href="#fb3_img_img_9c1213fc-8378-5753-8534-2edfd038a4de.gif" alt="Schematic illustration of coefficients of friction for airplane tires on a runway."/>
Figure 1.9 Coefficients of friction for airplane tires on a runway.
At the microscopic level, as in the surface of a wing, friction causes resistance and slows down the velocity of the air as it passes over it. The layer of air that is impacted by the friction of the wing, or any other surface of the aircraft, is referred to as the boundary layer.
Several factors are involved in determining friction effects on aircraft during takeoff and landing operations. Among these are runway surfacing material, condition of the runway, tire material and tread, and the amount of brake slippage. All of these variables determine a coefficient of friction μ (mu). The actual braking force, Fb, is the product of this coefficient μ (Greek symbol mu) and the normal force, N, between the tires and the runway (Eq. 1.14):
Figure 1.9 shows typical values of the coefficient of friction for various conditions. Note the value of μ for dry concrete is ~0.7 with ~10% wheel slip, while the μ on smooth, clear ice is ~0.2. This means that an airplane wheel rolling on smooth, clear ice will experience much lower friction (increased stopping distance) than a wheel rolling on dry concrete.
EXAMPLE
Calculate the braking force on dry concrete when the normal force (N) is 2000 lb.
SYMBOLS
| a | acceleration (ft/s2) |
|---|---|
| E | Energy (ft‐lb) |
| KE | Kinetic energy (ft‐lb) |
| PE | Potential energy (ft‐lb) |
| TE | Total energy (ft‐lb) |
| F | Force (lb) |
| F b | Braking force (lb) |
| g | Acceleration of gravity (ft/s2) |
| h | Height (ft) |
| HP | Horsepower |
| L | Moment arm (ft or in.) |
| m | Mass (slugs, lb‐s2/ft) |
| M | Moment (ft‐lb or in.‐lb) |
| N | Normal force (lb) |
| r | Radius (ft) |
| rpm | Revolutions per minute |
| s | Distance (ft) |
| T | Thrust (lb) |
| t | Time (second) |
| V | Velocity (ft/s) or (kts.) |
| Vf | Final velocity (ft/s) |
| V k | Velocity (kts.) |
| V i | Initial velocity (ft/s) |
| V t | Tangential (tip) speed (ft/s) |
| W | Weight (lb) |
| μ (mu) | Coefficient of friction (dimensionless) |
KEY TERMS
Acceleration
Area
Arm
Coefficient of friction
Centripetal force
Component vector
Energy
Equilibrium
Force
Friction
Kinetic energy
Laws of motion
Linear
Mass
Mechanical energy
Motion
Potential energy
Power
Pressure
Resultant vector
Rotational motion
Scalar quantity
Torque
Velocity
Vector quantity
Vector resolution
Work
PROBLEMS
Note: Answers to problems are given at the end of the book.
1 Convert 65 kts. to fps.
2 Convert 200 fps to kts.
3 Convert 35 kts. to fpm.
4 Convert 52 nm to sm.
5 An airplane weighs 16 000 lb. The local gravitational acceleration g is 32.2 fps2. What is the mass of the airplane?
6 The airplane in Problem 5 accelerates down the runway with a net forward force (thrust less drag) of 6000 lb. Find the acceleration of the airplane.
7 The airplane in Problem 6 starts from a brakes‐locked position on the runway. The