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Work done lifting an object underwater

  1. Sep 11, 2008 #1

    I have a question regarding work done lifting an object vertically upwards, under water.
    I am aware that work is done against hydrostatic pressure (which varies depending on a depth h from the surface), and that density of the fluid and the object may have a role in the calculation of the work done in lifting the object vertically upwards to the surface, from depth h.

    my question is, how is the work done generally calculated in this circumstance?
    Last edited: Sep 11, 2008
  2. jcsd
  3. Sep 11, 2008 #2


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    I would start ignoring everything but Archimedes' principle.
  4. Sep 11, 2008 #3


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    Work is generally defined as force times distance.

    The depth is the distance in this case (or however far you lift the object) and the force is the apparent weight of the submerged object (neglecting drag).

  5. Sep 12, 2008 #4

    Andy Resnick

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    That's a good question, and requires an involved answer.

    First, let's ignore viscosity- water has viscosity, which acts to oppose
    motion. We'll consider viscosity later.

    Consider a neutrally buoyant object within an inviscid liquid- it may
    have the same density as the fluid. Because it is neutrally buoyant,
    there is no gravitational acceleration, and no work is required to move
    the object.

    This may seem counterintuitive, but that's only because I started by
    ignoring viscosity and most people don't have a reference. Here's one-
    in orbit, astronauts can move about very easily: they are neutrally
    buoyant objects moving within an inviscid medium.

    So the total amount of work required to lift a nearly-neutrally buoyant
    object through a viscous medium has two terms, the gravitational term W
    = [itex]\Delta\rho V g \Delta h[/itex] and a drag force term, the force
    required to overcome viscous effects. This has to be written like W
    =[itex]\int f dl [/itex], because the viscous drag will depend on the
    path taken- straight up, zig-zag, whatever. The drag force f can be
    written simply as[itex] \mathbf{F}_d= {1 \over 2} \rho \mathbf{v}^2 C_d
    A [/itex] One simplification is to tow the body at constant speed, then you are left with a simple multiplication rather than an integration.

    To be honest, I didn't write the gravitational part correctly, I did not
    properly account for the spatial extent of the body, instead simply
    writing [itex]\Delta h[/itex].

    Edit: I have a tough time submitting this post... let me know if I left something unclear in the midst of my frustration.....
  6. Sep 12, 2008 #5


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    This is somewhat misleading, even though it is just an analogy. Just for clarification, astronauts are not neutrally buoyant they are weightless due to the absence of gravity. There is not really an inviscid (or viscid) medium in a vacuum – there’s nothing. Hence there is no buoyant force acting on them. They float due to the lack of a gravitational force acting on them. For an object to be neutrally buoyant, it implies that it is in a medium that exerts a buoyant force on it. Since there is no such medium in space (vacuum) they are not considered neutrally buoyant IMO.

    The gravitational term (i.e. the force required to lift the object, neglecting drag) is equal to the apparent weight of the object: [tex] W_{app} = mg - F_b = mg - \rho_{fluid}gV_{obj} [/tex]. If it is neutrally buoyant, then it is a special case where the apparent weight is 0. Hence, no work is done when lifting the submerged object (neglecting drag). If the object is not neutrally buoyant, the work is found by multiplying this (the apparent weight) times the height the object is lifted (again ignoring drag).

    If the object is lifted straight up, the drag force can just be summed with the apparent weight for a net force. If the distance the drag force is applied is different from the elevation change (i.e. it zigzags), the net work could be found by superposition.

  7. Sep 12, 2008 #6

    Andy Resnick

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    Yes and no- astronauts do move under the influence of gravity: they orbit. Astronauts within the space station move in an (approximately) inviscid medium- air. The difference is that instead of astronauts being isopycnic, the astronauts + space station + air all move together in free fall. But the physics is the same if I have a drop of fluid floating in air on the space station, or a drop of neutrally buoyant fluid in a Plateau tank- the fluid boundary is the same, etc.

    Right. My point was that the drag force is an essential component of the calculation, and depends of the path taken-it takes work to move an object around underwater, even if it follows a dh = 0 path. The work then is dissipated via viscosity.
  8. Sep 12, 2008 #7


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    I suppose I should have clarified what I meant by absence of gravity. If they are somewhere in space (orbiting or not) where the gravitational force and their inertial forces are equal they would experience zero gravity (weightlessness), or in other words, an absence of gravity. It is true, strictly speaking that gravity is still exerted on them, it just appears that it isn't since their inertial force is equal to the gravitational force (i.e. resultant force is 0).

    Of course if there is a medium (in this case air in a space station) there will be a buoyant force, but you omitted that condition from your original post.

  9. Sep 12, 2008 #8

    Andy Resnick

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    Yeah, I left that detail (and probably a few others) out of my original response. The central point to realize here is that even with air present, there is no buoyancy force if everything is in free fall. That is one major advantage of doing science on orbit- it's possible to use materials that normally sediment/stratify, and they don't on orbit.

    To be sure, there is a residual acceleration due to not being at the precise center of mass of the space station- and the Shuttle can fly in different orientations (gravity gradient, for example) to control this 2nd-order effect.
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