# Plane landing with retarding force

## 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.

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.

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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

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?

tiny-tim
Homework Helper
Ok, so then it would would be 1/2mv^2=Fb+μmgx?
yes, except you've missed out some brackets
then I would solve for x(t) by using v^2=(dx/dt)^2
uhh? 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:

where,

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

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

gneill
Mentor

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.

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

gneill
Mentor

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.

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?

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gneill
Mentor

@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.

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.

gneill
Mentor

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"?

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

gneill
Mentor

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?

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

gneill
Mentor

Happy to be of service

berkeman
Mentor