Free-Body Diagrams

Choosing the body is the whole decision

A free-body diagram answers one question: if this object were cut away from everything around it, what would the surroundings still be doing to it? The word free is literal. You draw the object detached - off the floor, off the pin, off the cable, away from the block beside it - and everything you removed reappears only as the force it was exerting.

That cut is a decision, and it is the decision that determines the answer you can get. Forces that cross the boundary you drew are external and appear on the diagram. Forces entirely inside the boundary are internal and do not appear at all. Isolate two blocks together and the push between them vanishes from the picture; isolate one of them and that same push becomes an arrow you can solve for. Neither choice is more correct - they answer different questions.

So work backwards from the unknown. Ask what you are trying to find, then choose the body whose boundary that unknown has to cross. If you want the force in a connection, your cut must pass through the connection. If you want the acceleration of an assembly and do not care about its internal forces, take the whole assembly and let the internal forces disappear. Practicing engineers do this at whatever scale suits the question: NASA's introductory aeronautics material treats a whole aircraft in flight as one body carrying weight, lift, thrust, and drag, with the distributed weight collected at the center of gravity - a single body chosen because the question was about the aircraft's overall motion, not about the load in any one wing spar.

The procedure, step by step

The sequence below is worth running mechanically until it becomes automatic.

1. Name the body. Write it down - "block B alone", "the beam plus both blocks". If you cannot name it in words, you have not chosen it.

1. Draw it isolated. Sketch the body by itself, not traced over the original picture. Drawing on top of the original is the single most reliable way to keep a support in the diagram that you were supposed to have removed. The sketch does not need to be pretty; for a particle problem a box or a dot is fine.

1. Walk the boundary. Go around the outline of the body. At every place where something was touching it, ask what that thing can push or pull with, and draw that force at the point of contact. This is the replace step, and doing it as a systematic lap around the perimeter is what stops you from forgetting the normal force or the friction force.

1. Add the body forces. Gravity does not act at a contact point; it acts throughout the body and is drawn as one arrow at the center of mass. Weight in the engineering sense is a force, measured in newtons, not a mass in kilograms; NIST's SI guide is explicit that the newton is the unit of the quantity weight, even though everyday speech uses "weight" for mass.

1. Choose and draw axes. Put the axes on the diagram, with the positive directions marked. Unstated axes are the source of most sign errors.

1. Label everything. Every arrow gets a name. Known values get their numbers and units; unknowns get symbols. Angles and the dimensions you will need for moments get labeled too.

1. Count. Count your unknowns and compare with the number of independent equations your chosen body will give you - two force equations in a two-dimensional particle problem, three (two force, one moment) for a rigid body in two dimensions. If you have more unknowns than equations, you need another body or another cut before you start algebra.

What each support permits, prevents, and costs

Replacing a support with a force is not guesswork. A support prevents certain motions and permits others, and it can only exert force in the directions it prevents motion. That correspondence gives a short standard inventory for two-dimensional problems, with the number of unknowns each one adds:

• Roller or rocker: prevents motion perpendicular to its surface, permits sliding along the surface and rotation. One unknown - a single force normal to the surface.

• Frictionless pin: prevents translation in both directions, permits rotation. Two unknowns - usually taken as x and y components, because the direction of the pin force is not known in advance. It contributes no couple-moment, which is exactly why the pin is drawn as two force components and not three.

• Fixed support (built-in, welded, cantilevered into a wall): prevents translation in both directions and rotation. Three unknowns - two force components plus a couple-moment.

• Cable, rope, or chain: can only pull, and only along its own line. One unknown, a tension whose direction is known. A negative cable tension is not a compression; it is a signal that the cable would have gone slack and the model no longer holds.

• Smooth surface: one unknown, normal to the surface. Add friction and you add a second unknown, tangent to the surface, whose direction opposes impending or actual relative sliding.

• Spring or two-force link: one unknown along its axis.

Counting unknowns this way gives you a cheap sanity check before any algebra. A beam held by a pin at one end and a roller at the other carries two plus one, that is three unknown reaction components, which is exactly the number of independent equilibrium equations available for a rigid body in two dimensions. That match is why the simply supported beam is solvable from statics alone.

Axes, signs, and what a negative answer means

Signs do not come from intuition; they come from the axes you drew. Once positive x and positive y are on the diagram, every force component's sign is determined, and your job is to be consistent rather than clever.

Choose axes to reduce work. On an inclined surface, aligning x with the slope and y with the normal means only the weight has to be resolved into components, instead of the normal force and friction both needing resolution. For a 25 kg crate on a 20 degree ramp, with the standard acceleration of free fall g_n = 9.806 65 m/s^2, the weight is 25 kg x 9.806 65 m/s^2 = 245.17 N. With rotated axes its components are 245.17 N x cos 20 deg = 230.4 N pressing into the surface and 245.17 N x sin 20 deg = 83.9 N down the slope. A student who leaves the axes horizontal and vertical and then writes the normal force as equal to the full 245 N is about 6.4 percent high, and the error is purely bookkeeping.

When you do not know which way an unknown force points, assume a direction, draw the arrow, and let the algebra decide. If the answer comes out negative, the magnitude is right and the assumed direction was wrong. That is a result, not a mistake - do not flip the arrow and re-solve, and above all do not quietly drop the minus sign. The one thing you must not do is change the assumed direction partway through, because the sign of every other term in the equation was written relative to it.

One body or several: where to cut

Because every part of a body in equilibrium is itself in equilibrium, you are free to isolate at any scale. That freedom is the main strategic lever you have.

Take the whole assembly as one body and the forces that connected its parts cancel in pairs. They are equal and opposite, they act on two different parts of the same body, and they contribute nothing to the sum of external forces. This is why a whole-system diagram is short: fewer arrows, fewer unknowns, and often a single equation for the acceleration or an overall reaction.

But those canceled forces are frequently exactly what you were asked for. The force in a pin, the tension in a cable, the push between two blocks, the internal load in a beam - none of these appears on the whole-system diagram, because it never crosses that boundary. To see one, you must draw a boundary that cuts through it. The moment your cut passes through a connection, the interaction that was invisible becomes an external force on each of the two new free bodies, pointing one way on one diagram and the opposite way on the other.

The practical routine follows: use the whole system first to get the global quantity cheaply, then cut where the unknown lives and use the value you just found. And keep a check in reserve - the third free body you did not need is an independent test of the two you used.

From diagram to equations

A finished diagram converts into equations mechanically. For each axis, sum the components of every arrow on the diagram, set the sum equal to zero if the body is in equilibrium or to mass times the acceleration component if it is not, and solve.

Two disciplines matter here. First, the mass-times-acceleration term is not a force and does not belong on the diagram. It lives on the other side of the equation. Drawing an ma arrow among the real forces double-counts it and produces equations that balance to nonsense. (There is an alternative convention, d'Alembert's, in which a fictitious inertial force of magnitude ma is deliberately added so the problem can be handled as a static balance - it is legitimate, but it is a different bookkeeping system and the two must never be mixed.) Second, take moments about a point that kills unknowns: summing moments about a pin removes both of that pin's force components from the equation at a stroke.

A worked instance: a 6.0 m beam, pin at A, roller at B, with a 1200 N load 2.0 m from A. Summing moments about A removes both pin components, leaving R_B x 6.0 m - 1200 N x 2.0 m = 0, so R_B = 400 N; then vertical equilibrium gives R_A = 1200 N - 400 N = 800 N. Checking moments about B independently gives 800 N x 6.0 m - 1200 N x 4.0 m = 0, which confirms the pair.

What the diagram assumes, and what this lesson is not

Every free-body diagram is a model, and the modeling assumptions are usually invisible until they fail. A rigid body does not deform, so the geometry you measured before loading is the geometry you use after - true enough for a steel bracket, badly wrong for a cable-stayed structure at large deflection. A frictionless pin transmits no moment, which is why it costs two unknowns and not three; a real pin that has seized transmits a moment your diagram denies. A smooth surface has no tangential force, which is exactly the assumption that fails the moment a crate stops sliding. A massless cable has the same tension throughout and pulls only along its line, which stops being true for a heavy chain hanging under its own weight. Naming the assumption you are making is part of drawing the diagram, not an afterthought.

This lesson is educational material. It teaches how to construct and read a free-body diagram; it is not engineering design guidance, and nothing here should be used to size, check, or approve a real structure, machine, lifting arrangement, or pressure-retaining part. Real design is performed by a licensed engineer working to the governing code or standard, with the load cases, load factors, and allowable values that code specifies.