Forces and Newton's Laws of a train

In summary, Cheddar's solution used 16320 N for part (a) and 26656 N for part (b). His problem is that he thinks the numbers are supposed to be divided by some amount, but he is unsure of how to do that. He recommends drawing free body force diagrams to determine the tension in the couplings.
  • #1
Cheddar
38
0

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.
 
Physics news on Phys.org
  • #2
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.)
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 

What is a force?

A force is a push or a pull on an object, which can cause it to accelerate or change its motion.

What are Newton's Laws of Motion?

Newton's Laws of Motion are three fundamental principles that explain how objects move and interact with each other. They were developed by Sir Isaac Newton in the 17th century and are still used today to understand the behavior of objects in motion.

How do forces affect a train?

Forces can affect a train in several ways. The force of the engine pulling the train forward is necessary to overcome the force of friction between the train and the tracks. Forces can also cause the train to accelerate, decelerate, or change direction.

What is the relationship between force and acceleration?

According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that the greater the force, the greater the acceleration, and the greater the mass, the smaller the acceleration.

How does Newton's Third Law of Motion apply to a train?

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In the case of a train, the force of the engine pushing the train forward is countered by the force of the tracks pushing back on the train, allowing it to move forward. Without this equal and opposite reaction, the train would not be able to move.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
537
  • Introductory Physics Homework Help
Replies
13
Views
959
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
787
Replies
44
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
669
Back
Top