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Chapter 6: Problem 45

calculate the ratio of the drag force on a passenger jet flying with a speed of \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a prop-driven transport flying at half the speed and half the altitude of the jet. At \(10 \mathrm{~km}\) the density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\), and at \(5.0 \mathrm{~km}\) it is \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\). Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient \(C\).

Short Answer

Expert verified
The ratio of the drag force on the jet to the prop plane is approximately 2.27.

Step by step solution

01

- Understand the Problem

Calculate the ratio of the drag forces on two airplanes with different speeds and altitudes. Use the drag equation: \[ F_D = \frac{1}{2} \rho v^2 C A \] where \begin{itemize} \rho is the air density, v is the velocity, C is the drag coefficient, and A is the cross-sectional area. ewline Given: \begin{itemize} \textbf{Passenger Jet:} \rho_1 = 0.38 \ \mathrm{kg} / \ \mathrm{m}^{3}, \ v_1 = 1000 \ \mathrm{km} / \ \mathrm{h} \ (or 277.78 \ \mathrm{m} / \ \mathrm{s}) \textbf{Prop Plane:} \rho_2 = 0.67 \ \mathrm{kg} / \ \mathrm{m}^{3}, \ v_2 = 500 \ \mathrm{km} / \ \mathrm{h} \ (or 138.89 \ \mathrm{m} / \ \mathrm{s})
02

- Calculate the Drag Force for the Jet

First, compute the drag force for the jet. \[ F_{D1} = \frac{1}{2} \rho_1 v_1^2 C A \] \[ F_{D1} = \frac{1}{2} \times 0.38 \ \mathrm{kg} / \ \mathrm{m}^{3} \times (277.78 \ \mathrm{m} / \ \mathrm{s})^2 \times C \times A \] \[ F_{D1} = 0.19 \times 77160.98 \times C \times A \] \[ F_{D1} = 14660.59 \times C \times A \]
03

- Calculate the Drag Force for the Prop Plane

Next, compute the drag force for the prop plane. \[ F_{D2} = \frac{1}{2} \rho_2 v_2^2 C A \] \[ F_{D2} = \frac{1}{2} \times 0.67 \ \mathrm{kg} / \ \mathrm{m}^{3} \times (138.89 \ \mathrm{m} / \ \mathrm{s})^2 \times C \times A \] \[ F_{D2} = 0.335 \times 19306.29 \times C \times A \] \[ F_{D2} = 6467.61 \times C \times A \]
04

- Calculate the Ratio of the Drag Forces

Finally, find the ratio of the drag forces by dividing the drag force of the jet by the drag force of the prop plane. \[ \text{Ratio} = \frac{F_{D1}}{F_{D2}} = \frac{14660.59 \times C \times A}{6467.61 \times C \times A} \] \[ \text{Ratio} = \frac{14660.59}{6467.61} \] \[ \text{Ratio} \approx 2.27 \]

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Key Concepts

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

drag equation
The drag equation is a fundamental principle in fluid dynamics. It helps us understand and calculate the force of resistance, called drag, experienced by objects moving through a fluid. The equation is given by: \[ F_D = \frac{1}{2} \rho v^2 C A \] This equation involves several variables: \( F_D \) is the drag force,\( \rho \) is the fluid density, \( v \) is the velocity of the object relative to the fluid, \( C \) is the drag coefficient, and \( A \) is the cross-sectional area. The drag force increases with the square of the velocity, making these calculations crucial for high-speed scenarios.
air density
Air density, represented as \( \rho \), is a measure of how much mass of air exists in a given volume. It varies with altitude, temperature, and pressure. At sea level, the standard density of air is approximately \( 1.225 \ \mathrm{kg/m^3} \). In the given exercise, the air density was \( 0.38 \ \mathrm{kg/m^3} \) at an altitude of 10 km and \( 0.67 \ \mathrm{kg/m^3} \) at 5 km. Lower air density at higher altitudes means less drag force for aircraft, which is crucial for fuel efficiency and speed.
velocity impact on drag
Velocity has a significant impact on drag force. As velocity increases, the drag force doesn't increase linearly but rather with the square of the velocity. This is described in the drag equation as \( v^2 \). For example, in our exercise: the jet flies at 1000 km/h (277.78 m/s), and the prop plane flies at 500 km/h (138.89 m/s). Despite the prop plane flying at half the speed, the ratio of drag forces ends up being around 2.27 because the jet's higher velocity greatly increases its drag force.
altitude effects on flight
Altitude affects flight primarily through changes in air density. Higher altitudes have lower air density, which decreases drag force. For instance, at 10 km altitude, the air density is 0.38 \( kg/m^3 \), while at 5 km, it's 0.67 \( kg/m^3 \). Thus, aircraft can fly faster and more efficiently at higher altitudes because reduced drag means less resistance and better fuel consumption. These factors are crucial for designing flight paths and optimizing fuel usage in aviation.

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Most popular questions from this chapter

A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. 6-75). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm}\). What is the magnitude of the force on the bolt from the rod?

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed \(1100 \mathrm{~N}\). The coefficient of static friction between the box and the floor is \(0.35 .\) (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. (a) What is the magnitude of the net force acting on the rider? (b) What is the magnitude of the force on the rider from the motorcycle?

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved \(30 \mathrm{~mm}\) horizontally.

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