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

1. May 11, 2014

### Nathanael

1. The problem statement, all variables and given/known data

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?

2. Relevant 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.

3. 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?

2. May 11, 2014

### Staff: Mentor

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.