As a kid I have seen several movies, cartoon etc. in which usually one guy just uncouple (unplugs (?), dunno) the rest of a running train and then runs off with the locomotive at full speed and escapes.
I just remembered that and find it ridiculous and I wondered, which kind of problems you would encounter by try doing that.
However I can imagine a following situation: two train wagon have two overlapping pieces of iron with a hole and one big iron stick is put in, so that they are interlocked. If one wagon moves now, there other will be pulled as well.The wholes are on top of each other, so there is a difference in altitude between them. That might be important.
Removing it, for example by pulling it out, seems at first easy, as there is no force, which presses it downwards apart from it's own weight, which seems to be a minor problem.
However I would also think, that you would have friction.
I think, that the iron stick is not perpendicular to the ground, as it has the two wagons pull it. So the angle might be slightly oblique and so the force pulling on the stick would be not parallel to the ground, hence you could split the force into two perpendicular parts, one parallel to the ground and one perpendicular to the ground.
However I wonder, if it possible to ignore friction, but keep the resistance of materials, so that one doesn't sink into one another. Similar to super fluids liquids.
So I assume pulling the iron out would be possible, if push it upwards from the downside.
I also wonder how such an object with no friction behaves, if you apply a really strong force on it.
I just try to imagine the whole situation and check my physical understanding of things. So feel free to correct me or add a thought.
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$\begingroup$ The friction force that you have to overcome is equal to the force transmitted from the locomotive to the rest of the train multiplied by the friction coefficient. The energy would therefore be the friction force multiplied by the length of the path you have to pull out. The power you need would be the energy divided by the time you need to pull it out. The force that is transmitted is (if the train is not accelerating) only the air and metal friction losses of the other wagons(that is what is pulling on it). It could be calculated if someone had the data for the coupling and friction. $\endgroup$– WalyKuCommented Dec 18, 2014 at 12:16
1 Answer
It depends on the coupler design, and wether the coupler is in tension or compression.
If you have a buffer-and-chain coupler there is simply no way you will uncouple the train when it is running without using explosives. Maybe a link and pin coupler can be released if you jiggle the throttle while extracting the pin.
If you have a Janey / SA3 / AAR / knuckle coupler it's possible. You can close the air brake valve and pull the release lever for the coupler. As soon as it is not loaded the coupler will release and the back part of the train will fall away, breaking the air hose connection (it's designed to do this) and stopping the back half.
The location of the air valve and release lever are usually not compatible with operating them while moving at 50km/h, there are exceptions but the movies usually ignore these details.
This exact process is used in reverse at train yards. Look up "Hump yard" in the usual place for more details.
