GATE 2021 AE – Question 63
An airplane of mass 5500 kg, initially in steady level flight at 225 m/s, starts a steady pull-up with curvature radius 775 m. CG is 20 cm ahead of wing center of pressure CP, and tail force point T is 550 cm aft of CP. Use g=9.81 m/s²; neglect tail drag and take tail force vertical. Thrust and drag are equal, opposite and collinear. Tail force is ___ kN (one decimal).
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Correct answer: 14.8 to 15.2
Explanation
**Pull-up manoeuvre.** At the start of a steady pull-up on a circular path of radius $r=775$ m at $V=225$ m/s, the normal acceleration is $V^2/r$. The net force normal to the flight path (lift of the wing plus tail force, minus the weight component) must supply it.
**Effective normal force** (the load the lifting surfaces must carry):
$$N=m\left(g+\frac{V^2}{r}\right)=5500\left(9.81+\frac{225^2}{775}\right)=5500\times(9.81+65.32)=413\,200\text{ N}.$$
**Moments about the wing centre of pressure (CP).** The wing lift $L$ acts at the CP, so it has no moment about the CP. The weight and the inertia force act at the CG, which is 0.2 m ahead of the CP, and together they equal the effective normal force $N$ (acting downward through the CG in the pull-up). The tail force $T$ acts downward at 5.5 m behind the CP.
Moments about the CP balance when
$$N\times0.2=T\times5.5\;\Rightarrow\;T=\frac{0.2\,N}{5.5}.$$
(The wing lift then satisfies $L-T=N$ for the force balance normal to the path.)
$$T=\frac{0.2\times413\,200}{5.5}=15\,026\text{ N}\approx\mathbf{15.0\ kN}\ (\text{downward}).$$
Official GATE 2021 answer key: https://gate2026.iitg.ac.in/doc/download/Answer_keys2021/ae_2021.pdf