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Statics · Module 3 of 11 · Beginner

Equilibrium of a Particle

If all forces meet at one point, the particle model is enough.

01

Readiness check

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).
0 or 1 weak itemsContinue with this module.
2 weak itemsRedo the Level 1 and 2 ladder of Force Vectors first.
3 or more weak itemsStep back to Module 2 and complete it fully before returning.
02

The core idea

At rest means the components cancel, in every direction at once.

ΣFx = 0ΣFy = 0

The 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 model works when: all forces are concurrent (cables, springs, pulleys, rings, hooks); size and rotation do not matter.
The model breaks down when: forces act at different points of an extended body; then moments matter, and you need Module 5.
The concept. Three concurrent forces in balance: the two cable tensions together supply exactly what the weight demands.
03

The method

1Look

What point do all the forces pass through?

2Simplify

Isolate that point as the particle.

3Draw

FBD: every contact force plus weight, with angles.

4Solve

ΣFx = 0 and ΣFy = 0; check signs and sense.

04

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.

Figure 1. Problem setup: a 20 kg light on a ring held by two cables at unequal angles.
Figure 2. Free-body diagram of the ring. Solution: T₁ = 143.6 N, T₂ = 175.9 N.
weightcable tensionsangles and axes
  1. ProblemFind both cable tensions for the system in Figure 1.
  2. Given / findm = 20 kg, θ₁ = 30°, θ₂ = 45°. Find T₁ and T₂.
  3. AssumptionsCables are massless and carry pure tension along their axes; the system is at rest; 2D.
  4. ModelFigure 2: FBD of the ring. Three forces, one point: particle equilibrium. W = 20 × 9.81 = 196.2 N.
  5. EquationsΣFx = 0: T₂ cos 45° − T₁ cos 30° = 0 ΣFy = 0: T₁ sin 30° + T₂ sin 45° − 196.2 = 0
  6. 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.
  7. 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.
  8. ConclusionSpecify both cables and anchors for at least 176 N working load, and note that making the cables shallower would increase the tensions sharply.
Result. T₁ = 143.6 N, T₂ = 175.9 N. Both checks pass.
05

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Assuming equal tensions with unequal anglesT₁ = 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 particleExtra 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 FBDEquations 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.
06

Practice ladder

Level 1 · Direct skill

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.

Level 2 · Mixed concept

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.

Level 3 · Independent problem

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.

Level 4 · Transfer to real engineering

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).

07

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here is my FBD described in words: [forces, angles]. Ask me three questions that would expose a missing or extra force."
"Check whether my equilibrium equations match the FBD I described; flag mismatches, don't solve."
"What are the tensions in this cable problem?" The FBD-to-equations habit is the entire skill.
"Draw the FBD for me." Drawing it yourself is the learning.

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.

Must include: photo and FBD side by side, the solved tensions, a reasonableness check, and one limitation of your model.
08

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.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRedraw the traffic-light FBD and equations from memory.
+3 daysSolve one new two-cable problem with your own numbers.
+7 daysMixed set: one particle problem plus one resultant problem (Module 2).
+30 daysRevisit your portfolio report: would you change the model?
09

Textbook mapping

ItemMapping
Main textbookR.C. Hibbeler, Engineering Mechanics: Statics, Chapter 3, Equilibrium of a Particle
Core sections3.1 Condition for Equilibrium · 3.2 The Free-Body Diagram · 3.3 Coplanar Force Systems · 3.4 Three-Dimensional Force Systems
Recommended problemsFundamental Problems F3-1 onward (partial solutions in the back). Master the coplanar set; the FBD habit you build here carries the whole course.
Skip on first pass3.4 (3D systems) can wait until you have finished 2D rigid-body equilibrium; then return with Cartesian vectors fresh.
Read nextChapter 4, sections 4.1 to 4.4 before opening Module 4.