# Drag force - dimensional analysis

I have tried the following example from the link: MIT OCW 8.012 PS1 It is about dimensional analysis.

Derive an expression for the drag force on a ball of radius $R$ and mass $M$ moving with velocity $v$ through a medium with mass density $\rho$.

I have tried the following: $F=R^a M^b v^c \rho^d$ Knowing the dimensions of force and the other quantities I get: $$M^1L^1 T^{-2}=L^aM^bL^cT^{-c}M^dL^{-3d}$$ This gives me a system of equations: $$1=b+d$$ $$1=a+c-3d$$ $$-2=-c$$ This gives $c=2$ thus the drag force must be proportional with $v^2$. Then I can put the $c=2$ into the second equation to get: $1=a+2-3d$ from which follows: $-1=a-3d$. Now I am stuck with a system of 2 equations: $$1=b+d$$ $$-1=a-3d$$ I know that this has not only one solution, so how to continue?

Is it even possible to derive an expression for the drag force? This example was given on a problem set. Is there only one correct solution or are there more possibilities?

I have thought of expressing $d$ and $a$ in terms of $b$ yielding: $d=1-b$ and $a=2-3b$ so a solution would be $F=R^{2-3b}M^bV^2\rho^{1-b}$. Could this be solved in a better way?

The way I like to phrase pretty much all of dimensional analysis is, "you can only take an arbitrary mathematical function of dimensionless parameters: mathematics doesn't directly deal in any other sorts of functions."

When you see $[[R]] = \text m, ~~[[M]] = \text {kg},~~[[v]] = \text{m/s},~~[[\rho]] = \text{kg}/{\text m^3}$ your first question needs to be, "are there any dimensionless combinations of my parameters, or are these all linearly independent?" In this case there is only one dimensionless parameter, $\rho R^3/m.$ You can indeed write exponents just the way you are doing to find a system of equations which "leaves some parameters unspecified".

Now take the dimensionful parameters and assemble them into any quantity $Q$ which has the appropriate dimensions to model some phenomenon $\Phi$, and take your dimensionless parameters $a_1, a_2, \dots a_n,$ and combine them together with an arbitrary function $f$:$$\Phi = Q~ f(a_1, a_2, \dots a_n) ~~~\text{for some } f.$$Functions are so generic that you have given a complete description of the system in that concise little blurb, and it is at the full generality that mathematics will allow.

So for example to get $\text{N} = \text{kg} ~\text m / \text s^2$ probably the easiest $Q$ is $M v^2 / R$. Now since there is only one dimensionless parameter, the drag force must be expressible as:$$F_d ~=~ \frac{M v^2}{R} \cdot~ f\left({\rho R^3 \over M}\right).$$ There are many potential solutions but they all tend to be quadratic in $v$ as that's the only place we can derive the fundamental unit of $\text s$ from.

However, if you roll viscosity into this mix, $[[\mu]] = \text {kg} / (\text m \cdot \text s),$ and a new parameter $a_2 = \rho ~ v ~ R / \mu$ emerges, part of a family of dimensionless parameters called Reynolds numbers. By using $f(\{a_i\}) = \alpha/a_2$ you can have drag linear in $v$ or cubic in $v$, for instance.

Now suppose that viscosity is not important and we discover that actually, the mass $M$ does not factor into the expression for drag force at all: then we know that the form of $f$ must be $k~a_1$ for some $k$, and $F_d = k~\rho~R^2 ~v^2.$ This dimensionless constant $k$ is part of a family of dimensionless parameters called drag coefficients.

Hopefully that gives you a sense of your options. You have to understand something more about the physical situation if you want to narrow down $f$.

You'll find a one parameter family of solutions, because you have 4 independent quantities while in this problem you have the 3 independent units for mass, length and time. In your solution, you can see that the freedom to choose the parameter b comes from the fact that the density of air divided by the density if the object is dimensionless.

To fix b requires bringing in some of the relevant physics back into the problem. E.g. you can argue that the ratio of the densities should not affect the air resistance, therefore the power to which you should raise that ratio should be zero to make this factor identical to 1.

Since drag is the resistance force of an object moving through a fluid (a fluid friction term), then you can make the physical argument for the value of $b$. This frictional force exists only at the boundary between the object and the fluid itself (a surface force), this means that the drag force must be independent of the mass of the object, thus $b=0$.

You then have \begin{align} 1&=0+d\\ -1&=a-3d \end{align} so clearly $d=1$ and now $a=2$. Your drag force law takes the form, $$F=k\rho R^2v^2$$ for some dimensionless constant $k$.