Mass Conversion 184B11

1. **State the problem:** We have five masses with different units and need to find the total mass in kilograms, pounds, and slugs. 2. **Given data:** - $m_1 = 500$ g - $m_2 = 800$ g$_e$ (grams-force equivalent) - $m_3 = 15$ poundals - $m_4 = 3$ lbf (pounds-force) - $m_5 = 0.10$ slug - Acceleration due to gravity $g = 30.5$ ft/s$^2$ 3. **Important conversions and formulas:** - $1$ slug $= 32.174$ lb$_m$ - $1$ lb$_m = 0.453592$ kg - $1$ lbf $= 1$ lb$_m \times g_{standard}$ where $g_{standard} = 32.174$ ft/s$^2$ - $1$ poundal $= 1$ lb$_m \times 1$ ft/s$^2$ - To convert force to mass: $m = \frac{F}{g}$ where $F$ is force in lb$_m$ ft/s$^2$ and $g$ is local gravity - $1$ g $= 0.001$ kg 4. **Convert each mass to lb$_m$ (pounds mass):** - $m_1$: $500$ g $= 500 \times 0.001 = 0.5$ kg Convert to lb$_m$: $$0.5 \text{ kg} \times \frac{1}{0.453592} = 1.10231 \text{ lb}_m$$ - $m_2$: $800$ g$_e$ means weight under standard gravity, so force $F = 800$ g$_e = 800 \times 0.00220462 = 1.7637$ lbf Convert force to mass at local gravity: $$m_2 = \frac{F}{g} = \frac{1.7637 \text{ lbf}}{30.5/32.174} = 1.7637 \times \frac{32.174}{30.5} = 1.863 \text{ lb}_m$$ - $m_3$: $15$ poundals Since $1$ poundal $= 1$ lb$_m \times 1$ ft/s$^2$, mass: $$m_3 = \frac{15 \text{ poundals}}{30.5} = 0.4918 \text{ lb}_m$$ - $m_4$: $3$ lbf Convert force to mass at local gravity: $$m_4 = \frac{3 \text{ lbf}}{30.5/32.174} = 3 \times \frac{32.174}{30.5} = 3.165 \text{ lb}_m$$ - $m_5$: $0.10$ slug Convert slug to lb$_m$: $$0.10 \times 32.174 = 3.2174 \text{ lb}_m$$ 5. **Sum all masses in lb$_m$:** $$m_{total} = 1.10231 + 1.863 + 0.4918 + 3.165 + 3.2174 = 9.8395 \text{ lb}_m$$ 6. **Convert total mass to kilograms:** $$9.8395 \times 0.453592 = 4.462 \text{ kg}$$ 7. **Convert total mass to slugs:** $$\frac{9.8395}{32.174} = 0.306 \text{ slugs}$$ **Final answers:** - (a) Total mass in kilograms: $4.462$ kg - (b) Total mass in pounds (lb$_m$): $9.84$ lb - Total mass in slugs: $0.306$ slugs