Train Wheels Problem: Solving for Force and Gravity | Homework Statement

[Jakecp]

Homework Statement

It is basicly on the image attached.

Homework Equations

where

is the force on the front tires,

is the force on the rear tires,

is the wheelbase,

is the distance from the center of mass (CM) to the rear wheels,

is the distance from the center of gravity to the front wheels (

+

= [PLAIN]https://upload.wikimedia.org/math/d/2/0/d20caec3b48a1eef164cb4ca81ba2587.png), https://upload.wikimedia.org/math/6/f/8/6f8f57715090da2632453988d9a1501b.png is the mass of the vehicle, and

is the gravity constant.

But i believe they are worthless because we do not have the measure for them.

The Attempt at a Solution

I think that the small wheel train is for passengers because what we need there is speed. With a small wheel we have the ability to turn it faster than in a cargo. Also i think is like a comparison with big trucks and mini cars. A big truck has massive wheels while a car has just little wheels. I think also the number of wheels affects it. And we can see that locomotive I has bigger wheels than locomotive 2.

My answer would be : A)

 

[Jakecp]

Also , traction affects? Bigger the wheel , bigger the traction ? Why? Is that true? or the smallest the wheel the bigger the traction? and why?

regards!

 

[haruspex]

Take note of where the shaft from the piston attaches to the wheels.

 

[Jakecp]

Take note of where the shaft from the piston attaches to the wheels.

I don't really get it , i have seen pistons in photos in google of freight trains and passenger trains. I see that in the locomotive they are the same.
Image of a freight train :
http://previews.123rf.com/images/bo...-closeup-of-wheels-and-boiler-Stock-Photo.jpg

Image of a passenger locomotive train :

http://c8.alamy.com/comp/DWCCB4/gnr...express-passenger-locomotive-built-DWCCB4.jpg

I see the wheel size difference but i don't know why , if you can explain me why would be awesome. I just noticed that there are more wheels in the freight than in the passenger and that there are more pistons in the freight one because of the wheels . But i don't see a position change of the pistons , i see them in the borders and that they way they are crossed in the freight train is not the same as in a passenger train .

 

[haruspex]

But i don't see a position change of the pistons

Look at where the shaft attaches to the wheel in relation to the radius of the wheel. Think about torque, and the relationship between angular velocity and linear velocity.

 

[Jakecp]

Look at where the shaft attaches to the wheel in relation to the radius of the wheel. Think about torque, and the relationship between angular velocity and linear velocity.

To be honest i have not seen angular nor linear velocity yet at school so i have no idea. I will do some research . I have observed that ( the shaft relation ) and you are right , but why is that? why is it better to put the shaft like that?

 

[haruspex]

To be honest i have not seen angular nor linear velocity yet at school so i have no idea. I will do some research . I have observed that ( the shaft relation ) and you are right , but why is that? why is it better to put the shaft like that?

Have you studied levers and mechanical advantage?

 

[Jakecp]

Have you studied levers and mechanical advantage?

Yes but why is it different , i mean , both need more MA right? , one needs to go faster but that does not means it needs less MA. What kind of lever is it? Type 1 , 2 or 3?

 

[haruspex]

Yes but why is it different , i mean , both need more MA right? , one needs to go faster but that does not means it needs less MA. What kind of lever is it? Type 1 , 2 or 3?

The categorisation of levers into those three types is a bit artificial. I think it is only taught in schools because it is easy to teach. The shaft/wheel/rail system does not fit neatly into it.
Let's label the centre of the wheel O, the point of attachment of the beam B, and the point of contact of the wheel with the rail R.
The power stroke from the piston occurs while B is below O. How would you assign the roles of fulcrum etc. to these three points?

It is more appropriate to consider the mechanical arrangement of the engine to be applying a couple to each wheel: during the power stroke, there is a roughly horizontal force from the axle at the centre of the wheel and a roughly equal and opposite force from the drive shaft where it attaches to the wheel. The result is a torque tending to rotate the wheel.
Correspondingly, the frictional force from the rail exerts an opposing torque.
How does the distance ratio OB:OR affect these torques?

 

[Jakecp]

The categorisation of levers into those three types is a bit artificial. I think it is only taught in schools because it is easy to teach. The shaft/wheel/rail system does not fit neatly into it.
Let's label the centre of the wheel O, the point of attachment of the beam B, and the point of contact of the wheel with the rail R.
The power stroke from the piston occurs while B is below O. How would you assign the roles of fulcrum etc. to these three points?

It is more appropriate to consider the mechanical arrangement of the engine to be applying a couple to each wheel: during the power stroke, there is a roughly horizontal force from the axle at the centre of the wheel and a roughly equal and opposite force from the drive shaft where it attaches to the wheel. The result is a torque tending to rotate the wheel.
Correspondingly, the frictional force from the rail exerts an opposing torque.
How does the distance ratio OB:OR affect these torques?

The fulcrum i believe would be B right? And the rest i have no idea. I have gathered data and will ask my teacher about this today.

 

[haruspex]

The fulcrum i believe would be B right? And the rest i have no idea. I have gathered data and will ask my teacher about this today.

The fulcrum is generally taken to be a point that is fixed. Which of the three points is (for the instant) stationary?

 

[Jakecp]

Yes , the fulcrum would be in the wheel. Because it would never move. In the side where the piston moves would be the input and near the fulcrum the outpput (near the wheel) and it would reverse as it rotates.

 

[haruspex]

Yes , the fulcrum would be in the wheel. Because it would never move. In the side where the piston moves would be the input and near the fulcrum the outpput (near the wheel) and it would reverse as it rotates.

Sorry, but I'm not at all sure what you are saying there.
My previous post was trying to get you to say that the fulcrum must be the point R, since that is stationary instantaneously. However, this leads to a bizarre result. This demonstrates the traditional type classification of levers has some issues.
As I mentioned, the engine exerts two horizontal forces on the wheel. The piston pushes backwards at point B. In doing so, the reaction pushes forwards on the piston, creating a forward force on the engine, and thus a forward force on the wheel axis. This combination of forward and backwards forces on the wheel generates a torque.
For the purposes of this question, think in terms of the reference frame of the engine. It is trying to push the rails backwards. Thinking this way, you can take the wheel centre O as the fulcrum and the rails (point R) as the load.

 

[Jakecp]

Newtons third law of action reaction . One force pushing forward and one backwards... Yes i get that point , But i do not understand why R is the fulcrum and not the center of the wheel. If the center of the wheel won't move also.

 

[Jakecp]

The fact is that i can't picture that in my mind but i will try to draw it and get some conclutions & will tell you my conclutions here later.

 

[haruspex]

Newtons third law of action reaction . One force pushing forward and one backwards... Yes i get that point , But i do not understand why R is the fulcrum and not the center of the wheel. If the center of the wheel won't move also.

No, the centre of the wheel must move, or the train would not advance.
But my point there was that taking R to be the fulcrum does not in fact work. I was trying to show that the simplistic view of levers is not general enough to deal with this case.
You can get the right answer by taking the centre of the wheel as the fulcrum, but it is far from obvious why that is the right role assignment and the other is wrong.

 

[Jakecp]

Oh ok , i understand your point now... You wanted to mean that this kind of levers didn't fit in the class 1,2,3 lever clasification.

 

[Jakecp]

In here traction and motion takes a big role because of Newton Second law of motion which says that acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).So we have :

F = MAAnd here Traction gets into action because with more traction you are able to move more mass. It takes a big role because in one case we need to get into motion a large quantity of weight and in the passenger train we just need to move it faster. Getting smaller and more wheels gives the engine the ability to rotate less times per wheel rotation than having less wheels and bigger wheels. This gives more traction because the engine produces more force without the need to rotate fast or more times.

We can compare this with trucks and cars. We see that trucks ( big trucks )have more wheels than cars and that their wheels are bigger.

Also , here the third law of Newton applies also . It states that for every action there must be a reaction . The action would be the wheel pushing the rail backwards and the rail pushing the train forward.

Taking this into comparison with big trucks , using the second law of Newton and seeing that traction takes a big role in this experiment i would say that the answer is (a) . In any other experiment we should take air resistance into account and also that in cargo trains there is the need to go backwards or do more stops which require the train to have a different kind of body. In this case we can not see any drastic or noticeable changes on the body because of air resistance or maneuverability .
BibliographyCsep10.phys.utk.edu,. (2015). Newton's Three Laws of Motion. Retrieved 20 November 2015, from http://csep10.phys.utk.edu/astr161/lect/history/Newton3laws.htmlPhysics Forums - The Fusion of Science and Community,. (2015). Train wheels problem. Retrieved 20 November 2015, from https://www.physicsforums.com/threads/train-wheels-problem.843739/reddit,. (2015). Passenger vs Freight locomotives? • /r/trains. Retrieved 20 November 2015, from http://www.reddit.com/r/trains/comments/30ezzx/passenger_vs_freight_locomotives/Teachertech.rice.edu,. (2015). Newton's 3 Laws of Motion. Retrieved 20 November 2015, from http://teachertech.rice.edu/Participants/louviere/Newton/law2.html

 

[haruspex]

Oh ok , i understand your point now... You wanted to mean that this kind of levers didn't fit in the class 1,2,3 lever clasification.

I mean that it is not clear how to fit it into that classification.
But it does turn out that if you settle on the wheel centre as the fulcrum, the shaft attachment as the applied force, and the rail (i.e. its frictional force on the wheel) as the load you will get the right answer. Stick with this, the answer is all about mechanical advantage and ratios! (Your post #18 is completely irrelevant.)
Study the two pictures, viewing the radius of the wheel from O to B to R as a lever. Look at the ratio of the distances OB, OR. Taking O as the fulcrum, what does this ratio say about mechanical advantage? How is the ratio different between the two trains?

 

[Jakecp]

That is my answer to the homework (#18) . Well , i think the train for freight has a better ratio because the wheels are smaller.In here : http://c8.alamy.com/comp/DWCCB4/gnr...express-passenger-locomotive-built-DWCCB4.jpg

OB distance is very small and OR is quite a big distance. Thinking of it as a lever it may have a MA under 1 . Because it would be like : Input to fulcrum distance : about 25 cm . (OB) Fulcrum to output about 100 cm (OR)so it would be ¼ or .25 .In here : https://c2.staticflickr.com/6/5024/5787221601_92b16728f0_b.jpg i would say that O B is of about 10 cm and O R is about twice as 0 B so it would be ½ . I don't know if i am or not correct.

 

[haruspex]

That is my answer to the homework (#18) . Well , i think the train for freight has a better ratio because the wheels are smaller.In here : http://c8.alamy.com/comp/DWCCB4/gnr...express-passenger-locomotive-built-DWCCB4.jpg

OB distance is very small and OR is quite a big distance. Thinking of it as a lever it may have a MA under 1 . Because it would be like : Input to fulcrum distance : about 25 cm . (OB) Fulcrum to output about 100 cm (OR)so it would be ¼ or .25 .In here : https://c2.staticflickr.com/6/5024/5787221601_92b16728f0_b.jpgi would say that O B is of about 10 cm and O R is about twice as 0 B so it would be ½ . I don't know if i am or not correct.

Yes, that looks roughly right, but to answer the question you were given, you need to be referring to the two pictures there.
There, A has OB:OR about 1:3 or 1:4, B has more like 2:3.
But what do you mean by a 'better' ratio. Which ratio will give the greater top speed? Which ratio will give the greater tractive force?

 

[Jakecp]

A has moré tractive force and b more speed

 

[Jakecp]

But why ?

 

[haruspex]

But why ?

The engines are essentially the same, so the pistons operate at the same rates. Suppose the max speed of the piston end is 30kph. If it attaches near the rim of the wheel, the max tangential speed of the wheel (and hence the max speed of the train) won't be much more. If it attaches only a thrid of the way from the centre of the wheel to the rim, the tangential speed at the rim can go up to 90kph.

 

[Jakecp]

Thanks a lot!

 

[Jakecp]

Basiclly the levers do not have mechanical advantage because they are MA < 1 right? I got it thanks! Because the maximum the piston can move is 30kph but that is not the maximum amount of force it can produce but just the speed. So it can be moving at 15kph with 150 Newtons of force and it would multiply by 3 ( the speed ) because of the 1:3 in the passenger train and the speed would be of 45kph while the piston is moving at 15kph right?.
Love physics! Thanks for your help.!

 

[haruspex]

Basiclly the levers do not have mechanical advantage because they are MA < 1 right? I got it thanks! Because the maximum the piston can move is 30kph but that is not the maximum amount of force it can produce but just the speed. So it can be moving at 15kph with 150 Newtons of force and it would multiply by 3 ( the speed ) because of the 1:3 in the passenger train and the speed would be of 45kph while the piston is moving at 15kph right?.
Love physics! Thanks for your help.!

"Mechanical advantage" does not have to be more than 1. It is just the ratio of output force over input force. The ratio of input and output speeds will be the other way around: you can arrange to get twice the force if you accept getting half the speed, or you can get twice the speed if you accept only getting half the force.
The basic laws are power in = power out, and power = force x speed.
See https://en.m.wikipedia.org/wiki/Mechanical_advantage

 

[Jakecp]

but , how is it possible for that to be efficient if you are losing the force initialy applied?

 

[haruspex]

but , how is it possible for that to be efficient if you are losing the force initialy applied?

There need not be much power lost. Inefficiency relates to loss of power, not loss of force.
The work done by a force is the magnitude of the force multiplied by the distance it advances. If the lever turns it into half the force but moving twice the distance, it does the same work.

 

[Jakecp]

You are right in that , but i don't get at 100% how does that happen? This : you can arrange to get twice the force if you accept getting half the speed, or you can get twice the speed if you accept only getting half the force.

 

[Jakecp]

I looked here https://en.m.wikipedia.org/wiki/Mechanical_advantage#Speed_ratio but it is just the MA

 

[haruspex]

You are right in that , but i don't get at 100% how does that happen? This : you can arrange to get twice the force if you accept getting half the speed, or you can get twice the speed if you accept only getting half the force.

Consider two of the front gears on a bicycle, one twice the diameter of the other. Uphill, in low gear, I apply a steady torque to the pedals and turn them at some constant rate. That exerts a force F on the chain and and makes it move at speed v. Cresting the hill, I change to the high gear, but find applying the same torque keeps the pedals rotating at the same rate as before. The force on the chain drops to F/2, but the chain moves twice as fast. The power transferred is Fv in both cases.