Locomotive tractive effort and draw-bar pull between cars

[paul9619]

Hi all,

Another teaser that I have been attempting is the following:

A Locomotive (A) pulls two tracks (B) and (C). The masses of A,B & C are 80, 25 and 20 tonne respectively. The train starts from rest and accelerates uniformally to cover 2KM in 8 minutes. Assume the resistance to motion of the locomotive and the trucks is 200N/tonne

1. Calculate the tractive effort developed by the locomotive A.

I have used the formula: Tractive Effort = Total resistance to motion (R) + Mass x Acceleration.

The total resistance to motion is 200/N tonne x (80+25+20) = 25KN
The mass is 125 Tonne
The acceleration I have as 0.01736111 m/s using s=ut+1/2at^2.

Therefore I have the tractive effort as 27.17K Newtons.

Is this right?

2. Calculate the draw-bar pull between A and B
3. Calculate the draw bar pull between B & C

I am struggling with 2 and 3. Could anyone give me any pointers??

 

[SGT]

1. Your reasoning is correct. I have not checked your numbers, but I suppose they are OK.
2. Remember that A must pull the combined masses of B and C.
3. The draw bar between B and C must pull the mass of C.

 

[Astronuc]

The acceleration of 0.01736111 m/s² is correct assuming that locomotive and trucks are accelerating uniformly.

As SGT mentioned, the drawbar between A and B has the masses and resistance of B and C pulling on it, and between B and C, the drawbar only pulls C.

 

[paul9619]

Ok taking this into consideration I have done the following.

Resistance to motion = 25 tonne x 200 N/tonne = 5000N for truck b
= 20 tonne x 200 N/tonne = 4000N for truck c

Therefore total resistance to motion is 9000N.

Force required to move B & C = Acceleration x mass = 781.25N

Therefore the pull between A and B is 9000N + 781.25 = 9781.25N

Does this seem right??

Obviously for the pull between B & C it will be the same precedure but just taking into account the resistance & motion of C.

 

[SGT]

It's right.