# Coffee filter lab

1. Apr 18, 2013

### keicee

If an object of mass M falls under the influence of gravity and a drag force Fdrag, we may write Newton’s Second Law as
Ma = Mg - Fdrag
Where a is the acceleration of the object and Mg is the weight of the object. The drag force has the general form:
Fdrag = bv^n

Where the drag coefficient b is a constant that depends on the shape of the object, and v is the velocity of the object. As an object falls from rest the velocity increases until the drag force and the weight are equal in magnitude. The acceleration then becomes zero and we have the following relation:
Mg = b(Vt)^n

In this lab, the mass ( M) will be comprised of coffee filters. We can then express the total mass in terms of the number of filters (N) and the mass of one filter (m):
M=N(m)
Now we can express a relationship between the number of filters and the terminal velocity.
N(m)g = b (Vt)^n or
N α (Vt)^n

Our goal in this lab is to find n for the drag force.

I need help with my lab report. I need to linearize this equation and i dont know how to do it

2. Apr 18, 2013

### Staff: Mentor

Welcome to the PF.

Why do you think you need to "linearize" that equation?

3. Apr 18, 2013

### keicee

to determine the value of n

4. Apr 18, 2013

### Staff: Mentor

What does the variable "t" represent? It doesn't seem like time enters into this problem...

5. Apr 18, 2013

### keicee

Vt stands for terminal velocity

6. Apr 18, 2013

### haruspex

Should that read N = α (Vt)^n?
As in, plot log(N) against Vt?

7. Apr 18, 2013

### collinsmark

One thing you already know is

I think you need to redo your equation from

Nmg = N α (Vt)n

to

Nmg = α (Vt)n.​

There is no need for that extra N on the right hand side of the equation. There is a single drag force involved (which is a function of the object's velocity). It is not separate drag forces being added together. So there is no need for that extra N. (I'm also assuming that the are different Vt values depending on the value of N; so Vt is a function of N.)

I assume that you have measured different Vt values for different N. Is that correct?

If so, you have two unknowns. Those unknowns are α and n. If you've taken at least two measurements, you can solve for the two unknowns. (Don't use linear algebra to solve for the two unknowns since the equations are not necessarily linear. But might be able to use substitution or some other method. [Graphing maybe, as haruspex suggested?])

8. Apr 19, 2013

### Basic_Physics

You can linearize your equation: Mg = b(Vt)^n
by taking the logarithm both sides, maybe change it first to

M = b/g (Vt)^n
or
M = a (Vt)^n where a = b/g