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Tractive effort and trains!!


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Hi nice forum, some very interesting threads! I'm a bit stuck with this mechanics question and hope someone can help me out!


Question: A train of mass 250 tonnes, moving with a uniform velocity of 63km/h along a horizontal track begins to climb up an inclined bank having a slope of 1 in 80. During the climb, the train exerts a constant tractive effort of 22000N, whilst the resistances to motion remain constant at 65N/tonne. Determine how far the train will climb up the bank before coming to a standstill?


I would be grateful for any advice, even on where to start!

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I think the train will travel about 1.5km up the bank before it stops, up to about 19m above the base of the grade.



T=tractive effort

B=net resistance=(65N/tonne)(mass of the train on the bank)

W=weight of the train on the bank



The resistance to the motion if the entire train was on the bank would be B=65N/tonne*(250tonne)=16,250N. That's less than the tractive effort T=22,000N. 1tonne=1000kg. The whole train weighs 250tonnes*g W=250000kg(9.8m/s^2)=2,450,000N. The force of the weight pulling the whole train back down the bank is W/sqr(1+80^2)=30,600N. If the train is going fast enough that all the cars make it onto the grade before it comes to a standstill then there will be a constant net force of F=22,000-16,250-30,600N= -24,850N. So it's going to be gradually slowing down. If you assume that this net force F is pretty much constant from the time the train enters the bank and goes up to where it stops then that's an acceleration of


Vo=63km/h=initial speed=17.5m/s


distance up the bank=1/2*a*t^2=1540m about 1.5km

or an elevation of about 1.5km/sqr(1+80^2)=19m


It's probably a bit more like 21m because the train starts out lighter on the grade, unless it's a windy day or there's wet leaves on the rails. Something like that.. It's getting late here so you might want to double check that. Cool problem. Good luck :cocktail:

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