## Pushing a railway car without spinning the wheels?

Hi everyone!

I'm struggling to get my head around a calculation having not done any physics studies for 10 years - My brain is a little rusty.

Railway trucks (railroad cars) are shunted around a yard using a tractor (Its a farm tractor on rubber tyres).

The rail itself is recessed into the concrete surface of the yard so that the steel wheels of the cars run on the rail and the tractor's rubber tyres run on the concrete.

The tractor is driven up to the rear of the car and simply pushes it via a steel attachment on the front.

Each truck weighs 80T (including the load) and up to 8 of them have to be pushed at once.

I have calculated the force required to push the load as follows:

F=N*Cr

F= force required
N= Vertical force of load (the weight)
Cr= Coefficient of friction

F=8*80*0.01 = 6.4T

We know from experience that the tractor can push this load without problems.

However, we would like to know how many trucks the tractor could push without breaking traction.

To do this I need some assistance!

I think I need to know:
Coefficient of friction between concrete and rubber
The weight of the tractor (4600kg)
The torque applied to the wheels
Diameter of the wheels (1.5M rear / 1.15m front)

The problem is that we dont know the wheel torque, just the flywheel torque. This can be done fairly easily if we know the gear and final drive ratio (The tractor manufacturer says they can provide that information) Until they've provided the information, lets assume a gear ratio of 1.5 and a final drive ratio of 4.5

The other issue is that the tractor is 4 wheel drive, has different size wheels front/rear and has a variable centre differential! For this calculation we're happy to assume a 60:40 torque split rear/front).

I've confused myself this morning so this may be really easy to calculate - I just cant see it!
 PhysOrg.com physics news on PhysOrg.com >> Study provides better understanding of water's freezing behavior at nanoscale>> Soft matter offers new ways to study how ordered materials arrange themselves>> Making quantum encryption practical
 Recognitions: Homework Help Some things to think about... The trucks are on wheels that turn? If so then the friction between the wheels and the ground won't resist the movement like that. The dominant retarding force will come from friction in the axles... this will depend on the speed so for a constant applied force the tractor+trucks accelerate to a maximum speed which may be less than the maximum speed of the tractor by itself. The maximum force the tractor can apply, though, will be it's weight multiplied by the coefficient of static friction between the tires and the concrete. It will have to oppose it's own losses as well as those in the trucks it pushes - when the applied force is equal to the total retarding forces, the trucks + tractor move at constant speed. While the applied force is higher - the whole will accelerate. Does the tractor push the trucks through their center if mass? If not, then the tractor is applying a torque to the trucks. Not all the force will go into horizontal motion.
 Thanks for your reply, Simon. The trucks are on standard rail wheels/axles that are free to turn when the brakes are released - The tractor is used to shunt them around in the yard so they can be loaded at a platform. Once loaded they're pushed out of the way into a siding area from where a railway locomotive takes them away. They only shunt the trucks at a little more than walking pace. The tractor is pushing on the trucks via their standard railway coupling (on the centre line of the truck, approx. 1.2m from the floor).

## Pushing a railway car without spinning the wheels?

It seems like you have two or three or four questions in your post:
1. How much hp does it take to push 1 railroad car down a track at a certain velocity.
Knowing my tractors hp output, how many railroad cars in total should my tractor be able to push.
2. Knowing the static coeficient of friction between my tractor tires and concrete, will the tractor be able to push that many cars or will the wheels spin?
3. Or, if the wheels do spin in 2. then knowing the static coeficient of friction between my tractor tires and concrete, how many cars can be pushed, before the wheels are just about to spin?
4. Can my tractor even spin its wheels on concrete?
 Recognitions: Homework Help There's also the issue of static friction in the axles of the rail cars, and rolling resistance in the axles and wheels of the rail cars. Getting the rail cars initially at rest to move will require the most force. Once they are moving, then you're dealing with kinetic friction in the axles (less than the static friction), and rolling resistances.

 Quote by rcgldr There's also the issue of static friction in the axles of the rail cars, and rolling resistance in the axles and wheels of the rail cars. Getting the rail cars initially at rest to move will require the most force. Once they are moving, then you're dealing with kinetic friction in the axles (less than the static friction), and rolling resistances.
If you have watched Thomas the Tank Engine (*) carefully you will have seen this effect mentioned. There is a certain amount of slack available in the U.K. style couplings on rail cars so that the engine can "pick up" one car at a time and get it moving before the coupling comes tight and the next car is picked up.

With U.S. style couplings I don't think that there's much slack, but as a train starts rolling you can definitely hear a kind of sequence of clanks as the couplings come tight from front to back.

(*) In the relevant episode a steam engine had broken its tie rods and had to work with less than the normal amount of traction. The remedy was to loosen up the couplings, providing more slack so that the relevant static friction would be on one car at a time.

 Quote by 256bits It seems like you have two or three or four questions in your post: 1. How much hp does it take to push 1 railroad car down a track at a certain velocity. Knowing my tractors hp output, how many railroad cars in total should my tractor be able to push. 2. Knowing the static coeficient of friction between my tractor tires and concrete, will the tractor be able to push that many cars or will the wheels spin? 3. Or, if the wheels do spin in 2. then knowing the static coeficient of friction between my tractor tires and concrete, how many cars can be pushed, before the wheels are just about to spin? 4. Can my tractor even spin its wheels on concrete?
I'm trying to answer all of the above.

Recognitions:
Homework Help
 Quote by Simon Bridge The dominant retarding force will come from friction in the axles... this will depend on the speed
Is that right? Why does it depend on speed when normally, e.g. pushing a box across a floor, it is considered not to depend on speed?
I would have thought that the friction in the axles would be a constant (for a given axle load), so steady state is reached when resistance * speed = power of tractor.
 Recognitions: Homework Help @Haruspex: That's what I'd have thought too but the resistance comes from a range of effects that tend to get clumped together and can vary with speed and other factors - like drag and lubricant temperature. There is a large flywheel bolted to the wall in the undergrad lab at the University of Auckland (actually there are six of them) that clearly demonstrates the speed-dependence of the losses on the system. It's a big iron cylinder mounted on lubricated bearings and students can apply various forces to it. Real life is quite messy. Notice that I did not call the dominant retarding force "friction".

Recognitions:
Homework Help
 Quote by Simon Bridge @Haruspex: That's what I'd have thought too but the resistance comes from a range of effects that tend to get clumped together and can vary with speed and other factors - like drag and lubricant temperature. There is a large flywheel bolted to the wall in the undergrad lab at the University of Auckland (actually there are six of them) that clearly demonstrates the speed-dependence of the losses on the system. It's a big iron cylinder mounted on lubricated bearings and students can apply various forces to it. Real life is quite messy. Notice that I did not call the dominant retarding force "friction".
OK, but I still think the power the tractor can develop will be the principal limiter, and speed = power / (constant) axle-friction-plus-rolling-resistance will give the first approximation.
 Recognitions: Homework Help Well - good point: you'll probably run out of tractor power before reaching any critical limit. Especially for a tractor routinely used for pushing railway carriages around. Would it be reasonable - considering that this situation involves low speed - to model the resistance as a direct proportion to the weight of the truck? for this kind of back-of-envelope? Fair enough I guess. I'm inclined to think this is the sort of thing you'd want to find out by experiment though. i.e. find out if the tractor wheels would slide in place by getting it to push against a wall... I don't think it'll get damaged... Aside: - when I was a freshman, the mechanics lecturer ("Prof. Orange" in case anyone else here was taught by him) was fond of using railway carriages as example moving objects. He'd draw a box on wheels and say "New Zealand railway carriage: completely frictionless!"
 I'm glad I'm not the only one who found this to be more puzzling that it seems. We know from experience that the tractor will spin it's wheels - On a few occasions the railway brakes have been left on and the tractor driver found he couldnt push the cars.

Recognitions:
Homework Help