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Extra force on a train problem

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Two long trains carrying coal are traveling in the same direction side by side on separate tracks. One train is moving at 40 ft/sec and the other at 50 ft/sec. In each coal car a man is shoveling coal and pitching it across to the neighboring train. The rate of coal transfer is 4 tons/min for each 100 feet of train length. This rate is the same for both trains. Find the extra force on each train per unit length caused by this mechanism.

    2. Relevant equations

    I think I will need: sum of Forces = Mass * dv/dt but I'm not certain



    3. The attempt at a solution
    I'm not sure exactly what to do. Would you just find the weight of coal added to each train or is there more to it. Does the momentum equation have any use here. I'm confused on how to start this. Any help would be appreciated!
     
  2. jcsd
  3. Oct 12, 2009 #2

    Delphi51

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    Yes, I think momentum is the best approach. The fast train will lose momentum as it exchanges fast-moving coal for slow-moving coal. If you can find out how much momentum is lost per second you can use
    dp/dt = m*dv/dt = F
     
  4. Oct 12, 2009 #3
    So this is what I've come up with...

    using F = d(mv)/dt = m*dv/dt + v*dm/dt => where dv/dt = 0, so F = v*dm/dt

    and dm/dt = 4 tons/min

    F(slower moving train)=d(mv)/dt = (4 tons/min)*(1 min/60sec)*(40ft/sec)/(100 ft of length)
    = 0.0267 tons-ft/s^2 per unit length = 53.4 lb ft/s^2 per unit length

    similarly for the faster moving train...

    with v = 50 ft/s => F(faster moving train) = 66.67 lb ft/s^2 per unit length

    How does this look? Did I miss something?
     
  5. Oct 12, 2009 #4

    Delphi51

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    My calcs are often wrong, but here is what I did for the fast train:
    dp = mv - mv = m(v1 - v2) since the mass doesn't change.
    = 4*2000/60 lb/s*(50-40) ft/s
    = 1333 lb*ft/s^2
    This is per 100 ft of train, so per foot it is 13.3 lb*ft/s^2.
     
  6. Oct 12, 2009 #5
    If that is for the fast train would the force on the slow moving train be the same( for example: the slower train has an "extra force" of +13.3 while the faster train has an "extra force" of -13.3)? Also, when it asked for the force on "each train," would they both be the same but have opposite signs? It seems like the momentum change would be the same with the slower train gaining the same amount of momentum that the fast train is losing. I'm not sure why my solution is wrong.
    Why wouldn't you just calculate the force for each train rather than find the difference (4*2000/60 lb/s*(50-40) ft/s). Sorry if I'm missing something very basic. Thanks very much for the help!
     
    Last edited: Oct 12, 2009
  7. Oct 12, 2009 #6
    I think I get it. If they were both traveling at the same velocity and all other things being the same, the "extra force" would be zero, but since the faster train is moving at 50 ft/s (10 ft/s faster than the slower one), the "extra force" is from this 10ft/s difference. For example, the coal from the fast train being thrown into the slow train creates an "extra" force of +13.3 lb*ft/s^2 and addes momentum to the slower train. If this is true then the coal from the slower train being thrown into the fast train creates an "extra" force of -13.3 lb*ft/s^2 and the faster train loses the same amount of momentum that the slower train gains.

    Does that sound correct? I think I might understand now.
     
  8. Oct 12, 2009 #7

    Delphi51

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    Sounds great!
     
  9. Oct 13, 2009 #8
    Thanks for the help!
     
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