Centripetal force - Train on a curved track

[madmax321]

Centripetal force -- Train on a curved track

Homework Statement

You have a circular arc of railroad track to lay on the line from Dubuque to Albuquerque. The arc of track is 30º and the turn radius for the arc is 425 meters. It has to keep the path steady and secure for an ALP-46
locomotive (m = 90,000 kg total; 22,500 kg per axle; 11,250 kg per wheel) at speed 36 m/sec.

Compute: the centripetal force that the rail must provide to one of the outside wheels for this turn at 36 m/sec.

Homework Equations

The Attempt at a Solution

Would this be the correct formula to use?
F = m v^2/R

11250 * 36*36 /425=34305.88

Max

 

[TSny]

Hi Madmax321.

I don't think that's the right answer. You might consider the following:

1. What is the magnitude of the total centripetal force on the locomotive?
2. How many outside wheels does the locomotive have?

 

[madmax321]

The question is for only one (1) wheel so the weight of the total train engine as well as how many wheels it has seems irrelevant since there is "11,250 kg per wheel" given in the problem

Max

 

[TSny]

Here's how I interpret the problem. I could be wrong.

The "11,250 kg per wheel" just means that when the locomotive sits at rest on the track, each wheel supports 11,250 kg out of the total 90,000 kg. However, when going around a curve, the statement of the problem implies that only the wheels on the outside of the curve provide centripetal force. (That makes sense if you think about how train wheels are shaped and how they sit on the track.) The wheels on the inside of the curve do not contribute to the centripetal force. So, you need to take into account that it's just the outside wheels that together must provide the total centripetal force for making the 90,000 kg of mass go around the turn.

 

[madmax321]

ahhh... So you would take the entire mass of the train, divide by 4 and use that mass for
F = mv2/R?

Max

 

[TSny]

ahhh... So you would take the entire mass of the train, divide by 4 and use that mass for
F = mv2/R?

Max

I think that's right. I prefer to use m = 90,000 kg in F = mv²/R to calculate the total centripetal force acting on the locomotive. Then divide that total centripetal force by the number of outside wheels to get the force per wheel. But either way yields the same answer.

 

[madmax321]

for V is it just, the speed 36 m/s or is it a vector say,
36cos(30)

 

[TSny]

Just the speed.

 

[madmax321]

thank you