Statics: Forces, Moments and Equilibrium
Free-body diagrams, resultants, moments, equilibrium, reactions and trusses.
The core ideas
Statics is the mechanics of bodies that are not accelerating. Almost every structure, machine frame, shaft and bracket is designed first as a static problem: find the forces, then check the stress (see Structures and Strength of Materials). The method never changes: draw the body free of its surroundings, replace each support by the forces and moments it exerts, and require that the forces and moments balance.
- A force is a vector: magnitude, direction and line of action (newtons, N).
- A moment is a force times the perpendicular distance from a point: M = F d (N·m). Its sign is counter-clockwise positive unless you choose otherwise, but you must keep one convention.
- A couple is two equal and opposite parallel forces: a pure moment with no resultant force that has the same value about every point.
- A body is in equilibrium when the resultant force and the resultant moment are both zero.
Free-body diagrams
A free-body diagram (FBD) isolates the body and shows every external force on it, including weight, applied loads and the reactions at supports. It is the most important step; most errors in statics are a missing or wrongly directed force on the FBD.
| Support | Reactions it provides |
|---|---|
| Roller or smooth surface | One force, perpendicular to the surface |
| Pin (hinge) | Two force components (x and y), no moment |
| Fixed (cantilever) support | Two force components and a moment |
| Cable or rope | One tension force along the cable, pulling only |
| Two-force member (pinned at both ends, no load between) | A force along the line joining the pins |
Forces and moments
M = F d⊥ = F r sin θ Rx = ΣFx, Ry = ΣFy xR = ΣM ÷ Ry
Worked example: moment and resultant
- A 200 N force acts at 30° to a spanner handle 350 mm long: only the perpendicular component turns the nut. M = 200 × sin 30° × 0.35 = 35.0 N·m. Pull at 90° and you get 70 N·m.
- Three forces act on a plate along the x axis: 40 N at 0° at x = 0, 30 N at 90° at x = 0.5 m and 20 N at 210° at x = 1.2 m. Rx = 22.68 N, Ry = 20.00 N, so R = 30.2 N at 41.4°. The moment about the origin is ΣM = 3.00 N·m, so the resultant acts along a line crossing the x axis at 0.150 m.
Equations of equilibrium
ΣFx = 0 ΣFy = 0 ΣMO = 0 (any point O)
If a body has more unknown reactions than equations it is statically indeterminate (for example a beam on three supports): equilibrium alone is not enough and the deflections must be considered as well, a topic covered in the beams guide.
Worked example: a loaded beam
Worked example: reactions and maximum moment
- Replace the 5 kN/m uniform load by its resultant, 5 × 6 = 30 kN acting at mid-span (3 m).
- ΣMA = 0: RB × 6 − 10 × 2 − 30 × 3 = 0, so RB = 18.33 kN.
- ΣFy = 0: RA + RB = 10 + 30, so RA = 21.67 kN.
- Shear just right of the point load is RA − P − w × 2 = 1.67 kN; it falls by 5 kN each meter and reaches zero at x = 2.33 m, where the bending moment is greatest: Mmax = 33.6 kN·m.
- Check with the other end: moments about B must also sum to zero (21.67 × 6 − 10 × 4 − 30 × 3 = 0.00). Always check a result with a second equation.
Worked example: a hanging sign
Worked example: two cables at different angles
- A 500 N sign hangs from two cables making 30° (left) and 60° (right) with the horizontal. Unknowns: T1 and T2.
- ΣFx = 0: T1 cos 30° = T2 cos 60°, so T2 = 1.732 T1.
- ΣFy = 0: T1 sin 30° + T2 sin 60° = 500, so T1 (0.5 + 1.5) = 500 and T1 = 250 N, T2 = 433 N. Check: residuals 0.000000 and 0.000000. The flatter cable carries less, but a very flat cable carries huge tension (at 5° each side the tension is W ÷ (2 sin 5°) = 2 868 N).
Trusses and two-force members
A truss is a set of straight members pinned at the joints and loaded only at the joints, so each member is a two-force member in pure tension or compression. The method of joints applies ΣFx = ΣFy = 0 to each joint in turn, starting where there are only two unknowns; the method of sections cuts through the truss and applies the three equations to one side. A planar truss is statically determinate if m + r = 2 j (members plus reaction components equal twice the joints). Numerical truss solving is in the Truss Calculator under Tools.
Common mistakes
- A missing force or a force with the wrong sense on the free-body diagram.
- Mixing the sign convention for moments halfway through.
- Using the component of a force along the wrong axis (check which angle is measured from which line).
- Replacing a distributed load by its resultant at the wrong point: for a triangular load the resultant acts at one third from the large end.
- Forgetting that a cable can only pull, and that a support reaction may come out negative (it acts the other way).
Related
- Next: Centroids, Area Moments and Friction. Back to the overview: Engineering Fundamentals.
- Structures and Strength of Materials for stress and deflection.
- The press frame case study uses these equations.
FAQ
Why take moments about a support?
Because the unknown reaction there has zero moment arm and drops out, leaving one unknown per equation.
What if a reaction comes out negative?
It acts opposite to the direction you assumed. Keep the sign in later equations.
When is a problem statically indeterminate?
When there are more unknown reactions or members than equilibrium equations; add deflection (compatibility) equations.
References
- Hibbeler, R. C., Engineering Mechanics: Statics, Pearson.
- Beer, F. P., Johnston, E. R., Vector Mechanics for Engineers: Statics, McGraw-Hill.
- Meriam, J. L., Kraige, L. G., Engineering Mechanics: Statics, Wiley.
Last reviewed: October 2026. Examples are for study; check any real structure to the code that governs it.