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Unit analysis - using measurements as variables - correct (?)

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Currently, I'm expected to find the side length of a square garden whose area is 25cm2. Of course, we're talking about a square here, and since the area of a rectangle is l * w, and, when talking about a square, l = w, the area of a square is S2. Of course, I know each side length is going to be 5cm - that's rather obvious. Instead, I'm asking as to whether I should treat the actual measurement, "cm", as a variable.


    25cm^2 = squareroot[25cm^2] * squareroot[25cm^2]
    25cm^2 = squareroot[25] * squareroot[25] * squareroot[cm^2] * squareroot[cm^2]
    25cm^2 = 5 * 5 * cm * cm
    25cm^2 = 25*cm^2
    25cm^2 = 25cm^2.)

    Please pay attention to how I'm treating cm as a variable - is it mathematically correct to do this? That's my question.

    If not, my question would be this:

    How do you algebraically find the final unit of measurement in which your final answer will be presented, in this question, of course.
    Last edited: Sep 23, 2011
  2. jcsd
  3. Sep 23, 2011 #2


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    No. "cm" is a unit of measurement. Length, which you called, l, is a variable. Width, which you called, w, is a varaible. Both the numbers, l and w, count centimeters.
  4. Sep 23, 2011 #3


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    Yes, in essence, a physical quantity consists of a number and a unit. The unit can be treated as something that is multiplied by the number (since you have 5 units of 1 cm). As a result, when you square the quantity, you square both the number and the unit. (The square of a product of two numbers is equal to the product of the square of those two numbers).

    (5 cm)2 = 52 cm2 = 25 cm2

    In reverse, it is as follows:

    (25 cm2)1/2 = (25)1/2 (cm2)1/2

    = 5 cm

    I guess that's basically what you wrote.
  5. Sep 23, 2011 #4
    Thanks a bunch!
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