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Could a person do this? -- Push a cruise ship by hand

  1. Oct 23, 2014 #1
    I was watching a program on the worlds largest cruise ship, the Oasis of the Seas. It displaces 243000 tons. They we're discussing the sophisticated maneuvering systems and I got to thinking.

    Let's say the ship is docked, within a foot of the dock and untethered. If I sat down with my legs dangling over the side of the dock, put my feet against the hull and pushed against the hull bracing my back against the pier, and pushed as hard as I could, would I be able to move the ship away from the dock?

    How much force would it take and would a human be able to do that? It might move slowly but would I be able to move it noticeably?

    How would you go about figuring that out?

    Tex
     
    Last edited: Oct 23, 2014
  2. jcsd
  3. Oct 23, 2014 #2
    I have no idea. My first thought is figure out what acceleration you define as perceptible and then apply newton's law to see what net force you would need. Then maybe figure out what the range of forces from the water and air pushing on the ship in various amounts are. If the force from your legs would just get lost in that value as "noise" then you would not be able to do it. If the force from your legs can rise above the typical forces, then you could do it.
     
  4. Oct 23, 2014 #3

    Nugatory

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    Newton's ##F=ma## will do the trick. You know the mass ##m##, you can come up with a pretty good estimate of the force ##F##, calculate the resulting acceleration.

    This is ignoring the effects of friction with the water - include these and the ship will act as if it's solidly aground instead of floating, no more likely to move than the average mountain.
     
  5. Oct 24, 2014 #4

    A.T.

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    That would be true for static friction. Can water exert something similar, that prevents any acceleration below a certain force?
     
  6. Oct 24, 2014 #5

    Borek

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    Let's separate theoretical answer from the practical one.

    In practice pretty small currents or winds (ones that will require sensitive sensors to be detected) will create forces that you are not able to overcome, just because they work on a huge surfaces.
     
  7. Oct 24, 2014 #6

    sophiecentaur

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    It also ignores the fact that a large mass of water needs to be moved as it sloshes round from one side of the ship to another. Its effective mass needs to be included in the "ma" thing. Even a light paddle would take quite a lot of pushing if it were as big as a ship's cross section.

    That said, it is possible to shift some pretty hefty barges on canals with the available power of just one person.
     
  8. Oct 24, 2014 #7

    NTW

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    In the absence of any perturbation, the acceleration would be F/m, where m will be just the mass of the ship. F will be the hydrodynamic resistance, that is a function of the cross-section of the ship perpendicular to its motion, the velocity squared, the density of the water, and some coefficient of hydrodynamic drag...
     
  9. Oct 24, 2014 #8
    So...help me out here please. If I pushed with 50 lbs of force against the hull, and assuming that water resistance or wind or waves don't play a factor, what distance of movement could I expect in 1 hour of pushing? Let's assume that floating in the water provides an almost frictionless surface.
     
  10. Oct 24, 2014 #9

    Bandersnatch

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    Just use the third equation:
    http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1
     
  11. Oct 24, 2014 #10

    NTW

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    It's an impossible situation, but that's easy to calculate. 50 lbs = 222 N. If the ship has a mass of 243000 metric tons = 2,43 * 108 kg

    a = 222/2,43 * 108 = 9,14 * 10-7 m/s2

    e = ½ * a * t2 = 0,5 * 9,14 * 10-7 * 36002 = 0,00165 m = 1,65 mm

    Sounds plausible... And that, with zero water resistance...
     
  12. Oct 24, 2014 #11

    sophiecentaur

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    I would agree that is possibly the case if the ship is out in the open water but, against the quay, the situation is different, surely. Water must be moved from the open sea to the gap in-between the hull and the wall. It has to be accelerated. (and a lot more than the acceleration of the ship am). I don't see how the acceleration of that water is less important than the acceleration of the boat itself. During the operation, there will be a difference in hydrostatic pressures on with sides of the ship because the level on the quayside will be lower. I realise we are trying to get as 'ideal' as possible but can you really ignore my point?
     
  13. Oct 24, 2014 #12

    NTW

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    That's always the case of an object moving in a fluid A force of drag appears, and opposes the movement. I have mentioned the equation used for drag in aerodynamics, but I feel pretty sure that it's generally valid for water also.

    I agree that the 'vacuum' formed between the hull and the quay may add some additional resistance, but not all quays are walls that reach the bottom. There are many pole-supported quays, and perhaps even cantilevered ones...
     
  14. Oct 24, 2014 #13

    A.T.

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    What do you mean by "In the absence of any perturbation"? Without the push by the human? Your statement only makes sense if hydrodynamic resistance is the only force acting.
     
  15. Oct 24, 2014 #14

    Borek

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    I think he refers to things I have mentioned - that is, no current and no wind (impossible in practice).
     
  16. Oct 24, 2014 #15

    NTW

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    Of course...
     
  17. Oct 24, 2014 #16

    A.T.

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    Then your statement was wrong. If the man still pushes, then hydrodynamic resistance is not the only force on the ship, and therefore it doesn't determine its acceleration in the way you claimed in post #7.
     
  18. Oct 24, 2014 #17

    mfb

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    We can safely use this as upper limit for a central push (it is a bit more if we push at one end). The additional amount of water moved and its velocity will depend on many factors, including the geometry of the walls and ground around the ship.
     
  19. Oct 24, 2014 #18

    NTW

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    There is a net force that produces the acceleration. The ship will only move if the push by the man is greater than the hydrodynamic resistance. But that hydrodynamic resistance is zero (because the ship's velocity is also zero) when the man starts pushing, As the speed builds up, the hydrodynamic resistance will grow, and if the man wishes to keep the acceleration constant, he will have to push harder and harder...

    In the case of zero hydrodynamic resistance, contemplated in post #9, things are quite different, and even more impossible...
     
  20. Oct 24, 2014 #19

    SteamKing

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    You must be careful here. Cruise vessels are advertised on the basis of their Gross or Net Tonnage, which is not the same as displacement tonnage.

    For example, the Oasis of the Seas has a Gross Tonnage of 225,282, but an actual displacement tonnage of about 100,000 metric tons. Gross tonnage is a measure of the internal volume of the vessel, and not necessarily its weight (or displacement).

    http://en.wikipedia.org/wiki/MS_Allure_of_the_Seas
     
  21. Oct 24, 2014 #20

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    I think it can't be even an upper limit when there exists hydrodynamic resistance.

    That distance of 1,65mm (one hour of pushing with a force of 50 lbs) was the result for a very unlikely case, namely, that the hydrodynamic resistance was zero... A weird condition stipulated by thetexan in post #8...
     
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