/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Calculate the magnitude of the d... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(C=0.75\).

Short Answer

Expert verified
The magnitude of the drag force is approximately \( 7770.43 \ \mathrm{N} \).

Step by step solution

01

Understand the Drag Force Formula

The formula for calculating the drag force \( F_d \) is given by \( F_d = \frac{1}{2} C \rho A v^2 \), where \( C \) is the drag coefficient, \( \rho \) is the density of the air, \( A \) is the cross-sectional area, and \( v \) is the velocity of the object.
02

Calculate the Cross-Sectional Area

The cross-sectional area \( A \) of the missile can be calculated using the formula for the area of a circle: \( A = \pi r^2 \). Given the diameter is \( 53 \ \mathrm{cm} = 0.53 \ \mathrm{m} \), the radius \( r \) is \( \frac{0.53}{2} \ \mathrm{m} = 0.265 \ \mathrm{m} \). Thus, \( A = \pi (0.265)^2 \).
03

Substitute Values into the Drag Force Formula

Now, substitute all known values into the drag force equation: \( F_d = \frac{1}{2} \times 0.75 \times 1.2 \ \mathrm{kg/m^3} \times \pi \times (0.265 \ \mathrm{m})^2 \times (250 \ \mathrm{m/s})^2 \).
04

Perform Calculations

Calculate the value of the cross-sectional area first: \( A = \pi (0.265)^2 \approx 0.2209 \ \mathrm{m^2} \). Then calculate the drag force: \[ F_d = \frac{1}{2} \times 0.75 \times 1.2 \times 0.2209 \times 250^2 = \frac{1}{2} \times 0.75 \times 1.2 \times 0.2209 \times 62500 \approx 7770.43 \ \mathrm{N} \].

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Key Concepts

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

Drag Coefficient
The drag coefficient, denoted as \( C \), is a dimensionless number that represents the drag per unit area of an object moving through a fluid, such as air or water. This concept is crucial because it helps quantify the resistance an object encounters when it moves through the air.
The drag coefficient does not remain constant and can vary depending on factors such as:
  • Shape of the object
  • Surface roughness
  • Flow conditions (laminar or turbulent)
The missile in our exercise has a drag coefficient of 0.75, which is typical for streamlined shapes, signifying moderate resistance. A lower \( C \) value indicates a more aerodynamic object, while a higher one suggests more resistance.
Understanding the significance of the drag coefficient can help improve the design and efficiency of various objects moving through air.
Air Density
Air density, represented by \( \rho \), is the mass per unit volume of air, typically measured in \( \text{kg/m}^3 \). It plays a significant role in determining how much drag force is exerted on an object.
Different factors influencing air density include:
  • Altitude
  • Temperature
  • Humidity
In our exercise, the air density is given as 1.2 \( \text{kg/m}^3 \) at low altitude—a relatively higher density due to denser air close to the earth’s surface.
Understanding air density is vital when calculating drag force, as it directly impacts the resistance experienced by moving objects.
Cross-Sectional Area
The cross-sectional area \( A \) is the surface area of the object facing the fluid flow. Calculating it is vital for understanding how much drag force an object faces. Typically, the cross-sectional area is directly dependent on the shape and size of the object.
For circular objects, like a missile's front, it is calculated using the formula \( A = \pi r^2 \). In our situation, with a radius of 0.265 meters, the cross-sectional area becomes approximately 0.2209 \( \text{m}^2 \).
The larger the cross-sectional area, the greater the drag force, as more surface area is exposed to the fluid flow. This concept is pivotal for designs in aerospace, automotive, and sports industries.
Velocity
Velocity \( v \) is the speed of an object in a particular direction. In the context of drag force calculation, it is a critical factor because the squared term \( v^2 \) in the drag force equation shows that even small increases in velocity result in a significantly larger drag force.
The missile discussed in the exercise is moving at 250 meters per second—a relatively high speed, which massively increases the drag force.
Key points about velocity impacting drag include:
  • Higher speed results in higher drag
  • Drag is proportional to the square of velocity
For engineers and designers, fully appreciating the impact of velocity is essential for optimizing aerodynamics and enhancing performance of vehicles and projectiles.

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Most popular questions from this chapter

A police officer in hot pursuit drives her car through a circular turn of radius \(300 \mathrm{~m}\) with a constant speed of \(80.0 \mathrm{~km} / \mathrm{h}\). Her mass is \(55.0 \mathrm{~kg}\). What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(30^{\circ}\), the box starts to slip, and it then slides \(2.5 \mathrm{~m}\) down the plank in \(4.0 \mathrm{~s}\) at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

A child weighing \(140 \mathrm{~N}\) sits at rest at the top of a playground slide that makes an angle of \(25^{\circ}\) with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of \(0.86 \mathrm{~m} / \mathrm{s}^{2}\) (down the slide, of course). (a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?

In about 1915, Henry Sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side (Fig. 6-21). Sincosky's mass was \(79 \mathrm{~kg}\). If the coefficient of static friction between hand and rafter was \(0.70\), what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers? (After suspending himself, Sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter. If you do not think Sincosky's grip was remarkable, try to repeat his stunt.)

ssM A car weighing \(10.7 \mathrm{kN}\) and traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) without negative lift attempts to round an unbanked curve with a radius of \(61.0 \mathrm{~m} .(\mathrm{a})\) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is \(0.350\), is the attempt at taking the curve successful?

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