3.1 Forces, Moments, Couples & Vectors

Forces, Moments, Couples & Vectors

Statics is the branch of mechanics that deals with bodies at rest or in uniform motion under balanced loads. For EASA Part-66 Module 2, you must treat force as a vector, compute moments and couples, and distinguish translational equilibrium (no net force) from rotational equilibrium (no net moment). These ideas underpin every free-body sketch of a spar, jack, hinge, or control surface.

Force as a Vector

A force is a push or pull that tends to change the motion of a body. In SI units the magnitude is measured in newtons (N). One newton is the force that gives a mass of 1 kg an acceleration of 1 m/s² (from Newton’s second law, F = ma). Because force has both magnitude and direction, it is a vector. Two forces of 100 N are not the same load if one acts vertically upward on a wing spar and the other acts aft along the fuselage.

Vectors are represented by arrows: length proportional to magnitude, arrowhead showing sense. On a free-body diagram (FBD) you draw every external force on the part of interest — weight, reaction, thrust, aerodynamic load, cable tension — and omit internal forces within a single rigid body. For airframe work this “free-body thinking” is how you decide whether a jack is overloaded, whether a bolt is in shear, or whether a control surface hinge moment is within limits.

Resolving Forces into Components

A force at an angle is usually resolved into rectangular components:

• F_x = F cos θ
• F_y = F sin θ

where θ is the angle from the x-axis. Components are scalars with assigned positive directions; you recombine them later if a resultant is needed. Example: a 500 N guy-wire tension at 30° above the horizontal has F_x = 500 cos 30° ≈ 433 N horizontal and F_y = 500 sin 30° = 250 N vertical. The horizontal component loads the attachment in shear or tension along the structure; the vertical component contributes to lift-off or compression on the support.

When several forces act at a point, the resultant is the vector sum of the components. For concurrent forces, ΣF_x and ΣF_y give the resultant’s components; magnitude R = √(ΣF_x² + ΣF_y²) and direction follows tan⁻¹(ΣF_y/ΣF_x) with careful quadrant checks.

Moments (Turning Effect)

A force that does not pass through a chosen point (pivot or axis) tends to rotate the body about that point. The moment (or torque) of a force about a point is:

M = F × d⊥

where d⊥ is the perpendicular distance from the point to the line of action of the force. Units are newton-metres (N·m). Sense is conventionally clockwise or anticlockwise; one sense is taken positive and the other negative so moments can be summed algebraically.

Worked example — hinge moment. A control surface is free to rotate about a hinge. A 40 N aerodynamic force acts on the surface 0.25 m behind the hinge line, perpendicular to the surface. The moment about the hinge is:

M = 40 N × 0.25 m = 10 N·m

If a balance tab produces a 25 N force 0.20 m ahead of the hinge (opposing sense), its moment is 25 × 0.20 = 5 N·m the other way. Net hinge moment is 10 − 5 = 5 N·m still tending to rotate the surface in the original sense. The pilot (or autopilot actuator) must supply that residual moment through the control run.

If the force is not perpendicular to a lever arm of length L, either use the perpendicular component of force (F sin φ or F cos φ as geometry requires) or the perpendicular distance from the pivot to the force’s line of action — both methods give the same M.

Couples

A couple is two equal, parallel, opposite forces whose lines of action do not coincide. The net force is zero, so a pure couple produces no translational acceleration, only a pure turning tendency. The moment of a couple is constant about every point in the plane and equals:

M_couple = F × s

where s is the perpendicular separation of the two lines of action. Example: two 100 N forces, opposite in sense, 0.40 m apart, form a couple of 40 N·m. Turning a control wheel, applying equal and opposite hand loads on a yoke, or applying equal lift forces of opposite sense on upper and lower surfaces of a tab can be modelled as couples.

Equilibrium of Translation vs Rotation

A rigid body is in static equilibrium only when both of the following hold:

1. Translational equilibrium: the vector sum of all forces is zero — ΣF_x = 0, ΣF_y = 0 (and ΣF_z = 0 in 3D). The body has no net tendency to accelerate linearly.
2. Rotational equilibrium: the algebraic sum of all moments about any point is zero — ΣM = 0. The body has no net tendency to angularly accelerate.

These are independent conditions. A body can have ΣF = 0 yet spin if an unbalanced couple remains; it can have ΣM = 0 about one point yet translate if a net force remains. Exam traps often offer a situation with balanced forces but an unbalanced moment (or the reverse) and ask whether the body is “in equilibrium.” Full equilibrium needs both.

Free-Body Thinking for Airframe Loads

Typical maintenance and design sketches:

• Aircraft on jacks: weight acts at the centre of gravity; jack reactions act at jacking points. Take moments about one jack to find the other reaction, then use ΣF_vertical = 0 for the remaining reaction. Never assume equal jack loads if the CG is not midway between jacks.
• Cantilever wing root: lift distributed along the span is replaced by an equivalent force and moment (or couple) at the root. The root fittings must carry both shear (force) and bending moment.
• Landing gear: ground reaction, side load, and drag during braking produce moments about attachment pins; bolts and lugs are sized for the resulting force and moment combination.
• Cable or pushrod control runs: tension is a force vector along the cable; idler and sector moments must balance for a stationary surface.

Worked example — jack reactions. An aircraft of weight 20 000 N rests on two main jacks 4.0 m apart. The CG is 1.5 m from the left jack (and therefore 2.5 m from the right). Taking moments about the left jack (anticlockwise positive for the right reaction):

ΣM_left = 0 ⇒ (R_right × 4.0) − (20 000 × 1.5) = 0
R_right = 30 000 / 4.0 = 7 500 N
ΣF_vertical = 0 ⇒ R_left + 7 500 − 20 000 = 0 ⇒ R_left = 12 500 N

Check: moments about the right jack give the same left reaction. The nearer jack always carries the larger share of weight.

Practical SI Habits

Always convert mass to weight when a force is required: weight W = mg with g ≈ 9.81 m/s² (often 9.81 or 10 m/s² in exam approximations). Moments must use perpendicular distance — using the slant length of a member without resolving the force is a common error. Label every force with magnitude, direction, and point of application on the FBD before writing equations. For Module 2, expect multi-choice questions that mix vector resolution, M = F × d⊥, couple recognition, and the two equilibrium statements.