Chris W wrote:
Owen Duffy wrote:


I was taught (in imperial units) to differentiate mass (pound) and
force (pound-force). That learning stood me well when we changed to SI
(metric) part way through school.


You were taught wrong. If you use pounds in a formula that wants mass
such as F=M*A you will get the wrong answer. So lets say you weigh 200
lbs on earth where A = 32 ft/sec^2. You can then calculate your mass by
solving for M = F/A or 200/32 = 6.25.

If you "weigh" 200 lb (no s at the end of unit symbols) on Earth, that
_is_ your mass, in normal usage in either the medical sciences or in
sports, which are of course the purposes for which we normally weigh
ourselves.

You can, of course, use those 200 pounds in the F = ma formula. For
example, if you accelerate those 200 pounds at 40 ft/s², the force is
8000 lb·ft/s², which is, of course, 8000 poundals, since a poundal is
the force which will accelerate a mass of one pound at a rate of 1
ft/s².


When you are doing physical calculations it is very important to use the
correct units, other wise you calculations are meaningless. Suppose you
want to know what you will weigh on the moon where the acceleration due
to gravity is 5.25 ft/sec^2. F = M*A if you use 200 for your mass you
get, 200 * 5.25 = 1050, that indicates you would weigh 1050 lbs on the
moon. Which is clearly wrong. Trying again with the correct units and
you get, 6.25 * 5.25 = 32.8, now that sounds more like what you would
weight on the moon.

Not at all. It indicates that you exert a force due to gravity of 1050
poundals (not lbs) on the moon. On Earth, you would exert a force of
somewhere in the neighborhood of 6410 poundals to 6450 poundals,
depending on your specific location.

Not only is it just as easy to use an unfamiliar unit for force as it
is to use some strange unit for mass, but the absolute
foot-pound-second system (which includes poundals) has been around
considerably longer than the gravitational foot-pound-second system
(which includes slugs), and rather than either of those systems, those
still using English units are more likely to use the engineering system
which includes both pounds and pounds force, but neither slugs nor
poundals. Since that system is not a completely coherent system, of
course, many of the formulas need to be adjusted with a g_c factor, a
dimensionless number equal to the ratio of the acceleration used to
define a pound-force to that used to define a poundal, or g_c = (32.174
ft/s²)/(1 ft/s²) = 32.174

In the non scientific world, where the metric unit KG is used for
weight, M=F*A works just fine if you put what you call "weight" in KG in
for M in the formula.

The symbol for kilograms is kg, not KG. There is nothing different
about the weight in the English units world, where the pound used for
this purposes is, by definition, exactly 0.45359237 kg.

Gene Nygaard