Question

An airplane (5500 kg) initiates a pull-up at 225 m/s with curvature radius 775 m. CG, CP, and tail point T are shown. Thrust and drag cancel. Tail force is vertical. Find the tail force (round to one decimal place). 

Hint:

In pull-up maneuvers, increased lift shifts the pitching moment; the tail force balances CP-CG moment.

Answer 1

Correct Answer: 14.8

Load factor: \[ n = 1 + \frac{V^2}{gR} = 1 + \frac{225^2}{9.81 \times 775} = 1 + 6.61 = 7.61 \] Total lift required: \[ L = nW = 7.61 \times (5500 \times 9.81) = 7.61 \times 53955 = 410{,}000\ \text{N} \] Let: Distance CG → CP = 0.20 m (nose-forward). Distance CG → Tail = 5.50 m (aft). Lift at CP creates nose-down moment about CG: \[ M_{CP} = L \times 0.20 = 410000 \times 0.20 = 82000\ \text{N·m} \] Tail force \(T\) must create opposite (nose-up) moment: \[ T \times 5.50 = 82000 \] \[ T = \frac{82000}{5.50} = 14909\ \text{N} \] Convert to kN: \[ T = 14.9\ \text{kN} \]