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Chapter 6: Problem 39

Calculate the ratio of the drag force on a jet flying at 1200 km/h at an altitude of 15 km to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is 0.38 kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.

Short Answer

Expert verified
The ratio of the drag forces is approximately 2.28.

Step by step solution

01

Identify Given Values

We have two scenarios to consider: a jet flying at 1200 km/h at an altitude of 15 km and a prop-driven transport flying at 600 km/h at an altitude of 5 km. We know the air density at 10 km altitude is 0.38 kg/m\(^3\) and at 5 km altitude is 0.67 kg/m\(^3\). The problem states that both airplanes have the same cross-sectional area and drag coefficient.
02

Recall the Drag Force Equation

The drag force \( F_d \) can be expressed using the equation:\[ F_d = \frac{1}{2} C \rho A v^2 \]where \( C \) is the drag coefficient, \( \rho \) is the air density, \( A \) is the cross-sectional area, and \( v \) is the velocity of the aircraft.
03

Calculate the Drag Force for the Jet

For the jet: using its speed (1200 km/h = 333.33 m/s) and assuming its altitude results in an air density of 0.38 kg/m\(^3\) (the value at 10 km, since we don't have the density at 15 km but assume it's similar), the drag force is:\[ F_{d, \text{jet}} = \frac{1}{2} C (0.38) A (333.33)^2 \]
04

Calculate the Drag Force for the Prop-driven Transport

For the prop-driven transport: using its speed (600 km/h = 166.67 m/s) and its air density at 5 km (0.67 kg/m\(^3\)), the drag force is:\[ F_{d, \text{prop}} = \frac{1}{2} C (0.67) A (166.67)^2 \]
05

Determine the Ratio of Drag Forces

The ratio of the drag force on the jet to the prop-driven transport is calculated by dividing the two expressions from Steps 3 and 4:\[\text{Ratio} = \frac{F_{d, \text{jet}}}{F_{d, \text{prop}}} = \frac{(0.38) (333.33)^2}{(0.67) (166.67)^2} \]
06

Simplify and Calculate the Ratio

Now calculate the ratio:\[\text{Ratio} = \frac{0.38 \times 333.33^2}{0.67 \times 166.67^2} = \frac{0.38 \times 111111.1}{0.67 \times 27777.8} = \frac{42222.218}{18518.526} \approx 2.28\]

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Key Concepts

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

Air Density
Air density is one of the most critical factors affecting drag force. It refers to the mass of air per unit volume, often denoted by the Greek letter \( \rho \). At higher altitudes, the air density decreases, which can significantly affect the drag force on an aircraft. In calculations involving drag force, air density is crucial because it determines how much air resistance a moving object will encounter.

  • Higher air density means more molecules of air in a given space, leading to greater air resistance and thus a higher drag force.
  • Lower air density, as found at higher altitudes, results in less air resistance and a lower drag force.

In our exercise, the air density values used are 0.38 kg/m\(^3\) at 10 km (assumed for 15 km altitude) and 0.67 kg/m\(^3\) at 5 km. These values help us understand how much resistance the planes face in the air.
Altitude
Altitude, or the height above sea level, directly influences air density and thus the drag force experienced by an aircraft. The higher the altitude, the lower the air density, since the atmospheric pressure decreases with altitude.

  • As altitude increases, the number of air molecules in a given volume decreases. This reduces the air resistance an aircraft faces.
  • A decreased air resistance means an aircraft can potentially fly faster without increasing its engines' power.

In this exercise, the jet operates at an altitude of 15 km, while the prop-driven transport flies at 5 km. These altitudes were key in determining the respective air densities and thus the drag forces on each type of aircraft.
Velocity
Velocity is the speed of an object in a specified direction, and it dramatically affects the drag force acting on an object. In the context of aerodynamics, velocity is crucial since the drag force increases with the square of the velocity. This means that if an aircraft doubles its speed, the drag force increases four times, assuming other factors remain constant.

  • The drag force is proportional to the square of the velocity, \( v^2 \). As such, even a slight increase in speed can greatly increase the drag.
  • In our example, the jet's velocity is 1,200 km/h, while the propeller-driven transport's velocity is 600 km/h鈥攁ffecting the ratio of the drag forces significantly.

This squared relationship underscores why speed control is critical in aircraft performance and efficiency management.
Cross-Sectional Area
The cross-sectional area is the portion of the aircraft's surface that is exposed to airflow during motion. It's an essential component in calculating the drag force because it determines how much of the aircraft is meeting air resistance directly.

  • A larger cross-sectional area results in a higher drag force, as more air collides with the aircraft.
  • Conversely, a smaller cross-sectional area tends to lower the drag force due to less air resistance.

In the problem, both aircraft are considered to have the same cross-sectional area. This assumption allows us to focus on the differences in air density and velocity to find how they affect the drag forces on their journey through the sky.

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Most popular questions from this chapter

A circular-motion addict of mass 80 kg rides a Ferris wheel around in a vertical circle of radius 12 m at a constant speed of 5.5 m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

The mysterious sliding stones. Along the remote Racetrack Playa in Death Valley, California, stones sometimes gouge out prominent trails in the desert floor, as if the stones had been migrating (Fig. 6-16). For years curiosity mounted about why the stones moved. One explanation was that strong winds during occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force must act on a 20 kg stone (a typical mass) to maintain the stone鈥檚 motion once a gust has started it moving? (Story continues with Problem 37.)

A worker pushes horizontally on a 35 kg crate with a force of magnitude 110 N.The coefficient of static friction between the crate and the floor is 0.37. (a) What is the value of fs,max under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker鈥檚 110 N push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15掳 above the horizontal. (a) If the coefficient of static friction is 0.65, what minimum force magnitude is required from the rope to start the crate moving? (b) If mk 0.35, what is the magnitude of the initial acceleration of the crate?

A 3.5 kg block is pushed along a horizontal floor by a force of magnitude 15 N at an angle 40 with the horizontal (Fig. 6-17). The coefficient of kinetic friction between the block and the floor is 0.25. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block鈥檚 acceleration.
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