- #1

200bhp

- 4

- 0

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.

Please consider the following scenario:

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 don't 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, let's 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 can't see it!