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Train moving along a bend

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data
    A train consists of a locomotive and five identical carriages, connected via massless ropes. Initially the train is moving at a speed V. At this speed, the tension in the rope, F between the locomotive and the first carriage exactly balances the resistive drag, so the train does not accelerate.

    1. Find the tension in all other ropes in terms of F.

    2. A single carriage has mass M and length L. The resistive force is given by kV, where k is a constant. If the train now moves along a bend of radius R, (R>>L) show that the velocity for which minimum force is exerted by the tracks on the central carriage is given by

    [itex]V_m = \frac{5 k L}{2M}[/itex]


    2. Relevant equations
    Centripetal acceleration
    [itex]a_c=\frac{v^2}{c}[/itex]

    Newton's Second Law

    3. The attempt at a solution

    1. Label the carriages 1 to 5. (1:closest to locomotive) Let Fi be the coupling between the ith an (i-1)th carriage. If I start from the last carriage, then Newton's second law will give: F5= MkV.

    Similarly for the second carriage, F4=F5+MkV=2MkV.

    Continuing all the way to the first carriage, I find F=5MkV

    So

    [itex]F_5=\frac{F}{5}[/itex]
    [itex]F_4=\frac{F}{4}[/itex]

    etc.

    2. The central carriage is the third one. At a velocity V, the drag on this carriage will be MkV. The other forces will be the tensions of the two connecting ropes. I'm confused whether to include a force due to the track? The resultant of the forces must be the centripetal force. Also I am unsure of how to include the length L in the equations.
     
    Last edited: Apr 24, 2013
  2. jcsd
  3. Apr 24, 2013 #2

    mfb

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    How did you get this? The equations in front of that are right, but this one is not.

    Why do you consider the central carriage only?
    That is the part you should minimize (I would expect that it is 0 at the minimum). I think it is implied that this force is acting sidewards only.
    That will be relevant for the direction of the forces.
     
  4. Apr 24, 2013 #3

    Andrew Mason

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    You don't have to use Newton's second law for the first part. You just need the first law - net force on each car is 0.

    The resistive force on each car is kV. So the total resistive force for all 5 (which is equal to F) is 5kV. ie. |F| - 5kV = 0.

    Apply that same concept to the next car (no. 2): |F2| - 4kV = 0.

    That allows you to get a value for F2 in terms of F. Repeat that for each car.

    AM
     
  5. Apr 24, 2013 #4
    Oh sorry, I wrote the wrong formula in haste. What I meant was

    [itex]F_5 = \frac{F}{5}[/itex]
    [itex]F_4 = \frac{2F}{5}[/itex]

    and so on.

    Also I noticed I didn't say that we have to find the velocity for which the rails exert minimum force on the central carriage. Sorry about the confusions.

    So in order to proceed, do I need to resolve the components of the tension and drag force along the radial direction? (I'm not sure how to calculate the tension in the ropes since the train is now accelerating)
     
  6. Apr 24, 2013 #5

    Andrew Mason

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    The car is accelerating in the radial direction, not in the tangential direction. So I think you have to assume the tension does not change.

    As you note, the two tensions on the middle car are not pulling in exactly the same direction so there is a radial component in the sum of those two tensions. Just calculate that radial tension. It will be a function of V, R and L. Compare that to the centripetal force that must be exerted on the car in order for it to prescribe the turn by subtracting the radial tension from the needed centripetal force.

    AM
     
  7. Apr 25, 2013 #6
    I'm unsure about how to get the radial component of the tension. Anyway here is my understanding of the geometry of the situation:

    etrwoy.png

    Is this correct? FT refers to the force due to the tracks. The centripetal force required by the car is

    [itex]\frac{MV^2}{R}[/itex]
     
  8. Apr 25, 2013 #7

    Andrew Mason

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    F3 and F4 are perpendicular to the respective radial vectors. Those radial vectors are separated by an angle (since L<<R, assume the angle, in radians, is L/R). The angle between the two forces is not ∏. You can see this if you draw the cars on either side.

    AM
     
  9. Apr 25, 2013 #8

    TSny

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    Make F3 and F4 perpendicular to R.
     
  10. Apr 25, 2013 #9
    But if the tensions are perpendicular to their radial vectors, how can they have a radial component?
     
  11. Apr 25, 2013 #10

    TSny

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    The external forces acting on the car determine the acceleration of the center of mass of the car. After making the force vectors F3 and F4 perpendicular to R at the ends of the car, imagine sliding the vectors to the center of mass. You can now think of the car as a single particle concentrated at the center of mass.
     
  12. Apr 25, 2013 #11

    Andrew Mason

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    When you add them together, do they sum to 0? If not, what is the direction of the resultant?

    AM
     
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