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Locomotive accelerates 34-car train

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A locomotive accelerates a 34-car train along a level track. Every car has a mass of 6.2 × 104 kg and is subject to a frictional force f = 165v, where the speed v is in meters per second and the force f is in newtons. At the instant when the speed of the train is 31 km/h, the magnitude of its acceleration is 0.11 m/s2. (a) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at 31 km/h?


    2. Relevant equations
    F=ma
    V was converted to m/s
    (Force of locomotive) = (34)(62000)(0.11) <-------total mass of the system includes all 34 cars
    (frictional force) = (0.11)(34) <-----since friction is acting on each individual car

    3. The attempt at a solution
    Okay, found an internet guru that said that: (Tension) = (Force of locomotive) + (frictional force)

    Using this equation, the tension was found. What I don't understand is. Why?

    First off, moving the friction force to the left of the equation using algebra we get that:
    (Tension) - (frictional force) = (Force of locomotive)
    What's confusing about this is that the Tension should be acting against the pull of the locomotive, which is what the friction is doing as well, so why aren't they added? Perhaps I am misunderstanding what forces are where and how they work?

    What am I missing? I've found the answer already, but I want to understand WHY.
    WHY.
    Y.
    Hahahah, seriously though, any help would be appreciated.
     
  2. jcsd
  3. Sep 24, 2014 #2

    haruspex

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    Where you've written 'force of locomotive', do you mean 'locomotive force'? If so, how are you defining that? You seem to be equating it with 'net force', i.e. the force which matches mass*acceleration.
    Tension acts in both directions. To the engine it is a retarding force; to the carriages it is a propulsive force.
     
  4. Sep 24, 2014 #3
    The reason you are having so much trouble with this problem is that you haven't drawn a free body diagram. Please show us your free body diagram, and then show us the force balance equation that you derive from this free body diagram.

    Chet
     
  5. Sep 25, 2014 #4
    How does one add a FBD? I'm new and haven't learned how to post pictures and such.
     
  6. Sep 25, 2014 #5
    I guess you start by figuring the force it takes to accelerate the dead mass of all the carriages at 0.11 m/s/s
    (f = m * a )
    Then add the total friction force at this speed.
    This = the total engine force available.
    By the way, is f = 165 v for each car, or all of them ?
     
  7. Sep 25, 2014 #6

    gneill

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    Create your diagram on your computer and save it in a file. Then on Physics Forums when you're editing a post hit the "UPLOAD A FILE" icon at the lower right of the editing pane. The rest should be intuitive.
     
  8. Sep 25, 2014 #7

    haruspex

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    In case what I posted earlier is not clear, your confusion seems to come from two different interpretations of "force of locomotive". In F = ma, F is the net force. You might validly call this the locomotive force, i.e., that force causing locomotion. In your first equations you seem to be interpreting "force of locomotive" as the net force.
    But later you say Tension should be acting against the pull of the locomotive, which is what the friction is doing as well, so why aren't they added? So now it seems you are referring to the force exerted by the 'locomotive', i.e. by the engine. This will be equal to the tension.
     
  9. Sep 26, 2014 #8
    Whats the mass of the locomotive ?
     
  10. Sep 26, 2014 #9

    haruspex

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    You don't need to know that to answer the questions posed.
     
  11. Sep 27, 2014 #10
    youre ignoring the mass of the locomotive then ?
     
  12. Sep 27, 2014 #11

    haruspex

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    No, we're not ignoring it, it just is not relevant to the question aked.
     
  13. Sep 29, 2014 #12
    So, summary :
    g assumed at 9.81 (m/s)/s
    Total mass (m) = 34 * 62,000 = 2,108,000 kg
    Force required to accelerate this mass @ 0.11 (m/s)/s
    = 2,108,000 * 0.11 = 231,880 N
    Force of friction = 34 * 165 * 8.6111 = 45,673.27 N
    Total driving force required (and tension in the coupling) = 277,553.27 N

    For equilibrium up the incline :
    277,553.27 = friction force + ( m * g * sine A )
    Transpose for A :
    A = inverse sine ( ( 277,555.27 - friction force ) / ( m * g ) )
    A = 0.6424 °
     
  14. Sep 29, 2014 #13
    Either I'm missing something here or this is a classic case of responders overthinking the problem!

    Going back to the original post, the question asked was: "What's confusing about this is that the Tension should be acting against the pull of the locomotive, which is what the friction is doing as well, so why aren't they added?"

    But you have already answered your own question, Fetch! You say the internet guru(!) said: (Tension) = (Force of locomotive) + (frictional force). Exactly! They are added.

    So think about what is giving rise to the tension in the coupling. The locomotive is pulling the train - let's say from right to left. The frictional force in the carriages is trying to prevent this and, in effect, 'pulling' from left to right. Thus the tension in the coupling is the sum of these two forces - exactly as your guru said!

    Quite why you then shuffled the equation round and caused yourself confusion I am not sure, but you don't need all the complicated equations, or the mass of the locomotive(!) to understand what I have just said. Do you? ;-)
     
  15. Sep 29, 2014 #14

    haruspex

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    Well, that's completely wrong.
    Tie a string to a hook on the wall and pull with force T. The string pulls back on you with force T. The string pulls on the hook with force T, and vice versa. The tension in the string is T, not 2T.
     
  16. Sep 29, 2014 #15
    You have signally failed to understand what I was saying - and probably confused the original poster completely! Where did I suggest that the tension was doubled?

    Let me try to put it in very simple terms: the locomotive is like a person on one end of a string; the frictional force is like someone else on the other end of the string, trying to hold it back. The tension in the string is therefore the SUM of the force exerted by the locomotive and that exerted by the frictional force trying to oppose it.

    If you can't accept this, consider the frictional effects gradually tending to zero. What is the effect on the tension in the coupling? It, too, reduces until it becomes equal to the force being exerted by the locomotive alone.

    So, I repeat: (Tension) = (Force of locomotive) + (frictional force). Just like the 'internet guru' said!
     
  17. Sep 29, 2014 #16

    haruspex

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    Still wrong.
    The force exerted by the engine (locomotive) is equal to the tension. The tension minus the frictional force on the carriages is the net force accelerating the carriages.
    If you apply your reasoning to my simple string and hook scenario you deduce that the tension is 2T.
    I suspect that the 'internet guru' was referring to "locomotive force", i.e. the net force producing an acceleration. This is a general concept, not specifically related to 'locomotives', as in railway engines.
     
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