Forces, equilibrium and free-body diagrams for QCE Engineering Unit 3

Syllabus dot point

“Apply the conditions of static equilibrium to determine unknown forces acting on a structure, using free-body diagrams and the resolution of forces into perpendicular components”

QCEEngineeringUnit 3: Civil structures6 min readUpdated 29 May 2026

Quick answer

A structure in static equilibrium has zero net force and zero net moment. Draw a free-body diagram showing every external force (loads, reactions, weights) as a vector, resolve each force into $x$ and $y$ components, then apply $\sum F_x = 0$ , $\sum F_y = 0$ and $\sum M = 0$ . Three equations let you solve for up to three unknown reactions. Choosing the pivot at a support eliminates that reaction from the moment equation, which is the fastest route to a clean answer.

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to treat a civil structure as a rigid body in static equilibrium and solve for the unknown forces acting on it. That means representing every force as a vector, drawing a clean free-body diagram (FBD), resolving forces into perpendicular components, and applying the three equilibrium conditions. This is the foundation of every Unit 3 calculation, from a simple bracket to a loaded truss.

The answer

Force as a vector

A force has magnitude (in newtons, $\text{N}$ ) and direction. Two forces can be added head-to-tail or, more usefully for calculation, split into perpendicular components. A force $F$ acting at angle $\theta$ to the horizontal has components:

F_x = F\cos\theta \qquad F_y = F\sin\theta

Working in components turns awkward vector geometry into ordinary arithmetic.

Free-body diagrams

A free-body diagram isolates one body and shows every external force acting on it. Internal forces between parts of the same body are excluded. For a civil structure you typically draw:

applied loads (the weight of traffic, wind, stored material)
the structure's own weight, acting at its centre of mass
support reactions (a pin gives a force in any direction, usually shown as $H$ and $V$ components; a roller gives a force perpendicular to its surface only)

A correct FBD is half the marks. Label each force, give its line of action, and show the angle of any inclined force.

The conditions of static equilibrium

A rigid body is in static equilibrium when it has no tendency to translate or rotate. In two dimensions this gives three independent equations:

\sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum M = 0

The moment (turning effect) of a force about a point is:

M = F \times d

where $d$ is the perpendicular distance from the pivot to the force's line of action. Anticlockwise is conventionally positive. Because you can take moments about any point, choosing the pivot at a support makes that support's reaction disappear from the equation (its moment arm is zero), leaving a single unknown.

A typical solution strategy

Draw the FBD and mark all known and unknown forces.
Resolve inclined forces into $x$ and $y$ components.
Take moments about one support to find the other reaction.
Use $\sum F_y = 0$ (and $\sum F_x = 0$ ) to find the remaining reactions.
Check that all three equations balance.

Beam with a point load

Find the vertical reactions on a simply supported beam of length $6\,\text{m}$ resting on a pin at A (left end) and a roller at B (right end). A vertical point load of $1200\,\text{N}$ acts $2\,\text{m}$ from A. The beam's own weight is neglected.

Step 1: Draw the free-body diagram and identify unknowns

The pin at A provides an unknown vertical reaction $R_A$ (and a horizontal reaction, which is zero here since there are no horizontal loads). The roller at B provides an unknown vertical reaction $R_B$ . The applied load acts downward at $2\,\text{m}$ from A and $4\,\text{m}$ from B. There are two unknowns, which the two equilibrium equations $\sum M = 0$ and $\sum F_y = 0$ can solve.

Step 2: Take moments about A to find $R_B$

Taking moments about A eliminates $R_A$ from the equation because its moment arm is zero (it acts at the pivot). This leaves a single equation in the single unknown $R_B$ :

\sum M_A = 0

R_B(6) - 1200(2) = 0

R_B = \frac{2400}{6} = 400\,\text{N}

The load tries to rotate the beam clockwise about A, so $R_B$ must act upward to resist that.

Step 3: Apply vertical force equilibrium to find $R_A$

With $R_B$ known, the sum of vertical forces gives $R_A$ directly. Upward forces equal the downward load:

\sum F_y = 0

R_A + R_B - 1200 = 0

R_A = 1200 - 400 = 800\,\text{N}

Step 4: Check by taking moments about B

As a verification, take moments about B. $R_A$ acts $6\,\text{m}$ from B (upward, anticlockwise) and the load acts $4\,\text{m}$ from B (downward, clockwise):

R_A(6) - 1200(4) = 800(6) - 4800 = 4800 - 4800 = 0 \checkmark

Final answer: The reactions are $R_A = 800\,\text{N}$ and $R_B = 400\,\text{N}$ . The larger reaction sits under the support nearer the load, which is the physically expected result.

Why this matters for civil structures

Every bridge pier, retaining wall and roof truss is designed by first assuming static equilibrium and solving for the forces the structure must carry. Those forces feed directly into the stress and material calculations later in Unit 3. Get the FBD wrong and every downstream number is wrong.

Common errors

Forgetting a force: Leaving the structure's own weight or a reaction off the FBD breaks equilibrium. List every contact point and every load before writing equations.
Mixing up moment arms: The distance $d$ is the perpendicular distance to the line of action, not the distance to the point where the force is drawn. For an inclined force, resolve into components first, then take moments of each component.
Sign errors: Pick one rotation direction as positive and keep it consistent. Mixing clockwise and anticlockwise positive within one equation is the most common arithmetic slip.
Treating a pin as a roller: A pin (or hinge) resists force in two directions and needs both $H$ and $V$ unknowns. A roller resists force in only one direction. Using the wrong support type gives the wrong number of unknowns.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A car with a mass of 1300 kg is parked on a driveway that makes an inclined angle of 25 degrees to the horizontal. a) Calculate the normal force acting on the car. b) Calculate the coefficient of static friction required between the car tyres and the driveway to prevent the car from rolling.

Show worked answer →

A parked car is in static equilibrium, so resolve the weight into components perpendicular and parallel to the incline (sum of forces = 0 in both directions). Four marks, two per part.

Part a (normal force). Perpendicular to the slope the forces balance, so F_N = mg cos(theta) = 1300 x 9.8 x cos 25 [1 mark for the method] = 11546 N, about 11.55 kN [1 mark].

Part b (coefficient of static friction). Parallel to the slope, friction must balance the down-slope weight component: F_s = mg sin(theta). Since F_s = mu_s x F_N, then mu_s = mg sin(theta) / (mg cos(theta)) = tan(theta) [1 mark for the method] = tan 25 = 0.47 [1 mark].

Note the tidy result: at the point of rolling, mu_s = tan(theta), independent of mass.

Practice4 marks

A parcel slides down an inclined ramp, which is at 20 degrees to the horizontal plane, at a constant velocity. The parcel experiences a normal force of 98.2 N. A coefficient of kinetic friction of 0.37 and a coefficient of static friction of 0.39 exist between the parcel and the ramp. Determine the mass of the parcel.

Show worked answer →

Constant velocity means zero acceleration, so the parcel is in equilibrium and the forces are balanced. Four marks.

Method using the normal force directly. Perpendicular to the ramp the forces balance, so the normal force equals the perpendicular component of weight: F_N = mg cos(theta) [1 mark for recognising forces are balanced perpendicular to the ramp].

Rearrange for mass: m = F_N / (g cos(theta)) [1 mark]. Substitute: m = 98.2 / (9.8 x cos 20) [1 mark] = 10.66 kg [1 mark].

A check using the parallel direction (down-slope weight equals kinetic friction, mg sin(theta) = mu_k F_N) gives the same mass; the static coefficient is a distractor because the parcel is already moving.

Practise this

Sources & how we know this

Engineering General senior syllabus 2025 — QCAA