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

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The discussion focuses on calculating the tension in the couplings between a 100-tonne engine and a 400-tonne train, given specific frictional forces and the engine's power output. Participants explore whether the train is accelerating, concluding that it is not, which suggests that the tension equals the frictional force acting on the train at 20,000N. They also discuss the relevance of the provided data, noting that it may serve to challenge students or prepare them for more complex problems. Calculations involving energy loss due to friction and kinetic energy at a constant speed are examined, emphasizing the relationship between power, force, and acceleration. Ultimately, the tension in the couplings can be determined by analyzing the forces acting on the system.
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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.
 
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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!
 
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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
 
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