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As I learned about force and weight, I was angered that the engineering handbooks never made an effort to state if a value was in ${lb}_{m}$ or ${lb}_{f}$. I finally learned why they never worried about it all.

It begins with Newton's equation $F=m\bullet a$, which is almost correct. As it turns out, the actual equation for our system that uses ${lb}_{f}$ and ${lb}_{m}$ is $F=\frac{m\bullet a}{{g}_{c}}$. Also, for years and years, I thought that ${g}_{c}$ was a pain-in-the-butt conversion factor that had to be memorized.

The value and units for ${g}_{c}$ do not need to be memorized. However, the phrase "one pound mass weighs one pound force, on earth, at sea level" needs to be memorized as does the actual form of Newton's equation. If units are included, the equation looks like:

$F\left({\mathrm{lb}}_{f}\right)=\frac{m\left({\mathrm{lb}}_{m}\right)\bullet a\left(\frac{\mathrm{ft}}{{s}^{2}}\right)}{{g}_{c}}$Now play the algebra game and solve for ${g}_{c}$:

${g}_{c}=\frac{m\left({\mathrm{lb}}_{m}\right)\bullet a\left(\frac{\mathrm{ft}}{{s}^{2}}\right)}{F\left({\mathrm{lb}}_{f}\right)}$ Now collect terms: ${g}_{c}=\frac{m\bullet a}{F}\bullet \frac{{\mathrm{lb}}_{m}\bullet \mathrm{ft}}{{\mathrm{lb}}_{f}\bullet {s}^{2}}$The numerical value of both "F" and "m" is unity. And, the numerical value for "a" is 32.174. Plug and chug:

${g}_{c}=32.174\bullet \frac{{\mathrm{lb}}_{m}\bullet \mathrm{ft}}{{\mathrm{lb}}_{f}\bullet {s}^{2}}$If we would just adopt SI, life would be so, very, much easier!