Experimental Derivation of the Drag Force

In summary, the conversation discusses the experimental derivation of the drag equation. By plotting data from dropping coffee filters with varying masses, it is found that the mass is proportional to the terminal velocity squared. To derive the drag equation, the known gravitational force must be equated to the unknown drag force and some algebraic manipulation is required. It is also suggested to vary the surface area to remove the dependency on area. The graph obtained shows that v^2 is proportional to mass, and from this, it is possible to determine the slope of the line, which includes the unknown drag coefficient (Cp). It is also noted that the drag force includes the air density (rho_air) and a dimensionless constant (Cd) that depends on the
  • #1
koko122
4
0
I am trying to experimentally derive the drag equation. I have dropped coffee filters with varying masses and determined their resulting terminal velocity. I plotted the data and found that the mass of the coffee filter is proportional to the terminal velocity squared. I was now wondering how I can use this to derive the drag equation, which states: F_D = .5CpAv2
 
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  • #2
At terminal velocity the known constant gravitational force is equal to the unknown drag force; equate them and do a bit of algebra and you'll have the ##v^2## dependency. You will also have to do some measurements with a constant mass and different surface areas to trade out the dependency on area.
 
  • #3
Would you be so kind to explain this a bit further. The graph I obtained is attached.
 

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  • #4
At terminal velocity the filter is not accelerating. So the forces acting on it must sum to zero. You know what the two forces are.
 
  • #5
That's obvious but from the graph that I have provided can I derive that mg=dv^2 where d is the drag constant? Or can I show that the proportionality constant is equal to g/d somehow?

The graph shows the following:
$$ {v_{\textrm{T}}}^{2} \propto m$$
and from this I want to show that:
$$\Sigma F=mg-D{v_{\textrm{T}}}^{2}=0$$
So that the drag force can be defined as following:
$$F_{drag}=Dv^{2}$$
 
  • #6
So...

mg = .5CpAv2

Your graph is for

v2 = mg/.5CpA

The standard equation for a straight line is..

y = slope*x + constant

but constant = 0

So in this case

y = v2
x = mass m

and the slope of the graph

= g/.5CpA

You know the slope.
You know the area A
You know g

That just leaves Cp.

Sorry had to edit a few typos in this post.
 
  • #7
koko122 said:
That's obvious but from the graph that I have provided can I derive that mg=dv^2 where d is the drag constant? Or can I show that the proportionality constant is equal to g/d somehow?

The graph shows the following:
$$ {v_{\textrm{T}}}^{2} \propto m$$
and from this I want to show that:
$$\Sigma F=mg-D{v_{\textrm{T}}}^{2}=0$$
I don't see how that would follow using only the graph. It follows from Newton's second law, and that ∑F=0 at a constant velocity.

See C. Watters post #6. Cp is the only unknown in the expression for the slope, so it can be calculated.
 
  • #8
Ya that makes sense. Would it be correct to infer from the graph that the drag force is equal to: $F_drag=Dv^2$, where D is just a general drag constant?
 
  • #9
I don't think you can call D a "general" drag constant because D includes A (the area of the object). So it's specific to that size of filter cone rather than cones in general.

D = .5CpA

Had to edit for typo again.
 
  • #10
CWatters said:
I don't think you can call D a "general" drag constant because D includes A (the area of the object). So it's specific to that size of filter cone rather than cones in general.

D = .5CpA

Had to edit for typo again.
And wouldn't Cp include the density of air as well? That has to be in the drag force expression somewhere.

From http://en.wikipedia.org/wiki/Drag_(physics)#Drag_at_high_velocity
we have
[tex]F_{drag}=\frac{1}{2} \ \rho_{air} \ C_d \ v^2[/tex]
where Cd is dimensionless -- and I'm pretty sure depends on the object's shape.
 

1. What is the drag force and how is it derived experimentally?

The drag force is a force that acts in the opposite direction of an object's motion through a fluid. It is derived experimentally by measuring the force acting on an object as it moves through the fluid at different velocities.

2. What factors affect the magnitude of the drag force?

The magnitude of the drag force is affected by the density and viscosity of the fluid, the size and shape of the object, and the velocity of the object.

3. How is the drag coefficient determined in experimental derivation of the drag force?

The drag coefficient, which is a dimensionless quantity that describes the relationship between the drag force, the object's frontal area, and the fluid density and velocity, is determined by conducting multiple experiments and analyzing the data to find the best fit for the drag force equation.

4. What are some common methods used in experimental derivation of the drag force?

Some common methods used in experimental derivation of the drag force include wind tunnel testing, water tank testing, and flow visualization techniques such as smoke or dye tracing.

5. How is the drag force used in real-world applications?

The drag force is an important factor in many real-world applications, such as designing aerodynamic vehicles, calculating the resistance of fluids on ships and submarines, and predicting the flight paths of projectiles. It is also used in sports, such as in determining the optimal shape and materials for sports equipment like golf balls and swimming caps.

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