Forces and Newton's Laws of a train

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Homework Help Overview

The problem involves a train consisting of 50 cars, each with a mass of 6800 kg, and an acceleration of 0.08 m/s². The goal is to determine the tension in the couplings between specific cars while ignoring friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for tension based on the mass and acceleration, with some questioning the correctness of the original poster's results. There is mention of the need to consider how many cars each coupling is responsible for pulling.

Discussion Status

There is an ongoing exploration of the problem, with participants providing insights into the implications of how cars are numbered and the resulting calculations for tension. Some guidance has been offered regarding the importance of free body diagrams and the interpretation of the problem setup.

Contextual Notes

Participants note potential confusion regarding the numbering of the train cars, which affects the calculations and interpretations of the tension in the couplings. There is a lack of explicit consensus on the correct approach due to differing assumptions about the car numbering.

Cheddar
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Homework Statement


A train consists of of 50 cars, each of which has a mass of 6800 kg. The train's acceleration = 0.08 m/s(squared). Ignore friction and determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars.

Homework Equations


Net Force = Tension = mass * acceleration

The Attempt at a Solution


Used the equation above I came to:
16320 N for part (a) and
26656 N for part (b).
My problem is, I'm thinking those answers are supposed to be divided by some amount since there are couplings between each car.
 
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Hi Cheddar,

Cheddar said:

Homework Statement


A train consists of of 50 cars, each of which has a mass of 6800 kg. The train's acceleration = 0.08 m/s(squared). Ignore friction and determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars.

Homework Equations


Net Force = Tension = mass * acceleration

The Attempt at a Solution


Used the equation above I came to:
16320 N for part (a) and
26656 N for part (b).

Those results don't sound right to me; how did you get those numbers? (Notice that in this problem the coupling in part a is responsible for pulling many more cars than the coupling in part b.)
 
Cheddar said:

Homework Statement


A train consists of of 50 cars, each of which has a mass of 6800 kg. The train's acceleration = 0.08 m/s(squared). Ignore friction and determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars.

Homework Equations


Net Force = Tension = mass * acceleration

The Attempt at a Solution


Used the equation above I came to:
16320 N for part (a) and
26656 N for part (b).
My problem is, I'm thinking those answers are supposed to be divided by some amount since there are couplings between each car.

Have you tried drawing free body force diagrams for the two cars?
 
This is what I did.:
Each car has a mass of 6800 kg.
Part (a) 6800 kg * 30 cars = 16320 N
Part (b) 6800 kg * 49 cars = 26656 N
 
I agree with your answers. Although, your equations above are confusing because you didn't show that you multiplied by the acceleration 0.08, but the numbers you wrote imply that you did muliply it.

Consider the last car on the train. It is pulling nothing. It is being pulled by the next to last car. So the coupler must be pulling only the mass of the last car multiplied by the acceleration. Now the third car is pulling on the two behind it. So it's coupler is pulling at 2*mass*accel, etc, etc all the way up to the engine which is providing the pull for the entire train of cars.
 
Cheddar said:
This is what I did.:
Each car has a mass of 6800 kg.
Part (a) 6800 kg * 30 cars = 16320 N
Part (b) 6800 kg * 49 cars = 26656 N

Okay, that's the right idea, but I think they mean the 1st car is the car right behind the engine, and the 50th car is the last car.

So that in part a, the coupling is pulling twenty cars, and in part b it is only pulling one car.
 
Chrisas said:
I agree with your answers. Although, your equations above are confusing because you didn't show that you multiplied by the acceleration 0.08, but the numbers you wrote imply that you did muliply it.

Consider the last car on the train. It is pulling nothing. It is being pulled by the next to last car. So the coupler must be pulling only the mass of the last car multiplied by the acceleration. Now the third car is pulling on the two behind it. So it's coupler is pulling at 2*mass*accel, etc, etc all the way up to the engine which is providing the pull for the entire train of cars.

Your second paragraph contradicts your first. If the second is correct (and it is), then the original answers must be wrong, because the tension in the last coupling must be less than that in any other coupling.
 
I don't think it's a contradiction because as alphy pointed out I was counting the cars from the back because that made Cheddar's answers correct. I was assuming that his actual problem drawing must have been numbered that way. However, if you number the cars the other way, then yes the answer changes.
 
Chrisas said:
I don't think it's a contradiction because as alphy pointed out I was counting the cars from the back because that made Cheddar's answers correct.

Sorry; I didn't realize that.
 
  • #10
No problem. Both the OP and I should have been specific on our counting reference. Sorry for the confusion.
 
  • #11
Thanks everyone. I was counting the opposite way I guess. There is no picture or explanation as to the numbering of the cars so I guess we really have no idea which answers are correct...
I'm using the one with the first car being behind the engine though. I think it makes more since. Thanks again.
 

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