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Distance travelled whilst decelerating

  1. Jun 24, 2013 #1
    1. The problem statement, all variables and given/known data
    A train of 80tonnes moves at a velocity of 70km/hr up to the bottom of a gradient of 8degrees
    The power is turned off and the train allowed to coast up the gradient before coming to rest,
    If the tractive resistance to motion is 70newtons per tonne, calculate the distance moved by the train up the gradient before coming to rest.

    2. Relevant equations
    Acceleration = Final velocity - initial velocity?
    Force = Mass x acceleration
    s=(u+v/2)t

    3. The attempt at a solution
    I have calculated the following to start with;
    Train mass = 80000kg
    Velocity = 19.4m/s
    tractive resistance = 0.07newtons per kg

    The problem again is not having a clue what formula(s) to use in which order etc

    I tried To use F=MA to calculate force then minus the tractive resistance
    then convert back to acceleration which i then get 19.33m/s, but that doesnt seem right at all
    Any ideas?

    Thanks,
     
  2. jcsd
  3. Jun 24, 2013 #2
    Using F = ma is correct.

    First you need to find out all the forces acting on the train.
     
  4. Jun 24, 2013 #3
    I calculated 1552000N for the force of the train moving,
    The only force I know from info provided is tractive resistance so,
    80000 x 0.07 = 5600N
    I guessed that the tractive resistance is subtracted from the train moving but didnt have a clue from where to go then, I cant find a formula that seems to contain the parameters I have
     
  5. Jun 24, 2013 #4
    Assume for a second there is no tractive resistance. Would the train be able to go up the gradient indefinitely?
     
  6. Jun 24, 2013 #5

    SteamKing

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    Acceleration is not just the difference in velocity. Remember, the units of acceleration are m/s^2.

    Think about the problem for a minute before worrying about what formulas to use. The train is merrily going on its way at constant velocity. When the train encounters the hill, the engineer closes the throttle and lets the train coast up the hill. When going up the hill, why does the train's velocity drop? Is it just because of the resistance to motion? Or is it also because when traveling up the hill, the train is being raised to a higher elevation?
     
  7. Jun 24, 2013 #6

    CWatters

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    How about drawing a diagram showing the forces acting on the train when it's on the slope?
     
  8. Jun 25, 2013 #7
    The train will go further if there was no tractive resistance, however, it wouldnt go on forever due to gravity?

    It is because it is traveling to a higher elevation, gravity comes in to play?
    My lecturer hinted that its the height distance I calculate, then use trigonometry to calculate the distance traveled up the gradient of 8degrees.

    I have now drawn one, I'm not sure quite sure if this is correct but below is my attempt,

    The train is moving at a force of, F = 80000 x 19.4 = 1552000N
    The tractive resistance force is F = 80000 x 0.07 = 5600N
    so we subtract tractive resistance force from moving force
    1552000-5600 = 1546400N
    Our acceleration now (or rather deceleration) with tractive resistance implemented is
    1546400N/80000 = 19.33m/s

    Then v^2 = u^2 + 2a x d
    so d = -u^2 / 2a

    distance = -19.33^2 / (2 x 9.81) = 19m (vertical distance?)
    Using trigonometry,
    Distance (on the gradient) = 19m/sin(8) = 136.5m

    That sounds a reasonable distance to me?

    If the above is correct, my next step is to
    "Prove my answer by an alternative method, compare and contrast the 2 methods."
    1 Method to be used is conservation of energy (as used) and the other is D'Alembert's principle which I have never done, I have just went through a powerpoint explaining the D'alemberts principle but it was far to complex for me. Is there a simplified explanation or could someone summarize the principle for me.

    All I gathered from it that it is used when there are multiple forces on a mass, which can be used where an object can move in xyz.
     
    Last edited: Jun 25, 2013
  9. Jun 25, 2013 #8

    haruspex

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    Right, so where is gravity in your equations?
    I don't see a diagram attached.
    The train isn't moving with a "force". It has momentum, given by mass * velocity, but the units would not be Newtons.
    It makes no sense to subtract a force from a momentum.
     
  10. Jun 25, 2013 #9
    Gravity is here: distance = -19.33^2 / (2 x 9.81) = 19m (vertical distance?) However, I presume this was incorrect,

    The diagram was drawn on paper, I was under the impression the diagram was for my benefit only.

    I see now how the train isn't moving with force seeing that there isn't any forces pulling or pushing it now, it is clearly momentum as you said.

    I see now that a force can't be taken from momentum like that (now it occurred to me that the train is under momentum not force), I have still not a clue how to work this one out :(

    Just a clue of what formulas I need to use would be a good start, Thanks again
     
  11. Jun 25, 2013 #10

    haruspex

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    Gravity and traction resistance both operate over the distance the train travels. This means you can't handle them separately. You need one equation that allows for both retarding forces.
    You wrote
    What are all the forces that will tend to retard the train?
    What is the component of each in the direction opposing the motion of the train?
    What do they add up to?
    From here there are two ways:
    - calculate the acceleration
    - use the work-energy theorem
     
  12. Jun 25, 2013 #11
    Using method one is faster: v2=u2+2as
    Kinematics equation to find s after finding acceleration thru F=ma.
    Tedious part is to find resultant force in the direction that retards the train first.
     
  13. Jun 26, 2013 #12
    I see what you mean now by what I wrote, I meant I drawn it and below was my calculation attempt

    The forces retarding the train would be gravity and tractive resistance,
    The tractive resistance is newtons/mass but I guess I would be needing to formulate this into some sort of negative acceleration?
    And gravity is basically an acceleration (or rather deceleration in this instance)

    But I dont have a clue what formulas I need, I have tried searching for formulas but I find ones that I cant seem to work into the question

    Thanks again,
     
  14. Jun 26, 2013 #13
    you need W= mg for weight
    and to calculate the frictional force.
    Calculate force in the direction of the retarding of the train through trigonometry then add them up to find total force slowing the train down. I thnk you can do the next part now.
    Formulas:W=mg
    F=ma
    v2=u2 + 2as
     
    Last edited: Jun 26, 2013
  15. Jun 26, 2013 #14
    So gravitational force slowing train down is 80000 x 9.81 = 784800N?
    I can't calculate the force retarding the train using trigonometry as I only have 1 angle, no distances yet?

    I know it seems like I am just being lazy here but I don't have a clue when it comes to mechanical physics, I don't have the ability to understand yet what affects what, what can add up and can't etc I need to have completed this coursework for this coming monday (and Im away for 2 days after tomorrow), Im starting to panic a little now so thats not helping,

    I really appreciate everyones help so far, I ideally need someone to dumb everything down for me :confused:
     
  16. Jun 26, 2013 #15
    Did this even though its not tested and i' m rushing to study for exams but oh well :D you helped me a bit there in circuits too

    ImageUploadedByTapatalk1372268488.093109.jpg
     
  17. Jun 26, 2013 #16
    Thanks, I only needed a hint on order of equations to be used and how to be implemented, that drawing is pretty much exactly what I drawn,

    However, I came to an answer that does not seem correct, my workings are below,

    Retarding force = 80000 x 9.81 / sin8 = 5639020.7N
    Friction Force = 80000 x 0.07 = 5600N
    Tot Retarding force = 5639020.7N+5600N = 5644620.7N

    F=MA so
    A = F/M
    so A = 5644620.7N/80000 = 70.6m/s

    V2 = u2 + 2as
    So;
    s = (v2 - u2) / 2a
    So;
    s = 02 - 19.42 / (2 x 70.6) = -2.665m

    I change -2.665m to positive 2.665m, because it is deceleration not acceleration.
    This however doesnt seem right, where have I gone wrong?
     
  18. Jun 26, 2013 #17
    What's the answer? Ill try to figure tht out (lazy to do checking on calculations)
    Seems like you didnt use exact values for initial speed
     
  19. Jun 26, 2013 #18
    I don't know the actual answer.
    the value I used for initial speed was 19.4m/s as you can see in this line
    s = 02 - 19.42 / (2 x 70.6) = -2.665m
     
  20. Jun 26, 2013 #19
    Why wouldnt it seem right? Hmm. You have to use 19.444 m/s though. More accurate
    Because its negative? Your acceleration should be negative btw. Unless you put deceleration
     
  21. Jun 26, 2013 #20
    Imagine, a train at 70km/hr going up 8degree gradient which isnt much really, and just lets off throttle, 2.6m isn't much for it to travel, I thought it would go a good few metres, however i must be wrong if the calculations are correct. Prehaps a train service provider could help me haha

    19.444 m/s is more accurate yes but that makes little difference to the answer, we are told working to 3sf is suffice.

    So would you say that is correct answer now?

    Next step is to work it out again but using D'Alemberts principle, know any good tutorials/websites that could teach me the principle? (bare in mind, I am very beginner at mechanical principles)
     
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