The full lecture, read side-by-side: every idea on the left comes alive in the live panel on the right as you scroll. Watch a door swing wide from its handle, a see-saw level out, a crane refuse to topple, a tightrope walker's pole save him, and a fan spin up to speed. Press ▶ for a continuous narrated tour.
1 — Torque: the turning effect of a force
Every object you have met so far was a particle that only slides. But a door, a wheel, a bottle cap also turn. A force applied off the axis produces a turning effect called torque (moment of force), symbol τ (tau).
Torqueτ = r F sin θ = (moment arm) × F
moment arm l = r sin θ = ⊥ distance from the axis (hinge) to the line of action
vector: τ = r × F · SI unit: N·m · anticlockwise positive
- θ = 90° — τ = rF, the maximum: push square-on to the door.
- θ = 0° — τ = 0: a push whose line of action runs through the hinge cannot swing it.
- Bigger r — bigger τ: that is exactly why the handle sits at the far edge, never beside the hinge.
worked — pushing a door
A 1.2 m wide door is pushed with 100 N at 90°. Compare a push at the handle (r = 1.1 m) with one near the hinge (r = 0.3 m).
handle: τ = 1.1×100 = 110 N·m · near hinge: τ = 0.3×100 = 30 N·m — same force, almost 4× the turn
2 — Second condition & the principle of moments
For rotational equilibrium the turning effects must cancel — the principle of moments:
Principle of momentsΣτ = 0 about any axis
Σ(clockwise torques) = Σ(anticlockwise torques)
for a see-saw: w₁d₁ = w₂d₂
On the right a 40 kg child sits close to the pivot while a lighter 25 kg child slides outward. The beam tips until the lighter child reaches the distance where the live Στ readout reads zero — lighter child, longer arm.
worked — balancing the see-saw
Aslam (40 kg) sits 1.5 m from the pivot. Where must Maryam (25 kg) sit to balance it? (g = 10 m/s²)
40×10×1.5 = 25×10×d → 600 = 250 d → d = 2.4 m from the pivot
Technique: take torques about the point where an unknown force acts — its moment arm is zero, so it vanishes from the equation.
3 — Why a longer arm wins
A stuck bolt will not move under a bare spanner. The mechanic never just pushes harder — he slips a pipe over the handle to make the arm longer. Since τ = rF sin θ, doubling r doubles the torque for the very same force.
Same force, longer armτ = r F sin θ
double r → double τ (with F, θ unchanged)
On the right, watch the bolt resist the short spanner, then break free once the green moment arm is extended by the pipe. This is the single trick that the whole chapter is built around.
worked — pipe extension
150 N at 90° on a 0.30 m spanner vs the same on a 0.60 m pipe-extended spanner?
short: τ = 0.30×150 = 45 N·m · extended: τ = 0.60×150 = 90 N·m — double the turn
4 — The two conditions of equilibrium
A body is in equilibrium when it has neither linear nor angular acceleration. That needs two conditions together:
First condition — no translationΣF = 0 → ΣFx = 0 and ΣFy = 0
Second condition — no rotationΣτ = 0 about any axis (Σclockwise = Σanticlockwise)
The tower crane on the right is the perfect example. A heavy load hangs far out on the jib; a counterweight sits on the short arm. When the counterweight's anticlockwise torque equals the load's clockwise torque, the arm is level and the whole crane stands — ΣF = 0 at the mast and Στ = 0 about the tower.
Static or dynamic: a body satisfying both conditions may be at rest (a parked crane) or moving uniformly (a paratrooper at terminal velocity) — both count as equilibrium.
5 — Couple: two forces, pure rotation
Turn a steering wheel: your two hands push with equal forces in opposite directions along different lines of action. Such a pair is a couple.
Moment of a coupleτ = F × d (one force × ⊥ distance between the two forces)
net force = F − F = 0 → no translation, pure rotation
same moment about every point — a couple has no unique pivot
worked — steering wheel
Two 25 N forces on opposite rims of a 40 cm wheel?
τ = F×d = 25 × 0.40 = 10 N·m · net force = 0
Everyday couples: turning a tap, winding a clock key, opening a bottle cap, a cyclist's two pedals — zero net force, pure turning.
6 — Centre of gravity & states of equilibrium
- Centre of gravity (CG) — the single point through which the body's total weight acts. In a uniform field (any school lab) CG = centre of mass.
A body stays standing only while the vertical line through its CG falls inside its base; it topples the instant that line crosses the edge.
| State | On tilting, CG… | Behaviour | Example |
|---|
| Stable | rises | returns | wide-base truck, book on a table |
| Unstable | falls | topples | pencil on its point, tall narrow stack |
| Neutral | stays level | rests where left | ball or cylinder on the floor |
On the right, a tightrope walker carries a long pole: it lowers and widens his effective CG so the weight line stays over the rope. The wide-base truck beside him keeps its weight line inside the wheels even when leaning. The two design rules: lower the CG and widen the base.
Truck loading (board favourite): stack heavy crates at the bottom — on a slope or sharp turn the weight line must stay inside the wheels' base, or the truck rolls.
7 — Angular displacement, velocity, acceleration & v = rω
- Angular displacement θ — angle swept, in radians: θ = s/r. 1 rev = 2π rad = 360°; 1 rad ≈ 57.3°.
- Angular velocity ω — ω = Δθ/Δt (rad/s); for f rev/s, ω = 2πf.
- Angular acceleration α — α = Δω/Δt (rad/s²).
Angular kinematics (constant α) & the linear linkωf = ωi + αt · θ = ωi t + ½αt² · 2αθ = ωf² − ωi²
s = rθ · v = rω · a = rα (θ in radians only)
The fan on the right starts from rest and speeds up: θ, ω and α all climb on the live readouts. Notice the blade-tip races at twice the speed of a point halfway in — same ω, double r, so v = rω doubles. That is why a long blade-tip whooshes while the hub barely moves.
worked — fan spin-up
Fan reaches 300 rpm from rest in 10 s. α? revolutions?
ωf = 2π(300/60) ≈ 31.4 rad/s → α = 3.14 rad/s² · θ = ½×3.14×100 = 157 rad = 25 rev
8 — Exam recap & numericals
numerical — door torque at an angle
100 N on a 1.2 m door at 30° to its plane, about the hinges?
τ = 1.2×100×sin30° = 60 N·m (90° would give 120 N·m)
numerical — braking wheel
60 rad/s to rest in 12 s. α and θ?
α = −5 rad/s² · θ = (0−3600)/(2×−5) = 360 rad ≈ 57.3 rev
- τ = rF sinθ = moment arm × force; max at 90°, zero through the axis; anticlockwise positive.
- Principle of moments: Σclockwise = Σanticlockwise; lighter load sits farther out.
- Longer arm wins: double r doubles τ for the same force (the pipe trick).
- Equilibrium = ΣF = 0 and Στ = 0; static or dynamic.
- Couple: τ = F·d; zero net force, pure rotation, same about every point.
- Stability: keep the weight line through the CG inside the base; lower CG, wider base.
- θ = s/r, ω = 2πf, α = Δω/Δt; angular equations mirror the linear set; v = rω, a = rα.