Understanding Forces in a Train System

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Homework Help Overview

The discussion revolves around understanding forces in a train system, specifically focusing on the interaction between the engine and the wagons. The original poster presents a scenario involving a train with a specific mass and acceleration, seeking to determine the force exerted by one wagon on another.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of Newton's laws to the train system, questioning how to calculate the forces between the wagons and the engine. There are discussions about the forces acting on different parts of the train and the implications of Newton's third law.

Discussion Status

The discussion is ongoing, with participants raising questions about the relationships between forces and masses in the system. Some have provided insights into the equations needed for analysis, while others express confusion about specific aspects of the problem, indicating a productive exploration of the topic.

Contextual Notes

Participants are grappling with the complexities of force interactions in a multi-body system, including the need to consider different forces acting on the engine and wagons. There is an emphasis on understanding the implications of Newton's laws in this context.

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There is something that I not able to understand about force. I would like to tell it to you with an example.
There is a train consisting of 1 engine of 8000kg and 5 wagons of 2000 kg each
The acceleration of the train is 2 m/s2

How would you tell what force is exerted by wagon 1 on wagon 2 and why?
 
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AakashPandita said:
There is something that I not able to understand about force. I would like to tell it to you with an example.
There is a train consisting of 1 engine of 8000kg and 5 wagons of 2000 kg each
The acceleration of the train is 2 m/s2

How would you tell what force is exerted by wagon 1 on wagon 2 and why?

1<->2<->3<->4<->5->

F exerted to the ultimate wagon on the line

look at it as a whole

Find acceleration of the train and then look at 1<->2 ( Forces act equal but in different directions in each case )

What do you guess the acceleration of the first wagon be related to overall acceleration.

Good Luck.
 
Actually what i did not understand was...
we need to find force by 1 on 2...
but 1 is connected to engine and 2 is connected to wagon 3,4, and 5.
And according to Newton's 3rd law,
the force exerted by engine and wagon1 should be equal to the force exerted by wagons 2,3,4,5 in opp. direction
i.e,
ma=-ma
(8000+2000) x 2 should be equal to -(2000 x 4) x 2
but it is not so...

it is difficult to understand my question but if you understand...please tell me the reason...where am i getting wrong?
 
AakashPandita said:
Actually what i did not understand was...
we need to find force by 1 on 2...
but 1 is connected to engine and 2 is connected to wagon 3,4, and 5.
And according to Newton's 3rd law,
the force exerted by engine and wagon1 should be equal to the force exerted by wagons 2,3,4,5 in opp. direction
i.e,
ma=-ma
(8000+2000) x 2 should be equal to -(2000 x 4) x 2
but it is not so...

it is difficult to understand my question but if you understand...please tell me the reason...where am i getting wrong?

Mtotal * a = what is it?
 
Hi Aakash! :smile:
AakashPandita said:
… according to Newton's 3rd law,
the force exerted by engine and wagon1 should be equal to the force exerted by wagons 2,3,4,5 in opp. direction

That's right!

You have correctly identified the two bodies whose equations of motion you need to consider. :smile:
i.e,
ma=-ma
(8000+2000) x 2 should be equal to -(2000 x 4) x 2

No, you need two equations, one for each body.

For the rear body, yes, it's F = mreara.

But for the front body, it's force-from-the-engine - F = mfronta.

So how do you find force-from-the-engine ? :wink:​
 
First of all, For the front body what is "F" that you mentioned?
And...If the equation of the rear was right then shouldn't it be equal to (m front a)?
And why can' t the equation for the front be written as the product of its mass and a?
Why is it wrong?
 
stallionx said:
1<->2<->3<->4<->5->

F exerted to the ultimate wagon on the line

look at it as a whole

Find acceleration of the train and then look at 1<->2 ( Forces act equal but in different directions in each case )

What do you guess the acceleration of the first wagon be related to overall acceleration.

Good Luck.

Look at the interaction between 4 and 5

1,2,3,4(4 masses ) -> some force

that some force <-5-> Force total

Drawing FBD's what do you arrive at ?
 
i understand your point.
but i don't find the answer to my question which is in the 3 rd post of this forum.
 
Hi Aakash! :smile:
AakashPandita said:
First of all, For the front body what is "F" that you mentioned?

For both bodies, F is the equal and opposite forces between the front and back parts.

Now take a closer look at the question …
There is a train consisting of 1 engine of 8000kg and 5 wagons of 2000 kg each
The acceleration of the train is 2 m/s2

i] what is meant by the word "engine"?

ii] why is the whole train accelerating?? :wink:
 
  • #10
engine is pulling all the wagons, i.e, applying a force on them and so it is accelerating.
 
  • #11
please tell me what mass should we multiply with acceleration...mass of wagon1+ that of engine......or mass of wagon2+3+4+5.
 
  • #12
Hi Aakash! :smile:

(just got up :zzz: …)
AakashPandita said:
engine is pulling all the wagons, i.e, applying a force on them and so it is accelerating.

Exactly …

the question is telling you (without actually saying so) that there is an extra force (call it E).​

So how are you going to include E in your equations? :smile:
AakashPandita said:
please tell me what mass should we multiply with acceleration...mass of wagon1+ that of engine......or mass of wagon2+3+4+5.

You always use the mass of whatever "body" your equation is about.

And you need a separate equation for each body. :wink:
 
  • #13
i don't understand what you are trying to say.
 
  • #14
which part? :confused:
 
  • #15
the answer for my question( which you have quoted)
 
  • #16
AakashPandita said:
please tell me what mass should we multiply with acceleration...mass of wagon1+ that of engine......or mass of wagon2+3+4+5.
To apply Newton's 2nd law:
Pick the body (or group of bodies) you want to consider as your system.
Identify all the forces acting on that body.
Find the net force on the body.
Then apply ΣF = ma, where m is the mass of whatever 'body' you are analyzing.
 
  • #17
what i do not understand is if wagon 1 is exerting force on the 2nd what mass should be multiplied with acceleration...
engine+wagon1...or mass of wagon2+3+4+5
 
  • #18
AakashPandita said:
what i do not understand is if wagon 1 is exerting force on the 2nd what mass should be multiplied with acceleration...
engine+wagon1...or mass of wagon2+3+4+5
First tell me:
What forces act on 'engine+wagon1'?
What forces act on 'wagon2+3+4+5'?
 
  • #19
Another hint if you need it: What's producing the acceleration of the entire train (including all the cars)?
 
  • #20
the force acting in both the cases should have the same magnitude... shouldn't it?
 
  • #21
AakashPandita said:
the force acting in both the cases should have the same magnitude... shouldn't it?
When you say 'the force' what force do you mean? There may be multiple forces involved.

Of course the force that wagon 1 exerts on wagon 2 is equal in magnitude to the force that wagon 2 exerts on wagon 1.
 
  • #22
the force means the force that is applied by the engine...
 
  • #23
AakashPandita said:
the force means the force that is applied by the engine...
Well, that force acts on the engine, right?

Again:
What forces act on 'engine+wagon1'?
What forces act on 'wagon2+3+4+5'?
 
  • #24
the force that acts on engine+wagon1 should be =wagon2+3+4+5 x acceleration
and
the force that acts on wagon 2+3+4+5 should be=engine+wagon1 x acceleration

am I right?
 
  • #25
AakashPandita said:
the force that acts on engine+wagon1 should be =wagon2+3+4+5 x acceleration
and
the force that acts on wagon 2+3+4+5 should be=engine+wagon1 x acceleration

am I right?
No. There are two forces acting on 'engine + wagon 1': One from the track, which propels the entire train, and one from its contact with wagon 2.

But how many forces act on 'wagon 2+3+4+5' ?
 
  • #26
only from engine+wagon1.
 
  • #27
AakashPandita said:
only from engine+wagon1.
Right. The only force on 'wagon 2+3+4+5' is the force from wagon 1. Since you know the acceleration, can you figure out what that force must be using Newton's 2nd law?
 
  • #28
yes...i understood that.
But i have another problem.
the force exerted by engine+wagon1 should be equal to the reaction force exerted by wagon2+3+4+5...in the opp. direction.
i.e,
-(8000+2000 x a) = a x 2000 x 4
As "a" is same.
10000 should be equal to 8000...but it is not so...tell me how this is not a violation of Newton's third law.
 
  • #29
Aakash, as you now know, there is one force, F on the back part, and two forces, F and E (or rather -F and E) on the front part.

So your equations for the two parts will look different.

What are they? :smile:​
 
  • #30
AakashPandita said:
yes...i understood that.
But i have another problem.
the force exerted by engine+wagon1 should be equal to the reaction force exerted by wagon2+3+4+5...in the opp. direction.
So far, so good. The force that those two pieces exert on each other is equal and opposite.
i.e,
-(8000+2000 x a) = a x 2000 x 4
As "a" is same.
10000 should be equal to 8000...but it is not so...tell me how this is not a violation of Newton's third law.
Careful in how you apply Newton's 2nd law. The force you need is the net force:
ΣF = m*a

When dealing with the 'engine + wagon 1' piece, there are multiple forces involved. Since the net force on each piece will be different, you cannot expect 'm*a' to be the same for each.
 

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