 2014 #3

    utkarshakash

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    I have edited my original post. Please see it.
     
  5. Feb 6, 2014 #4

    BvU

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    Your editing confuses. What happened to the momentum balance you mentioned?
    In the "a = .." expression I see one term with the dimension of force and one term with no dimension at all. What is this ?
     
  6. Feb 6, 2014 #5

    utkarshakash

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    Sorry. I think that my entire expression for a is wrong.
    The net force on the container should be equal to rate of change of momentum.
    Momentum = [itex]m(t) v(t) \\ (m_i - \delta m) v(t) [/itex]

    where [itex]\delta m = ρS(√2gh)Δt [/itex].

    Now If I differentiate it wrt t,

    [itex]m(t) dv/dt + v(t) dm/dt \\ (m_i - \delta m) a + v(t) (\rho V - \rho S \sqrt{2gh} ) [/itex]

    But frictional force also acts. Thus
    [itex] m(t) a = \mu m(t) g - (m_i - \delta m) a + v(t) (\rho V - \rho S \sqrt{2gh} )[/itex]

    Do you think I'm going right or is there anything that needs to be corrected?
     
  7. Feb 6, 2014 #6

    BvU

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    Mind you, they ask for the initial acceleration. Perhaps a force balance is already enough. Just a hunch. You are concentrating on the container momentum. Anything else in the scene that carries momentum ?
     
  8. Feb 6, 2014 #7

    BvU

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    Something else:
    My suspicion is that S1 ##\ne## S2 but h1 = h2 and V1 = V2. Wrong ? S1 = S2 is also a resonable possibility: they want to make it easy for us.
     
  9. Feb 6, 2014 #8

    utkarshakash

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    The water which is coming out. You said about force balance. If force is balanced, how is the container accelerating?
     
  10. Feb 6, 2014 #9

    BvU

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    One goes one way, the other goes the other way. Total momentum conserved.
    Compare with accelerating rockets in deep space: shoot away some exhaust gas and you accelerate in the other direction.
     
  11. Feb 6, 2014 #10

    utkarshakash

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    But frictional force also acts on the system. So how is the momentum conserved? Also If I assume that it is conserved, how will it give me initial acceleration?
     
  12. Feb 7, 2014 #11

    BvU

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    That's the nice thing about his exercise: they ask for the initial acceleration. Situation ##t < 0##: friction is zero. Situation ##t \ge 0##: max friction given by ##\mu F_N = \mu S_1h/V * \rho V \ g= \mu g h S_1##

    Now the other side. Chap who makes the hole gets blown away by a big fat water jet (unless he stands aside). Fat enough to move the container, they say. I had big fun looking at the Jetlev videos. You know speed and mass (rate, that is!), so momentum there is known.
    (subtle: from the OP I have to guess that speed is ##\sqrt{2gh}##, not ##\sqrt 2 gh##...)

    Now assume the container accelerates; write down the momentum balance. Make up your mind about ## S_1 = S_2## or ## S_1 \ne S_2##. First case is a piece of cake, second leaves you with a condition on ## S_2/ S_1## (because a can't be <0). I find the latter more realistic, but who cares.
     
    Last edited: Feb 7, 2014
  13. Feb 7, 2014 #12

    utkarshakash

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    Sorry for this but I'm having a hard time understanding your post. You are saying about momentum balance. Now let's say the container gains a speed v. The water coming out of the hole has a speed [itex]\sqrt{2gh}[/itex]. But what about the masses? We are talking about the situation at t=0. So water just starts flowing out and no appreciable change in mass of container is observed. Really confused :confused:
     
  14. Feb 7, 2014 #13

    BvU

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    What is the momentum of the water coming out of the hole? If you can't tell it in kg m/s, then express it in kg m/s per second.....
     
  15. Feb 7, 2014 #14

    utkarshakash

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    [itex]m\sqrt{2gh}[/itex]

    But my question is what do you substitute for mass? Since the situation is at time t=0 no water has come out of the hole yet. If, suppose, Δt time has passed, then mass of the water flown out the hole is [itex]\rho S_1 Δt[/itex].
     
    Last edited: Feb 7, 2014
  16. Feb 8, 2014 #15

    haruspex

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    Quite so. The force is the rate of change of momentum. How much momentum is added to the jet of water in time Δt?
     
  17. Feb 8, 2014 #16

    utkarshakash

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    [itex] \rho S_1 Δt \sqrt{2gh} [/itex]
     
  18. Feb 8, 2014 #17

    BvU

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    Hold it. In time ##\Delta t## the mass of the water coming out can not be [itex]\rho S_1 Δt[/itex] because that does not have the dimension of mass. Nor can [itex] \rho S_1 Δt \sqrt{2gh} [/itex] represent a momentum.

    What is missing ? .... Right ! Fill it in and you have ##\Delta p## (##p## = momentum). Divide by ##\Delta t##, and you have ##F={dp\over dt}##. Don't worry about the acceleration phase of the water coming out: it's so fast that it can be ignored.
     
  19. Feb 8, 2014 #18

    utkarshakash

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    Sorry. The mass should be [itex]\rho S_1 \sqrt{2gh} Δ t [/itex].
    The momentum should be [itex]\rho S_1 (2gh) Δ t [/itex]

    [itex]F = \rho S_1 (2gh) [/itex]
     
  20. Feb 8, 2014 #19

    BvU

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    So you're done. Right?
     
  21. Feb 9, 2014 #20

    utkarshakash

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    No. If I write F=ma then
    [itex] a = \dfrac{\sqrt{2gh}}{Δ t} [/itex]

    But this is not the answer.
     
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