Monorail problem with coefficient of fiction and front and rear wheel drive

[wallism]

Hi, this problem’s been bugging me for ages. I’m sure it’s just a simple misunderstanding somewhere on my behalf but I just can’t seem to get my head round it.

Homework Statement

A monorail car with front wheel drive and mass of 5 tonnes is suspended from two wheels 10m apart as shown. Its centre of mass G is midway between the wheels and 4m below the rail.

Starting from rest, the train is designed to achieve a maximum speed of 20km/h in 40s with uniform acceleration. Calculate the minimum coefficient of friction between the front wheel and the rail to achieve this.

Part 2
Determine the acceleration that could be achieved with the same coefficient of friction if rear wheel drive was used, and hence the time to reach 200km/hr.

The answers given are 0.254 and 49.1

I've attached a photo of the diagram.

Homework Equations

F=ma
F=uR
V=at

The Attempt at a Solution

1. Firstly calculate the acceleration in m/s^2
To do this I converted the speed
200 km/hr is equivalent to 200x10^3 m/hr
Divide into seconds 200x10^3/3600 = 55.5 m/s
Then divide by 40s to get the acceleration needed (v/t=a)so a = 1.38ms^-2 or 25/18

2. Calculate the force needed to accelerate the mass
F=ma, m=5000kg, a=1.38ms^-2
5000 x 1.38 = 6944.4N or 62500/9

3. Work out the coefficient of friction.

a) Work out the reaction force of the front wheel on the beam, using moments and resolving forces, giving you 24525N (2500kg x 9.81)

b) Use F=uR
Calculated F needed as 6944.4, and R as 24525N

Rearrange formula F/R=u
6944.4/24525 = 0.283 which is wrong!?

Surely the centre of mass G must affect it somehow but I don’t understand how to incorporate that. I can’t do the second part of the question because of the reason above. If the centre of mass was off centre then I could understand how the coefficient of friction and acceleration between front and back would differ.

Any help would be greatly appreciated.

 

[gneill]

Consider that the force that drives the monorail forward is not acting through its center of mass. Rather, the friction force of the tires on the rail acts horizontally above and to the front (or rear) of the center of mass. There will be components, moments, etc., to consider. What center would you choose for moments affecting the normal force acting on the drive wheel? What forces produce those moments?

 

[wallism]

Finally! I’ve cracked the first part. Thank you very much for your help.

I’m still a little unsure as to how to do the second part though. It seems straight forward enough after doing the first bit. It’s basically doing the same thing but backwards.

The bit I’m unsure about is finding the reaction force and friction.

I’m starting off by taking moments around the bottom left to find out the reaction forces. I need a value of friction to start off with. Surely I can’t just use the same value as friction for the first part? So that leaves me with two unknowns in my equation.

Starting with sum of moments clockwise should equal 0
(R2 being the right hand support, F being friction)

5G - 10(R2) + 4F = 0

If I do use F from part one it gives the wrong answer (50.3 seconds), obvious because the reaction force of the rear wheel will be different from the front so the friction will differ.

Alternatively use the friction from part one to find R2 with moments then resolve to find R1. Then use R1 to find a new friction (F = u R, been given u and have already calculated R), which gives a new friction of 5523.79N. Use this new friction in the old equation to find new values of R2 and R1.

5G - 10(R2) + 4F = 0

Use the new R1 of 22315.484N to get an acceleration of 1.1336 and a time of 49.0 seconds. It’s close to the real answer but I’m pretty sure it’s not a correct way to do it and that there’s a far better and more logical approach.

Thanks Again

 

[PeterO]

Finally! I’ve cracked the first part. Thank you very much for your help.

I’m still a little unsure as to how to do the second part though. It seems straight forward enough after doing the first bit. It’s basically doing the same thing but backwards.

The bit I’m unsure about is finding the reaction force and friction.

I’m starting off by taking moments around the bottom left to find out the reaction forces. I need a value of friction to start off with. Surely I can’t just use the same value as friction for the first part? So that leaves me with two unknowns in my equation.

Starting with sum of moments clockwise should equal 0
(R2 being the right hand support, F being friction)

5G - 10(R2) + 4F = 0

If I do use F from part one it gives the wrong answer (50.3 seconds), obvious because the reaction force of the rear wheel will be different from the front so the friction will differ.

Alternatively use the friction from part one to find R2 with moments then resolve to find R1. Then use R1 to find a new friction (F = u R, been given u and have already calculated R), which gives a new friction of 5523.79N. Use this new friction in the old equation to find new values of R2 and R1.

5G - 10(R2) + 4F = 0

Use the new R1 of 22315.484N to get an acceleration of 1.1336 and a time of 49.0 seconds. It’s close to the real answer but I’m pretty sure it’s not a correct way to do it and that there’s a far better and more logical approach.

Thanks Again

Is there any chance they mean "rear wheel drive as well" ?

 

[gneill]

Develop an expression for the normal force on the drive wheel given some acceleration "a" of the monorail. It'll be the sum of half the weight of the monorail and the (upward) force due to the moment about the front wheel.

The frictional force F_f should equal $\mu\;F_n$, and $F_f = m\;a$. You should be able to solve for a.

 

[NascentOxygen]

Is there any chance they mean "rear wheel drive as well" ?

I doubt it. That would make a third case.

Consider a motorcycle. With rear-wheel drive, it can do wheelstands under high acceleration. With the front wheel off the ground, this means all of the weight is being carried by the rear wheel, so this allows it to grip better and you can accelerate faster. Up to that point, the harder you accelerate, the better it grips. But with the front tyre not touching the road, it means you lose steering.

If your motorcycle was front-wheel drive, it could not do wheelstands. During acceleration the weight is constantly shared by both wheels, so by bearing less weight the driven wheel now won't get as good a grip. But the cycle is more controllable, since you all the time have both wheels on the road.

 

[infolil]

Hello everybody
I have a problem of monorail but I can't solve it.
Please could you help me?
I have a problem to solve as follows:

A monorail system consists of a car suspended on a horizontal cable.
The car is driven by one of the wheels. If the coefficient of friction
between the driving wheel and the cable is µ=0.5.

Determine:
1. which of the 2 wheels will cause a movement of the car?
2. calculate the maximum acceleration.

thank you.

 

[infolil]

How I can know the driving wheel?
The monorail car is suspended from two wheels with distance L. Its centre of mass G is midway between the wheels and 0.4L below the rail.

 

[gneill]

How I can know the driving wheel?
The monorail car is suspended from two wheels with distance L. Its centre of mass G is midway between the wheels and 0.4L below the rail.

You can't know the driving wheel. Your problem doesn't provide enough information to determine it.

Take a look at the problem as posed in this thread. There's much more information provided for that version of the problem.

 

[infolil]

Thank you
This thread is very interessant but we haven't the same problem.
I attached the statement . May be, this will more clear but I haven't other information.

 

[gneill]

Thank you
This thread is very interessant but we haven't the same problem.
I attached the statement . May be, this will more clear but I haven't other information.

Ah! That's different! The wording is asking you to determine which would be the better wheel for driving the car. You'll have to determine which wheel would allow for the most acceleration without the drive wheel slipping. It will involve writing equations for moments about different points in order to determine the normal forces that act between the wheel and cable for the two situations.

 

[infolil]

Thanks

I tried. Is that fair and I'm in the right direction?

 

[gneill]

I think you've got the right ideas. If I recall correctly from when I looked at this thread back in August, I found that front wheel drive was the better choice.

 

[infolil]

Thank you for your effort.
I finished the problem.
Best regards

 

[moram]

i know its an old thread but could anybody post a correct solution to this question, as i am trying to do the same one and have done the same method as the OP and got the same wrong answer of 0.283 and can't see how to get the correct answer. thanks

 

[SteamKing]

Sorry. At PF, it's a policy not to encourage users to resurrect old posts. If you have a problem, post it in its own thread. You can always reference any previous threads in your new post.

It's also a big policy in the HW forums that we don't post solutions to any problems. Show your work and tell us where you are having a problem. We will make suggestions to help you work past your difficulties, but you have to work out your own solutions.