Aeronautical Engineering

NSWEngineering StudiesSyllabus dot point

Engineering mechanics: How are the four forces of flight (lift, weight, thrust, drag) balanced in steady level flight, climb and descent?

Identify the four forces of flight, apply equilibrium conditions to steady level flight, climbs and descents, and calculate net force and acceleration during accelerated phases

A focused answer to the HSC Engineering Studies Aeronautical Engineering dot point on the four forces of flight. Lift, weight, thrust and drag in steady level flight, balance in climb and descent, and worked HSC-style past exam questions.

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What this dot point is asking

NESA wants you to identify the four forces of flight, apply equilibrium in steady level flight, climb and descent, and find net force and acceleration when the forces are unbalanced.

The answer

The four forces

  • Lift (L). Aerodynamic force on the wings perpendicular to the airflow. Always pointing approximately upward in level flight.
  • Weight (W). Gravitational force on the aircraft, W=mgW = mg, directed toward the centre of the Earth.
  • Thrust (T). Propulsive force from the engines, along the engine axis (approximately aligned with the flight direction for cruise).
  • Drag (D). Aerodynamic resistance, parallel to the airflow and opposite to the flight direction.

Steady level flight

In steady level flight (constant altitude, constant airspeed, no acceleration), both pairs of forces balance:

L=WT=DL = W \qquad T = D

A typical airliner cruises with L/D15L/D \approx 15 to 20. Higher L/DL/D means lower fuel burn per kilometre.

Steady climb

In a climb at angle θ\theta at constant airspeed, the forces along the flight path and perpendicular to it must each balance:

Along the flight path:

T=D+WsinθT = D + W \sin\theta

Perpendicular to the flight path:

L=WcosθL = W \cos\theta

In a climb, lift is less than weight (because the weight component perpendicular to the flight path is Wcosθ<WW\cos\theta < W), and thrust is more than cruise thrust (to lift the aircraft against gravity).

Steady descent

In a descent at angle θ\theta at constant airspeed:

T+Wsinθ=DL=WcosθT + W \sin\theta = D \qquad L = W \cos\theta

If thrust is zero (glide), the descent angle is determined by the lift-to-drag ratio:

tanθ=DL=1L/D\tan\theta = \frac{D}{L} = \frac{1}{L/D}

A glider with L/D=40L/D = 40 descends at tan1(1/40)=1.4\tan^{-1}(1/40) = 1.4 degrees, travelling 40 m horizontally per 1 m of altitude lost.

Accelerated flight

If forces are not balanced, Newton's second law gives:

Fnet=maF_{\text{net}} = ma

For takeoff acceleration along the runway: TDμ(WL)=maT - D - \mu (W - L) = ma, where μ\mu is the rolling friction coefficient and WLW - L is the normal force on the gear (which decreases as lift builds up with airspeed).

Australian context

Qantas operates Boeing 737s on domestic routes and Boeing 787, Airbus A330 and Airbus A380 on international routes. The new Boeing 787-9 has a quoted cruise L/DL/D of about 21 (composite wing, advanced aerofoil design). The Royal Australian Air Force operates Boeing F/A-18F Super Hornets and Lockheed F-35A Lightning II, which trade L/DL/D for supersonic capability and manoeuvrability.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2021 HSC style5 marksA Boeing 737 of mass 70 tonnes is flying in straight and level cruise at 800 km/h. (a) Calculate the lift force required. (b) If the engines produce a combined thrust of 45 kN, calculate the drag force. (c) During a 5 degree climb at constant airspeed, calculate the new thrust required (assume drag is unchanged).
Show worked answer →

(a) Lift in level flight. In steady level flight, lift equals weight.

L=W=mg=70,000×9.81=6.87×105 N=687 kNL = W = mg = 70{,}000 \times 9.81 = 6.87 \times 10^5 \text{ N} = 687 \text{ kN}

(b) Drag in level flight. In steady (constant velocity) flight, thrust equals drag.

D=T=45 kND = T = 45 \text{ kN}

The lift-to-drag ratio is L/D=687/45=15.3L / D = 687 / 45 = 15.3, typical for a swept-wing airliner at cruise.

(c) Thrust required in climb. In a steady climb at angle θ\theta, the forces along the flight direction are thrust forward, drag rearward and the component of weight rearward WsinθW \sin\theta. For constant airspeed:

T=D+WsinθT = D + W \sin\theta

T=45,000+687,000×sin5°=45,000+59,900=104,900 N105 kNT = 45{,}000 + 687{,}000 \times \sin 5° = 45{,}000 + 59{,}900 = 104{,}900 \text{ N} \approx 105 \text{ kN}

Climb requires more than twice the cruise thrust because the engine must overcome both drag and the rearward component of weight.

Markers reward (1) lift equals weight in level flight, (2) thrust equals drag at constant speed, (3) decomposition of weight into components along and perpendicular to the flight path in the climb case, and (4) units throughout.

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