Discussion
Two engines at \( 7.0 \) m produce \( 95 \) kN each. If the failed engine's windmilling drag is \( 12 \) kN at the same arm, the total asymmetric moment is
Answer: A
Correct answer: 749 kN m
Both the live engine's thrust and the failed engine's drag act at the same arm and in the same rotational sense, so they add:
\[ N = (95 + 12) \times 7.0 = 107 \times 7.0 = 749\ \mathrm{kN\,m} \]
The windmilling drag is easy to overlook and adds nearly 13 per cent here. It is largest when the failed engine is windmilling with a high bypass fan turning in the flow, which is precisely the case on the aircraft where the arms are longest.
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