How to Draw a Free Body Diagram in 6 Steps

The same six steps work on every problem, from a block on a table to a loaded bridge truss. Follow them in order and you will not miss a force.

Weight, then one force for every contact you removed. That is the whole checklist.

1 — Isolate the body

Decide what you are analysing and cut it away from everything touching it. Draw it as a simple shape: a box for a block, a line for a beam, a dot for a particle. Do not draw the ramp, the rope or the wall — each is about to become a force.

Objects that move together, like stacked blocks, can be treated as one body. Isolate them separately only when you need the contact force between them.

2 — Add the weight

One arrow, labelled W or mg, straight down from the centre of gravity. In introductory problems it is the only force that acts without touching.

3 — Turn each contact into a force

Walk the boundary of the body. Everywhere something was touching it, draw what that contact provides:

• Surface → normal force N, perpendicular to it, plus friction f along it if the surface is rough.
• Rope or cable → tension T, pulling away from the body along the rope. A rope can only pull.
• Spring → force along the spring axis, pushing if compressed and pulling if stretched.
• Support → a roller gives one reaction, a pin gives two, a fixed support gives two plus a moment.

4 — Add the applied loads

Whatever the problem states: pushes, pulls, distributed loads, applied moments. Put each at the point where it acts, with its magnitude and angle.

5 — Choose the axes

Draw a small x–y pair beside the diagram. On an incline, tilt it so x runs along the slope. That splits the weight into mg sin θ and mg cos θ and leaves every other force on an axis, which is most of the algebra done before you have written anything.

6 — Check before you solve

• Every arrow starts on the body, and every arrow is labelled.
• Count: one force per contact, plus weight. More arrows than that means you invented one.
• No “force of motion” and no centrifugal force. Motion is what forces cause, not a force.
• In equilibrium, could these arrows plausibly cancel? If they all point the same way, one is missing.

Worked example: block on a rough incline

1. Isolate — a tilted box, no ramp.
2. Weight mg straight down.
3. One contact, the slope: normal N perpendicular to it, friction f along it, pointing up-slope because the block tends to slide down.
4. No applied loads in this problem.
5. Axes tilted along the slope.
6. Check: three arrows, one contact plus weight. Correct.

The equations drop straight out: N = mg cos θ perpendicular to the slope, and f = mg sin θ along it while the block stays put.

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