9702

CIE AS Physics

Topics 4.1, 4.2 & 5 · 2025–2027 syllabus

Cambridge International AS Level · 9702

Turning effects, equilibrium, and work, energy & power

A concise teaching page for 4.1 Turning effects of forces, 4.2 Equilibrium of forces, and all of Topic 5 Work, energy and power, followed by interactive flashcards. Definitions and equations use the wording and symbols of the 2025–2027 syllabus. These energy and moment formulae are not printed on the exam data sheet — you must recall them.

4.1 Turning effects 4.2 Equilibrium 5.1 Energy conservation 5.2 EP and EK
4.1 Turning effects of forcesCentre of gravity, moment, couple and torque 4.2 Equilibrium of forcesPrinciple of moments, equilibrium, vector triangle 5 Work, energy and powerWork, conservation, efficiency, power, mgΔh and ½mv²

4.1 Turning effects of forces

You should be able to:
  1. understand that the weight of an object may be taken as acting at a single point known as its centre of gravity
  2. define and apply the moment of a force
  3. understand that a couple is a pair of forces that acts to produce rotation only
  4. define and apply the torque of a couple

Centre of gravity

Definition — memorise this

The centre of gravity of an object is the point at which its entire weight may be taken as acting.

For calculations, you can replace the object’s distributed weight with a single arrow W = mg drawn downward from the centre of gravity (C.G.). For a uniform, regular object the C.G. lies at the geometric centre. For an irregular object it can be found experimentally by hanging the object from two different points and seeing where the vertical plumb lines cross.

In a uniform gravitational field (near the Earth’s surface) the centre of gravity coincides with the centre of mass.

Weight acts at the centre of gravity FIG. 1 Weight acts at the centre of gravity C.G. W = mg uniform rod, length L C.G. at L/2 W
Fig. 1 For an irregular object the centre of gravity may not look “central”. For a uniform rod it is at the midpoint. In both cases the whole weight acts there.

Moment of a force

Definition — memorise this

The moment of a force about a point is the product of the force and the perpendicular distance from that point to the line of action of the force.

moment = F × d⊥ unit: N m  ·  direction: clockwise or anticlockwise

The line of action is the infinite straight line along which the force points. The distance you need is not necessarily the distance to the point of application — it is the shortest (perpendicular) gap from the pivot to that line.

If a force F is applied at a distance r from the pivot, and the angle between the force and the line from the pivot to the point of application is θ, then

moment = F r sin θ because  d⊥ = r sin θ

A force through the pivot (d⊥ = 0), or a force parallel to the line from the pivot (θ = 0° or 180°), has zero moment: it does not tend to turn the object.

Moment equals force times perpendicular distance FIG. 2 Moment = force × perpendicular distance to the line of action F d⊥ pivot clockwise (a) force already perpendicular F d⊥ = r sin θ pivot r θ (b) force at an angle — use d⊥, not r
Fig. 2 (a) When the force is already perpendicular to the rod, d⊥ is just the distance along the rod. (b) When the force is at an angle, drop a perpendicular from the pivot onto the line of action. moment = F r sin θ.
Worked example

Moment with a force at an angle

A force of 40 N is applied to a door at 0.80 m from the hinge. The force is at 30° to the door, in the plane of the door’s swing. Calculate the moment about the hinge.

The perpendicular distance from the hinge to the line of action is d⊥ = 0.80 × sin 30° = 0.40 m.

moment = F × d⊥ = 40 × 0.40 = 16 N m (tending to open the door).

Exam trap

The unit of moment is N m, never the joule. A joule is also 1 N m, but a joule is a unit of energy. A moment is a turning effect, not a form of energy — writing “J” loses the mark.

Couples and the torque of a couple

Definitions — memorise these

A couple is a pair of forces that acts to produce rotation only. The two forces are equal in magnitude, opposite in direction, and have parallel lines of action that are separated (they are not in the same straight line).

The torque of a couple is the product of one of the forces and the perpendicular distance between the two lines of action.

torque of a couple = F × d unit: N m  ·  same about every point

Because the forces are equal and opposite, the resultant force is zero, so the object’s centre of mass does not accelerate. There is still a turning effect, which is why a couple produces rotation only. A useful property: the torque of a given couple is the same no matter which point you take moments about.

Examples: turning a steering wheel, a screwdriver, or the two hands on a tap.

A couple produces rotation only FIG. 3 A couple: equal, opposite, parallel forces, separated by distance d F F d torque = F d resultant force = 0 rotation only
Fig. 3 The two forces cancel as a resultant force, so there is no translational acceleration. The perpendicular separation d means they do not cancel as moments, so the object turns. Torque = F d.

4.2 Equilibrium of forces

You should be able to:
  1. state and apply the principle of moments
  2. understand that, when there is no resultant force and no resultant torque, a system is in equilibrium
  3. use a vector triangle to represent coplanar forces in equilibrium

The principle of moments

Definition — memorise this

The principle of moments: for a system in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.

Σ clockwise moments = Σ anticlockwise moments about the same point

You may take moments about any convenient point. A standard trick is to choose a point on the line of action of an unknown force (often a support reaction). That force then has zero moment, and it drops out of the equation.

Principle of moments applied to a beam FIG. 4 Uniform beam in equilibrium — moments about the pivot A B pivot W 180 N C.G. R 0.80 m 0.40 m 2.4 m
Fig. 4 The worked example below. Taking moments about the pivot eliminates the unknown reaction R, because R acts through the pivot so its moment is zero.
Worked example

A uniform beam on a pivot

A uniform beam AB has length 2.4 m and weight 180 N. It rests horizontally on a pivot 0.80 m from A, with a load W at end A, and is in equilibrium. Find W and the support reaction R.

The beam’s weight acts at its centre, 1.2 m from A, so 0.40 m to the right of the pivot.

Take moments about the pivot.

Anticlockwise: W × 0.80.   Clockwise: 180 × 0.40 = 72 N m.

Principle of moments: W × 0.80 = 72 ⇒ W = 90 N.

No resultant force (vertical): R = 90 + 180 = 270 N upward.

The two conditions for equilibrium

Statement — memorise this

A system is in equilibrium when there is no resultant force and no resultant torque (moment).

Translational equilibrium

No resultant force: ΣFx = 0 and ΣFy = 0. The centre of mass does not accelerate (it is at rest or moves at constant velocity).

Rotational equilibrium

No resultant torque: Σ moments = 0 about any point. The object does not have an angular acceleration (it is not starting to spin, or it rotates at a constant rate).

Both conditions must hold. A couple is a reminder that “no resultant force” is not enough on its own — a couple has zero resultant force but is not in equilibrium, because there is a resultant torque.

The vector triangle

If three coplanar forces act on a system that is in equilibrium, the forces can be represented by a closed vector triangle. Draw the three force vectors tip-to-tail. Because the resultant is zero, the tip of the last vector meets the tail of the first — the triangle closes.

Each side is parallel to one of the forces and has a length proportional to that force’s magnitude. You can then use scale drawing, or trigonometry (sine or cosine rule) on the triangle, to find unknown forces or angles.

For more than three coplanar forces in equilibrium, the same idea gives a closed vector polygon.

Three coplanar forces in equilibrium form a closed vector triangle FIG. 5 A lamp held by two strings, and the closed vector triangle of its three forces T₁ T₂ W = 12.0 N 30° (a) forces on the lamp (b) closed vector triangle W 12.0 N T₁ = 20.8 N T₂ = 24.0 N 30°
Fig. 5 A 12.0 N lamp hangs from a horizontal string and a string at 30° to the horizontal. The three forces form a closed right-angled triangle. From the triangle: T₂ = 12.0 / sin 30° = 24.0 N, and T₁ = 24.0 × cos 30° = 20.8 N.
Exam tip

If you are asked to “draw a vector triangle”, do it as a separate diagram, not just the free-body arrows on the object. Label each side with the force’s name and, if known, its magnitude. Mark the angles. State the scale if it is a scale drawing.

5 Work, energy and power

The syllabus assumes you already know the forms of energy and energy transfers from IGCSE / O Level. Topic 5 is in two parts: 5.1 Energy conservation (work, conservation, efficiency, power) and 5.2 Gravitational potential energy and kinetic energy.

5.1 Energy conservation

Work done

Definition — memorise this

Work done by a force is the product of the force and the displacement in the direction of the force.

W = F s cos θ unit: J  ·  1 J = 1 N m  ·  θ is the angle between F and s

Work is a scalar. If the force is parallel to the displacement, θ = 0° and W = F s. If the force is perpendicular to the displacement (θ = 90°), cos 90° = 0 so W = 0 — a useful fact for circular motion and for the normal contact force on a level surface. A force opposite to the motion (θ = 180°) does negative work: the object loses energy.

Work done equals F s cos theta FIG. 6 Only the component of force along the displacement does work crate s F F cos θ θ W = F s cos θ θ = angle between F and s If θ = 0°, W = F s If θ = 90°, W = 0
Fig. 6 The vertical component F sin θ does no work if the crate moves horizontally. Only F cos θ, the component along s, contributes: W = (F cos θ) s = F s cos θ.
Worked example

Work done at an angle

A crate is pulled 6.0 m along a floor by a force of 50 N at 25° to the horizontal. Calculate the work done by this force.

W = F s cos θ = 50 × 6.0 × cos 25° = 272 J (3 significant figures).

Principle of conservation of energy

Definition — memorise this

The principle of conservation of energy: energy cannot be created or destroyed. It can be transferred from one form to another, but the total energy of a closed system remains constant.

In a mechanics problem this usually means that a loss in one form equals a gain in others. If there is no air resistance and no friction, gravitational potential energy and kinetic energy simply swap. If resistive forces act, some energy is transferred to thermal energy (internal energy of the object and the surroundings). The total is still constant; it is just no longer all mechanical.

Efficiency

Definition — memorise this

The efficiency of a system is the ratio of useful energy output from the system to the total energy input. The same ratio can be written with power, because power is energy per unit time.

efficiency = useful energy output / total energy input or   useful power output / total power input

Efficiency is a ratio: a number between 0 and 1. It is often written as a percentage (multiply by 100%). Real machines always have efficiency less than 1, because some input energy is wasted, usually as thermal energy.

Sankey diagram showing useful and wasted energy FIG. 7 Efficiency is useful output divided by total input total energy input useful energy output wasted (usually thermal) efficiency = 0.70 or 70%
Fig. 7 A Sankey diagram is a useful picture of energy flow. The width of each arrow represents the amount of energy. Here the useful output is 70% of the input.
Worked example

Efficiency from power

A motor has an input power of 400 W and a useful output power of 280 W. Find its efficiency.

efficiency = useful power output / total power input = 280 / 400 = 0.70 (or 70%).

Power

Definition — memorise this

Power is the work done per unit time (the rate of doing work, or the rate of energy transfer).

P = W / t unit: W  ·  1 W = 1 J s−1
Derivation you can be asked for — P = Fv

If a constant force F moves an object a distance s in the direction of the force in time t,

P = W / t = (F s) / t = F (s / t) = F v,

where v is the speed in the direction of the force. If F and v are at an angle θ, then P = F v cos θ.

A boat moving at constant speed, power equals Fv FIG. 8 At constant speed, driving force equals drag, and P = Fv F (driving force) v P = F v same direction constant speed: F = drag
Fig. 8 A boat travelling at constant speed. The driving force does work against drag. Instantaneous power output is P = Fv.
Worked example

Power at constant speed

A boat’s engine provides a driving force of 1.5 kN at a constant speed of 8.0 m s−1. Calculate the output power.

P = F v = 1.5 × 103 × 8.0 = 1.2 × 104 W (12 kW).

5.2 Gravitational potential energy and kinetic energy

Gravitational potential energy

Formula — recall and use

In a uniform gravitational field (near the Earth’s surface), the change in gravitational potential energy is

ΔEP = mgΔh unit: J  ·  g = 9.81 m s−2 (from the data sheet)
Derivation from W = F s — required by the syllabus

To raise a mass m slowly through a vertical height Δh at constant speed, the applied upward force equals the weight mg.

Work done by this force: W = F s = mg × Δh.

That work is stored as gravitational potential energy, so ΔEP = mg Δh.

This holds only in a uniform field, where g is constant. (A different formula, not needed at AS, is used at large distances from the Earth.)

Raising a mass through height delta h FIG. 9 Work done in raising a mass is stored as gravitational potential energy start mass m Δh F = mg mg W = F s = mg Δh ΔEₚ = mg Δh uniform gravitational field only
Fig. 9 Raising the mass at constant speed means the applied force equals the weight. The work done by that force equals the gain in EP.
Worked example

Calculating ΔEP

A student of mass 55 kg climbs a vertical height of 12 m. Calculate the gain in gravitational potential energy. Take g = 9.81 m s−2.

ΔEP = mg Δh = 55 × 9.81 × 12 = 6.47 × 103 J.

Kinetic energy

EK = ½ m v2 unit: J  ·  must be recalled — not on the data sheet
Derivation from the equations of motion — required by the syllabus

An object of mass m starts from rest (u = 0) and is accelerated by a constant force F through a displacement s, reaching speed v.

Newton’s second law: F = ma.

Equation of motion: v2 = u2 + 2as = 2as, so s = v2 / 2a.

Work done by the force: W = F s = (ma)(v2 / 2a) = ½ m v2.

This work equals the kinetic energy gained, so EK = ½ m v2.

If the object already has speed u, the same steps give W = ½ m v2 − ½ m u2 = ΔEK. This is the work–energy idea: the work done by the resultant force equals the change in kinetic energy.

Falling object converts gravitational potential energy to kinetic energy FIG. 10 If resistive forces are negligible, a loss in Eₚ equals a gain in Eₖ 1 Eₚ = mg Δh Eₖ = 0 2 Eₚ smaller Eₖ larger 3 Eₚ = 0 Eₖ = mg Δh = ½mv² Δh no resistance: mg Δh = ½ m v² v = √(2g Δh) with resistance: mg Δh = ½ m v² + thermal total energy still conserved
Fig. 10 A falling object. Without air resistance the loss in EP equals the gain in EK. With air resistance the object is slower than √(2gΔh); the “missing” energy has become thermal energy.
Worked example

Energy transfer, with and without resistance

An apple of mass 0.15 kg falls 3.0 m from rest. Take g = 9.81 m s−2.

(a) If air resistance is negligible, find its speed just before it lands.

(b) If it actually lands at 6.0 m s−1, how much energy has been transferred to thermal energy?

(a) Loss in EP = gain in EK:

mg Δh = ½ m v2 ⇒ v = √(2g Δh) = √(2 × 9.81 × 3.0) = 7.67 m s−1.

(b) Loss in EP = 0.15 × 9.81 × 3.0 = 4.41 J.

Gain in EK = ½ × 0.15 × (6.0)2 = 2.70 J.

Energy transferred to thermal = 4.41 − 2.70 = 1.71 J.

Formulae to recall

None of the equations below appear on the AS data sheet. You must remember them, including the derivations marked ★.

QuantityEquationNotes
Moment of a forcemoment = F d⊥ = F r sin θUnit N m, not J. d⊥ is to the line of action.
Torque of a coupletorque = F dd is the perpendicular separation of the two forces. Same about any point.
Principle of momentsΣ clockwise = Σ anticlockwiseFor a system in equilibrium, about any one point.
EquilibriumΣF = 0  and  Σ torque = 0Both conditions are required.
Work doneW = F s cos θUnit J. θ between force and displacement.
Efficiencyuseful output / total inputUse energy or power. A ratio, or × 100%.
PowerP = W/t   and   P = F v ★Unit W. Derive P = Fv from W = Fs.
Gravitational potential energyΔEP = mgΔh ★Uniform field only. Derive from W = Fs.
Kinetic energyEK = ½ m v2 ★Derive from the equations of motion.
Before the exam

Use g = 9.81 m s−2 unless the question gives another value. State the point you are taking moments about. Always use the perpendicular distance. For energy, check whether resistive forces are mentioned — if they are, mg Δh is not equal to ½mv2.

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