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

[Dennis240]

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

 

[haruspex]

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.

0.499% grade

Think carefully what that means. It is not an angle.

Sin(0.00499)

((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?)

 

[Dennis240]

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?

 

[haruspex]

Yes, grade is not an angle

So why are you taking sin and cos of it?

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.

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.

 

[Dennis240]

ok
thank you

 

[Dennis240]

Force Normal to the inclined surface (Fn)
Fn = mass*gravity*Cos(theta)

 

[haruspex]

Fn = mass*gravity*Cos(theta)

Right, and which of those terms is missing from:

((82500*2*0.453592)*Cos(0.00499)*0.05)/1000

 

[Dennis240]

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?

 

[Dennis240]

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.

 

[haruspex]

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, μ_s. Consequently the maximum frictional force would be μ_sN, where N is the normal force.

would the rolling resistance be substituted in for gravity

It's another coefficient, so you would just use that in place of μ_s.
But if the problem as given to you has .05 then I guess you have to use that. Where is the question from?

 

[Dennis240]

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.

 

[haruspex]

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.

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 μ_s. That is the standard variable for static friction coefficient.
At https://en.wikipedia.org/wiki/Rolling_resistance it is named C_rr.

 

[Dennis240]

Understood, will do.