Calculate the moment of a force about a point and apply moment equilibrium together with force equilibrium to determine the support reactions of a simply supported structure

What this dot point is asking

QCAA wants you to compute the turning effect of a force, the moment, and use it together with force balance to find the unknown reaction forces at the supports of a structure. Reactions are the forces the ground or supports push back with to hold the structure still. You cannot analyse a truss member or a beam section until you know them, so this is the gateway calculation of Unit 3.

The answer

The moment of a force

A force applied at a distance from a pivot tends to rotate the body. That turning effect is the moment:

Here $F$ is the force and $d$ is the perpendicular distance from the pivot to the line of action of the force. The perpendicular distance matters: a force whose line passes through the pivot has zero moment because $d = 0$. Moments are measured in newton metres (N m) and carry a sense, conventionally taken as positive anticlockwise or positive clockwise as long as you stay consistent.

The principle of moments

For a body in rotational equilibrium, the total clockwise moment about any point equals the total anticlockwise moment, which is the same as saying:

This is the third equilibrium condition, alongside $\sum F_x = 0$ and $\sum F_y = 0$. The power of the moment equation is that you can take moments about any point you like. Choosing a point where an unknown force acts makes that force drop out (its $d$ is zero), leaving fewer unknowns.

Supports and the reactions they provide

Real structures rest on supports that supply reaction forces:

• A roller support can only push perpendicular to its surface, so it gives one reaction (usually vertical).
• A pin (hinge) support can push in any direction, so it gives two components, vertical and horizontal.
• A fixed (built-in) support resists translation and rotation, giving two force components and a moment.

A simply supported beam has a pin at one end and a roller at the other. That is three reaction components for three equilibrium equations, so the structure is statically determinate and solvable by hand.

The strategy for finding reactions

1. Draw the free-body diagram with all loads and the unknown reactions.
2. Take moments about one support. Its own reaction has zero moment there, so the equation contains only the other reaction and the loads. Solve it.
3. Use $\sum F_y = 0$ to find the remaining vertical reaction.
4. Use $\sum F_x = 0$ for any horizontal reaction.

Why this matters for civil structures

Reactions are the starting point for everything that follows. The method of joints needs them before any truss member can be solved, and a shear-force and bending-moment diagram is built directly from the reactions and loads. A wrong reaction silently corrupts every later number, so the moment-about-a-support technique and its built-in check are core skills for the Unit 3 data test and external exam.