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Homework Help: Tractive force

  1. Apr 27, 2008 #1
    Hi everyone, just found this place and think its a brill idea that people can get extra help outside of class! Hope one or two of you can assist me with my problem!


    1. The problem statement, all variables and given/known data

    3) (a) Define the terms Work and Power. Derive an equation relating the power developed by an engine and the tractive force it produces. Note: ‘derive’ means show all working! [4]

    The resistance to the motion of a car, Fr, of mass 2000 kg is proportional to its speed (i.e. Fr=kV). With the engine working at 72 kW the car can attain a maximum speed of 12 m/s when travelling up a straight hill inclined at and angle  to the horizontal when sin = 0.1.

    (b) Calculate the value of the resistive force and show that the value of k is
    333.3 Ns/m. [6]
    (c) Find the greatest speed at which the car could travel down the road with the engine working at 72 kW.

    2. Relevant equations


    3. The attempt at a solution

    Ive had a good look at this, and the only relationship that i am able to come up with in relation to both the work and power definitions is basically just


    I know that this statement is correct, but i just dont think that it is the correct answer for the question. And it seems that without this its not very easy to progess onto the next parts of the question.
    Being quite honest, im not entirely sure what tractive force actually is, ive had a good look around google and not come up with much, and especially nothing in the way of a defintion.

    So basically im stumped

    Hope you can help

  2. jcsd
  3. Apr 28, 2008 #2
    a tractive force is an antiquated term for force pulling something, specifically regarding a locomotive (or an engine in general in this problem).
    The mathematical definitions of work and power are the equations that you gave (a more elegant definition of work might be 'change in energy' (Ef - Ei) ); perhaps they're looking for a verbal definition?
  4. Apr 28, 2008 #3


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    Welcome to PF!

    Hi farso! Welcome to PF! :smile:

    Yes … as lzkelley says, "tractive" just means pulling … they use it a lot on Star Trek!

    Your equation for power seems fine, except that they ask you to "show all working" … and what working is there in a definnition?

    I think perhaps they want you to put the units in, to show that you know whether it's newtons or millie newtons or whatever. :confused:

    Are you ok now on (b) and (c)? :smile:
  5. Apr 28, 2008 #4
    Thats great thank you! I'll include units and also a verbal definition of what ive shown to make it clear.

    Quite honestly, im having trouble finding a way of working out resistive force, seems like the term isnt used very widely?
    Naturally i know that F=mg, but I cant make the link between that and Fr=kV or the equation i previously wrote.

    Thanks again for your helps!
  6. Apr 28, 2008 #5
    when the car/train/thing reaches its maximum velocity - what happens to its acceleration? What does that tell you about the net force? what does that tell you about the individual forces?
  7. Apr 29, 2008 #6
    I assumed instant acceleration to max velocity, and i know that they are all in equilibrium, im just having real difficulties proving that coefficent f is 333.33... Have you got any hints on how this might be proved?

  8. Apr 29, 2008 #7


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    Hi farso! :smile:

    How it got there (the acceleration) doesn't matter …

    Anyway, show us what figure you got, and how you got it! :smile:
  9. Apr 29, 2008 #8
    Ok then, work so far...

    a) work = force*distance
    power = work/time
    therefore power=(force*displacement)/time
    as displacement/time = velocity
    power = force*velocity

    tractive force = power/velocity

    Ive tried calculating each force acting upon the vehicle if the surface was frictionless.
    theta = 5.74deg
    in its components,
    wx=mg sin(theta) = 1962.28J
    wy=-mg cos(theta) = -19521.63
    tractive force= 6000(not sure on units)

    I also know the resistive force is supposed to equal the tractive force while it is not accelerating, and the power remains constant.

    Im certainly missing something, but it feels like ive tried everything.
  10. Apr 29, 2008 #9


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    Hi farso! :smile:

    Yes … you're missing an equation! :wink:

    Both the resistive and the tractive force are along the slope, so calculate the forces along the slope.

    That'll be tractive force = resistive force + component of weight.

    Which is … ? :smile:
  11. Apr 29, 2008 #10
    Ok, I think i may have got this now thanks to tiny tim!

    Tractive force id calculated to 6000 using tf=p/v
    the resistive force was calculated to nearly 4000 using fr=kV
    and the weight of the vehicle is naturally 2000kg

    Therefore using your equation, this proves k to be 333.3

    Please tell me im correct on this... I could nearly cry over this assingment ;)
  12. Apr 29, 2008 #11


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    Hi farso! :smile:

    Yes, you've certainly got the right result :smile: … but I think you've fiddled it … :rolleyes:

    You haven't converted the kg into N, and you haven't used the slope.

    (btw, what letter did you use for the angle? it came out as a square on my computer)

    Try again! :smile:
  13. Apr 29, 2008 #12
    Sorry, i copied and pasted and it was theta before that.

    Still cant get this working right, im using wx=mg sin(theta) to get component force but it still falls short. Im really sorry if you think im being slightly dumb here, I just dont get this problem at all and its always the way that anything that I dont understand tends to come up in the exams!!
  14. Apr 29, 2008 #13


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    Hi farso! :smile:

    (Copy this θ in future … it's unicode, so it should show up fine on any computer … I got it from http://www.barzilai.org/math_sym.htm. :smile: )

    If you used mg sinθ, you should have got 2000*9.81*0.1, which gets you back to about 2000, which is ok. :smile:
  15. Apr 29, 2008 #14
    Ah, theres my problem! I did asin(0.1) to try and actually figure out the angle which θ related to. Thanks very much for all your help!
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