Distance travelled whilst decelerating

[gary32]

Homework Statement

A train of 80tonnes moves at a velocity of 70km/hr up to the bottom of a gradient of 8degrees
The power is turned off and the train allowed to coast up the gradient before coming to rest,
If the tractive resistance to motion is 70Newtons per tonne, calculate the distance moved by the train up the gradient before coming to rest.

Homework Equations

Acceleration = Final velocity - initial velocity?
Force = Mass x acceleration
s=(u+v/2)t

The Attempt at a Solution

I have calculated the following to start with;
Train mass = 80000kg
Velocity = 19.4m/s
tractive resistance = 0.07Newtons per kg

The problem again is not having a clue what formula(s) to use in which order etc

I tried To use F=MA to calculate force then minus the tractive resistance
then convert back to acceleration which i then get 19.33m/s, but that doesn't seem right at all
Any ideas?

Thanks,

 

[voko]

Using F = ma is correct.

First you need to find out all the forces acting on the train.

 

[gary32]

Using F = ma is correct.

First you need to find out all the forces acting on the train.

I calculated 1552000N for the force of the train moving,
The only force I know from info provided is tractive resistance so,
80000 x 0.07 = 5600N
I guessed that the tractive resistance is subtracted from the train moving but didnt have a clue from where to go then, I can't find a formula that seems to contain the parameters I have

 

[voko]

Assume for a second there is no tractive resistance. Would the train be able to go up the gradient indefinitely?

 

[SteamKing]

Acceleration is not just the difference in velocity. Remember, the units of acceleration are m/s^2.

Think about the problem for a minute before worrying about what formulas to use. The train is merrily going on its way at constant velocity. When the train encounters the hill, the engineer closes the throttle and let's the train coast up the hill. When going up the hill, why does the train's velocity drop? Is it just because of the resistance to motion? Or is it also because when traveling up the hill, the train is being raised to a higher elevation?

 

[CWatters]

How about drawing a diagram showing the forces acting on the train when it's on the slope?

 

[gary32]

Assume for a second there is no tractive resistance. Would the train be able to go up the gradient indefinitely?

The train will go further if there was no tractive resistance, however, it wouldn't go on forever due to gravity?

Acceleration is not just the difference in velocity. Remember, the units of acceleration are m/s^2.

Think about the problem for a minute before worrying about what formulas to use. The train is merrily going on its way at constant velocity. When the train encounters the hill, the engineer closes the throttle and let's the train coast up the hill. When going up the hill, why does the train's velocity drop? Is it just because of the resistance to motion? Or is it also because when traveling up the hill, the train is being raised to a higher elevation?

It is because it is traveling to a higher elevation, gravity comes into play?
My lecturer hinted that its the height distance I calculate, then use trigonometry to calculate the distance traveled up the gradient of 8degrees.

How about drawing a diagram showing the forces acting on the train when it's on the slope?

I have now drawn one, I'm not sure quite sure if this is correct but below is my attempt,

The train is moving at a force of, F = 80000 x 19.4 = 1552000N
The tractive resistance force is F = 80000 x 0.07 = 5600N
so we subtract tractive resistance force from moving force
1552000-5600 = 1546400N
Our acceleration now (or rather deceleration) with tractive resistance implemented is
1546400N/80000 = 19.33m/s

Then v^2 = u^2 + 2a x d
so d = -u^2 / 2a

distance = -19.33^2 / (2 x 9.81) = 19m (vertical distance?)
Using trigonometry,
Distance (on the gradient) = 19m/sin(8) = 136.5m

That sounds a reasonable distance to me?

If the above is correct, my next step is to
"Prove my answer by an alternative method, compare and contrast the 2 methods."
1 Method to be used is conservation of energy (as used) and the other is D'Alembert's principle which I have never done, I have just went through a powerpoint explaining the D'alemberts principle but it was far to complex for me. Is there a simplified explanation or could someone summarize the principle for me.

All I gathered from it that it is used when there are multiple forces on a mass, which can be used where an object can move in xyz.

 

[haruspex]

The train will go further if there was no tractive resistance, however, it wouldn't go on forever due to gravity?

Right, so where is gravity in your equations?

I have now drawn one, I'm not sure quite sure if this is correct but below is my attempt,

I don't see a diagram attached.

The train is moving at a force of, F = 80000 x 19.4 = 1552000N

The train isn't moving with a "force". It has momentum, given by mass * velocity, but the units would not be Newtons.

The tractive resistance force is F = 80000 x 0.07 = 5600N
so we subtract tractive resistance force from moving force
1552000-5600 = 1546400N

It makes no sense to subtract a force from a momentum.

 

[gary32]

Right, so where is gravity in your equations?

I don't see a diagram attached.

The train isn't moving with a "force". It has momentum, given by mass * velocity, but the units would not be Newtons.

It makes no sense to subtract a force from a momentum.

Gravity is here: distance = -19.33^2 / (2 x 9.81) = 19m (vertical distance?) However, I presume this was incorrect,

The diagram was drawn on paper, I was under the impression the diagram was for my benefit only.

I see now how the train isn't moving with force seeing that there isn't any forces pulling or pushing it now, it is clearly momentum as you said.

I see now that a force can't be taken from momentum like that (now it occurred to me that the train is under momentum not force), I have still not a clue how to work this one out :(

Just a clue of what formulas I need to use would be a good start, Thanks again

 

[haruspex]

Gravity is here: distance = -19.33^2 / (2 x 9.81) = 19m (vertical distance?) However, I presume this was incorrect,

Gravity and traction resistance both operate over the distance the train travels. This means you can't handle them separately. You need one equation that allows for both retarding forces.

The diagram was drawn on paper, I was under the impression the diagram was for my benefit only.

You wrote

I have now drawn one ... below is my attempt

I see now how the train isn't moving with force seeing that there isn't any forces pulling or pushing it now, it is clearly momentum as you said.

I see now that a force can't be taken from momentum like that (now it occurred to me that the train is under momentum not force), I have still not a clue how to work this one out :(

Just a clue of what formulas I need to use would be a good start, Thanks again

What are all the forces that will tend to retard the train?
What is the component of each in the direction opposing the motion of the train?
What do they add up to?
From here there are two ways:
- calculate the acceleration
- use the work-energy theorem

 

[vadevalor]

Using method one is faster: v2=u2+2as
Kinematics equation to find s after finding acceleration thru F=ma.
Tedious part is to find resultant force in the direction that retards the train first.

 

[gary32]

Gravity and traction resistance both operate over the distance the train travels. This means you can't handle them separately. You need one equation that allows for both retarding forces.

You wrote


What are all the forces that will tend to retard the train?
What is the component of each in the direction opposing the motion of the train?
What do they add up to?
From here there are two ways:
- calculate the acceleration
- use the work-energy theorem

I see what you mean now by what I wrote, I meant I drawn it and below was my calculation attempt

The forces retarding the train would be gravity and tractive resistance,
The tractive resistance is Newtons/mass but I guess I would be needing to formulate this into some sort of negative acceleration?
And gravity is basically an acceleration (or rather deceleration in this instance)

But I don't have a clue what formulas I need, I have tried searching for formulas but I find ones that I can't seem to work into the question

Thanks again,

 

[vadevalor]

you need W= mg for weight
and to calculate the frictional force.
Calculate force in the direction of the retarding of the train through trigonometry then add them up to find total force slowing the train down. I thnk you can do the next part now.
Formulas:W=mg
F=ma
v2=u2 + 2as

 

[gary32]

you need W= mg for weight
and to calculate the frictional force.
Calculate force in the direction of the retarding of the train through trigonometry then add them up to find total force slowing the train down. I thnk you can do the next part now.
Formulas:W=mg
F=ma
v2=u2 + 2as

So gravitational force slowing train down is 80000 x 9.81 = 784800N?
I can't calculate the force retarding the train using trigonometry as I only have 1 angle, no distances yet?

I know it seems like I am just being lazy here but I don't have a clue when it comes to mechanical physics, I don't have the ability to understand yet what affects what, what can add up and can't etc I need to have completed this coursework for this coming monday (and I am away for 2 days after tomorrow), I am starting to panic a little now so that's not helping,

I really appreciate everyones help so far, I ideally need someone to dumb everything down for me

 

[vadevalor]

Did this even though its not tested and i' m rushing to study for exams but oh well :D you helped me a bit there in circuits too

 

[gary32]

Did this even though its not tested and i' m rushing to study for exams but oh well :D you helped me a bit there in circuits too

View attachment 59879

Thanks, I only needed a hint on order of equations to be used and how to be implemented, that drawing is pretty much exactly what I drawn,

However, I came to an answer that does not seem correct, my workings are below,

Retarding force = 80000 x 9.81 / sin8 = 5639020.7N
Friction Force = 80000 x 0.07 = 5600N
Tot Retarding force = 5639020.7N+5600N = 5644620.7N

F=MA so
A = F/M
so A = 5644620.7N/80000 = 70.6m/s

V2 = u2 + 2as
So;
s = (v2 - u2) / 2a
So;
s = 0² - 19.4² / (2 x 70.6) = -2.665m

I change -2.665m to positive 2.665m, because it is deceleration not acceleration.
This however doesn't seem right, where have I gone wrong?

 

[vadevalor]

What's the answer? Ill try to figure tht out (lazy to do checking on calculations)
Seems like you didnt use exact values for initial speed

 

[gary32]

What's the answer? Ill try to figure tht out (lazy to do checking on calculations)
Seems like you didnt use exact values for initial speed

I don't know the actual answer.
the value I used for initial speed was 19.4m/s as you can see in this line
s = 0² - 19.4² / (2 x 70.6) = -2.665m

 

[vadevalor]

Why wouldn't it seem right? Hmm. You have to use 19.444 m/s though. More accurate
Because its negative? Your acceleration should be negative btw. Unless you put deceleration

 

[gary32]

Why wouldn't it seem right? Hmm. You have to use 19.444 m/s though. More accurate

Imagine, a train at 70km/hr going up 8degree gradient which isn't much really, and just let's off throttle, 2.6m isn't much for it to travel, I thought it would go a good few metres, however i must be wrong if the calculations are correct. Prehaps a train service provider could help me haha

19.444 m/s is more accurate yes but that makes little difference to the answer, we are told working to 3sf is suffice.

So would you say that is correct answer now?

Next step is to work it out again but using D'Alemberts principle, know any good tutorials/websites that could teach me the principle? (bare in mind, I am very beginner at mechanical principles)

 

[vadevalor]

Considering the train to be of 80 thousand kg i think that should be it. I don't know what's that principle lol, but i thought your own school tutorials and notes will suffice?

 

[gary32]

Considering the train to be of 80 thousand kg i think that should be it. I don't know what's that principle lol, but i thought your own school tutorials and notes will suffice?

We only cover Pass criteria stuff during our lectures, this is merit criteria so i have not covered this.
So in short, I don't have notes, that's why I don't have much of an idea
The idea is kind of to encourage students to learn to be able to go out there own way and teach themselves which is slightly odd

 

[haruspex]

The forces retarding the train would be gravity and tractive resistance,
The tractive resistance is Newtons/mass but I guess I would be needing to formulate this into some sort of negative acceleration?

It's a force, so Newtons.

And gravity is basically an acceleration (or rather deceleration in this instance)

Again, it's a force. Acceleration is the consequence of that force (to the extent that the force is unopposed).
Seems like you have no idea about free body diagrams.
There are three forces acting on the train: gravity, the normal force N from the track (i.e. acting at right angles to the track), and the tractive resistance T, acting parallel to the track.
The trick is to 'resolve' the forces into their components in two directions at right angles. In this case, I would choose parallel and perpendicular to the track.
The tractive resistance is easy: that all acts parallel to the track so has no component in the other direction.
Similarly, the normal force is all in the perpendicular direction.
Gravity acts straight down. Given that the track is at some angle θ to the horizontal, it resolves into mg cos(θ) perpendicular to the track and mg sin(θ) parallel to it. (Is this really new to you? You would not be expected to be able to solve questions like this without having been taught how to resolve forces.)
Now we can consider summing the forces in the two directions, and that the sum, F, = ma, where a is the acceleration in that direction.
We are only interested in acceleration parallel to the track, so we have F = T+mg sin(θ).
See if you can take it from there.

 

[gary32]

It's a force, so Newtons.
Again, it's a force. Acceleration is the consequence of that force (to the extent that the force is unopposed).
Seems like you have no idea about free body diagrams.
There are three forces acting on the train: gravity, the normal force N from the track (i.e. acting at right angles to the track), and the tractive resistance T, acting parallel to the track.
The trick is to 'resolve' the forces into their components in two directions at right angles. In this case, I would choose parallel and perpendicular to the track.
The tractive resistance is easy: that all acts parallel to the track so has no component in the other direction.
Similarly, the normal force is all in the perpendicular direction.
Gravity acts straight down. Given that the track is at some angle θ to the horizontal, it resolves into mg cos(θ) perpendicular to the track and mg sin(θ) parallel to it. (Is this really new to you? You would not be expected to be able to solve questions like this without having been taught how to resolve forces.)
Now we can consider summing the forces in the two directions, and that the sum, F, = ma, where a is the acceleration in that direction.
We are only interested in acceleration parallel to the track, so we have F = T+mg sin(θ).
See if you can take it from there.

I have attached my FBD,
I honestly have not been taught this particular thing using free body diagrams etc, I have also not seen some formula needed.
I have done F=MA and the distance/acceleration/velocity formula though, however, putting them altogether is a totally different thing to me as I am not confident with merging the formula like I am with electrical formulas for instance.

I have however came to a reasonable answer.

Tractive force = 0.07 x 80000 = 5600N
Normal force = 80000 x 9.81 x sin8 = 109223N
Total Retarding Force = 5600N + 109223N = 114823N

A = F/M
A = 114823/80000 = -1.435m/s (negative for deceleration)

s = (0² - 19.4²) / (2 x -1.435) = 131m

How does that look?
Or is it cos 8 in the normal force calculation?
I feel like I'm being a pain, please assure me I am not haha

Thanks!

 

[voko]

You could solve this using energy and work.

 

[haruspex]

A = 114823/80000 = -1.435m/s (negative for deceleration)

Watch the units: -1.435m/s²

s = (0² - 19.4²) / (2 x -1.435) = 131m

Bingo.

 

[gary32]

You could solve this using energy and work.

Please elaborate, I was supposed to be working the problem out using Conservation of energy (which now looking at it, I have not done silly me I can't remember if I mentioned. What would be the difference in calculations and what method have I used?)

After calculating my answer using Conservation of energy, I was then supposed to use D'Alemberts Principle which I have never covered to work it out again and then compare the 2 methods of calculation

Watch the units: -1.435m/s²

Bingo.

Thanks about the heads up on the units, my lecturer goes nuts haha I always forget, but would the following be true then; Velocity isn't the same as acceleration so the units are difference. Velocity measured in m/s and acceleration measured in m/s²

Also as above if you could contribute, thanks :)

 

[voko]

What is total energy at the bottom? What is it where the train stops? What is the difference and where has it gone?

 

[gary32]

What is total energy at the bottom? What is it where the train stops? What is the difference and where has it gone?

Potential energy = mass x gravity x distance?
So, Potential energy = 80000 x 9.81 x ?
Energy where train stops is
Potential energy = 80000 x 9.81 x ?
Sin8 probably goes into either of them formulas some how and then transpose them both to find distances?

 

[voko]

#29 did not answer any of the questions in #28. Potential energy is relevant, but the question was about total energy.

 

[gary32]

#29 did not answer any of the questions in #28. Potential energy is relevant, but the question was about total energy.

I think I've got it,

KE = 0.5 x Mass x Velocity²
KE = 15054400J

So that is the amount of energy dissipated when it comes to a stop so
height = energy / m x g
height = 15054400 / 80000 x 9.81 = 19.18m
using trig,
hyp (distance) = opp (height) / sin(8)
distance = 19.18/sin(8) = 137.8m

NOW, I know I have not implemented tractive resistance here, I am not sure where I would implement it?
What is Newtons/kg? I kind of figure if I implemented it with an acceleration/velocity it would be word done?

 

[voko]

You have computed the height assuming no resistance. But there is resistance. So the train performs some WORK to overcome it. How are the initial total energy and the final total energy are interrelated with work?

 

[gary32]

You have computed the height assuming no resistance. But there is resistance. So the train performs some WORK to overcome it. How are the initial total energy and the final total energy are interrelated with work?

Initial energy + Work energy = Final total energy?

What is Newtons per kg physically? to me, Newtons is a force, as in weight, as in, mass x gravity and then kg is a mass so to me i see it as mass x gravity / mass which is just gravity so its confusing.
How can I formulate N/Kg into work energy, I have searched the internet and cannot find a formula that uses Newtons mass and energy in it

 

[voko]

Initial energy + Work energy = Final total energy?

Correct.

What is Newtons per kg physically? to me, Newtons is a force, as in weight, as in, mass x gravity and then kg is a mass so to me i see it as mass x gravity / mass which is just gravity so its confusing.

You were able to use the coefficient to find the tractive resistance force earlier, why are you confused now?

How do you compute work when you know the force?

 

[gary32]

Correct.
You were able to use the coefficient to find the tractive resistance force earlier, why are you confused now?

How do you compute work when you know the force?

I calculated tractive resistance force as F = 80000 x 0.07 = 5600N
That is just a force, not work done?
I know work done is Force x Distance, but I don't have an actual distance other than the one I calculate in the end (and tractive resistance needs to be implemented before you work out the distance right?)

EDIT: Unless I am being plain stupid and Work done = Force x Some angle (sin8)?

 

[voko]

I calculated tractive resistance force as F = 80000 x 0.07 = 5600N
That is just a force, not work done?

Correct.

I know work done is Force x Distance

Correct, too.

but I don't have an actual distance other than the one I calculate in the end

So make it some unknown variable, and solve for it.

 

[gary32]

So make it some unknown variable, and solve for it.

You've lost me now

Please give me a clue or example,

 

[voko]

If force is F, and displacement is s, what is the work of the force?

If the displacement is s along the gradient angle a, what is the elevation in the end? And what is the corresponding potential energy?

 

[gary32]

If force is F, and displacement is s, what is the work of the force?

If the displacement is s along the gradient angle a, what is the elevation in the end? And what is the corresponding potential energy?

Work of the force is F x displacement?

If displacement is s along the gradient a,
The end elevation is Sin8 x s
Corresponding potential energy is 15054400J?

I still am lost, honestly, I am probably as dumb as it gets when it comes to mechanics, you need to dumb things down to my level

 

[voko]

Work of the force is F x displacement?

Use the symbol $s$ for displacement.

If displacement is s along the gradient a,
The end elevation is Sin8 x s
Corresponding potential energy is 15054400J?

This cannot be a number, because the end elevation is $s \cdot \sin a$, where $s$ is unknown. So you have to write that down as a formula.

 

[gary32]

Use the symbol $s$ for displacement.



This cannot be a number, because the end elevation is $s \cdot \sin a$, where $s$ is unknown. So you have to write that down as a formula.

So what now? Also, what was the principle I used to initially work the problem out?

 

[voko]

So what now?

Write the work-energy equation in terms of known and unknown variables, and solve it.

Also, what was the principle I used to initially work the problem out?

Not sure whether there is a particular name to it. FBD analysis and first principles of motion, perhaps.

 

[gary32]

Write the work-energy equation in terms of known and unknown variables, and solve it.

Not sure whether there is a particular name to it. FBD analysis and first principles of motion, perhaps.

There are 2 unknown variables (from what I can see) so I can't see a way for solving it,
I will have to not submit this work and not get the mark as I am away until the day it needs submitting.

Thanks for everyones help, it has been appreciated.

 

[voko]

Which two unknowns do you think you have here?

 

[gary32]

Which two unknowns do you think you have here?

Work done is unknown because that's what I am trying to find and s appears to be unknown, i know you've said s sin but that means nothing to me. Of course this is solvable but I don't have a clue how

 

[voko]

Work depends on s. Just write that down as a formula. Then plug that formula into the energy-work equation. That will eliminate the work as an unknown.

 

[gary32]

Work depends on s. Just write that down as a formula. Then plug that formula into the energy-work equation. That will eliminate the work as an unknown.

Work = F x s
Energy-work equation = F d = (1/2 mv²) - (1/2 mv²)

Energy-work equation = F d = (1/2 x 80000 x 0²) - (1/2 x 80000 x 19.4²)

= -15054400J

d = W/F
d = -15054400J/what force?

 

[voko]

Work = F x s
Energy-work equation = F d = (1/2 mv²) - (1/2 mv²)

This is not what you had in #33.