Physics dimensional analysis problem?

Question

When an object falls through the air, there is a drag force that depends on the product of the surface area of the object and the square of its velocity. Fair = CAv^2. Where C is a constant. Determine the dimensions of C.

I know that Force is (kg)(m)/(s^2).
Area = m^2
v^2 = m^2/s^2

I have to find what units C has to have in order for both sides of the equation to be dimensionally correct.

Answer 1

kg/ ( m^3)

kgm/ s^2 = C * m^2*(m^2/s^2)
sec cancel..........kg*m = C* ( m^4)
so C = kg / ( m^3)

Answer 2

You can minimize a lot of tedious re-writing by using L, M and T for the dimensions.

Then [F] = LM/T^2, where [...] means "the dimensions of... ".

hence [F] = [C][A][v]^2, that is

LM/T^2 = [C]*L^2*L^2/T^2, that is [C] = M / L^3.

In other words, C has the dimensions of a DENSITY.

Only now do you need to insert your chosen system of UNITS. In your chosen system that means that:

C is given in units of kg /m^3. QED

Live long and prosper.

P.S. There is a great deal of confusion between the concept of PHYSICAL DIMENSIONS and the UNITS that you choose to measure them in. That confusion is present in your question, where you first ask about the dimensions of C, but then switch to say that you have to find its units. That may well reflect the teaching to which you've been exposed.

By using quantities like L. M, and T in a dimensional analysis, as above, you not only shorten what you need to write (compactness in algebra is a significant virtue). You are also concentrating on the ESSENCE of what you are doing, namely that lengths, masses and time are DIFFERENT kinds of things. At that stage you simply don't CARE what units you will measure them in. Your preferred choice of units is a question that can be temporarily left aside, to avoid unnecessary clutter if nothing else.

In addition, what I wrote out carefully, above, can be further streamlined, once you're used to thinking and writing in this way.

You can simply write F = CAv^2 so that with essential dimensional substitutions,

LM/T^2 = C*L^2*L^2/T^2, that is C = M / L^3.

In other words, simply let the presence of L, M and/or T let you know that this is to be read as a DIMENSIONAL calculation. That means that you can drop all the "[...]" brackets that I was careful to use the first time around.

That can CONSIDERABLY speed up the use of dimensional considerations.

Here's another example. We know that the force of gravity satisfies:

F = G m1*m2 / r^2.

So what are the dimensions of G? Using the speeded up method:

LM/T^2 = G M^2 / L^2, so G = L^3 / (M T^2).

In your preferred units, that means that G is measured in units of m^3 kg^(-1) s^(-2).

The same result can be found by equating the inward acceleration of something in a circular orbit, v^2 / r , to the inward gravitational acceleration it must have, Gm / r^2. (In fact recalling that Gm / r^2 IS an acceleration is actually the simplest way of all to obtain it, without having to recall the dimensions of force F. Try it!)

Answer 3

[kg][m][s^2] = [C][m]^2[m]^2[s]^-2

Let the unit of C be [kg^x ] [m^y][s^z]

Therefore,

Writing the equation
[kg^1][m^1][s^2] = [kg^x] [m^y] [s^z] [m^2] [m^2] [s^-2]

The right hand side reduces to

[kg^x] [m^(y +4)][s^ (z- 2)

Comparing the powers of kg, m and s on both sides,
x =1, y+4 =1 and z-2 =2

x=1, y = -3 and z= 4

The unit of C is [kg^1] [m^ (- 3)] [s^ (4)]

{ Note; When an equation is dimensionally correct, then it follows that the equation will be correct in its units also}

Answer 4

Let's start by reversing sides.

C*A*v^2 = F
C*m^2*m^2/s^2 = Kg*m/s^2

combine exponents to get

C*m^4/s^2 = Kg*m/s^2

multiply both sides by s^2, and get

C*m^4 = Kg*m

divide both sides by m^4, and get

C = Kg/m^3

It's a measure of density

Answer 5

[kg][m][s]^2 = [C][m]^2[m]^2[s]^-2
[kg][s]^2 = [C][m]^3[s]^-2 (after working with [m])
[kg][s]^4 = [C][m]^3 (after working with [s])
[kg][s]^4[m]^-3 = [C] (after bringing [m]^3 to the left hand side)
So units of C are kg s^-4 m^-3