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Chapter 5: Problem 40

The lift and drag measured on an aerodynamic body mounted at a five -degree angle of attack in a wind tunnel are 100 and \(145 \mathrm{lb}\), respectively. Calculate the corresponding normal and axial forces.

Short Answer

Expert verified
Normal force is 112.26 lb, and axial force is 135.74 lb.

Step by step solution

01

Understanding the Problem

We need to calculate the normal ( F_n ) and axial ( F_a ) forces based on the given lift and drag forces when the body is at a 5-degree angle of attack. The lift force ( L ) is perpendicular to the direction of airflow, and the drag force ( D ) is parallel to the direction of airflow.
02

Using Trigonometry

The normal force (F_n) is the force component perpendicular to the body, and the axial force (F_a) is the force component along the body.To find these, use trigonometry based on the angle of attack ( heta = 5^ ext{掳}):- The equations are: \[ F_n = L imes ext{cos}( heta) + D imes ext{sin}( heta) \]\[ F_a = -L imes ext{sin}( heta) + D imes ext{cos}( heta) \]
03

Calculating Normal Force

Substitute the given values into the formula for normal force:\[L = 100 \text{ lb}, D = 145 \text{ lb}, \theta = 5^ ext{掳}\]\[F_n = 100 \cdot \cos(5^ ext{掳}) + 145 \cdot \sin(5^ ext{掳})\]Calculating using trigonometric values: \[ F_n = 100 \cdot 0.9962 + 145 \cdot 0.0872 \approx 99.62 + 12.64 = 112.26 \text{ lb}\]
04

Calculating Axial Force

Now substitute the values into the formula for axial force:\[F_a = -100 \cdot \sin(5^ ext{掳}) + 145 \cdot \cos(5^ ext{掳})\]\[F_a = -100 \cdot 0.0872 + 145 \cdot 0.9962 \approx -8.72 + 144.46 = 135.74 \text{ lb}\]
05

Verification of Results

Check both results for any calculation errors again:- Normal Force \[ F_n = 112.26 \text{ lb}\]- Axial Force \[ F_a = 135.74 \text{ lb}\]Both are calculated correctly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lift and Drag Forces
The concepts of lift and drag are fundamental in understanding how objects move through a fluid, such as air. Lift is the force that acts perpendicular to the direction of the oncoming flow. Imagine a plane's wing, designed to create lift, allowing it to rise into the sky by manipulating the airflow.
It occurs mainly due to the difference in pressure on the upper and lower surfaces of a wing.
On the other hand, drag is a force parallel to the direction of the airflow. It's what tries to slow down an object moving through a fluid. Think about the resistance you feel when sticking your hand out of a moving car window. This resistance is drag.
Understanding these forces is crucial for designing and optimizing vehicles, aircraft, and even sports equipment. They must be balanced correctly for efficient movement.
Angle of Attack
The angle of attack is a vital parameter in aerodynamics. It is the angle between the chord line of a wing or airfoil and the oncoming air or relative wind.
Simply put, it's the tilt of the wing in the airflow. When this angle increases, the lift force generally increases -- to a point. However, after reaching a critical angle, the lift can dramatically decrease, leading the wing to stall. Monitoring and adjusting the angle of attack ensures that the maximum efficient lift is generated during flight.
This angle is particularly important during takeoff and landing of an aircraft, where different aerodynamic needs must be met.
Normal and Axial Forces
Understanding normal and axial forces helps in decomposing the effects acting on an object in airflow. Normal force is a component that is perpendicular to the surface of the body in the airflow, while axial force is aligned with the body's axis in the direction of the airflow.
To imagine this, consider the forces acting on the wing of an airplane which is tilted upwards. The lift generates a normal force contributing to the upward movement, whereas the drag generates an axial force opposing forward movement.
Using the angle of attack, these forces can be calculated using trigonometry. By resolving the lift and drag forces into their normal and axial components, you can better understand the aerodynamic performance of an airfoil.

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Most popular questions from this chapter

Consider a finite wing at an angle of attack of \(6^{\circ}\). The normal and axial force coefficients are \(0.8\) and \(0.06\), respectively. Calculate the corresponding lift and drag coefficients. What comparison can you make between the lift and normal force coefficients?

Consider a wing in a high-speed wind tunnel. At a point on the wing, the velocity is \(850 \mathrm{ft} / \mathrm{s}\). If the test-section flow is at a velocity of \(780 \mathrm{ft} / \mathrm{s}\), with a pressure and temperature of \(1 \mathrm{~atm}\) and \(505^{\circ} \mathrm{R}\), respectively, calculate the pressure coefficient at the point.

The airfoil on the Lockheed F-104 straight-wing supersonic fighter is a thin, symmetric airfoil with a thickness ratio of \(3.5\) percent. Consider this airfoil in a flow at an angle of attack of \(5^{\circ}\). The incompressible lift coefficient for the airfoil is given approximately by \(c_{1}=2 \pi \alpha\), where \(\alpha\) is the angle of attack in radians. Estimate the airfoil lift coefficient for \((a) M=0.2,(b) M=0.7\), and \((c) M=2.0\).

Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is \(7780 \mathrm{~N}\), the wing area is \(16.6 \mathrm{~m}^{2}\), and the maximum lift coefficient is \(2.1\) with flaps down, calculate the stalling speed at sea level.

Consider a rectangular wing with a NACA 0009 airfoil section spanning the test section of a wind tunnel. The test-section airflow conditions are standard sea level with a velocity of \(120 \mathrm{mi} / \mathrm{h}\). The wing is at an angle of attack of \(4^{\circ}\), and the wind tunnel force balance measures a lift of \(29.5 \mathrm{lb}\). What is the area of the wing?
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