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

• ChrisBaker8
In summary: Frictional force = 21,000N Therefore, velocity = power/frictional force = 3530KJ/21KJ = 168.1m/sIn summary, the question asks to find the tension in the couplings between the train and engine at the instant when the train is traveling at a speed of 80km/h and the engine is exerting a power of 4000kW. The train has a mass of 400 tonnes and experiences a frictional force of 20,000N, while the engine has a mass of 100 tonnes and experiences a frictional force of 1,000N. It is assumed that the train is not accelerating at this speed. The driving force
ChrisBaker8

## 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.

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?

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

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?

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

power = force * velocity

Question 1:

## What is tension and why is it important to calculate it in couplings of a 100T engine and 400T train?

Tension is the force that pulls or stretches an object. It is important to calculate tension in couplings of a 100T engine and 400T train because it determines the amount of force that is exerted on the couplings and ensures the safe operation of the train.

Question 2:

## What factors affect the tension in couplings of a 100T engine and 400T train?

The tension in couplings of a 100T engine and 400T train is affected by the weight of the train, the speed at which it is moving, the angle of the couplings, and the resistance of the tracks.

Question 3:

## How is tension calculated in couplings of a 100T engine and 400T train?

The tension in couplings of a 100T engine and 400T train can be calculated using the formula T = F x d, where T is the tension, F is the force applied to the couplings, and d is the distance between the couplings. This formula takes into account the factors that affect tension mentioned in the previous question.

Question 4:

## What are some potential problems that can arise from incorrect tension in couplings of a 100T engine and 400T train?

Incorrect tension in couplings of a 100T engine and 400T train can lead to couplings breaking, which can cause accidents or delays in train operations. It can also put excessive strain on the engine and other components of the train, leading to potential damage and breakdowns.

Question 5:

## How can tension in couplings of a 100T engine and 400T train be adjusted and maintained?

Tension in couplings of a 100T engine and 400T train can be adjusted by changing the force applied to the couplings or by adjusting the distance between the couplings. It is important to regularly check and maintain the tension in couplings to ensure safe and efficient operation of the train.

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