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Chapter 3: Problem 334

The 200 -kg glider \(B\) is being towed by airplane \(A\) which is flying horizontally with a constant speed of \(220 \mathrm{km} / \mathrm{h}\). The tow cable has a length \(r=60 \mathrm{m}\) and may be assumed to form a straight line. The glider is gaining altitude and when \(\theta\) reaches \(15^{\circ}\) the angle is increasing at the constant rate \(\dot{\theta}=\) 5 deg/s. At the same time the tension in the tow cable is \(1520 \mathrm{N}\) for this position. Calculate the aerodynamic lift \(L\) and \(\operatorname{drag} D\) acting on the glider.

Short Answer

Expert verified
Lift \(L = 2365 \text{ N}\) and Drag \(D = 1466 \text{ N}\).

Step by step solution

01

Convert Values to Appropriate Units

First, we need to convert the speed of the airplane and the rate of change of angle to SI units. The speed of the airplane is given as \(220 \text{ km/h}\). Converting this to meters per second, we get:\[\text{Speed} = \frac{220 \times 1000}{3600} \approx 61.11 \text{ m/s}\]The rate of angle change \(\dot{\theta} = 5 \text{ degrees/s}\) is converted to radians per second:\[\dot{\theta} = 5 \times \frac{\pi}{180} \approx 0.0873 \text{ rad/s}\]
02

Calculate Horizontal and Vertical Components of Velocity

We decompose the velocity of the glider into horizontal and vertical components. Using the definition of tangent and angle \(\theta = 15^\circ\):\[\text{Horizontal velocity } v_x = v \cdot \cos(\theta)\]\[\text{Vertical velocity } v_y = r \cdot \dot{\theta} = 60 \cdot 0.0873 \approx 5.24 \text{ m/s}\]For the horizontal part, as the towline forms the angle \(\theta\):\[v_x = 61.11 \cdot \cos(15^{\circ}) \approx 59.01 \text{ m/s}\]
03

Apply Forces Equation in Horizontal Direction

In the horizontal direction, the tension \(T\) provides the only horizontal force to overcome the drag \(D\).\[T \cdot \cos(\theta) = D\]Given \(T = 1520 \text{ N}\) and \(\theta = 15^\circ\):\[D = 1520 \cdot \cos(15^{\circ}) \approx 1466 \text{ N}\]
04

Apply Forces Equation in Vertical Direction

Similarly, in the vertical direction, the lift \(L\) and the vertical component of tension counteract the weight \(W\) of the glider.\[L = T \cdot \sin(\theta) + W\]Where the weight \(W = mg = 200 \cdot 9.81 = 1962 \text{ N}\).\[L = 1520 \cdot \sin(15^{\circ}) + 1962 \approx 2365 \text{ N}\]

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Key Concepts

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

Dynamics
Dynamics is the study of forces and their impact on motion. In the context of the glider and airplane system, dynamics involves understanding how different forces interact to affect the glider's movement.
Key dynamics concepts include:
  • Velocity and acceleration: The glider's velocity must be broken down into horizontal and vertical components to analyze its motion accurately.
  • Force balance: Ensures that the forces acting on the glider add up to influence its motion as predicted by Newton's laws.
In this problem, dynamics was used to determine how the glider's lift and drag forces influence its ascent while being towed at a certain angle.
Aerodynamics
Aerodynamics involves the study of the motion of air and the forces it exerts on objects. For a glider, aerodynamic principles are crucial because they determine how the air's movement influences its lift and drag. Aerodynamic forces are essential for a glider's flight dynamics. They are typically split into two main components:
  • Lift: The upward force that counteracts gravity.
  • Drag: The resistive force that acts opposite to the direction of movement.
In the given scenario, calculating the lift and drag involves understanding how air interacts with the glider at the specified towing angle and rate of climb.
Glider mechanics
Glider mechanics deals with the principles that enable a glider to maintain flight without its own power source. It involves understanding the interaction between the glider's structure and the aerodynamic forces acting upon it.
In this exercise:
  • The glider is being towed, which influences how it gains altitude.
  • The tow cable's length and angle adjustment play a significant role in directing its flight path.
  • By analyzing these interactions, we can determine the glider's position and control inputs needed to maintain its glide and ascent.
Essentially, glider mechanics help in ensuring stable and efficient flight when only aerodynamic forces and towing forces are in play.
Force analysis
Force analysis is a crucial part of engineering mechanics that involves breaking down forces to understand their components and impact on structures or vehicles.
In this problem, key steps in force analysis include:
  • Resolving tension into horizontal and vertical components. This helps in calculating the drag and lift acting on the glider.
  • Understanding the equilibrium position of forces where the upwards lift and downwards gravitational force balance each other.
  • Using trigonometric functions like sine and cosine to relate the angle of tow with the components of forces.
This helps us see the interplay between different mechanical forces and their role in enabling the glider's intended motion at a constant angle rate.

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Most popular questions from this chapter

A projectile is launched from the north pole with an initial vertical velocity \(v_{0} .\) What value of \(v_{0}\) will result in a maximum altitude of \(R / 3 ?\) Neglect aerodynamic drag and use \(g=9.825 \mathrm{m} / \mathrm{s}^{2}\) as the surface-level acceleration due to gravity.

Each tire on the 1350 -kg car can support a maximum friction force parallel to the road surface of 2500 N. This force limit is nearly constant over all possible rectilinear and curvilinear car motions and is attainable only if the car does not skid. Under this maximum braking, determine the total stopping distance \(s\) if the brakes are first applied at point \(A\) when the car speed is \(25 \mathrm{m} / \mathrm{s}\) and if the car follows the centerline of the road.

The quarter-circular slotted arm \(O A\) is rotating about a horizontal axis through point \(O\) with a constant counterclockwise angular velocity \(\Omega=\) 7 rad/sec. The 0.1 -lb particle \(P\) is epoxied to the arm at the position \(\beta=60^{\circ} .\) Determine the tangential force \(F\) parallel to the slot which the epoxy must support so that the particle does not move along the slot. The value of \(R=1.4 \mathrm{ft}\).

Two steel balls of the same diameter are connected by a rigid bar of negligible mass as shown and are dropped in the horizontal position from a height of \(150 \mathrm{mm}\) above the heavy steel and brass base plates. If the coefficient of restitution between the ball and the steel base is 0.6 and that between the other ball and the brass base is \(0.4,\) determine the angular velocity \(\omega\) of the bar immediately after impact. Assume that the two impacts are simultaneous.

Just after launch from the earth, the space-shuttle orbiter is in the \(37 \times 137-\mathrm{mi}\) orbit shown. At the apogee point \(A,\) its speed is \(17,290 \mathrm{mi} / \mathrm{hr}\). If nothing were done to modify the orbit, what would be its speed at the perigee \(P ?\) Neglect aerodynamic drag. (Note that the normal practice is to add speed at \(A,\) which raises the perigee altitude to a value that is well above the bulk of the atmosphere.
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