Distance and velocity from non-constant force, time, mass?

1. Nov 2, 2013

Darkbound

Helloo guuuys I've been often reading stuff from this forum and the time for me to get involved has finally came, I am faced with the following problem:

1. The problem statement, all variables and given/known data
An airliner makes an emergency landing with its nose wheel locked in a position perpendicular to its normal rolling position. The forces acting to stop the airliner arise from friction due to the wheels and from the breaking effort of the engines in reverse thrust mode. The force of the engine on the plane is constant, Fengine = -F0. The sum of the horizontal forces on the airliner can be written as F(t)=-F0+(t/ts-1)*F1, from touchdown at time t=0s to the final stop time at ts=28s the mass of the plane is M=80tonnes, F0=262kN, F1=41kN Neglect all air drag and friction forces, except the one stated in the problem.

a) Find the speed of the plain at touchdown
b) horizontal acceleration at time a(ts) a(0)
c) what distance s does the plane go between touchdown and the final stop time ts
d) What work do the engines during the reverese thrust mode during the landing
e) How much heat energy is absorbed by the wheels during the emergency landing?

Now for a) I am thinking about conservation of momentum, q2-q1= ∫F*dt integrated over the time interval 0->28 which will lead me to the equation v=F*t/m is this correct? I haven't tried inputing the numbers since I have limited submissions and I've noticed that in physics we can often be wrong even that something seems to be right :)

for b) taking the initial velocity which we find in a), is it as simple as (for t=0) a=(v2-v1)/(t2-t1) and since we know that v2 is 0 and t1 is 0 we get a=-v1/t2 and I'm being asked for the absolute value so it turns out a positive number v1/t2 ? What about the acceleration at time = 28s since they are saying that the airlaner stops after 28 seconds shouldnt the acceleration be 0? (i tried 0 its wrong answer)..

for c) well i know the time i know the velocity.. s=v*t?

d) Work = F*s and since the force is negative then I'll get a negative answer for work?

e) Now this is the one that I don't know how to approach, assuming that I am right about all the rest (which I'm pretty sure I'm not :D )

Last edited: Nov 2, 2013
2. Nov 2, 2013

cepheid

Staff Emeritus
Welcome to PF,

You are correct that the total impulse J = ∫F(t)dt is equal to the change in momentum. However, the equation v = F*t/m is totally wrong. It would only be true that J = F*t IF the force were constant. But the force is not constant, it is a function of time. That is why you have to *integrate* to find the impulse (as opposed to just multiplying). So, integrate the function F(t) that you've been given in order to find the change in momentum. Since you know that the final momentum is zero, the change in momentum tells you the initial momentum.

No, of course not. You should know that this equation for acceleration only applies if the acceleration is constant. In this problem, the acceleration is not constant, because the force is not constant. It is a function of time: F(t). Given the force F(t), how would you find the acceleration a(t)?

First of all, being "stopped" means that v = 0. It does not necessarily mean that a = 0. In any case, as I said above, what you need to do here is find the function a(t). Then you'll be able to find the acceleration at the two times by plugging in t = 0 s or t = 28 s into that function.

Are you being serious here? s = vt only applies when the speed is constant. In this problem, the speed is not constant, because there is a net force, therefore the plane is accelerating.

Well, actually, W = Fs only applies for a constant force. In this problem, the force changes as a function of time, and therefore as a function of distance travelled after landing. So, to compute the work you'd have to compute the integral W = ∫F(s)ds. However, you don't have to worry about that here, because you know the change in kinetic energy, so you can use the work-energy theorem to figure out the work done. And yes, of course the work done will be negative since the force opposes the motion.

Work done by friction reduces the kinetic energy of the object, and that energy is turned into heat. So, to answer this question, just find the work done by the part of the force that is due to friction (not the part that is from the engine).

3. Nov 2, 2013

Darkbound

Hahahaha I actually laughed at some point(at myself) when reading what you said, I totally failed here with that constant non-constant :)

Ookay so velocity, I find the force that's left after 28 seconds which basicly means that all the friction force is gone according to the equation they gave me, and I am left with -262000 N that is F(28), then -m*v1=-F(28)dt integrated from 0 to 28, gives v1=91.7 m/s ?

acceleration... would be that whole equation that I am given divided by m, a(t)=F(t)/m for a(0)=-3.7875 m/s^2, a(28)=-3.275 m/s^2

I am not entirely sure how to find the distance in that case... i know x(t)=x0+v*t+1/2*a*t^2 but that's for constant acceleration so I guess that I can't use it here, or if I put my a(t) in there it will work?

d) well ye then it would be W=F(28)*s*dt integrated over 0->28 and for e) W=Ff*s*dt 0->28

Am I on the right track?

Edit: About the distance, I have a value for a(28) if I integrate it twice that should give me the distance?

4. Nov 2, 2013

cepheid

Staff Emeritus
Why are you integrating F(28)? That is just the value of the force at a single instant. Integrate F(t) from t = 0 to t = 28.

Haven't checked the numbers, but yes, a(t) = F(t) / m

Nope. Remember that a = dv/dt, and v = dx/dt. So, to get v(t), you'd have to integrate a(t) once with respect to time. To get x(t), you'd have to integrate v(t) with respect to time. Net result: to get x(t) from a(t), you have to integrate a(t) twice with respect to time. The equation you posted above is what you get if you do that integration for a(t) = a = const. You have to derive a new equation by doing this integration for the particular a(t) that you have.

No, that is just wrong. As I said before, just use the work-energy theorem. No integration involved.

Again, a(28) is a number. You have to integrate the FUNCTION a(t) twice, with limits of integration from t = 0 to t = 28. I don't understand why you insist on plugging in 28???

5. Nov 2, 2013

Darkbound

Okay here are my equations/calculations, I hope I got it right this time

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Last edited: Nov 2, 2013
6. Nov 2, 2013

cepheid

Staff Emeritus
Things look okay except for the second last step. Again, the friction force isn't constant, so you can't just say that the work done by friction is force*distance. To do this, you need to do the following:

According to the work-energy theorem, the TOTAL work done is equal to the change in kinetic energy. So if you calculate the change in KE of the plane between 0 and 28 s, you know how much total work is done. You also know how much work is done by the engines, because this is easy to compute (since the engine force is constant). You calculated this above. Obviously the difference between the TOTAL work, and the work done by the engines, must be the work done by friction.

Last edited: Nov 2, 2013
7. Nov 2, 2013

Jamestoomer

How would the calculated work done by friction transform into the heat energy? Should it be thought of as a 100% conversion.

8. Nov 2, 2013

Darkbound

so Wtotal=KE2-KE1 , Wfriction=Wtotal-Wengines , Wtotal=-KE1 since we have no kinetic energy at time 28, Wtotal=-1/2*80000*(98.875)^2=-391050625 Wfriction=-391050625-(-725347000)=334296375 I somehow don't believe that this is true, how is it even possible that the total work is less than the engines work alone?

Edit: my distance and engines work is also wrong I just submited the answers... accelerations and velocity are correct I have one submission left, where did I make a mistake?

9. Nov 2, 2013

cepheid

Staff Emeritus
Hello Jamestoomer. That is what is to be assumed for simplicity. Of cousre some of the kinetic energy of the plane is going to go into sound, and light, and work done to deform the plane, and kinetic energy of pieces of shrapnel that fly off the landing gear etc. But mostly, friction converts the overall ordered motion of the object into disordered random motions of the individual atoms in the lattice that makes up the metal in the plane. This disordered random motion of individual atoms or molecules corresponds to an increase in temperature of the material. Temperature is a statistical measure of the average kinetic energy due these disordred/random motions at the atomic level.

Hmm. For distance: your first integral needs to be an INdefinite integral, not a definite one. That way it produces a v(t) function, which you then integrate from 0 to 28 to get the change in x position. If the first integral is definite, then this produces a number for the total change in speed, which is not what you want to integrate to get x. Does that make sense?

As far as the work goes: all I can say is that you must have an arithmetic error somewhere. The total work done by the two forces should greater (in absolute value) than the work done by friction alone. It's probably just because your distance was wrong that your Wengine turned out wrong.

10. Nov 2, 2013

Darkbound

Thanks I got the correct distance with Indefinite and then Definite integration :)

11. Nov 2, 2013

speed01

@Darkbound could you please elaborate how did you got the heat absorbed and work engines??????

12. Nov 3, 2013

wanglikun08

Hi, guys. Is the magnitude of friction work equal to heat absorbed by the wheels?
I calculated the total work and the engine work, then the difference delta E between them should be friction work. I also calculated the friction work Ef alone with integration. Then I can see that delta E = Ef, but when I submit the answer, it's wrong.

13. Nov 3, 2013

cepheid

Staff Emeritus
Welcome to PF,

We can help you if you show your work. Post your two calculations. There must simply be an error with one of them.

14. Nov 3, 2013

wanglikun08

For the total work, E = -(1/2)*m*v0^2 = -3.8553*10^8 J
the engine work, W = -F0*s = -3.6606*10^8 J
so delta E = E - W = -1.947*10^7 J.

For the integration,
F/m = -dv/dt, =>
v(t) = -∫(F/m)dt
= -∫[-(F0+F1)/m + F1*t/(m*ts)]dt
= -(1/m)*[-(F0+F1)*t + (1/2)*F1*(t^2)/ts]
so the friction work is
Ef = ∫(F+F0)ds
= ∫(F+F0)*vdt
= -(1/m)*∫(-F1 + F1*t/ts)*[-(F0+F1)*t + (1/2)*F1*(t^2)/ts]dt
= -(1/m){(1/2)*F1*(F0+F1)*t^2 - [F1^2 + 2*F1*(F0+F1)*t^3]/(6*ts) + (1/8)*F1^2*t^4/ts^2]
let t = ts = 28, then,
Ef = -1.947*10^7 J
=================================
should have written it in the formula editor, but I'm lazy so... sorry

15. Nov 4, 2013

abialterego

Hi everyone. Thanks for all the explanations, this topic helped me a lot.

I still dont' get the last question right.

I tried getting the total work (-1/2m*v^2) and substract the work of the engines (which i got right) but i got the answer wrong.

The complete question said:
(e) How much heat energy is absorbed by the wheels during the emergency landing?
(magnitude in Joules; the force due to wheels is (F(t)+F0))

So because of this what I tought I could do was:
integrate ( (F(t)+F0) * s ) from 0 to 28, but it was also wrong.

Any help would be appreciated.

16. Nov 4, 2013

cepheid

Staff Emeritus
Welcome to PF abialterego,

Work is the integral of force with respect to distance

You are integrating your force with respect to time, with an extra factor of s in there. Whatever this gives you, it is certainly not the work done.

The easiest way to find the work done by friction in this problem is as follows:

1. Using the work-energy theorem, find the total work done by finding the change in kinetic energy
2. Find the work done by the constant engine force (easy).
3. Subtract 2 from 1. The difference must be the work done by friction.

17. Nov 4, 2013

kbingo

Hi and thanks everybody for all the help.
Actually that what wanglikun08 did, also I believe that's also abialterego first method. I did the same - computing the initial energy from the touchdown speed (m*v^2/2) and subtracting the engine energy (and that number is correct), but the answer was wrong. It might be another catch in this question but I'm not getting it...

18. Nov 4, 2013

abialterego

Thank you cepheid, I can see why the second method was completely wrong. I will keep thinking of this. We must be missing something.

19. Nov 4, 2013

cepheid

Staff Emeritus
Wow, how many of you guys are working on this? This should be fairly straightforward. Can you post the answers that you are getting?

20. Nov 6, 2013

bb18

I was working on this, my mistake and probably everyone else is when integrating dv/dt I forgot to put v0 on the equation. By luck the error on distance is within accepted range but the error didn't get accepted when calculating heat energy.