If you're like me, you are so baffled by physics problems involving British units that you convert everything to SI to avoid having to deal with pounds and feet. Here is an attempt at explaining the concept of pounds force, pounds mass, and acceleration in somewhat easier terms.

Physics majors: let me know if I mess this up! I do not guarantee this information against errors.

Let's start with a basic review of some kinematics. You are probably familiar with Newton's claim that:

F=ma(1)

We generally do not give this equation much thought because it works so well, but to illustrate an important point, let's annotate the formula a bit:

F(force, N)=m(mass, kg) ⋅a(acceleration, m/s^{2})(2)

Equation 2 shows us that Newton's second law relates three properties of
a body:

Remember that this formula works equally well in either direction; a body can react to a force by accelerating, or it can exert a force under acceleration. There is really no difference between these two phenomena, but it helps to think of it this way when considering the problem with British units.

By this principle, whenever a body is near a source of gravity, as
is the case on the surface of the Earth, that body experiences
an acceleration that gives it its *weight*. We can rewrite Newton's
second law as:

the special case in which W is the object'sW=mg(3)

We see from this equation that the units are
*consistent*, that is, if you multiply kilograms by meters over
seconds squared, the resulting unit is the kg⋅m/s^{2}, or Newton.

We should also note from equation 3 that the metric system distinguishes the
*weight* of a body from the body's *mass*. Weight is an expression
of the force the object applies on a scale or balance, while mass refers
to the quantity of matter making up that object.
When a physics professor
asks you your *weight*, he expects you to reply in Newtons rather
than in kilograms.

Suppose you stand on a purist
metric scale and discover your *weight*, that
is, the force you exert on the scale, to be 785 Newtons. Knowing that **g**
is about 9.8 m/s^{2}, you can use equation 1 to determine that
your *mass* to be 80 kilograms.

Of course, this is not how you typically determine your weight with
a bathroom scale calibrated in pounds.
You may be familiar with the British distinction between *pounds mass*
(lbm) and *pounds force* (lbf);
one unit refers to mass, and the other to force.
However, unlike the metric units of mass and force, *pounds force and
pounds mass are equivalent!* That means that you cannot use Newton's
second law with British units, such as:

F(force, lbf)=m(mass, lbm) ⋅a(acceleration, feet/s^{2})(incorrect!)(4)

Why not? The reason is a fatal flaw in the British unit system.
Recall the conclusion from equation 3 that weight and mass are two entirely
different properties. Without gravity, bodies are weightless, but still
have mass and are still governed by Newton's second law. Thus, the weight of
an object depends not only on mass, but on the local gravity acting on that
mass. On Earth, one pound of mass has a weight of one pound of force. This
is analogous to a **g** of 1, which would be terribly convenient, but
*this is not correct*. The quantity of **g** is dictated by the mass
of the Earth and cannot be arbitrarily assigned. It can only be measured or
computed from measured quantities. The key to understanding the failure of
the British system is to note that *the relationship between mass and weight
is fixed and cannot be assigned.*

Fine, then, but how do we overcome a flaw in our system of units? We begin
by eliminating some definitions that were erroneously made in defiance to
Newton. As noted above, we can arbitrarily choose a unit of mass or of
force, but not both. Here, we will forget about mass and consider the pound
to be a unit of force. (The reason for this should hopefully be apparent;
when we speak of pounds,
most of the time, we are concerned with the *weight* of things, not
their mass.)
Returing, then, to Newton's second law, in terms of weight,

W(force, lbf)=m(mass, ??) ⋅a(acceleration, feet/s^{2})(5)

Note that we have no idea what the unit for mass is at this point.
Since we can convert meters to feet, we can conclude that the value of
**g** at the Earth's surface is about 32.2 ft/s^{2}. Now we know
the quantities and units of every value in equation 5 except for *m*.
Let's examine a hypothetical body having a weight of one pound:

1 lb = m ⋅ 32.1 ft/s^{2}(6)

This gives us the quantity of *m*
Keeping in mind that we must obey algebra, we multiply to ensure
that *m* is a unit quantity:

1 ?? ⋅ 32.1 ft/s^{2}= 32.1 lbf(7)

We now know that a body with a true mass of 1 has a weight of 32.1 pounds here on Earth. All that's missing is a name for our unit of true mass. Are you ready?

The family of the engineer who conceived the slug (abbreviated sg) have requested that his identity and place of burial be kept secret, for their own protection. Unfortunately, since the engineering community is yet unwilling to give up the British system, even after its inherent ambiguity was instrumental in a multi-million-dollar failure in the NASA Mars program, we are stuck with it, at least for now. Just like the Newton, the slug can be derived from other units:

1 sg ⋅ 32.1 ft/sYou should think of the slug as^{2}= 32.1 lb

1 lb = 1 sg⋅ft/s^{2}(8)

Also remember equation 8. It gives us the ability to confirm our results when making calculations involving slugs.

One last note: the slug exists only as a bridge to allow you to perform kinematics calculations with British units. At no time should you be asked to provide a mass in slugs, or be given a mass quantity in slugs. Slugs should appear only within calculations.

Here's an example of an easy problem involving British units:

Q: What work is necessary to raise a pound of mass one foot?

A:This question calls for the computation of energy from mass and height. (You should know the answer without doing the math!) The relevant formula is:Since we are given theU = mgh(9)weightof the object, and require itsmass, we need to convert pounds mass to slugs. Remember that a quantity in pounds mass is still a measure of weight! We therefore divide by 32.1 to find a mass of 0.0312 sg. A mental check confirms that we want to divide; we are lifting a relatively small weight compared to that of a slug. Plugging in the quantities,U= (0.0312 sg) (32.1 ft/s^{2}) (1 ft)(10)The units of

uare of particular interest:Rewriting equation 10 and referring back to equation 8 confirms the answer we suspected, and in addition, demonstrates that our system of units is now consistent, a very important step in error-checking our answer. note the resemblance in derived units of the Newton to the pound:U= sg ⋅ ft^{2}/s^{2}(1 ft)(10)U= 1 sg⋅ft/s^{2}⋅ ftU= 1 lb⋅ft(11)## Example 1 conclusion

Pounds are a measure of weight, not mass. When given the weight of an object in pounds, if you need the mass, divide by 32.1 to obtain the mass in slugs.Remember that a slug is quite a bit. Your result should be a much smaller number.

Here's one more example:

Q: How much mass (in pounds) can be lifted one foot by one foot-pound of energy?

A:The answer is obvious, but again, let's do the math anyway. Rewrite equation 9 and plug in the numbers:1 lb⋅ft =We can easily solve form(32.1 ft/s^{2}) (1 ft)(12)m, which we know to be a mass quantity in keeping with equation 9. Pay attention to how the units work out in slugs, not pounds.m= 0.0312 lb⋅ft⋅s^{2}/ft^{2}

m= 0.0312 lb ⋅ s^{2}/ftm= 0.0312 sg⋅ft/s^{2}⋅ s^{2}/ftm= 0.0312 sg(13)But remember, the question asks for weight, albeit in an indirect and sneaky manner. We therefore need to convert slugs into pounds. Since there are many pounds to a slug, we multiply for a final answer of 1 lbm. We can refer back to equation 5 to confirm the weight of a slug on Earth:

W= (0.0312 sg) (32.1 ft/s^{2}) = 1 lbf(14)## Example 2 conclusion

When asked for mass in pounds, you must consider the difference between true mass and mass represented by pounds. When pounds are used, gravity is "built-in." Multiply by 32.1 whenever you must convert true mass to mass in pounds.

Hopefully, you now have a better understanding of how the British
system is used in physics. You should be equally comfortable solving
kinematics problems in SI or British units provided you always remember
these points:

The physics community, and most engineers on the rest of the planet, have abandoned the British unit system in the interest of safety and economy. Educators should refrain from teaching the British system other than as a historical curiosity. Until that happens, good luck solving those problems!

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