Statics · Module 3 of 11
Equilibrium of a Particle
If all forces meet at one point, the particle model is enough.
Readiness check
Learning objectives
- State the equilibrium conditions for a particle in 2D and in 3D.
- Solve a free-body diagram for two unknown cable tensions.
- Test whether an assumed cable angle agrees with the forces it implies.
From Module 2. Tick only what you can do closed-notes.
- Resolve any 2D force into x and y components with correct signs.
- Recombine components into a magnitude and direction.
- Convert a mass into a weight force (W = mg).
- Solve two linear equations with two unknowns.
- Build a unit vector between two points (needed for the 3D section).
The core idea
At rest means the components cancel, in every direction at once.
ΣFx = 0ΣFy = 0The free-body diagram (FBD) is the contract: every force that touches the particle goes on it, nothing else. Two equations in 2D, three in 3D, so count your unknowns before solving.
The method
What point do all the forces pass through?
Isolate that point as the particle.
FBD: every contact force plus weight, with angles.
ΣFx = 0 and ΣFy = 0; check signs and sense.
Worked example: the hanging traffic light
A 20 kg traffic light hangs from a ring supported by two cables: cable 1 at 30° above horizontal (left side), cable 2 at 45° above horizontal (right side). Find both cable tensions.
- ProblemFind both cable tensions for the system in Figure 1.
- Given / findm = 20 kg, θ₁ = 30°, θ₂ = 45°. Find T₁ and T₂.
- AssumptionsCables are massless and carry pure tension along their axes; the system is at rest; 2D.
- ModelFigure 2: FBD of the ring. Three forces, one point: particle equilibrium. W = 20 × 9.81 = 196.2 N.
- EquationsΣFx = 0: T₂ cos 45° − T₁ cos 30° = 0 ΣFy = 0: T₁ sin 30° + T₂ sin 45° − 196.2 = 0
- SolveFrom x: T₂ = T₁ (cos 30°/cos 45°) = 1.225 T₁. Substitute into y: T₁(0.5) + 1.225 T₁(0.7071) = 196.2, so 1.366 T₁ = 196.2, giving T₁ = 143.6 N and T₂ = 175.9 N.
- CheckHorizontal: 175.9 cos 45° = 124.4 N = 143.6 cos 30°. Both tensions positive (cables can only pull). The steeper cable (45°) carries more; it does more of the vertical lifting.
- ConclusionSpecify both cables and anchors for at least 176 N working load, and note that making the cables shallower would increase the tensions sharply.
Worked example 2: diagnose a cable-angle assumption
Problem. Two identical cables support a 200 N light. Each cable is 30° above horizontal. A student divides the load by two and reports 100 N tension in each cable.
- Draw the particleThe horizontal components cancel by symmetry, but only the vertical components support the light.
- Test the proposed answerTwo 100 N tensions provide 2(100)sin30° = 100 N upward, only half the required support.
- Apply equilibrium2T sin30° - 200 N = 0, so T = 200 N in each cable.
- InterpretShallow cables need greater tension because only part of each tension acts vertically.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Assuming equal tensions with unequal angles | T₁ = T₂ = 98.1 N in the example above | "Are the cable angles actually symmetric?" | Tensions are equal only for mirror-symmetric geometry. Otherwise solve both equations. |
| Forces on the FBD that do not touch the particle | Extra unknowns; unsolvable system | "Does this force physically act on this point?" | Only contact forces on the particle plus its weight. Re-isolate. |
| Missing weight on the FBD | Equations balance trivially; tensions come out zero | "What is pulling the system down in the first place?" | Gravity acts on every mass. Add W = mg before anything else. |
| A cable in "compression" | Negative tension in the result | "Can a rope push?" | Negative cable tension means the model is wrong; recheck geometry or directions. |
Practice ladder
A 50 kg crate hangs from two cables, each at 60° above horizontal, symmetric about vertical. Find the tension in each cable.
Show answer
By symmetry T₁ = T₂ = T. ΣFy: 2T sin 60° = 50 × 9.81 = 490.5, so T = 283.2 N.
Repeat the worked example but with θ₁ = 20° and θ₂ = 60°. Before solving, predict which cable carries more.
Show answer
T₂ = T₁(cos 20°/cos 60°) = 1.879 T₁. ΣFy: T₁(0.342 + 1.627) = 196.2, so T₁ = 99.7 N and T₂ = 187.3 N. The steeper cable carries more, as predicted.
A spring (stiffness k = 500 N/m, unstretched length 0.4 m) hangs vertically above a 10 kg mass and is stretched to a length of 0.6 m. Is the spring alone enough to hold the mass?
Show answer
Spring force F = ks = 500 × (0.6 − 0.4) = 100 N upward. Weight = 98.1 N. Vertical balance: 100 N ≥ 98.1 N: yes, barely (1.9 N margin, about 2%). An engineer would reject this margin and specify a stiffer spring or more stretch.
Design the cable angles for a 15 kg hanging planter between two walls 3 m apart, where each cable's safe working load is 250 N. Find the minimum cable angle above horizontal, then state the cable lengths.
Show answer
Symmetric: T = W/(2 sin θ) ≤ 250, so sin θ ≥ 147.2/500 = 0.294, giving θ ≥ 17.1°. Choose a round, safer θ = 30°: each cable spans 1.5 m horizontally, so length = 1.5/cos 30° = 1.73 m and T = 147.2 N (41% margin).
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Find one real hanging system (sign, lamp, plant, gym rings). Photograph it, measure or estimate the geometry, draw the FBD, compute the tensions, and write a half-page report: assumptions, model, numbers, one design recommendation.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What are the equilibrium conditions for a particle in 2D? In 3D?
2D: ΣFx = 0 and ΣFy = 0 (two equations, at most two unknowns). 3D: add ΣFz = 0.
2. What belongs on a free-body diagram, and what never does?
Every external force acting on the isolated particle (contact forces plus weight). Never: forces the particle exerts on other things, or internal forces.
3. What force law governs a linear spring?
F = ks, where s is the stretch or compression from the unstretched length.
4. What does a frictionless pulley change about a cable, and what does it not change?
It changes the cable's direction; it does not change the tension magnitude.
5. Why do cable tensions grow as the cables get closer to horizontal?
Only the vertical component T sin θ supports the weight; as θ approaches zero, T = W/(2 sin θ) grows without bound.