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I am trying to understand how to calculate the necessary force to start moving a train and I was wondering about the roles of rolling and static friction into the equation. From several sources I read that the rolling friction coefficient is about 0.0015 and the static friction ranges between 0.35 and 0.50 for most trains.

Let's say there is a train with 5 train cars, each weighs 100,000lb. As per my understanding, when a locomotive pulls a train, it does it one car at a time and that is achieved by the joints between the cars being able to expand and contract to alleviate the load on the locomotive when starting to move or when coming to a stop.

Question 1: My current understanding is that the locomotive needs to overcome the static friction of one car at a time, and the rolling friction of the cars that are already in motion. Is this correct?

Question 2: if we assume we have a 5 train car, 0% grade, no other friction forces are acting, how do we calculate the maximum necessary force to get the whole train moving by moving one car at a time?

Based on my current understanding, this is what I came up with:

F = Fs + Fr (where Fs is force necessary to get the 5th car moving, and Fr is the accumulated force necessary to keep the first 4 cars moving)

F = Cs*M + 4 * Cr * M (where Cs is coefficient for static friction, M is the car's mass, and Cr is coefficient for rolling friction)

F = 0.35 * 100,000 + 0.0015 * 100,000 * 4 = 35,000 + 600 = 35,600lbf

Once the whole train is moving, the force necessary to keep it moving is:

F = 0.0015 * 100,000 * 5 = 750lbf

Then say we want to move it at 10mph, the total horsepower needed would be:

Horsepower = 750 * 5280 * 10 / 3600 = 11,000 ft-lbf/s = 20HP

Question 3: how to calculate the total power necessary to bring this train from 0mph to 10mph over a distance of 500ft? How do I factor in the inertia into the equation?

Everyone
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  • Given a choice (not always possible), the engineer will always elect to take out the slack in the train because many cargos don't like the jerk as the stationary car lurches forward (passengers are particularly sensitive). It is also harder on all the couplers, which can snap causing a delay. – Jon Custer Jan 05 '24 at 14:30
  • @JonCuster and that is the effect of Newton's third law, right? Once it starts moving, the force that was exerted through the moving cars will no longer be necessary and an opposite and equal force is then instantly exerted on all those cars causing them to jerk. – Everyone Jan 05 '24 at 14:40
  • The jerk is from the coupling between a now-moving car and a still-stationary car hitting its mechanical stops and the stationary car goes from zero to sixty (well, not sixty) in a fraction of a second. The engineer would prefer to gently accelerate the whole train, having no slack at all between cars. – Jon Custer Jan 05 '24 at 14:43
  • @JonCuster aah that makes sense. Either way, if the slack is there, does the math above check out? Or is it not correct? – Everyone Jan 05 '24 at 14:44
  • My answer at https://physics.stackexchange.com/questions/378312/how-does-a-man-move-a-train-with-his-teeth may be relevant – akhmeteli Jan 05 '24 at 15:12

1 Answers1

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To question one, yes your reasoning here is correct.

To question two, the frictional force is given by $F=C_s\cdot N$ for static friction and $F=C_r\cdot N$ for rolling respectively, where $N$ is the normal force. The normal force is such that $N=M\cdot g$, where $g$ is the magnitude of the gravitational acceleration given as $32ft/s^2$ in the imperial units. So your calculations look ok except for you do not include the weight of your cars, only their mass. Be sure to bring in the factor $g$, e.g. $F_s=C_s\cdot M\cdot 32ft/s^2$ and similarly for the rolling term.

To question three, you have not given enough information to calculate it, e.g. you can calculate the work required via the work energy theorem, however, you do not specify the time over which the acceleration will be accomplished. Obviously getting the train from zero to ten miles per hour in three seconds takes alot more power then getting it there over an hour. So if you knew how fast you wanted to accomplish the job, then the by the work energy theorem the work done is equal to the change in kinetic energy which is in this case: ${1\over 2}MV^2$, so multiply by the factor of time to get the horsepower, keeping in mind all the correct units.

Albertus Magnus
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