Calculating Tension in Couplings of 100T Engine & 400T Train

  • Thread starter Thread starter ChrisBaker8
  • Start date Start date
  • Tags Tags
    Engine Train
Click For Summary

Homework Help Overview

The problem involves calculating the tension in the couplings between a 100-tonne engine and a 400-tonne train, considering various forces including friction and power exerted by the engine while moving at a constant speed of 80 km/h.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of the provided data, questioning whether the train is accelerating and how that affects the tension calculation. There are attempts to relate power, friction, and energy to the problem.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants suggest that the train's constant speed implies a specific tension value, while others consider the implications of the engine's power and friction forces. There is no explicit consensus on the approach to take.

Contextual Notes

Participants note the potential for additional context or follow-up questions that may not be provided, which could influence the interpretation of the problem. The exact wording of the question is highlighted as potentially ambiguous regarding acceleration.

ChrisBaker8
Messages
23
Reaction score
0

Homework Statement



Engine: 100 tonnes, with 1,000N frictional force
Train: 400 tonnes, with 20,000N frictional force

Speed = 80km/hour
Engine is exerting a power of 4000kW

Find the tension in the couplings between train and engine

The Attempt at a Solution



I'm not sure where to start, I thought of finding coefficient of friction but it didn't play out.
 
Physics news on Phys.org
Draw a diagram with the forces for a train moving at constant speed
 
Engine:
-981,000N vertically both up and down
-Not sure what force is to right (forward). Calculate from power?
-T+1,000N to left

Train:
-3,924,000N vertically both up and down
-T to right
-20,000N to left

wait, if the train isn't accelerating, does that mean T = 20,000N?

if it is, why would they provide all the other data?
 
Last edited:
ChrisBaker8 said:
wait, if the train isn't accelerating, does that mean T = 20,000N?
Exactly

if it is, why would they provide all the other data?
It could be there is another part to the question you haven't been given, or it could just be to confuse you!
 
Last edited:
That's weird, the question is worth 11 marks!

Okay, here's the exact wording:

An engine of mass 100 tonne pulls a train of mass 400 tonne along a horizontal track. There is a horizontal frictional force of size 1 kN acting on the engine and a horizontal frictional force of size 20 kN acting on the train. Find the tension in the couplings between the engine and the train at the instant when the speed of the train is 80 km/h and the engine is exerting a power of 4000 kW.

That doesn't indicate it's accelerating when it's at 80km/h, does it?
 
Correct it's not accelating.
What i meant was that in the advanced class there might be a part 2 to the question that says, eg. now the train is moving up a 10% slope, find the maximum speed for the given engine power.
 
No, there's no follow up question.

Are you sure it's not accelerating? Can I work out the driving force of the Engine from the power it exerts?

Maybe the engine is accelerating and I have to work out what the tension is as ir reaches 80km/h, as it does specify the instant[/i] it is traveling at 80km/h
 
Yes the point was that quite often a textbook or a professor will reuse the easy part of a more advanced question in an easier class.
So it's not unusual for their to be extra information given that you don't need.
It's good practice in spotting which bits of data are relavent for the problem you are studying
 
so, is it possible to calculate acceleration from the power the engine is exerting?
 
  • #10
Yes,
The energy lost to friction is friction_force*distance
So with the speed you can work out the rate of energy lost to friction = power.
Then with the mass you can work out the change in kinetic energy for the power that's left
 
  • #11
80km/h = 22.222222...m/s

so at this speed, the engine is losing (22.222222... x 1,000) joules per second

=-22,222 J/s
=-22,222 W

So that leaves 3,977,778 W

The kinetic energy of the engine at this speed is 24,691,358 J [ii]

so, do I add to [ii], calculate the new speed, and take difference between the speeds to be the acceleration?
 
  • #12
21,000N of friction
At 80km/h it's doing 22.2m/s or 1m in 0.045s

So to move 1m with 21,000N of friction takes 21KJ of energy, in 0.045s = 21,000/0.045 = 470KW
That means the engine at full power has 4000-470KW spare = 3530KW

At 80km/h it has a kinetic energy of = 1/2 mv^2 = 0.5 * 500,000kg * 22.2^2 = 123MJ
In one second it can supply 3530KJ extra energy so KE goes to 123,000+3530 KJ
an then working back with 1/2 mv^2 you can get the new v^2
 
  • #13
power = force * velocity
 

Similar threads

Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
10
Views
4K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K