WJEC WJEC 2018-style practice: A uniform beam of weight 120\,N and length 3.0\,m rests horizontally on a pivot 1.0\,m from its left end. A load is hung from the left end to keep the beam horizontal. Calculate the load required.

Take moments about the pivot so the unknown pivot reaction does not appear in the equation. The weight of a uniform beam acts at its centre, which is 1.5\,m from the left end, so it is 0.5\,m to the right of the pivot. This produces a clockwise moment. Clockwise moment from the weight: 120 × 0.5 = 60\,N m. The load W at the left end is 1.0\,m to the left of the pivot, producing an anticlockwise moment W × 1.0. For equilibrium the moments balance: W × 1.0 = 60, so W = 60\,N. Markers reward locating the beam weight at the centre, using perpendicular distances, and applying the principle of moments about the pivot.

WJEC WJEC 2022-style practice: Show that the equation v^2 = u^2 + 2ax is homogeneous with respect to base units.

Homogeneity means every term must have the same base units. Velocity has units m s^-1, so v^2 and u^2 both have units (m s^-1)^2 = m^2\,s^-2. For the final term, acceleration a has units m s^-2 and displacement x has units m, so 2ax has units m s^-2 × m = m^2\,s^-2 (the 2 is a dimensionless constant). Every term reduces to m^2\,s^-2, so the equation is homogeneous. Markers reward showing the base units of each term explicitly and concluding they match.

WalesPhysicsSyllabus dot point

What is the basic toolkit of units, vectors and forces that the rest of physics is built on?

SI units and prefixes, homogeneity of equations, scalars and vectors, resolving and adding vectors, density, and the equilibrium of coplanar forces including moments.

A focused answer to WJEC A-Level Physics Unit 1 basic physics, covering SI base units and prefixes, checking homogeneity of equations, scalars and vectors, resolving and adding vectors, density, and the equilibrium of coplanar forces using moments.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

WJEC wants you to use SI base units and prefixes, check that an equation is homogeneous, distinguish scalars from vectors, add and resolve vectors, use density, and apply the conditions for equilibrium of coplanar forces, including taking moments. This dot point is the toolkit the rest of the AS and A-level course relies on, so the techniques here reappear inside almost every later calculation.

The answer

SI units and homogeneity

The seven SI base units include the metre (m), kilogram (kg), second (s), ampere (A) and kelvin (K). Every other unit is derived from these, for example the newton is $\text{kg m s}^{-2}$ and the joule is $\text{kg m}^2\,\text{s}^{-2}$ . WJEC also expects fluency with the standard prefixes from pico ( $10^{-12}$ ) through to giga ( $10^{9}$ ), used when quoting quantities such as a $470\,\mu\text{F}$ capacitor or a $2.4\,\text{GHz}$ signal.

Scalars and vectors

A scalar has magnitude only (mass, energy, speed). A vector has magnitude and direction (displacement, velocity, force). Vectors are added by scale drawing (tip to tail) or by components. To resolve a vector at angle $\theta$ to the horizontal, use $F_x = F\cos\theta$ and $F_y = F\sin\theta$ . The magnitude of a resultant from two perpendicular components is $F = \sqrt{F_x^2 + F_y^2}$ , with direction $\theta = \tan^{-1}(F_y / F_x)$ .

Density

Equilibrium and moments

A body is in equilibrium when two conditions both hold: the resultant force is zero and the resultant moment about any point is zero. The moment of a force is $F \times d$ , the force multiplied by the perpendicular distance from the line of action to the pivot. The principle of moments states that for equilibrium the total clockwise moment equals the total anticlockwise moment about the same point.

Resolving a force on a slope

A box of weight $50\,\text{N}$ sits on a slope inclined at $25^{\circ}$ to the horizontal. Find the components of the weight along and perpendicular to the slope.

step Set up the geometry

The weight acts vertically downward. The angle between the weight and the line perpendicular to the slope equals the slope angle, $25^{\circ}$ .

step Component perpendicular to the slope

$W_{\perp} = W\cos 25^{\circ} = 50 \times 0.906 = 45.3\,\text{N}$ . This is the force pressing the box onto the surface.

step Component along the slope

$W_{\parallel} = W\sin 25^{\circ} = 50 \times 0.423 = 21.1\,\text{N}$ . This is the force pulling the box down the slope, which friction or an applied force must balance for equilibrium.

Examples in context

Example 1. A diving board. A $600\,\text{N}$ diver stands at the free end of a board $2.0\,\text{m}$ from a supporting roller, with the board bolted down $0.5\,\text{m}$ behind the roller. Taking moments about the roller, the diver gives a clockwise moment of $600 \times 2.0 = 1200\,\text{N m}$ , so the bolt must supply an anticlockwise moment of $1200\,\text{N m}$ , meaning a downward hold-down force of $1200 / 0.5 = 2400\,\text{N}$ . This is why diving boards are anchored so firmly.

Example 2. Checking a wrong formula. A student writes $E = mgh^2$ for gravitational potential energy. The right-hand side has units $\text{kg} \times \text{m s}^{-2} \times \text{m}^2 = \text{kg m}^3\,\text{s}^{-2}$ , whereas energy is $\text{kg m}^2\,\text{s}^{-2}$ . The units do not match, so the formula is wrong without any further work. A homogeneity check catches the error instantly.

Try this

Q1. A $40\,\text{N}$ force acts at $30^\circ$ above the horizontal. Find its horizontal and vertical components. [2 marks]

Cue. $F_x = 40\cos30^\circ \approx 34.6\,\text{N}$ , $F_y = 40\sin30^\circ = 20\,\text{N}$ .

Q2. State the two conditions for a body to be in equilibrium under coplanar forces. [2 marks]

Cue. Resultant force zero; resultant moment about any point zero.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20184 marksA uniform beam of weight

120\,\text{N}

and length

3.0\,\text{m}

rests horizontally on a pivot

1.0\,\text{m}

from its left end. A load is hung from the left end to keep the beam horizontal. Calculate the load required.

Show worked answer →

Take moments about the pivot so the unknown pivot reaction does not appear in the equation.

The weight of a uniform beam acts at its centre, which is $1.5\,\text{m}$ from the left end, so it is $0.5\,\text{m}$ to the right of the pivot. This produces a clockwise moment.

Clockwise moment from the weight: $120 \times 0.5 = 60\,\text{N m}$ .

The load $W$ at the left end is $1.0\,\text{m}$ to the left of the pivot, producing an anticlockwise moment $W \times 1.0$ .

For equilibrium the moments balance: $W \times 1.0 = 60$ , so $W = 60\,\text{N}$ .

Markers reward locating the beam weight at the centre, using perpendicular distances, and applying the principle of moments about the pivot.

WJEC 20223 marksShow that the equation

v^2 = u^2 + 2ax

is homogeneous with respect to base units.

Show worked answer →

Homogeneity means every term must have the same base units. Velocity has units $\text{m s}^{-1}$ , so $v^2$ and $u^2$ both have units $(\text{m s}^{-1})^2 = \text{m}^2\,\text{s}^{-2}$ .

For the final term, acceleration $a$ has units $\text{m s}^{-2}$ and displacement $x$ has units $\text{m}$ , so $2ax$ has units $\text{m s}^{-2} \times \text{m} = \text{m}^2\,\text{s}^{-2}$ (the $2$ is a dimensionless constant).

Every term reduces to $\text{m}^2\,\text{s}^{-2}$ , so the equation is homogeneous. Markers reward showing the base units of each term explicitly and concluding they match.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)