lugger Posted September 24, 2006 Report Share Posted September 24, 2006 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! Quote Link to comment Share on other sites More sharing options...
max4236 Posted September 25, 2006 Report Share Posted September 25, 2006 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. ma=T-B-Wsin@T=tractive effortB=net resistance=(65N/tonne)(mass of the train on the bank)W=weight of the train on the banksin@=1/sqr(1+80^2) 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 a=-24,850N/250,000kg=-.0994m/s^2Vo=63km/h=initial speed=17.5m/st=(V-Vo)/a=(0-17.5)/-/.0994=176secondsdistance up the bank=1/2*a*t^2=1540m about 1.5kmor 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: Quote Link to comment Share on other sites More sharing options...
lugger Posted September 25, 2006 Author Report Share Posted September 25, 2006 Thankyou for your help, i understand it much better. I missed the topic at college and was really starting to struggle!! Its actually starting to make sense now:) Quote Link to comment Share on other sites More sharing options...
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