# Homework Help: Banked curve and tilted train.

1. Apr 20, 2007

### JoshuaR

1. Tilting trains such as the Acela, which runs from Washinton to new York to Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force Fs (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 125mi/h on a curved section of track banked at an angle theta=8 and with a rated speed of 75 mi/h determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt (phi=0) (b) teh required angle of tilt phi if the passenger is to feel no side force.
(There is a diagram in the book of a slight bank (8 degrees) and a train on it with trucks tilting another phi degrees).

2. N=mgcos@ F=ma=mv^2/r = mgcos@sin@ (the horizontal component of the normal)

3. So from rated speed and above equations I got a radius of 2726.59 ft. Theoretically, however, another method for getting the radius would be to use a=v^2/R where a is found from arctan(a/g)=8 degrees rather than gcos@sin@=v^2/r This comes out to a radius of 2673.79 ft. Why do these numbers differ? Am I making a wrong assumption somewhere in one of the two methods? The problem is with the first method I get the answer from the book eventually by... a=v^2/r (using the radius found from the first method) and then calculating (a-gcos@sin@)/(gcos@)=coefficient of friction=Fs side force thingy. So using this method a would be 12.327ft/s*s and the answer matches the back of the book of 0.247W. However, if you use the arctan(a/g) method you get an answer more like .255. Why are these numbers different?
Part (b): The most simple solution is arctan(a/g)=sum of theta and phi, where phi is easily solved for as 13.3 degrees, matching the back of the book. But this is only if you use the second method initially of arctan(a/g) to find the radius! If you go by the other radius, hence acceleration, the numbers are slightly off. Worse, you can't use the first method at all here. I assumed mgcos@sin@=mv^2/r still, changing @ in this case to be the sum of theta and phi. This method, however, yields an angle of 17 degrees. (sin2@=2v^2/(gr))Definitely incorrect. Why does this method fail in this case?
As you can see. I've solved the problem, but I had to use different rationale for each part of it to match book answers. I'm unsure as to why these to rationales didn't match up. Thanks!
Josh

2. Apr 20, 2007

### AlephZero

For the 75 mi/h = 110 ft/sec case:

The vertical acceleration is mg

The horizontal acceleration is mv^2/r

To get no side force relative to the tilted train the resultant must be 8 degrees from the vertical which gives v^2/(rg) = tan 8

Taking g = 32.2 ft/sec^2 I get r = 2673.79 ft.

I can't follow your argument in the first case without a diagram, but it must be wrong because it gives a different answer. I can't see where your "cos theta sin theta" term came from.

3. Apr 20, 2007

### JoshuaR

If the normal component of mg is mgcos@, then the horizontal component of the normal force would be the sin of the normal force, mgcos@sin@. Or (mgsin2@)/2. Is that incorrect? The problem is I used those numbers to reach the answer to part (a), but used the answers as you have described to reach part (b). Interesting conundrum. Thanks for the help.

4. Apr 20, 2007

### AlephZero

Yes that is correct.

But there is also the tangential component of mg = -mg sin theta, and the horizontal component of the tangential force is -mg sin theta cos theta.

The resultant horizontal force from the weight is the sum of those two components, which = 0. It must = 0, because resolving a force into components doesn't change the force, and we agree the weight acts vertically.

5. Apr 20, 2007

### JoshuaR

Okay, so the weight acts vertically, leaving no room for a horizontal component. But the normal force acts perpendicular to the plane, even if it is only the equal and opposite of one component of gravity. This is the only force that could possibly provide the centripetal acceleration of going around the curve, right? If I can't use this, what can I use for the centripetal acceleration?
I have v^2/r = mgcos@sin@
Is the right hand of that equation wrong?

6. Apr 20, 2007

### AlephZero

No, the force to provide the centripetal acceleration comes from the force of the track on the train wheels. That can have components perpendicular to the track and parallel to it (sideways).

If the passengers don't feel any "sideways" force, the force on the wheels must be perpendicular to the track. You can resolve that into a vertical component to balance the weight mg, and a horizontal component to give the centripetal force mv^2/r.

Resolving vertically, to balance the weight mg, the magnitude of the force at the wheels must be mg/cos theta.

The centripetal component is then (mg/cos theta) sin theta = mg tan theta.

7. Apr 20, 2007

### JoshuaR

Ah, of course. I'm a bit rusty on my physics/brain. Now that I see it, it becomes instantly recognizable. Before, I was just thinking the normal force must equal mgcos@. I remember now. Thanks!

Edit: Of course, by remember, I mean I understand again...