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Acceleration and a Train

  1. Jun 9, 2005 #1
    Hi,
    Question:
    A train locomotive is putting two cars of the same mass behind it. Determine the ratio of the tension in the coupling between the locomotive and the first car(FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train.

    Work:
    F=ma
    Car 1:FN- M1G=m1*a
    Car 2:FN- M2G=m2*a

    2FN- M1G- M2G=m1*a+m2*a

    Answer:a=(-g(m1+m2))/(m1+m2))

    Is this correct?

    Thank You
     
  2. jcsd
  3. Jun 9, 2005 #2

    siddharth

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    Homework Helper
    Gold Member

    How did you get those two equations? The acceleration of the cars are in the horizontal direction (ie, x-direction). But gravity acts in the vertical direction (ie, y-direction).
    Also what about the Tension in the ropes? The question asks for the ratio of the Tension in the ropes (or couplings), not the acceleration of the cars.

    So, to approach any problem of this type, the first step is to draw a Force diagram. Draw seperate Force diagrams for each car and clearly mark the various forces acting on the cars.
    Now apply newton's second law separately on each car.
    What is the relation between the accelerations of the two cars and why?
    From this you will be able to calculate the ratio of the tensions.
     
  4. Jun 10, 2005 #3
    Hi,
    I did like you said and made a free body diagram. I made a table of my cars:

    Car2
    x Y
    Fn 0 Fn
    Fg 0 -mg
    T2 T2 0

    Car 1
    x Y
    Fn 0 Fn
    Fg 0 -mg
    T2 -T2 0
    T1 T1

    Car
    x Y
    Fn 0 Fn
    Fg 0 -mg
    T1 -T1 0

    Then from the table I calculated the x and y components of the tensions, using newton's second law FT=ma

    Car 2
    F=T2=ma,since a=a
    F=Fn=mg, since a=o

    Car 1
    F=T2=ma,since a=a
    F=Fn=mg, since a=o

    Car
    F=-T1=ma,since a=a
    F=Fn=mg, since a=o

    From here I'm kind of lost on how to calculate a ratio of these components to help me arrive at a answer of a 2 to 1 ratio.

    Please help and Thank you
     
    Last edited: Jun 10, 2005
  5. Jun 10, 2005 #4
    The vertical components all cancel, you need not worry about those. What you need to worry about are the horizontal forces, namely the tension force pulling each cart. This is basically the n-block string problem.

    The force acting on the LAST cart is just the tension in the second coupling. The forrces on the second cart are the tension in the couplings on either side of it, and the force on the entire system is coming from the engine.

    Use Newton's 2nd Law on all 3 to fidn expressions for the tension.
     
  6. Jun 11, 2005 #5

    siddharth

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    Homework Helper
    Gold Member

    Look at the picture I attached to see the Free body diagram for this problem.

    Apply newton's second law for CAR 1 in the x-direction. What are the forces acting on it? T1 acts to the right and T2 acts to the left. So the net force acting on it is T1-T2.

    By newton's second law,
    T1-T2=ma - Equation I

    Apply newton's second law for CAR 2 in the x-direction. What are the forces acting on it? T2 acts to the right.
    Therefore, by newton's second law
    T2=ma - Equation II

    From this you should be able to find the ratio of the Tensions.
    If you have a problem with any of the steps, do post and tell what the problem is.
     

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