Orientation · Module 5 of 10
Forces, Moments, and Equilibrium
Structures and machines carry forces, and staying still is a balance of them. This first look at statics introduces forces as vectors, the turning effect of a moment, and the two conditions for equilibrium.
Readiness check
This module is a first taste of statics. Tick what you can do comfortably.
- Multiply a force by a distance.
- Recall that a force has size and direction.
- Recall that balanced things do not accelerate.
- Divide a load evenly between two supports.
- Read a simple diagram of arrows.
The core idea
A force is a vector, with magnitude and direction. Its moment about a point is the force times the perpendicular lever arm, the turning effect. A body sits in equilibrium when the forces balance and the moments balance, the two conditions that let us solve for unknown supports.
moment M = F × d (perpendicular)equilibrium: ΣF = 0equilibrium: ΣM = 0Statics is the study of things that do not accelerate, which includes almost every structure and every machine part at rest or moving steadily. Its language is force. A force is a vector: it has a magnitude and a direction, and several forces combine into a resultant by adding their components. A force can also turn a body, and that turning effect is the moment, equal to the force times the perpendicular distance from the pivot to the line of action, the lever arm. A longer arm gives more moment for the same force, which is why a wrench has a handle. A body is in equilibrium when two conditions hold at once: the forces sum to zero, so it does not accelerate, and the moments about any point sum to zero, so it does not spin. These two statements, written as ΣF = 0 and ΣM = 0, are enough to find unknown support forces, like the reactions under a loaded beam. The tool that makes this reliable is the free body diagram, a sketch of the object alone with every force acting on it drawn in. This is the analytical core the Statics and Mechanics of Materials courses expand.
The skills, taught in order
Five skills open the door to statics.
5.1 Forces as vectors
A force has magnitude and direction, so it is a vector. It can be split into perpendicular components, which is how forces at angles are handled: work with the components, then recombine.
5.2 The resultant of several forces
Several forces on a body combine into one resultant by adding their components separately along each axis. The resultant is the single force that would have the same effect.
5.3 The moment of a force
A moment is the turning effect of a force, equal to the force times the perpendicular lever arm to the pivot. The perpendicular distance is what matters, not the straight-line distance to where the force is applied.
5.4 Equilibrium
A body in equilibrium satisfies ΣF = 0 and ΣM = 0: forces balance so it does not accelerate, and moments balance so it does not rotate. These two conditions solve for unknown forces.
| Condition | Meaning | Prevents |
|---|---|---|
| ΣF = 0 | forces balance | linear acceleration |
| ΣM = 0 | moments balance | rotation |
Both conditions must hold together. One without the other is not equilibrium.
5.5 Free body diagrams
A free body diagram isolates the object and draws every external force on it, including supports and weight. It is the single most important habit in statics, because a correct diagram makes the equations almost write themselves.
Engineering connection: finding the reactions under a bridge beam, or the forces in a bracket, always starts with a free body diagram and the two equilibrium conditions.
Worked example 1: the moment of a force
A worker pushes with 200 N at the end of a wrench, 0.3 m from the bolt, perpendicular to the handle. Find the moment about the bolt.
- ProblemFind the moment about the bolt in Figure 1.
- Given / findForce 200 N, perpendicular arm 0.3 m. Find the moment.
- AssumptionsThe force acts perpendicular to the handle, so d is the full length.
- ModelM = F × d, with d the perpendicular lever arm.
- EquationsM = 200 × 0.3
- SolveM = 60 N·m.
- CheckUnits N × m give N·m, a moment; doubling the handle would double the moment, as expected.
- ConclusionThe wrench applies 60 N·m to the bolt; a longer handle would loosen a tighter bolt for the same push.
Worked example 2: reactions on a beam
A beam 2 m long rests on a support at each end. An 800 N load sits at the centre. Find the force carried by each support.
- ProblemFind each support force for the beam in Figure 2.
- Given / findSpan 2 m, central load 800 N. Find the two reactions.
- AssumptionsWeightless rigid beam, simple supports, load at the exact centre.
- ModelEquilibrium: ΣM about one end = 0, then ΣF = 0.
- EquationsR × 2 = 800 × 1Rleft + Rright = 800
- SolveMoments about the left end: Rright × 2 = 800 × 1, so Rright = 400 N, and Rleft = 400 N.
- Check400 + 400 = 800 N, so the vertical forces balance, and symmetry agrees.
- ConclusionEach support carries 400 N; moving the load off-centre would shift more onto the nearer support.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Wrong lever arm | Moment too large | "Is d perpendicular to the force?" | Use the perpendicular distance to the line of action. |
| Forgetting moments | Forces balance but body still rotates | "Did I check ΣM = 0?" | Equilibrium needs both ΣF = 0 and ΣM = 0. |
| Incomplete free body diagram | A missing support or weight | "Have I drawn every force?" | Include all external forces, including weight. |
| Ignoring direction | Adding forces as plain numbers | "Which way does each point?" | Forces are vectors; add by components. |
Practice ladder
A force of 150 N acts perpendicular to a lever 0.4 m from the pivot. Find the moment.
Show answer
M = 150 × 0.4 = 60 N·m.
A 3 m beam on two end supports carries a 900 N load at its centre. Find each reaction.
Show answer
By symmetry, each support carries 900 / 2 = 450 N.
On a lever, an effort of 100 N acts 0.8 m from the pivot. What load at 0.2 m on the other side does it balance?
Show answer
Balance of moments: 100 × 0.8 = L × 0.2, so L = 80 / 0.2 = 400 N.
Draw a free body diagram of a ladder leaning against a smooth wall, and list every force on it.
What good work looks like
The ladder's weight acting down at its centre, a normal force from the wall pushing horizontally, and at the ground both a normal force up and a friction force inward. A good answer names all four and notes that equilibrium of forces and moments would solve for them.
Working with AI, and proving it yourself
Use AI as a guide, not an oracle
Portfolio task
Pick a simple loaded object, draw its free body diagram, and use equilibrium to find one unknown force.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What is a moment?
The turning effect of a force, equal to force times perpendicular lever arm.
2. State the two equilibrium conditions.
ΣF = 0 and ΣM = 0.
3. Why is a force a vector?
It has both a magnitude and a direction.
4. What is a free body diagram?
A sketch of the object alone with every external force drawn on it.
5. Central load on two equal supports?
Each support carries half the load.
Textbook mapping
This module follows Wickert and Lewis, An Introduction to Mechanical Engineering, 3rd edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| Forces and resultants | Wickert and Lewis, Sections 4.2 and 4.3 |
| Moment of a force | Wickert and Lewis, Section 4.4, Moment of a Force |
| Equilibrium and free body diagrams | Wickert and Lewis, Section 4.5, Equilibrium of Forces and Moments |
Section numbers refer to Wickert and Lewis, 3rd edition. Any edition with the same chapter titles is equivalent for study.