Tractive Force Train

In summary: Whichever way it is, we would need to know either the acceleration, or the time, for the calculation. The 90000 N for the engine and the 65000 N for car B to accelerate them to 27.78 m/s would be sensible. We could find the acceleration from the force, mass, and acceleration equations, but if we are finding the force needing to decelerate car B, we would need to know the coefficient of friction for car B's wheels.In summary, the conversation discusses the necessary tractive force and distance traveled for a subway train to reach a speed of 100km/h in 1 minute, as well as the distance and time required to slow down the train to
  • #1
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Homework Statement



What is the tractive force F necessary between subway car A and the track to reach a speed of 100km/h in 1 minute? What distance was traveled during this time? At this point the brakes are fully applied on the wheels of car A in order to slow the train to 40km/h. The brakes are not applied on the wheels of car B and C. Knowing the coefficient of kinetic friction is 0.30 between the wheels and the track. Determine a) the distance required to slow the train b) time required to slow train c) force in each coupling

Ma= 25000 kg, Mb = 40000 kg, Mc = 25000 kg


Homework Equations



m1vx1+ Fdt = m2vx2

100km/h = 27.78 m/s


The Attempt at a Solution




Complete train

0 + F(60) = 90000(27.78)

F = 41670 N

Coupling force between A & B

0 + F(60) = 65000(27.78)

F = 30095 N

Coupling between B & C

0 + F(60) = 25000(27.78)

F = 11575 N

Question do I need to apply the frictional force over this impulse??

For distance over a minute

27.78 m/s * 60 =1668.8 m?

To find part A & B

40km/h = 11.11m/s
(Using work done)

W = Fd, work = change in PE


1/2 mv^2 = 1/2 90000(27.78 -11.11)^2 =12505000 J

d = w/f (force =ukN = 90000(9.81)0.3 = 264870

d = 12.5X10^6/264870 = 47.2 m *part A*

b) 90000 -264870(T) =90000(11.11)

t = 5.66s
 
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  • #2
The wording is confusing. Is it correct to assume that car A is the engine pulling B and C?
I don't see how you can include the friction force of the other cars in the first part because you aren't given any information about the rolling friction (nothing to do with the kinetic friction coefficient given).
For distance over a minute
27.78 m/s * 60 =1668.8 m?
Something wrong here; the average velocity over that minute would be less than 27.78.
1/2 mv^2 = 1/2 90000(27.78 -11.11)^2 =12505000 J
d = w/f (force =ukN = 90000(9.81)0.3 = 264870
d = 12.5X10^6/264870 = 47.2 m *part A*
I got different answers for (a) and (b) working them out from a force, acceleration approach - getting the time first (over 20 seconds). In your second line, I think the 90000 should be replaced by just the mass of car A; only its friction force is working to stop the train. The kinetic energy should be the change between the energy at 27.78 m/s and the energy at 11.11 m/s.

The coupling forces asked for in (c) more likely apply to the slowing down than the speeding up, though it isn't clear in the question.
 

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