The four forces of flight: HSC Engineering Studies Aeronautical Engineering

Syllabus dot point

“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”

5 min readUpdated 2 Jun 2026

Quick answer

In steady level flight, lift equals weight and thrust equals drag. In a steady climb at angle $\theta$ , lift equals $W\cos\theta$ and thrust equals $D + W\sin\theta$ , so the engines do additional work to gain altitude. Unbalanced forces give acceleration through $F = ma$ .

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What this dot point is asking
The answer
Exam-style questions
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In plain English

Think of an aircraft cruising at constant speed and altitude like a cyclist riding steadily along a flat road. Gravity pulls the bike and rider down (weight); the road pushes back up with exactly the same force so they neither sink nor float (lift). Pedalling pushes the bike forward (thrust); wind resistance and rolling friction push back exactly as hard (drag). As long as all four pushes and pulls cancel out in pairs, the cyclist keeps going at a constant speed in a straight line. Climb a hill (climb at an angle) and the road no longer pushes straight up against gravity; some of gravity now pulls the rider backward down the slope too, so the legs (engines) have to push harder just to keep the same speed. Coast down a hill with no pedalling (a power-off glide) and gravity itself supplies the forward push, with the descent angle set entirely by how efficiently the bike rolls versus how much air resistance it fights, exactly the ratio engineers call lift-to-drag.

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, $W = mg$ , 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 = W \qquad T = D

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

Equilibrium conditions for each flight phase

Steady level flight: $L = W$ and $T = D$ (zero net force, constant velocity).
Steady climb at angle $\theta$ : $T = D + W\sin\theta$ (along the flight path) and $L = W\cos\theta$ (perpendicular to it).
Steady descent at angle $\theta$ : $T + W\sin\theta = D$ , and $L = W\cos\theta$ . If $T = 0$ (glide), $\tan\theta = D/L = 1/(L/D)$ .
Unbalanced forces (any phase): $F_{\text{net}} = ma$ , from Newton's second law.

Steady climb

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

Along the flight path:

T = D + W \sin\theta

Perpendicular to the flight path:

L = W \cos\theta

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

Steady descent

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

T + 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:

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

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

Accelerated flight

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

F_{\text{net}} = ma

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

What each force depends on

Two of the four forces are quantified by standard relationships that recur in HSC questions:

L = \tfrac{1}{2} \rho v^2 C_L A \qquad D = \tfrac{1}{2} \rho v^2 C_D A

where $\rho$ is air density, $v$ is true airspeed, $A$ is wing planform area, and $C_L$ and $C_D$ are the lift and drag coefficients (set by aerofoil shape and angle of attack). Both forces rise with the square of airspeed, which is why a small drop in speed near the stall causes a large loss of lift, and why drag (and fuel burn) climbs steeply at high cruise speeds. Air density falls with altitude, so the same indicated airspeed produces less lift higher up, one reason airliners cruise at carefully chosen altitudes that balance reduced drag against available engine thrust.

Why high lift-to-drag ratio matters

The lift-to-drag ratio $L/D$ is the single most important measure of aerodynamic efficiency. Because $L = W$ in level flight, the thrust needed is $T = D = W \div (L/D)$ : a higher $L/D$ means less thrust, less fuel and longer range for the same weight. Designers raise $L/D$ with high-aspect-ratio wings, smooth laminar-flow surfaces, winglets to reduce induced drag, and clean aerofoil sections. This is why a sailplane ( $L/D$ around 40 to 60) glides so far while a fighter optimised for speed and manoeuvre accepts a much lower $L/D$ .

An illustrative aircraft of weight 600 kN with drag held at 50 kN shows how quickly the required thrust grows with climb angle, because the $W\sin\theta$ term dwarfs the roughly-unchanged drag term:

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 $L/D$ 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 $L/D$ for supersonic capability and manoeuvrability.

Worked example

A 350-tonne Boeing 747 cruises at a quoted $L/D$ of 17. Check the engine thrust needed.

Step 1: Find the required lift

In level flight lift equals weight: $L = mg = 350{,}000 \times 9.81 = 3.43$ MN.

Step 2: Find the drag (and thrust)

At constant speed thrust equals drag: $D = L / 17 = 3.43 / 17 = 0.202$ MN = 202 kN.

Step 3: Check against the engines

Four engines at about 50 kN cruise thrust each total 200 kN, consistent with the 202 kN required.

Answering equilibrium questions fast

First state the FLIGHT PHASE (level, climb or descent) before writing any equation; that single word tells you whether $L = W$ or $L = W\cos\theta$ . Then resolve forces ALONG and PERPENDICULAR to the flight path, not vertically and horizontally, whenever the flight path is not level. If the question gives a climb or descent angle, expect a $\sin\theta$ or $\cos\theta$ somewhere; if it gives thrust and asks for acceleration, expect Newton's second law ( $F_{\text{net}} = ma$ ) rather than an equilibrium equation, since an accelerating aircraft is, by definition, not in equilibrium.

Common traps

Saying lift = weight in a climb: Only in level flight. In a climb at angle $\theta$ , lift is $W \cos\theta < W$ .
Ignoring the weight component along the flight path in a climb: The extra thrust during climb is mainly to overcome this component, not the drag.
Treating thrust and drag as always equal: Only at constant speed. During acceleration (takeoff, climbout, descent), the net force is non-zero.
Forgetting to decompose forces for climb angles: Always use $\sin\theta$ and $\cos\theta$ components when the flight path is not horizontal.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2021 HSC-style5 marks

A 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 \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 \text{ kN}

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

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

T = D + W \sin\theta

T = 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.

HSC 2020-style3 marks

Sketch a free-body diagram of an aircraft in steady level flight and explain the equilibrium conditions that apply.

Show worked answer →

The diagram shows four arrows from the aircraft centre of gravity: lift up, weight down, thrust forward, drag rearward. In steady level flight there is no acceleration, so the aircraft is in equilibrium: the vertical forces balance ( $L = W$ ) and the horizontal forces balance ( $T = D$ ). Markers award marks for a correctly labelled four-force diagram and for stating both equilibrium equations with the reason that constant velocity means zero net force.

HSC 2023-style6 marks

A light aircraft of mass 1200 kg accelerates along a runway. The engine produces a constant thrust of 4.0 kN, aerodynamic drag at lift-off speed is 0.8 kN, and rolling resistance averages 0.3 kN during the ground roll. Determine the acceleration of the aircraft and the runway distance needed to reach a lift-off speed of 30 m/s from rest.

Show worked answer →

Net accelerating force.

F_{\text{net}} = T - D - F_r = 4000 - 800 - 300 = 2900 \text{ N}

Acceleration (Newton's second law).

a = \frac{F_{\text{net}}}{m} = \frac{2900}{1200} = 2.42 \text{ m/s}^2

Runway distance. Using $v^2 = u^2 + 2as$ with $u = 0$ :

s = \frac{v^2}{2a} = \frac{30^2}{2 \times 2.42} = \frac{900}{4.84} = 186 \text{ m}

The aircraft needs about 186 m of runway to reach lift-off speed. Markers reward the net force as thrust minus both resistances, the acceleration from $F = ma$ , the correct equation of motion, and a final distance with units. Noting that drag and rolling resistance actually vary with speed (so this is an average estimate) earns the engineering-reasoning mark.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

A light aircraft has a mass of 900 kg and is flying in steady, level flight. Calculate (a) the lift force required, and (b) the drag force if the engine produces 1.1 kN of thrust.

Show worked solution →

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

L = W = mg = 900 \times 9.81 = 8829\ \text{N} \approx 8.83\ \text{kN}

(b) Drag. At constant speed, thrust equals drag.

D = T = 1.1\ \text{kN}

Marking criteria: 1 mark for $L = W$ , 1 mark for the correct numeric lift value with units, 1 mark for $D = T$ with the correct value.

foundation3 marks

Sketch and label the four forces acting on an aircraft in steady level flight, and state which pairs are equal in magnitude.

Show worked solution →

A correct diagram shows lift acting upward, weight acting downward, thrust acting forward (in the direction of flight) and drag acting rearward, all drawn from (or near) the aircraft's centre of gravity.

In steady level flight, the vertical pair balances ( $L = W$ ) and the horizontal pair balances ( $T = D$ ), because the aircraft has zero net force and therefore zero acceleration (constant velocity).

Marking criteria: 1 mark for four correctly labelled and directed force arrows, 1 mark for stating $L = W$ , 1 mark for stating $T = D$ with the reason (constant velocity implies zero net force).

core4 marks

An aircraft of mass 45 tonnes climbs at a constant angle of 6 degrees at constant airspeed. Drag at this airspeed is 38 kN. Calculate (a) the lift required and (b) the thrust required.

Show worked solution →

(a) Lift perpendicular to the flight path.

W = mg = 45{,}000 \times 9.81 = 4.41 \times 10^5\ \text{N} = 441\ \text{kN}

L = W\cos\theta = 441 \times \cos 6° = 441 \times 0.9945 = 438.6\ \text{kN}

(b) Thrust along the flight path.

T = D + W\sin\theta = 38 + 441 \times \sin 6° = 38 + 46.1 = 84.1\ \text{kN}

Marking criteria: 1 mark for correctly finding $W$ , 1 mark for $L = W\cos\theta$ with correct value, 1 mark for $T = D + W\sin\theta$ , 1 mark for the correct final thrust with units.

core4 marks

A glider with a lift-to-drag ratio of 32 is released from a tow at 800 m altitude with the engine (there is none) providing no thrust. Calculate (a) the glide angle, and (b) the horizontal distance it can travel before landing.

Show worked solution →

(a) Glide angle. With no thrust, $\tan\theta = 1/(L/D)$ .

\tan\theta = \frac{1}{32} = 0.03125 \quad \Rightarrow \quad \theta = \tan^{-1}(0.03125) = 1.79°

(b) Horizontal distance. Since $\tan\theta = \text{altitude} / \text{horizontal distance}$ :

\text{distance} = \frac{800}{\tan\theta} = \frac{800}{0.03125} = 25{,}600\ \text{m} = 25.6\ \text{km}

Marking criteria: 1 mark for $\tan\theta = 1/(L/D)$ , 1 mark for the correct angle, 1 mark for correctly relating altitude and distance via the tangent, 1 mark for the correct final distance with units.

core5 marks

The table below gives the thrust required for an aircraft of weight 600 kN at constant airspeed, climbing at different angles theta, with drag held at 50 kN. | theta (deg) | 0 | 6 | 12 | | Thrust required (kN) | 50 | 112.7 | 174.7 | Using the pattern in the table, (a) state the relationship between thrust required and climb angle, and (b) predict the thrust required at theta = 9 degrees.

Show worked solution →

(a) Relationship. $T = D + W\sin\theta$ , so as theta increases, thrust required increases because more of the aircraft's weight must be supported by the forward thrust component (via the increasing $\sin\theta$ term), not because drag itself changes.

(b) Prediction at 9 degrees.

T = 50 + 600 \times \sin 9° = 50 + 600 \times 0.1564 = 50 + 93.8 = 143.8\ \text{kN}

This sits between the 112.7 kN at 6 degrees and 174.7 kN at 12 degrees given in the table, consistent with the non-linear but increasing trend of $\sin\theta$ .

Marking criteria: 1 mark for correctly identifying $T = D + W\sin\theta$ as the governing relationship, 1 mark for explaining the physical reason (supporting weight along the flight path), 2 marks for correct substitution and calculation, 1 mark for the answer sitting sensibly between the table values.

exam6 marks

Explain why the thrust required for a steady climb increases much faster than the thrust required merely to overcome drag, and discuss why this places a practical limit on the maximum sustainable climb angle of a commercial airliner.

Show worked solution →

This is a 6-mark EXPLAIN/DISCUSS: markers reward a physically reasoned account linking the equilibrium equations to a real engineering limit, not just a restatement of the formula.

Core physics. In a steady climb at angle theta, $T = D + W\sin\theta$ . Drag itself changes only modestly with climb angle (since airspeed and hence $\tfrac{1}{2}\rho v^2 C_D A$ are held roughly constant), but the $W\sin\theta$ term grows with the sine of the climb angle and is scaled by the FULL weight of the aircraft, which for a large airliner is far larger than typical cruise drag. Because weight is roughly an order of magnitude larger than drag for a modern airliner (L/D of 15 to 20), even a modest climb angle adds a thrust demand comparable to or larger than the drag term itself.

The practical limit. Engine thrust is finite and roughly constant with altitude and speed (within limits). Once $D + W\sin\theta$ approaches the maximum available thrust $T_{max}$ , the aircraft cannot climb any faster at that weight and airspeed; the maximum sustainable climb angle occurs when $T_{max} = D + W\sin\theta_{max}$ , i.e. $\sin\theta_{max} = (T_{max} - D)/W$ . Since $W$ appears in the denominator, a heavier aircraft (more fuel, more passengers and cargo) has a lower maximum climb angle for the same engines, which is why airliners climb more steeply just after takeoff (light, near-empty fuel tanks relative to landing weight for very long routes) and why maximum climb angle decreases as an aircraft approaches its maximum takeoff weight.

Marker's note: top-band answers (1) correctly identify that drag is roughly unchanged while the $W\sin\theta$ term dominates the increase, (2) quantify or at least reason about the relative sizes of $D$ and $W$ for a real airliner, (3) connect this to a finite maximum thrust to derive the existence of a maximum climb angle, and (4) give a real consequence (weight-dependence of climb performance) rather than stopping at the formula.

exam5 marks

A twin-engine aircraft of mass 1500 kg begins its takeoff roll. Each engine produces 3.2 kN of thrust, aerodynamic drag averages 0.6 kN over the roll, and rolling resistance averages 0.4 kN. Determine the acceleration and the time needed to reach a lift-off speed of 28 m/s from rest.

Show worked solution →

Net accelerating force (two engines).

F_{\text{net}} = 2 \times 3.2 - 0.6 - 0.4 = 6.4 - 1.0 = 5.4\ \text{kN} = 5400\ \text{N}

Acceleration.

a = \frac{F_{\text{net}}}{m} = \frac{5400}{1500} = 3.6\ \text{m/s}^2

Time to lift-off speed. Using $v = u + at$ with $u = 0$ :

t = \frac{v}{a} = \frac{28}{3.6} = 7.78\ \text{s}

Marking criteria: 1 mark for correctly combining both engines' thrust, 1 mark for the net force subtracting both resistances, 1 mark for $a = F_{\text{net}}/m$ with correct value, 1 mark for the correct equation of motion, 1 mark for the correct final time with units.

Practise this

Sources & how we know this

Engineering Studies Stage 6 Syllabus (2011) — NESA
Engineering Studies HSC exam papers — NESA