Do the rail cars move on a 0.499% grade without brakes or external force?

  • Thread starter Dennis240
  • Start date
In summary, the rail cars cannot roll under their own weight. The static friction between wheel and rail does not stop the wheel from rotating. The force normal to the inclined surface is Fp, while the force parallel to the inclined surface is Fp. The force holding the railcar in place is Fw.
  • #1
Dennis240
8
0

Homework Statement


two rail cars are coupled together, each weigh 82500 lbs. They are on tracks at a 0.499% grade. each rail car has 2 trucks of 2 wheels each, for a total of 16 wheels. The assumed force of friction for contact with the wheel and rail is 0.05 (very low). If the cars have no brakes or anything holding them do they roll under their own weight?

Homework Equations


Frc = (total mass of cars)*9.81*Sin(angle of grade)
Ffr = (total mass cars)*Assumed Friction Factor*Cos(angle of grade)

The Attempt at a Solution


Force of rail cars down the slope - Frc - (kN) -> ((82500*2*0.453592)*9.81*Sin(0.00499))/1000= 3.66 kN
Force of Friction - Ffr - (kN) -> ((82500*2*0.453592)*Cos(0.00499)*0.05)/1000=3.74 kN
 
Physics news on Phys.org
  • #2
Dennis240 said:
The assumed force of friction for contact with the wheel and rail is 0.05
They cannot mean that. They must mean rolling resistance, not friction. The static friction between wheel and rail does not stop the wheel from rotating.
Dennis240 said:
0.499% grade
Think carefully what that means. It is not an angle.
Dennis240 said:
Sin(0.00499)

Dennis240 said:
((82500*2*0.453592)*Cos(0.00499)*0.05)/1000=3.74 kN
You left something out.
(But are you sure the coefficient is 0.05, not 0.005?)
 
  • #3
Sorry, I did not see this come in.
Yes, grade is not an angle, and when using Excel I had to remember defaults to Radians - this grade works out to about 0.28 degrees.
I went looking and found a railway document that indicated 0.05 was under extreme conditions the lowest. Since the assignment was to determine if hte rail cars would move, I presumed the lowest coefficient of friction steel rail on steel wheel would yield the most conservative result.
Left out? Hmm, not sure I can easily address that one. Hint?
 
  • #4
Dennis240 said:
Yes, grade is not an angle
So why are you taking sin and cos of it?
Dennis240 said:
found a railway document that indicated 0.05 was under extreme conditions the lowest.
As I wrote, coefficient of friction is irrelevant. The brakes are off, so the wheels will roll, not slide. What matters is the rolling resistance. This arises from various sources, including the elastic properties of the rail and axial friction. Rolling resistance of a railroad wheel on its rail would typically be much lower, like 0.001.
Dennis240 said:
Left out?
It might become apparent if you take it in smaller steps. What will the normal force be?

I urge you to adopt the style of keeping everything symbolic, not plugging in any numbers until the final step. Where it gives you numbers, like the mass of each car, ignore that and just write a variable, like m.
This has many advantages in readability, in finding errors, in sanity checking; and in many cases, including this problem, you find that a lot of terms cancel out, so you can avoid some redundant calculation and the inaccuracies that can introduce.
 
  • #5
ok
thank you
 
  • #6
Force Normal to the inclined surface (Fn)
Fn = mass*gravity*Cos(theta)
 
  • #7
Dennis240 said:
Fn = mass*gravity*Cos(theta)
Right, and which of those terms is missing from:
Dennis240 said:
((82500*2*0.453592)*Cos(0.00499)*0.05)/1000
 
  • #8
Odd, I was on line last night, looking for your response and did not see this come in. Hmmm.
Ahh, so you're saying that I incorrectly used the assumed friction force in place of gravity.

Ok, got that. That will give the Force Normal (or perpendicular) to the inclined surface (the rail).

The Force Parallel to the inclined surface (Fp) appears to be correct.
Fp = Mass*gravity*Sin(theta)

If I have the Fp, you indicated above, there needs to be a force holding the wheels from turning (Fw) - involving the rolling resistance - You indicated, "Rolling resistance of a railroad wheel on its rail would typically be much lower, like 0.001." So, would the rolling resistance be substituted in for gravity in the Fp equation yielding Fw? If that is true, then the rail car would always have the ability / desire to roll, as the rolling resistance is far less. Or did I miss something again, another turn along the path?
 
  • #9
Yes, now understand the step missed - that the force down the incline needed to be calculated first, then the force holding the railcar in place (keeping the wheels from turning) needed to be calculated.
 
  • #10
Dennis240 said:
used the assumed friction force
I know you called it the force of friction in post #1, but the 0.05 is clearly a coefficient, not a force, so I assumed you meant coefficient of static friction, μs. Consequently the maximum frictional force would be μsN, where N is the normal force.
Dennis240 said:
would the rolling resistance be substituted in for gravity
It's another coefficient, so you would just use that in place of μs.
But if the problem as given to you has .05 then I guess you have to use that. Where is the question from?
 
  • #11
We were not given the 0.05, but instructed to research it and justify the value. The 0.05 was found in Wikipedia - Adhesion Railway article. I forgot to check the engineeringtoolbox site.
So, the Normal force is as indicated above (Fn or N) [or mass*gravity*Cos(theta)] and the Maximum frictional force to overcome to get the pair of cars to roll would be [as you stated above] μsN or μs(mass*gravity*Cos(theta)). If I followed your path correctly.

We had an upper level engineering intern attend class and they provided the professor with some examples that were not focused on a block on a plane scenario.
 
  • #12
Dennis240 said:
We were not given the 0.05, but instructed to research it and justify the value.
In that case I suggest you use the .001 value and describe it as coefficient of rolling resistance.
Dennis240 said:
So, the Normal force is as indicated above (Fn or N) [or mass*gravity*Cos(theta)] and the Maximum frictional force to overcome to get the pair of cars to roll would be [as you stated above] μsN or μs(mass*gravity*Cos(theta)).
Yes, except that you should not call it μs. That is the standard variable for static friction coefficient.
At https://en.wikipedia.org/wiki/Rolling_resistance it is named Crr.
 
  • #13
Understood, will do.
 

1. How does the railcar move?

The railcar moves by utilizing the force of a locomotive or other power source, which pulls or pushes the railcar along the tracks. The wheels of the railcar are also designed to move smoothly along the tracks, allowing for efficient movement.

2. What makes the railcar move?

The railcar is typically powered by a locomotive, which can use electricity, diesel, or steam to generate the necessary force to move the railcar. In some cases, the railcar may also be pushed or pulled by another vehicle or equipment.

3. Can the railcar move without a locomotive?

No, the railcar requires a power source such as a locomotive to move along the tracks. The design and weight of the railcar make it impossible to move on its own without an external force.

4. What is the maximum speed of a railcar?

The maximum speed of a railcar can vary depending on several factors, including the type and condition of the tracks, the power source, and the weight and design of the railcar itself. However, most railcars can reach speeds of up to 70 miles per hour.

5. How does the railcar stay on the tracks?

The railcar stays on the tracks through the use of wheels that are specifically designed to fit on the tracks and keep the railcar stable. The tracks also have a specific shape and design, including the use of rails and ties, to ensure that the railcar stays in place while moving.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Replies
6
Views
842
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
11K
  • Introductory Physics Homework Help
Replies
10
Views
7K
Back
Top