2
$\begingroup$

If I were to solve the equation:

$$d=\frac{a^3}{cb^2}$$

and values were given as so: $a=9.7\ \mathrm m$, $b=4.2\ \mathrm s$, and $c=69\ \mathrm{m/s}$.

Why would it be wrong for me to use dimensional analysis first?

Using dimensional analysis you would simplify: $\mathrm{m^3/((m/s)(s^2))}$ to $\mathrm{m^2/s}$.

That is right as far as I know, but when I plug in the values I get $d=.07\ \mathrm{m^2/s}$, that is I plug it in as $(9.7\ \mathrm m^2)/(69\ 4.2\ \mathrm s)$ since I already did dimensional analysis.

However, when I plug in the values as $((9.7^3)(\mathrm m^3))/((69\mathrm m/\mathrm s)((4.2^2)(\mathrm s^2)$ I get the right answer of $d=.75\ \mathrm{m^2/s}$.

I know my dimensional analysis is correct but why is it that I cannot simplify the original equation before plugging in the values (as in I have to plug it in as $9.7^3\ \mathrm m^3$ instead of $9.7\ \mathrm m^2$)?

So essentially in this question I should think of it as for example if I take the equation $5x^2$, then $5x^2$ is equal to $25x^2$? Is that how I should think of it?

$\endgroup$
  • $\begingroup$ The exponents apply both to the units and to the numbers. $\endgroup$ – Javier Jan 5 '18 at 19:49
  • $\begingroup$ Oh right, I did forget about that rule, but can you explain why the exponent applies to both the units and numbers (specifically why does it apply to the unit)? $\endgroup$ – dawnwall Jan 5 '18 at 19:51
  • $\begingroup$ Imagine if you were to evaluate (ab)^2, where a is a constant representing the value or magnitude and b is representing the unit. You would get a^2 * b^2 instead of just a*b^2 (like what you have shown in your question). $\endgroup$ – Kane Billiot Jan 5 '18 at 23:02
  • $\begingroup$ That makes a lot of sense. I just want to clarify for the example given (ab)^2 it follows the power of a product rule, and in my question it does not. cb^2 would just be c*b^2 and as with a^3 you apply the product rule with the units, that is it becomes: 9.7^3(m^3) according to the power of a product rule. Thanks so much that was helpful $\endgroup$ – dawnwall Jan 6 '18 at 2:57
0
$\begingroup$

If you treat units like algebraic, multiplicative factors in every sense, you cannot go wrong. All rules of algebra apply. Realizing this makes matters more simple in every sense.

Just treat units as you would numbers.

$\endgroup$
  • $\begingroup$ Thank you for your answer, I will take that into consideration from now on. My only question left is wjy units are treated algebraically? $\endgroup$ – dawnwall Jan 5 '18 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.