Plane landing with retarding force

  • Thread starter Thread starter kraigandrews
  • Start date Start date
  • Tags Tags
    Force Plane
Click For Summary

Homework Help Overview

The problem involves a light plane making an emergency landing on a short runway, where it lands at a certain speed and subsequently drags a sandbag after snagging a cable. The forces acting on the plane include friction and a retarding force from the brakes, and the objective is to determine how far the plane travels before coming to a stop.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the plane and the setup of differential equations to describe the motion. There are attempts to apply the work-energy theorem and kinematic equations, with some questioning the integration process and the resulting functions.

Discussion Status

There is ongoing exploration of different approaches, including the use of conservation of momentum and energy. Some participants have provided guidance on applying kinematic equations, while others are questioning the assumptions made regarding the collision between the plane and the sandbag.

Contextual Notes

Participants are navigating through various interpretations of the problem, including the nature of the collision and the energy considerations involved. There is a noted lack of consensus on the correct approach, with some expressing uncertainty about the setup of equations and the role of different forces.

kraigandrews
Messages
105
Reaction score
0

Homework Statement


A light plane (mass = M) makes an emergency landing on a short runway. With its engine off, it lands at speed v0. A hook on the plane snags a cable attached to a sandbag (mass = m) and drags the sandbag along. The coefficient of friction between the sandbag and the runway is μ, and the plane's brakes give a retarding force of Fb. How far will the plane go before it stops?
Data: v0 = 43.0 m/s; M = 839 kg; m = 97 kg; μ = 0.32; Fb = 1208 N.


Homework Equations


F=mdv/dt
F=dp/dt=mdv/dt


The Attempt at a Solution


-Fb-μmg=(M+m)dv/dt
Im pretty sure these are the correct forces, however, the solution to the diff eq does not yield a function (i.e. e^-(something) that goes to zero) that will give a stopping point. My thought is I am missing a part of the equation or have to incorporate momentum.
Thanks.
 
Physics news on Phys.org
Notice that the left hand side if the equation you determined is constant:

-Fb-μmg=(M+m)dv/dt

You can now integrate this to find v(t) or use constant acceleration kinematic equations.
 
That was just my guess at it, however, it cannot be correct because when you integrate for v, then again for x you get a parabolic function which does not go to zero, hence it does not stop, so it is wrong. Suggestions for fixing this?
 
Plane making an emergency landing!

Homework Statement


A light plane (mass = M) makes an emergency landing on a short runway. With its engine off, it lands at speed v0. A hook on the plane snags a cable attached to a sandbag (mass = m) and drags the sandbag along. The coefficient of friction between the sandbag and the runway is μ, and the plane's brakes give a retarding force of Fb. How far will the plane go before it stops?
Data: v0 = 43.0 m/s; M = 839 kg; m = 97 kg; μ = 0.32; Fb = 1208 N


Homework Equations


F=dp/dt


The Attempt at a Solution


not sure how to set this up I know the forces are:
μmg=Ff
(M+m)g=N
Fb=retarding force
and initial momentum = Mv0
other than that I have no idea how to set up the differential equation to obtain V(t)
suggestions?
 


Ok, so then it would would be 1/2mv^2=Fb+μmgx?
then I would solve for x(t) by using v^2=(dx/dt)^2
is this correct?
 
kraigandrews said:
Ok, so then it would would be 1/2mv^2=Fb+μmgx?

yes, except you've missed out some brackets :redface:
then I would solve for x(t) by using v^2=(dx/dt)^2

uhh? :confused: v is given

get some sleep! :zzz:​
 
OK, we have established it is constant acceleration and we need to find the distance before the plane stops. The solution to your equations already exist.

Could we apply this constant acceleration kinematic equation:

vf^2 = vo^2 + 2ad

where,

vf = final velocity = 0
vo = initial velocity (given)
a = acceleration = -Fb - μmg / (M + m)
d = distance

solve for 'd' and note the sign of 'a' is negative.
 


When the plane hooks the sandbag, it sounds like an inelastic collision to me. This occurs immediately before the dragging and braking... so, re-evaluate the remaining KE.
 
  • #10


Care to elaborate at all? Because the conservation approach is not working unless I am missing something?
 
  • #11


Plane has initial velocity. Grabs sandbag. That's an inelastic collision, so use conservation of momentum to work out the new velocity of the plane+sandbag. Plane+sandbag has kinetic energy. Go from there with conservation of energy and work-energy.
 
  • #12


Kraigandrews, if you solve it and get it right (eventually), could you post what answer it is? I'd like to know if I got it right :)

@gneill: it's not really an inelastic collision, is it? Sounds like you're thinking of an inelastic collision as one where the objects stick to each other, which is true most of the time, but the real definition of an inelastic collision is where energy isn't conserved. And I'm pretty sure energy is conserved when the plane is just snagging a sandbag cable ... unlike a bullet slamming into a block of wood, or two cars crashing together, to quote other common conservation-of-momentum-but-not-energy problems / inelastic collision problems. Because, if energy is lost in the plane snagging the sandbag, where does the energy go?
 
Last edited:
  • #13


janrs said:
@gneill: it's not really an inelastic collision, is it? Sounds like you're thinking of an inelastic collision as one where the objects stick to each other, which is true most of the time, but the real definition of an inelastic collision is where energy isn't conserved. And I'm pretty sure energy is conserved when the plane is just snagging a sandbag cable ... unlike a bullet slamming into a block of wood, or two cars crashing together, to quote other common conservation-of-momentum-but-not-energy problems / inelastic collision problems. Because, if energy is lost in the plane snagging the sandbag, where does the energy go?

Plane grabs sandbag. Plane and sandbag move as one thereafter. That's an inelastic collision, no matter what color the plane is. Where the energy goes is for the crash inspectors to worry over. :smile:
 
  • #14


Ok, so I have tried this several ways and have not been able to get it using energy:

.5M(v0^2)=(Fb+(μ(M+m)g))x solving for x
also
.5M(v0^2)=(Fb+(μ(m)g))x solving for x
and some other variations of that.

so I'm really not sure what to do from here.
 
  • #15


kraigandrews said:
Ok, so I have tried this several ways and have not been able to get it using energy:

.5M(v0^2)=(Fb+(μ(M+m)g))x solving for x
also
.5M(v0^2)=(Fb+(μ(m)g))x solving for x
and some other variations of that.

so I'm really not sure what to do from here.

Question: How will you know when you've "got it"?
 
  • #16


It's homework set online, you enter the answer, tells you if its right or wrong.
 
  • #17


Did you give any thought to my suggestion that before the plane and sandbag start braking the velocity, that there is an inelastic collision to consider?
 
  • #18


Ok finally got, your approach was correct, I just was forgetting to neglect the mass of the plane when calculating the work done by friction. Thank you very much
 
  • #19


Happy to be of service :smile:
 
  • #20
(Two threads merged. Please do not multiple-post here.)
 

Similar threads

Replies
1
Views
5K
Replies
6
Views
3K
Replies
16
Views
3K
Replies
11
Views
3K
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 22 ·
Replies
22
Views
3K