Landing a 747 with only friction

[armolinasf]

Homework Statement

I had this problem floating around in my head and when I worked it out I was a little surprised. A 747 is coming in for a landing and the only force opposing its motion is the force of friction between the ground and its tires how much runway is needed for the plane to come to a complete stop?

landing speed: 270 kph = 75 m/s
coefficient of kinetic friction (f) = .7
mass = 390,000 kg

The Attempt at a Solution

I apply the work energy theorem and solve for distance (d)

.5mv^2=fmgd

d=v^2/(2fg) ==> (75)^2/(.7*9.8)= 409m

This is far shorter than the minimum runway needed to land a 747 and I'm ignoring so many other forces. Could someone provide some insight into why this answer is incorrect (or right?) Thanks

 

[MisterX]

Usually, planes are not landed with the majority of breaking force coming from the tires skidding. The wheels typically would roughly match the ground speed, and the a plane would be slowed via breaks on the wheels and aerodynamic forces. There are limits to the amounts of mechanical forces it is advisable to put on the landing gear, and how much force the breaks would be able to exert on the wheels, as well as limits for break heating; the landing gear could catch fire if too much breaking is attempted.

 

[Fightfish]

Hmm...one possible reason would be that your answer assumes that N = mg immediately upon touching down. It is likely that N is less than mg at the start of the landing due to lift forces.

 

[ehild]

The coefficient of kinetic friction you used is valid for sliding friction, when the Boeing were sliding on its belly. But it has got wheels and they roll. Rolling friction is in the range of 0.01.

ehild

 

[asteriatic]

I would like to first acknowledge that this is a hypothetical scenario and not something that would realistically occur in an actual landing of a 747. However, for the sake of theoretical discussion, I will provide a response to this content.

Based on the information provided, the solution you have calculated is technically correct. However, it is important to note that the scenario described is highly unrealistic and does not take into account many important factors that would affect the landing of a 747.

Firstly, the coefficient of friction given (0.7) is likely an underestimate for the actual friction between the tires and the ground. In reality, the friction between the tires and the runway would be much higher due to the weight and speed of the aircraft, as well as the type of tires and runway surface. This would result in a shorter stopping distance than what was calculated.

Additionally, the solution does not take into account the aerodynamic drag and lift forces acting on the aircraft during landing. These forces would also contribute to the deceleration of the aircraft and would need to be considered in the calculation of the stopping distance.

Furthermore, the scenario does not account for any braking systems or methods that would be used in a real landing. A 747 has powerful braking systems, including thrust reversers and spoilers, which would greatly aid in reducing the stopping distance.

In conclusion, while the solution calculated may be technically correct, it is not a realistic representation of a 747 landing and does not take into account important factors that would greatly affect the stopping distance. As a scientist, it is important to consider all relevant factors and limitations when conducting calculations and making conclusions.