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

[thetexan]

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

 

[ModusPwnd]

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.

 

[Nugatory]

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.

 

[A.T.]

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.

That would be true for static friction. Can water exert something similar, that prevents any acceleration below a certain force?

 

[Borek]

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.

 

[sophiecentaur]

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.

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.

 

[NTW]

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.

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...

 

[thetexan]

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.

 

[Bandersnatch]

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.

Just use the third equation:
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1

 

[NTW]

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.

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...

 

[sophiecentaur]

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...

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 realize we are trying to get as 'ideal' as possible but can you really ignore my point?

 

[NTW]

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 realize we are trying to get as 'ideal' as possible but can you really ignore my point?

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...

 

[A.T.]

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

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.

 

[Borek]

What do you mean by "In the absence of any perturbation"?

I think he refers to things I have mentioned - that is, no current and no wind (impossible in practice).

 

[NTW]

I think he refers to things I have mentioned - that is, no current and no wind (impossible in practice).

Of course...

 

[A.T.]

Of course...

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.

 

[mfb]

1,65 mm

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.

 

[NTW]

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.

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...

 

[SteamKing]

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.

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

 

[NTW]

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.

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...

 

[mfb]

How can you get faster motion if you increase resistance?

 

[willem2]

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

This is wrong. You multiplied with 36002 and not with (3600)^2. The real answer is 5.92 metres.

 

[thetexan]

5.92 meters in one hour is substantial! I would not have guessed it would be that much. That's about 10 centimeters per minute or one centimeter every 6 seconds. That would be an observable movement.

 

[NTW]

This is wrong. You multiplied with 36002 and not with (3600)^2. The real answer is 5.92 metres.

Right. It was a mistake with my calculator...

 

[mfb]

5.92 meters is also more than the typical length of a human. Okay, so a human can move the ship - in the absence of wind and water currents.

 

[NTW]

5.92 meters is also more than the typical length of a human. Okay, so a human can move the ship - in the absence of wind and water currents.

And of hydrodynamic resistance, nothing less. Just like if the ship glided on a frictionless surface...

 

[Simon Wilson]

There is no threshold force that needs to be overcome to move a ship in the absence of wind or current. In fact it is remarkably easy for an unassisted person to move a large ship. This can be explained in terms of kinetic energy (E) and momentum.
Consider a ship with a mass of 20,000 tons. If the ship is given a velocity of 0.5 inch per second then its energy, E = 1/2 mv2 is about 1,000 joules. A thousand joules is a very modest amount of energy. It is the energy expended by a 120-pound man climbing up a 6-foot-high flight of stairs.
At 0.5 inch per second the ship’s momentum, mass x velocity = 2 x 105 Newton-seconds.
The 120-pound man can impart this to the ship by applying his full weight for 400 seconds. If he moves the ship by standing with full weight on one of the mooring lines, he will have descended by 6 feet by the time the ship is moving at 0.5 inches per second.
Actually when a ship is set in motion, a comparable mass of water is also set in motion at a comparable speed.
Consequently the kinetic energy and momentum calculated above have been underestimated by a factor of two or so. However, the main conclusion stands: an unaided person can easily move a ship.
The ship will move. Fluid forces don’t have a limiting static friction. We can think of these frictional fluid forces as being directly proportional to the speed of the ship. They are close to zero when the speed is close to zero, and so on.

 

[sophiecentaur]

Great to see some numbers in a question like this.

 

[Uridium]

In order for the ship to move forward water molecules at the leading edge of the ship would have to be separated, is there enough of a bond to prevent motion starting below an initial starting energy level?

 

[DaveC426913]

In order for the ship to move forward water molecules at the leading edge of the ship would have to be separated, is there enough of a bond to prevent motion starting below an initial starting energy level?

You are simply re-wording the property of viscosity of water. Yes, it is taken into account.

Note, BTW, that moving the ship bow-first would be the easiest way to do it, since it presents the smallest area and that area is streamlined, thus providing the least resistance (after all, it is the very core of ship design: to facilitate forward motion, and resist lateral motion).

It would be much harder to move a ship broadside - you are, in your parlance, "separating" far more water molecules, and making them move a lot larger distance - and yet as explained above, it can certainly be done.