# Can you express mass in other dimensional units?

I'm just started a Physics I course, and while I've paid attention, I'm stuck on one of the first problems:

Express mass ($M$) in terms of acceleration($a$), density($D$), area($A$), and time($t$).

In the first class, we learned that everything in physics has either no unit, or a unit of mass, length, or time and we applied dimensional analysis to some simple problems. The problem I'm having is trying to convert between the 'fundamental' units. Can you dimensionally reduce an equation to only have mass on one side:

$${\rm M} = {\rm L}^\alpha {\rm T}^\beta$$

Where $\alpha$ and $\beta$ are some exponents? If so, could I have some hints as to how to do this? I know some equations that involve some of these quantities ( $F = Ma$, ${\rm density} = {\rm mass}/{\rm volume}$). Otherwise, am I interpreting this question correctly?

• No you can't, that is why it's fundamental. – Kyle Kanos Aug 19 '15 at 21:11
• "mass, length or volume" are you sure that third one should be volume? seems like an odd selection to me. – Kyle Oman Aug 19 '15 at 21:50
• @KyleOman, I'm sorry. It's time. I edited the question – Ben Aug 19 '15 at 22:22

When you are given acceleration, density, area and time, you can indeed find an expression for mass in terms of these. Here is how you go about it:

Make a table of the units that occur in each, and their exponents:

    L  M   T
a   1     -2
D  -3  1
A   2
t          1


As you can see, you need to use D (density) as the only one that contains mass. But when you use it, you bring length along (with an exponent of -3). Now you can simply look up and down your table for a factor that will eliminate length. A will do that.

I will leave it as an exercise for you to see what coefficients A and D need to have in the expression

$m = D^\alpha A^\beta$

Note that you could write the answer as "any set of coefficients that solves" four equations with three unknowns; this means that there is more than one solution - infintely many, in fact. For example, $at^2$ has units of length, so $D\cdot(at^2)^3$ is one of many combinations that yields mass - but the expression I wrote earlier is simpler. And any product of powers of $at^2$ and $A$ also has dimensions of length (to some power) and could therefore be combined with density to give mass.

As Kyle Kanos stated earlier, mass is a fundamental unit; that doesn't mean you can't express it in terms of derived units, although it's slightly unusual to do so - but certainly not unheard of. Dimensional analysis is very powerful - it's well worth getting to know it better.

And if you are actually looking for $\alpha$ and $\beta$ in the expression you gave in your question - that is indeed impossible. Since neither length nor time contain "mass" in their units, there is no way to make mass appear by multiplying them together. That's how "fundamental units" work.

• Thanks! Does Area really have a L power of -2? It seems like it would be positive. – Ben Aug 19 '15 at 22:33
• @ben Sorry - should be plus 2. I got carried away by the -3 in the line before. – Floris Aug 19 '15 at 22:36

regarding the first half: Density is Mass/Volume (not the other way around), so you need a Volume.. You could achieve that by a sufficient power of the area or find yourself a length-scale from acceleration and time

regarding the second half: that's the whole point of dimensional analysis, a mass-like quantity is a mass-like quantity and must come from one.

• Thanks. I edited the question. Also, thanks for the clarification on the second part. I didn't appreciate that Density was a mass-like quality enough. – Ben Aug 19 '15 at 22:41