Did the book make a mistake? Air friction and Drag-Force problem

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SUMMARY

The discussion centers on a physics problem regarding the calculation of air drag force on a 20 kg stone with a cross-sectional area of 0.040 m² and a drag coefficient of 0.80, using Eq. 6-14 from "Fundamentals of Physics 9th ed." by Resnick and Halliday. The calculated wind speed required to maintain the stone's motion is approximately 320 km/hr. However, there is confusion regarding whether this speed represents the wind speed relative to the ground or the stone's speed relative to the wind, highlighting a potential misunderstanding of the problem's requirements.

PREREQUISITES
  • Understanding of drag force equations, specifically Eq. 6-14: Drag-Force = (C*ρ*A*v²)*(1/2)
  • Knowledge of fluid dynamics concepts, particularly drag coefficient and air density
  • Familiarity with basic physics principles regarding motion and forces
  • Ability to interpret physics problems and equations from textbooks
NEXT STEPS
  • Review the concept of drag force and its dependence on relative velocity
  • Study the effects of air density on drag force calculations
  • Explore the relationship between wind speed and object motion in fluid dynamics
  • Examine similar physics problems involving drag force and motion to solidify understanding
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Students studying physics, educators teaching fluid dynamics, and anyone interested in understanding the principles of drag force and motion in relation to air resistance.

Nathanael
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Homework Statement



Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vertical cross-sectional area of 0.040 m^2 and has a drag coefficient C of 0.80. Take the air density to be 1.21 kg/m^3, and the coefficient of kinetic friction to be 0.80. (a) In kilometers per hour, what wind speed V along the ground is needed to maintain the stone’s motion once it has started moving?

Homework Equations



So called "Eq. 6-14" is: Drag-Force = (C*ρ*A*v^2)*(1/2)

C is "the drag coefficient" A is the "effective cross sectional area" ρ is the density and v is velocity.

The Attempt at a Solution



I've solved the problem and my answer is right, according to the book. The reason I ask this question is because I don't actually think it's the right answer and I want help clearing up my misunderstanding.

The answer is about 320 km/hr but I think the actual depends on another unspecified variable (the speed that the stone has relative to the ground.)

It says in the chapter that this problem is from (Ch.6; Fundamentals of Physics 9th ed.; Resnick/Hallidy):
"the magnitude of the drag force is related to the relative speed v" (between the fluid and body)

Doesn't this mean that the answer of 320 km/hr is not actually the speed of the wind "along the ground" as the problem specifically says to find, but instead is the speed of the wind relative to the speed of the block?


Did they just make a careless mistake or am I missing something here?
 
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You're right. But the minimum wind speed would be expected to be that which rolls the pebble with minimal speed, i.e., the speed of the pebble barely exceeding 0 m/s.

So the answer seems acceptable. :smile:
 
You should state the problem completely or we can't be of much help to you.
 

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