OxfordAQA International AS and A-level Physics: Vectors and moments (9630) – Revision Notes

Condensed notes for OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, sections 3.2.1 (Scalars and vectors) and 3.2.2 (Moments). This is shared International AS and A-level content. For full explanations and longer worked examples, use the study guide. Then test yourself with the practice questions.

Quick links: Physics hub, printable checklist, free 10-minute diagnostics, Motion and Newton’s laws guide.

Where g is needed, use g = 9.81 N kg⁻¹.

Key definitions

Term	Meaning
Scalar	Quantity fully described by its size
Vector	Quantity needing both size and direction
Resultant	Single vector equivalent to the sum of several
Equilibrium	Net force on the object is zero
Moment of a force	Force × shortest distance between the point and the force’s line of action
Couple	Two coplanar forces, equal in size, opposite in direction, acting along different lines
Moment of a couple	One force × perpendicular separation of the two lines
Centre of mass	Point through which the whole weight can be taken to act

Scalars and vectors (3.2.1)

Scalar	Vector
distance	displacement
speed	velocity
mass	weight, force
	acceleration

• Distance is path length. Displacement is the shortest separation between start and finish, with a direction.
• Speed = distance/time. Velocity = displacement/time.
• Mass does not depend on place. Weight = mass × gravitational field strength and points towards the planet’s centre.

Adding vectors

Case	Method
Two vectors at 90°	R = √(a² + b²); angle from tan θ = opposite/adjacent
Two vectors at angle θ, tail to tail	R² = a² + b² + 2ab cos θ
Several vectors, any angles	Resolve into perpendicular components, add each axis, recombine
Scale drawing	Choose a scale, draw tip to tail, join start to end, measure

Resolving F at θ to the horizontal: horizontal = F cos θ, vertical = F sin θ.

Method: add vectors by components

1. Choose two perpendicular axes.
2. Resolve every vector along them.
3. Add the components on each axis, keeping signs.
4. R = √(Rx² + Ry²); angle from tan θ = Ry/Rx. State what the angle is measured from.

Method: scale drawing

1. State the scale (for example 1 cm : 5 N).
2. Draw vector one at its measured angle.
3. Draw vector two from the tip of vector one.
4. Join start to finish; measure length and angle.

Worked reminder: 12 N east plus 5 N north gives R = √(144 + 25) = 13 N at tan⁻¹(5/12) = 22.6° north of east.

Equilibrium of coplanar forces at a point (3.2.1)

• Resultant force = 0.
• Two forces: equal, opposite, in line.
• Three forces: they form a closed triangle when drawn tip to tail, and components sum to zero along any direction.
• An object at rest or moving at constant velocity is in equilibrium. The converse does not hold: rest is only one case.

Method: three forces at a point

1. Draw a free-body diagram with every force and angle marked.
2. Choose axes to put as many forces as possible on an axis (along and perpendicular to a slope, or horizontal and vertical).
3. Write the sum of components = 0 along each axis.
4. Solve. Check by confirming that the force triangle closes.

Worked reminder: a 200 N weight hangs from two equal ropes, each 60° to the horizontal. 2T sin 60° = 200, so T = 115 N.

Moments (3.2.2)

• moment = F × d, where d is perpendicular to the line of action. Unit: N m.
• For a force at angle θ to a rod of length L: moment = F L sin θ.
• Say clockwise or anticlockwise.

Principle of moments

When a body is in equilibrium, total clockwise moment = total anticlockwise moment, about any one point.

Method: beam in equilibrium

1. Mark all forces; put a uniform beam’s weight at its midpoint.
2. Take moments about a point that an unknown force passes through, which eliminates it.
3. Equate clockwise and anticlockwise moments.
4. Resolve vertically (and horizontally if needed) for the other unknowns.

Worked reminder: a uniform 2.0 m rod of mass 3.0 kg is pivoted 0.40 m from its left end. What mass m hung on the left end balances it? The rod’s weight acts 0.60 m to the right of the pivot, so m × 0.40 = 3.0 × 0.60 and m = 4.5 kg. The pivot force is not needed, because it passes through the pivot.

Couples (3.2.2)

• Equal in size, opposite in direction, on parallel but separate lines.
• The net force is zero, so the object turns without accelerating sideways.
• Moment of couple = F × perpendicular distance between the lines.
• Examples: turning a tap with finger and thumb, a steering wheel turned with both hands, a wheel brace.

Worked reminder: 18 N forces 0.25 m apart give a couple of 18 × 0.25 = 4.5 N m.

Centre of mass (3.2.2)

• The point where the weight can be taken to act.
• A uniform regular solid has its centre of mass at its centre (middle of a rod, cube or sphere).
• It can lie outside the material, for example the centre of a ring.
• For a loaded rod, take moments about the balance point: moments from each side are equal.
• Two point masses on a light rod: the centre of mass is closer to the larger mass. For 2.0 kg and 6.0 kg placed 0.80 m apart, the distance from the 6.0 kg mass is 2.0 × 0.80 / 8.0 = 0.20 m.
• If a body hangs freely from a point, its centre of mass lies vertically below that point once it settles.

Must-know distinctions

• Distance is not displacement; speed is not velocity.
• Mass is a scalar in kg; weight is a force in N.
• A couple has zero resultant force; an unbalanced single force has a resultant force.
• Equilibrium means zero resultant force, not zero velocity.
• Moment of a force uses distance to one line from a point; moment of a couple uses distance between two lines.

Quick self-test

1. Distinguish a scalar from a vector, giving one example of each.
2. Forces of 6.0 N and 8.0 N act at a point at 90° to each other. Find the resultant and its angle from the 8.0 N force.
3. A force of 120 N acts at 25° above the horizontal. Find its horizontal and vertical components.
4. State two conditions that must hold for three coplanar forces at a point to be in equilibrium.
5. A box slides across a floor at constant velocity under a horizontal push of 45 N. What is the friction force? Is the box in equilibrium?
6. Find the moment of a 35 N force acting perpendicular to a lever 0.40 m from the pivot.
7. Two forces of 18 N, equal and opposite, act on a disc with their lines 0.25 m apart. Find the moment of the couple.
8. A uniform seesaw balances at its centre. A 30 kg child sits 1.6 m from the pivot. Where must a 24 kg child sit to balance?
9. A 40 N push on a door acts 0.80 m from the hinge, at 65° to the plane of the door. Find the moment about the hinge.
10. Where is the centre of mass of a uniform cube?

Answers

1. A scalar has size only (for example mass); a vector also has a direction (for example force).
2. R = √(6.0² + 8.0²) = 10 N at tan⁻¹(6.0/8.0) = 36.9° from the 8.0 N force.
3. Horizontal = 120 cos 25° = 109 N; vertical = 120 sin 25° = 50.7 N.
4. The resultant force is zero (components sum to zero in any direction), and drawn tip to tail the forces form a closed triangle.
5. Friction is 45 N, opposite to the push. Yes: constant velocity means zero resultant force.
6. 35 × 0.40 = 14 N m.
7. 18 × 0.25 = 4.5 N m.
8. 30 × 1.6 = 24 × d, so d = 2.0 m on the other side of the pivot.
9. d = 0.80 sin 65° = 0.725 m; moment = 40 × 0.725 = 29.0 N m.
10. At its geometric centre, where the diagonals meet.

Where marks are usually lost

• Using the full length of a lever as the distance when the force is angled to it.
• Writing “moment = force × distance” without saying perpendicular distance.
• Quoting a resultant size with no direction, or an angle with no stated reference line.
• Using cos θ for the vertical component when θ is measured from the horizontal.
• Omitting the weight of a uniform beam, or placing it at the wrong point.
• Taking moments about one point on one side and a different point on the other.
• Forgetting that a force through the pivot has zero moment, so it can be ignored.
• Treating a couple as if it produced a net force.
• Stating that equilibrium needs the object to be at rest.
• Leaving out units, or giving N m as a force.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, sections 3.2.1 and 3.2.2, published by OxfordAQA.