Units in Physics Calculations

The key idea is to think of the number and the units of any one quantity as being multiplied by each other. Thus 6 feet can be thought of as (6)(feet). Then you just use the commutative and associative laws of algebra. For example:

Addition and Subtraction

Here you will use the associative law, factoring out the units, which must already be the same in order to do that; for example:

6 ft + 2 ft  =  (6)(ft) + (2)(ft)     ;

             =  [(6) + (2)](ft)       ;

             =  (8)(ft)               ;

             =  8 ft                  .

It makes no sense to add things that are of different kinds (a time plus a distance does not yield an interesting sum!), and if the same kind of thing is measured in different units, one or both must be converted so that the factoring-out can proceed.

Multiplication

Here you will use the commutative law, grouping the units together and the numbers together; for example:

6 ft * 2 ft  =  (6)(ft) * (2)(ft)         ;

             =  [(6) * (2)][(ft)(ft)]     ;

             =  (12)(ft2)                 ;

             =  12 ft2                    .

Division

Here also, you will use the commutative law, grouping the units together and the numbers together; for example:

6 ft / 2 s  =  (6)(ft) / (2)(s)          ;

            =  [(6) / (2)][(ft)/(s)]     ;

            =  (3)(ft/s)                 ;

            =  3 ft/s                    .

Unit Conversion

Here you will use what the mathematicians call the "multiplicative identity":

• you can multiply anything by 1 without changing its value: a * 1 = a;

• if a = b, then a/b = 1 and b/a = 1.

Thus, using the fact that 1 yd = 3 ft, we can write as follows:

1  =  (1 yd)/(3 ft)     ,

and we can also write

1  =  (3 ft)/(1 yd)     .

Which of these two forms of 1 will be useful to multiply by depends on the situation. Remember, when adding two things of the same kind that have been measured in different units, you must first convert the units to match, and then you will be able to use the associative law to factor out the common units and add the numbers.

6 yd + 2 ft  =  (6)(yd) + (2)(ft)                               ;

             =  [(6)(yd) * (1)] + (2)(ft)                       ;

             =  [(6)(yd) * (3 ft)/(1 yd)] + (2)(ft)             ;

             =  [(6) * (3)/(1)  (yd) * (ft)/(yd)] + (2)(ft)     ;

             =  [(18) (ft)] + (2)(ft)                           ;

             =  [(18) + (2)] (ft)                               ;

             =  (20) (ft)                                       ;

             =  20 ft                                           .

Problems

1. Use the fact that 1.00 inch = 2.54 * 10^-2 m (this was once upon a time an experimental result, with some reasonable number of significant figures, but now is exact: it is the legal definition of an inch in the U.S.) to calculate the height in meters of a person who is 6 ft 6.5 inches tall.

2. There are 640 acres in a square mile, and 5,280 feet in one mile. How many acres are covered by a standard U.S. college football field, including the end zones (120 yards long and 160 feet wide)?

3. A particular computer operating system has a time-out parameter that is specified as 2 microfortnights (programmers sometimes get away with their little jokes). How long is that in seconds?

Dick Piccard revised this file (http://oak.cats.ohiou.edu/~piccard/phys251/units.html) on March 25, 2007.

Please E-Mail comments or suggestions to "piccard@ohio.edu".