Is the Thrust Problem on an Airbus A320 Jetliner Unsolvable?

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Homework Help Overview

The problem involves calculating the thrust of an Airbus A320 jetliner given its takeoff mass, takeoff speed, and time to reach that speed, while considering rolling friction. The original poster attempts to derive the thrust using the equations of motion and the formula for rolling friction but encounters difficulties due to the lack of a specific coefficient of rolling friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of thrust based on acceleration and friction, questioning the necessity of the coefficient of rolling friction. Some suggest that the problem cannot be solved without this coefficient, while others propose alternative interpretations of the problem setup.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered guidance on the need for additional equations or information, while others debate the validity of the provided formulas and assumptions. There is no explicit consensus on whether the problem is solvable as presented.

Contextual Notes

Participants note that the coefficient of rolling friction is not provided, leading to uncertainty about the problem's solvability. There are references to typical values for this coefficient, but the necessity of a specific value versus a general expression is also under discussion.

acspin
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Homework Statement


An airbus A320 jetliner has a takeoff mass of 75,000 kg. It reaches its takeoff speed of 82 m/s in 35 s. What is the thrust of the engines? You can neglect air resistance but not rolling friction. (The force of rolling friction is given by Fr = [tex]\mu[/tex]r N


Homework Equations



Fr = [tex]\mu[/tex]r N
V = V0 + at
[tex]\Sigma[/tex]F = ma
T = thrust

The Attempt at a Solution



The first thing i did was solve for the acceleration. I got 2.343 m/s2.

From here I set

[tex]\Sigma[/tex]F = T - [tex]\mu[/tex]r N = ma

Thus:

T = ma + [tex]\mu[/tex]r mg

Then

T = m (a + [tex]\mu[/tex]r g)

So:

T = 75,000kg (2.343 m/s2 + [tex]\mu[/tex]r9.8 m/s2)

I don't know what the heck to do from here. There is no way I can calculate the coefficient of rolling friction, and it is not given. So I have 2 unknowns and 1 equation. Does this problem need more info?
 
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The formula for rolling friction is not correct.
 
Really? That's what was given for the rolling friction formula...

I tried doing this:

Fr = murN

Fr = murmg

ma = mur mg

mur = a/g

Would that be correct for the coefficient of friction?

I'm just not sure that I'm able to set Fnet equal to the Fr
 
No, that's not correct. You should consider the forces acting on the airplane wheel assuming the wheel rolls without slipping. You will need one more equation for the rotational motion of the wheel. However, since you are not given any information about the wheel, you should ultimately set its mass to be zero and analyze what effect does this have on the forces acting on the wheel when the wheel itself is accelerating.
 
Since you are not given the coefficient of rolling friction, you'll have to look it up in a table somewhere, otherwise, you can't solve the problem. The coefficient of rolling friction for rubber on concrete is about 0.02 for a car.
 
The "coefficient of rolling friction" is not given because you don't need it.
 
According to the definition of rolling friction I'm used to, which I believe agrees with the one given on Wikipedia, I believe this problem can't be solved numerically without knowing the coefficient of rolling friction. Dickfore, if you're claiming otherwise, let's see some math.
 
Dickfore said:
No, that's not correct. You should consider the forces acting on the airplane wheel assuming the wheel rolls without slipping. You will need one more equation for the rotational motion of the wheel. However, since you are not given any information about the wheel, you should ultimately set its mass to be zero and analyze what effect does this have on the forces acting on the wheel when the wheel itself is accelerating.

What are you going on about? The wheels mass is immaterial in this pretend problem where air drag in not even a factor. And the equation for the rolling friction is certainly correct despite your objections. You might like to start a thread on the complete dynamics of takeoff thrust.
 
We do need the coefficient of rolling friction to solve the problem. My professor asked me to look at a certain page in my textbook, and there it says:

The Coefficient of Rolling Friction is the horizontal force needed for constant speed on a flat surface, divided by the upward normal force exerted by the surface.

Typical values of mur for rubber tires on concrete are 0.01 - 0.02 As Phantom Jay explained.


I think I'm going to use 0.02. This is dimensionless correct?

Thanks for all the help!
 
  • #10
acspin said:
I think I'm going to use 0.02. This is dimensionless correct?
Yes, the coefficient of rolling friction is dimensionless.

However, why do you need a specific value? What is wrong with expressing the answer in terms of the coefficient of friction.


Here's a completely different problem to illustrate what I am talking about. Suppose you are told that an object starts from rest and moves a distance of d while undergoing a constant acceleration a. How long did it take to move that distance?

Just because you don't the specific value for d or a does not mean the problem is unsolvable. The solution is given by

[tex]t=\sqrt{\frac {2d}{a}}[/tex]
 
  • #11
True, but in my experience, the fact that the problem gives specific values for mass, velocity, and time strongly suggests that a numeric solution is desired. If the intent was to express the answer in terms of [itex]\mu_r[/itex], I would expect to see phrasing like "Find the thrust as a function of the coefficient of rolling friction."
 
  • #12
Good point. Here he has a range of values for [itex]\mu_r[/itex], so he can find a range for the thrust that corresponds to this range for the friction coefficient.
 

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