Moments, Couples and Equilibrium

Spec mapping: OCR H556 Module 3.2 — Forces in action (moment of a force; torque of a couple; conditions for equilibrium of coplanar forces; centre of mass and centre of gravity). Refer to the official OCR H556 specification document for exact wording.

So far in this course we have treated bodies as point masses. A point mass has position and momentum but no extent — it cannot rotate. Real objects, from a paperback book on a desk to the Tyne Bridge or a wind-turbine blade, are extended in space, and forces applied to them can cause rotation as well as translation. To describe rotation we need the moment of a force; to handle the equilibrium of an extended body we need a second condition on top of $\Sigma\vec{F} = 0$ . This lesson develops that toolkit and applies it to beams, ladders, cranes, signposts and the centre-of-mass practical that anchors OCR PAG 4.

Key Definition: The moment of a force about a chosen pivot is the product of the magnitude of the force and the perpendicular distance from the pivot to the line of action of the force. It is a measure of the force's tendency to cause angular acceleration about that pivot. Its SI unit is the newton metre (N m).

Moment of a Force

For a force of magnitude $F$ acting at a perpendicular distance $x$ from a chosen pivot, the moment $\tau$ (sometimes written $M$ ) is:

$\tau = F\,x$

If the force is applied along a lever arm of length $d$ at an angle $\theta$ to that arm, only the component of $F$ perpendicular to $d$ produces a moment about the pivot:

$\tau = F\,d\sin\theta$

Equivalently, $\tau = F \times (d\sin\theta)$ — the perpendicular distance from the pivot to the line of action of $\vec{F}$ is $d\sin\theta$ . Both formulae are different ways of computing the same number; choose whichever matches the information you have.

SI unit: newton metre (N m). This has the same base units (kg m $^2$ s $^{-2}$ ) as the joule, but moments and energy are conceptually distinct. By convention, the joule is reserved for scalar energy and N m for vector-like rotational quantities — never abbreviate N m as J.
Sense: every moment has a sense of rotation, clockwise (CW) or anticlockwise (ACW). State it explicitly; OCR mark schemes routinely penalise unstated rotation senses.
Choice of pivot: the value of an individual moment depends on the pivot chosen, but for a body in equilibrium the net moment is zero about every point. This freedom is the most powerful problem-solving lever in statics.

Worked Example 1 — Spanner Turning a Nut

A mechanic applies a $60$ N force perpendicular to a $20$ cm spanner to loosen a wheel nut. Find the moment about the nut.

$\tau = F\,x = 60 \times 0.20 = 12\;\text{N m (anticlockwise, viewed from the spanner side)}$

Now suppose the mechanic instead pushes at $70°$ to the spanner shaft. Only the perpendicular component contributes:

$\tau = F\,d\sin\theta = 60 \times 0.20 \times \sin 70° = 60 \times 0.20 \times 0.940 = 11.3\;\text{N m}$

A force applied along the spanner ( $\theta = 0$ ) would give $\sin\theta = 0$ and zero moment — the force would only stretch the spanner, not turn the nut. This is why one instinctively pulls perpendicular to a lever for maximum turning effect, and why the moment is sometimes called the torque arm of the force.

Visualising Moment Geometry

The geometry has a useful dual reading. Either treat $F$ as the full force and use the perpendicular distance $d\sin\theta$ from the pivot to the line of action, or treat $d$ as the full lever arm and use the perpendicular component $F\sin\theta$ of the force. They give the same moment.

Principle of Moments

A rigid body is in rotational equilibrium about a chosen pivot if, and only if:

$\sum \tau_{\text{CW}} = \sum \tau_{\text{ACW}}$

equivalently $\sum \tau = 0$ if we adopt a sign convention (e.g. ACW positive). This statement is the principle of moments. It is exactly the rotational analogue of $\sum \vec{F} = 0$ for translational equilibrium.

A crucial subtlety: if the principle of moments holds about one point, and the body is also in translational equilibrium, it holds about every point. The choice of pivot is therefore arbitrary for a body already in equilibrium. We exploit this by pivoting at the point where the largest unknown force acts — that force then contributes zero moment, leaving fewer unknowns to solve for.

Worked Example 2 — Seesaw

A $30$ kg child sits $1.5$ m from the pivot of a uniform seesaw. Where must a $45$ kg child sit on the opposite side to balance it?

Taking moments about the pivot (the seesaw's own weight contributes zero moment because it acts at the pivot for a uniform plank):

$45 \times g \times x = 30 \times g \times 1.5$

$g$ cancels, so:

$45 x = 45 \Rightarrow x = 1.0\;\text{m}$

The heavier child sits closer to the pivot. Only the mass ratio matters — the result is independent of $g$ , which is why seesaws balance the same way on the Moon.

Couples and Torque

A couple is a pair of equal-magnitude, opposite-direction forces whose lines of action do not coincide. The two forces have zero vector sum, so they cause no translation. However they have a non-zero net moment about every point in the plane, equal in magnitude to:

$\tau_{\text{couple}} = F\,d$

where $d$ is the perpendicular distance between the two lines of action. This is the torque of the couple.

That the moment is independent of the chosen pivot is the defining feature of a couple. Try it for any point you like — the contributions from the two forces always add to $F d$ with the same rotational sense.

Familiar examples:

Steering wheel: you push on one rim with one hand and pull on the other rim with the other — equal and opposite forces, separated by the wheel's diameter, producing rotation without translation.
Screwdriver: the cross-grip of a screwdriver delivers a pure couple to the screw head.
Magnetic compass needle in a field: equal-and-opposite forces on the two poles (a magnetic dipole couple) rotate the needle.
Tightening a bolt with a torque wrench: the wrench is calibrated to apply a specified torque about the bolt's axis.

"Torque" and "moment of a couple" are usually synonymous at A-Level. Both are measured in N m. Engineers often quote engine torque in N m to characterise the rotational driving moment delivered to the crankshaft.

Full Conditions for Static Equilibrium

A rigid body is in complete (static) equilibrium if both of:

Translational equilibrium: $\sum \vec{F} = 0$ — no net force, so no linear acceleration.
Rotational equilibrium: $\sum \tau = 0$ about any (and therefore every) point — no net moment, so no angular acceleration.

Both conditions are independent and must be checked separately. A pure couple has $\sum \vec{F} = 0$ but $\sum \tau \neq 0$ — it accelerates angularly without translating. A single off-axis force has $\sum \vec{F} \neq 0$ and $\sum \tau \neq 0$ about most points — it accelerates both ways. Static equilibrium requires both to vanish simultaneously.

flowchart TD
  A[Extended rigid body] --> B[Identify every force on the body]
  B --> C[Draw a labelled free-body diagram]
  C --> D{Both equilibrium conditions to check}
  D --> E[Sum F equals 0: translational]
  D --> F[Sum tau equals 0: rotational]
  E --> G[Resolve into x and y components]
  F --> H[Choose smart pivot to eliminate an unknown]
  G --> I[Solve simultaneous equations]
  H --> I

Worked Example 3 — Loaded Beam on Two Supports

A uniform horizontal beam of length $4.0$ m and mass $20$ kg rests on two vertical supports: support A at the left end and support B located $3.0$ m from A. A $30$ kg load is placed on the beam $1.0$ m from A. Find the reaction forces at A and B.

Let $R_A$ and $R_B$ be the upward reactions at A and B respectively.

Translational equilibrium (vertical):

$R_A + R_B = (m_{\text{beam}} + m_{\text{load}})\,g = (20 + 30) \times 9.81 = 490.5\;\text{N}\quad(1)$

Rotational equilibrium — moments about A (eliminates $R_A$ ):

Clockwise moments about A:

Load: $30 \times 9.81 \times 1.0 = 294.3$ N m
Beam's own weight, acting at the beam's centre (uniform beam ⇒ centre of mass at $2.0$ m from A): $20 \times 9.81 \times 2.0 = 392.4$ N m

Anticlockwise moment about A:

$R_B$ at $3.0$ m: $R_B \times 3.0$

Principle of moments: $R_B \times 3.0 = 294.3 + 392.4 = 686.7$ ⇒ $R_B = 228.9$ N.

From (1): $R_A = 490.5 - 228.9 = 261.6$ N.

Sanity check — take moments about B and re-derive $R_A$ :

ACW about B: $R_A \times 3.0$ . CW about B: load at $(3.0 - 1.0) = 2.0$ m gives $294.3$ N m; beam weight at $(3.0 - 2.0) = 1.0$ m gives $196.2$ N m. (The overhanging part of the beam beyond B, from $3.0$ m to $4.0$ m, is to the right of B and contributes an ACW moment; but for a uniform beam the centre of mass is at $2.0$ m, not at the geometric centre of the supported portion — so we use the centre of mass of the whole beam.) Re-checking: beam weight acts at $2.0$ m from A, which is $1.0$ m to the left of B, giving a CW moment about B of $20 \times 9.81 \times 1.0 = 196.2$ N m. Total CW about B = $294.3 + 196.2 = 490.5$ N m. So $R_A \times 3.0 = 490.5$ ⇒ $R_A = 163.5$ N — which contradicts the first calculation. Where is the discrepancy?

Re-examining: the overhanging $1.0$ m of beam beyond B carries part of the beam's mass to the right of B, contributing an ACW moment about B that we omitted. For a uniform beam, the centre of mass is at the geometric centre ( $2.0$ m from A), but when taking moments about B we should treat the whole beam weight as acting at the centre of mass, $1.0$ m to the left of B — and that is what we did. The remaining error must be arithmetical. Let us redo it more carefully using the symmetry that the principle of moments is consistent: if $R_B = 228.9$ N is correct and the vertical-equilibrium equation is correct, then $R_A = 261.6$ N must also be correct. Taking moments about B should confirm this.

Moments about B (ACW positive, treating B at position $3.0$ m from A on the beam):

$R_A$ acts at A, $3.0$ m to the left of B, force upward ⇒ moment about B is $R_A \times 3.0$ ACW.
Load at $1.0$ m from A is at $(3.0 - 1.0) = 2.0$ m to the left of B, weight downward ⇒ moment $30 \times 9.81 \times 2.0 = 588.6$ N m, clockwise.
Beam weight at $2.0$ m from A is at $(3.0 - 2.0) = 1.0$ m to the left of B ⇒ moment $20 \times 9.81 \times 1.0 = 196.2$ N m, clockwise.

Equilibrium: $R_A \times 3.0 = 588.6 + 196.2 = 784.8 \Rightarrow R_A = 261.6$ N. ✓

The earlier slip was using the load moment as $294.3$ N m instead of $588.6$ N m — the load is $2.0$ m to the left of B, not $1.0$ m. Habit of taking moments twice catches exactly this kind of mistake; OCR examiners reward the check.

Worked Example 4 — Ladder Against a Wall

A uniform ladder of length $5.0$ m and mass $15$ kg rests against a smooth vertical wall at $70°$ to the horizontal. A $75$ kg climber stands $3.5$ m up the ladder (measured along its length). The ground is rough. Find the horizontal force from the wall and the friction force on the floor. (Smooth wall means only a horizontal contact force — no friction at the top.)

Let:

$W_L = 15 \times 9.81 = 147.2$ N (ladder weight, acting at midpoint $2.5$ m along).
$W_P = 75 \times 9.81 = 735.8$ N (climber weight at $3.5$ m along).
$N_w$ = horizontal reaction from the smooth wall (top of ladder).
$N_f$ = upward normal from floor; $F_f$ = horizontal friction from floor.

Vertical equilibrium: $N_f = W_L + W_P = 883$ N.

Horizontal equilibrium: $F_f = N_w$ .

Take moments about the foot of the ladder (eliminates both floor forces):

CW moments about foot (weights pulling ladder over):

Ladder weight × horizontal distance from foot to its centre of mass: $W_L \times 2.5\cos 70° = 147.2 \times 2.5 \times 0.342 = 125.8$ N m.
Climber weight × horizontal distance: $W_P \times 3.5\cos 70° = 735.8 \times 3.5 \times 0.342 = 880.7$ N m.

ACW moment about foot (wall holding the top in):

$N_w \times 5.0\sin 70° = N_w \times 5.0 \times 0.940 = 4.70\,N_w$ N m.

Setting CW = ACW: $4.70\,N_w = 125.8 + 880.7 = 1006.5$ ⇒ $N_w = 214$ N.

So $F_f = N_w = 214$ N and $N_f = 883$ N. The ratio $F_f / N_f = 214/883 = 0.24$ is the minimum coefficient of static friction needed to prevent slipping. A typical rough concrete floor has $\mu_s \approx 0.6$ , so the ladder is comfortably safe — but if the climber went higher up, or the ladder were steeper, $F_f / N_f$ would rise and the ladder would eventually slip.

This is also why ladder leaning angles are quoted as the "4-to-1 rule" (foot 1 m out from the wall for every 4 m of vertical rise, giving $\arctan(4/1) \approx 76°$ ) — chosen to keep $F_f/N_f$ well below typical $\mu_s$ values.

Centre of Mass and Centre of Gravity

The centre of mass (CoM) of a body is the unique point at which the body's mass can be considered to be concentrated for purposes of translational motion. The centre of gravity is the point through which the resultant of all gravitational forces on the body acts. In a uniform gravitational field the two coincide; in a non-uniform field (e.g. a tower so tall that $g$ differs significantly between bottom and top) they can differ slightly. At A-Level, treat the two as synonymous.

Moments, Couples and Equilibrium

Moments, Couples and Equilibrium

Moment of a Force

Worked Example 1 — Spanner Turning a Nut

Visualising Moment Geometry

Principle of Moments

Worked Example 2 — Seesaw

Couples and Torque

Full Conditions for Static Equilibrium

Worked Example 3 — Loaded Beam on Two Supports

Worked Example 4 — Ladder Against a Wall

Centre of Mass and Centre of Gravity

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