Physics for ME · Module 5 of 16 · Short and practical
Free-Body Diagrams and Equilibrium Preview
One body, isolated, with every force on it and nothing else. This small habit is the single biggest predictor of success in Engineering Statics.
Readiness check
From Module 4. Tick only what you can do closed-notes.
- Name the six force models and their direction rules.
- Solve ΣFy = 0 for an unknown normal force.
- Split weight into slope components.
- Distinguish forces on a body from forces by it.
- Declare and keep a sign convention.
The core idea
Isolate the body. Draw every force on it. Nothing else exists.
ΣFx = 0 · ΣFy = 0 (at rest)The recipe never changes: pick the body, replace each touch with its modeled force, add weight, choose axes, write the balance. Equilibrium (this module's preview) means the sums vanish; Statics spends a whole course on what that implies for machines and structures.
The skills, taught in order
The free-body diagram is one habit done in a fixed order. Build it the same way every time and equilibrium problems stop being puzzles and become bookkeeping.
5.1 The isolation step
Draw the boundary first: decide exactly which body you are analysing and cut it free from everything else. Every other object then becomes nothing but a force crossing that boundary.
5.2 Force inventory
Walk the boundary and add one arrow per physical interaction: a normal force wherever a surface touches, friction along it, tension along each rope, plus the one long-range force, weight, acting at the centre of gravity. If an arrow cannot name its source, it does not belong.
5.3 Axes that fit the problem
Choose axes that put the most forces directly on them. On a slope, take one axis along the surface and one perpendicular to it, so that only the weight needs splitting (Module 2's module, reused).
5.4 Equilibrium of a particle
If the forces all act through one point and the body does not accelerate, equilibrium is two scalar equations: ΣFx = 0 and ΣFy = 0. Laid tip to tail, the force arrows close into a polygon.
Micro-example. A 50 N sign on two symmetric cords: ΣFx = 0 makes the horizontal pulls equal and opposite, and ΣFy = 0 makes their vertical parts add to 50 N.
5.5 Knowns, unknowns, and counting
A particle FBD in two dimensions gives exactly two equations, so it can resolve two unknowns (say a tension and an angle, or two reactions). Counting unknowns against equations tells you, before any algebra, whether the problem is solvable as stated.
5.6 The handoff to Statics
Real parts are extended bodies, and forces act at different points, so a body can have zero net force yet still tend to rotate. Statics adds the second condition, ΣM = 0 (the torques, or moments, about any point sum to zero), giving a third equation per body in 2D. Worked Example 2 is a first taste; rigid-body equilibrium develops it fully.
| Body model | Equilibrium conditions (2D) | Unknowns it can solve |
|---|---|---|
| Particle (forces meet at a point) | ΣFx = 0, ΣFy = 0 | 2 |
| Rigid body (forces at different points) | ΣFx = 0, ΣFy = 0, ΣM = 0 | 3 |
Engineering connection: the direct gateway to Statics Module 3 and rigid-body equilibrium. The two-equation habit becomes force balance first, then full rigid-body balance.
Worked example 1: the toolbox on the ramp
A 15 kg toolbox rests on a 20° loading ramp without sliding. Find the friction force and the normal force holding it, and the minimum friction coefficient this requires.
- ProblemFind F, N, and the minimum μs for the toolbox in Figure 1.
- Given / findm = 15 kg, θ = 20°, at rest. W = 147.2 N.
- AssumptionsRigid box, dry contact, no other touches.
- ModelFigure 2 with axes along and normal to the ramp: only the weight needs splitting.
- EquationsAlong: F − W sin 20° = 0 Normal: N − W cos 20° = 0
- SolveF = 147.2 × 0.342 = 50.3 N up the slope; N = 147.2 × 0.940 = 138.3 N. Holding requires μs ≥ F/N = tan 20° = 0.364.
- Check√(50.3² + 138.3²) = 147.2 N: the components reassemble the weight. The μ condition matches the Statics tan-θ slip rule exactly.
- ConclusionThree arrows and two sums answered a real question: rubber on wood (μ ≈ 0.6) holds, smooth steel on steel (μ ≈ 0.3) does not. Statics Module 3 starts from precisely this diagram.
Worked example 2: reactions on a loaded plank
A uniform plank 4.0 m long weighs 200 N and rests on two supports: A at the left end and B at the right end. A 700 N person stands 3.0 m from the left. Find the upward reaction at each support. This is a rigid body, so a moment balance enters for the first time.
- ProblemFind the support reactions RA and RB for the loaded plank in Figure 3.
- Given / findPlank L = 4.0 m, weight 200 N at its centre (2.0 m); a 700 N person at 3.0 m; supports A at 0 and B at 4.0 m. Find RA and RB.
- AssumptionsRigid plank, supports give vertical reactions only, all forces vertical and in one plane.
- ModelRigid-body equilibrium: forces balance and moments balance. Take moments about A so that RA drops out and RB falls straight out.
- EquationsΣMA = 0: RB(4.0) − 200(2.0) − 700(3.0) = 0 ΣFy = 0: RA + RB = 900
- SolveRB = (400 + 2100) / 4.0 = 625 N. Then RA = 900 − 625 = 275 N.
- CheckTake moments about B instead: RA(4.0) = 200(2.0) + 700(1.0) = 1100, so RA = 275 N, the same value. Equilibrium holds about every point, which is the free check.
- ConclusionA moment equation about a well-chosen point isolates one unknown at a time. The person standing nearer B loads it more (625 N versus 275 N), exactly as intuition expects. This one torque balance is the idea Statics extends to beams, frames, and supports.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Forces by the body drawn on it | The box's push on the ramp appears on the box's FBD | "Does this force act on my chosen body?" | Only forces on the body. Its pushes on the world belong to other diagrams. |
| Arrows without sources | "Equilibrium force" or "balancing force" invented | "What touch or field creates this arrow?" | Every arrow is gravity or a named contact. No source, no arrow. |
| Axes fighting the geometry | Every force needing decomposition on a simple slope | "Could one axis lie along the surface?" | Align axes with the constraint; then only gravity splits. |
| Normal force drawn vertical on a slope | N pointing up instead of perpendicular to the ramp | "What direction does this surface push?" | Surfaces push perpendicular to themselves. Always. |
Practice ladder
Draw the FBD of a 6 kg picture hanging from two symmetric wires at 60° above horizontal, and find each tension.
Show answer
Three arrows: W = 58.9 N down, two tensions up-and-out. 2T sin 60° = 58.9 gives T = 34.0 N.
For the worked-example toolbox, the ramp angle is slowly raised. At what angle does it let go if μs = 0.50, and what changes on the FBD as θ grows?
Show answer
Slip at tan θ = 0.50: θ = 26.6°. On the FBD nothing changes but proportions: W sin θ grows, W cos θ (and with it the friction ceiling μN) shrinks: the diagram explains the failure before the algebra does.
A 40 kg engine hangs from a chain; a worker pulls it sideways with a horizontal rope until the chain leans 15° from vertical. Draw the FBD of the hook and find both the rope force and the chain tension.
Show answer
Chain tension T along the chain, rope force P horizontal, W = 392.4 N down. Vertical: T cos 15° = 392.4, T = 406.3 N. Horizontal: P = T sin 15° = 105.1 N. Leaning the chain 15° costs 14 N of extra tension and needs a 105 N pull.
Photograph three real supported objects (a hanging sign, a propped ladder, a shelf bracket). For each, draw the FBD with sources named, mark knowns and unknowns, and state which Statics tool (force balance alone, or moments too) finishing it would need.
What good work looks like
Three clean diagrams, every arrow sourced, unknown counts stated, and the correct call on which need moment equations (the ladder and bracket do; the sign may not).
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Build a ten-diagram FBD album: ten everyday objects (cup on table, hanging plant, leaning broom, braking bicycle, and six of your own), each with the isolated body, sourced arrows, and a one-line equilibrium statement. This album is your entry ticket to Statics.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What belongs on a free-body diagram, and what never does?
Every external force acting on the isolated body (contacts plus gravity). Never: forces the body exerts elsewhere, or internal forces.
2. What does equilibrium mean for a particle?
ΣFx = 0 and ΣFy = 0: the arrows close into a polygon.
3. On a slope of angle θ, what are the weight components and the no-slip condition?
W sin θ along, W cos θ normal; holding requires μs ≥ tan θ.
4. Why choose slope-aligned axes?
So the constraint forces (N, F) lie on the axes and only gravity needs decomposition: less algebra, fewer sign errors.
5. What does Statics add to this module's picture?
Extended bodies: forces act at points, so moments must also balance (ΣM = 0), and supports get reaction models.