The KC-10 Extender tanker aircraft of the United States Air Force is used to refuel other planes in flight. The Extender can carry 365,000 lb of jet fuel, which can be transferred to another aircraft through a boom that temporarily connects the two planes.

(a) Express the mass of the fuel in the units of slugs and LBM.

(b) Express the mass and weight of the fuel in the SI.

Solution:

Approdch:-

we will calculate the merss $\mathrm{m}$, weight w, and the gravitational acceleration

$ g=32.2 \mathrm{ft} / \mathrm{s}^2 \\ $.

The expression $w=m g$,

we will then Convert from slugs to lbm Using the Conversion factor 1 slug

$ =32.1 \neq 4 \mathrm{~J} \mathrm{bm} \\ $

$ 1 \text { slug }=14.59 \mathrm{~kg}\\ $

Solution:- (a) we first determent the mass of the fuel in the unit of slugs:

$ \begin{aligned} m &=\frac{w}{g} \\ m &=\frac{3.65 \times 10^5 \mathrm{lb}}{32.2 \mathrm{ft}^2 \mathrm{~/s}^2} \\ &=1.234 \times 10^4 \frac{\mathrm{lb} \mathrm{s}^2}{\mathrm{ft}} \\ m &=1.134 \times 10^4 \mathrm{slugs} \end{aligned} $

(b) we Convert the mass quantity $1.134 \times 10^4$ slugs into the units of $\mathrm{kg}$ in the SI:

$ \begin{aligned}\\ m &=\left(1.134 \times 10^4 \text { slugg }\right)\left(24.59 \frac{\mathrm{kg}}{5 \mathrm{lug}}\right) \\\\ &=1.655 \times 10^5(5 \text { lugs })\left(\frac{\mathrm{kg}}{5 \operatorname{lug}}\right) \\\\ &=1.655 \times 10^5 \mathrm{~kg}\\ \end{aligned}\\ $

we first write $ m=165.5 \times 10^3 \mathrm{~kg}\\ $

since kilo prefix alrealy implies of $ 10^3 \mathrm{~g} \\ $

$ m=165.5 \times 20^6 \mathrm{~g} \\ $

The fuel weight in the SI is:

$ \begin{aligned}\\ w &=\left(1.655 \times 10^5 \mathrm{~kg}\right)\left(9.81 \mathrm{~ms}^{-2}\right)\\ \\ &=1.62 \times 10^6 \frac{\mathrm{kg} . \mathrm{m}}{\mathrm{s}^2} \\\\ &=1.62 \times 10^6 \mathrm{~N}\\ \end{aligned}\\ $

Discussion:-

To double-check the coalition of weight in the SI, By using the Conversion factor

$ 1 \mathrm{lb}=4.448 \mathrm{~N} \\ $

$ \begin{aligned}\\ w &=\left(3.65 \times 10^5 1 b\right)\left(4.448 \mathrm{NJb}^{-1}\right) \\\\ &=1.62 \times 10^6(1 b)\left(\frac{\mathrm{N}}{\mathrm{lb}}\right) \\\\ w &=1.62 \times 10^6 \mathrm{~N} \\ \end{aligned} \\ $