Classical Mechanics

Free Body Diagram: How to Draw One Step by Step

Definition

A free body diagram is a simplified sketch that shows one chosen object as a dot or box, with every external force acting on it drawn as a labelled arrow. It strips away the surroundings, so the vector sum of those arrows gives the net force — which equals mass times acceleration.

You already do this without noticing. Push a stalled car and you feel exactly three things that matter: your push, the ground dragging back, and the car’s stubborn weight pressing down. Your brain quietly discards the paint, the passengers, the shopping in the boot.

A free body diagram is that instinct, made rigorous. Get it right and a page of confusing physics collapses into two short equations. Get it wrong — one stray arrow, one missing contact — and every number after it is wrong too.

What Is a Free Body Diagram?

A free body diagram is a drawing of a single object showing only the external forces acting on it, with each force drawn as an arrow starting at the object. Nothing else appears: no surfaces, no ropes, no neighbouring blocks.

The word free is doing real work here. You are freeing the body from its surroundings, then replacing every removed contact with the force it was exerting. The floor vanishes and becomes an upward arrow. The rope vanishes and becomes a pull along its old direction.

Why bother? Because forces obey vector arithmetic, and vector arithmetic needs a clean list. As soon as the picture contains a wall, a pulley and three blocks, you cannot tell which forces belong to which object.

One body, one diagram

This is the rule beginners break most often. If a problem has two blocks, it has two free body diagrams — never one crowded picture with arrows going everywhere.

The payoff is that each diagram gives you its own equation. Two diagrams, two equations, two unknowns: the system solves.

The Free Body Diagram Formula

Every free body diagram exists to feed one equation — Newton’s second law, applied separately to each axis.

ΣF = ma

In practice you never use it in that vector form. You split it the moment you have chosen axes:

ΣFx = max and ΣFy = may
  • ΣF — the net force, the vector sum of every arrow on the diagram, in newtons (N)
  • ΣFx, ΣFy — the net force along the x and y axes, in newtons (N)
  • m — the mass of the body you isolated, in kilograms (kg)
  • a — the acceleration of that body, in metres per second squared (m/s²)
  • ax, ay — the components of that acceleration along each axis, in m/s²

Two details decide whether your answer is right. First, ΣF means sum: forces pointing along the negative axis enter with a minus sign. Second, m is the mass of the isolated body only — not the whole system, unless you deliberately drew the whole system as one body.

Once the diagram is done and the components are listed, the arithmetic is mechanical, and you can check any single-force result against our Newton’s Second Law Calculator before committing it to a full solution.

How to Draw a Free Body Diagram in 5 Steps

Draw a free body diagram by isolating one body, reducing it to a dot, adding one arrow for gravity and one for each physical contact, labelling every arrow, then choosing axes and resolving. These five steps work for every mechanics problem you will meet.

Step 1 — Pick one body and circle it

Decide what you are analysing before you draw anything. Literally draw a loop around it in the original picture — this single habit prevents most stray-arrow errors.

Step 2 — Redraw it as a dot

Shape and size rarely matter for force problems, so shrink the object to a point and put that point at the origin of your axes. All arrows will now start from this dot.

Step 3 — Add gravity, then walk the boundary

Weight always acts, so draw it first, straight down. Then trace around the object’s surface and ask at every point: is something touching me here? Each contact earns exactly one force — or two if the surface is rough, since a rough surface gives both a normal force and friction.

Step 4 — Draw and label every arrow

Each arrow starts at the dot and points the way the force actually acts. Length should roughly reflect size: if you know the object is not sinking through the floor, draw N about as long as W.

Label with symbols, not numbers — N, W, T, f. Numbers come later, and symbols keep the algebra honest.

Step 5 — Choose axes, then resolve

Pick axes that put as many arrows as possible directly on an axis. Any arrow left at an angle gets split into components, and only then do you write ΣFx and ΣFy.

1. The real situation 2. The free body diagram crate rope 30° A rough floor, a crate, a rope x y N W = mg T f The crate becomes a dot. Only forces ON it survive.

The five steps in action: a real crate on a rough floor becomes four labelled arrows on a point.

Free Body Diagram Lab

The 4 Forces That Never Belong on a Free Body Diagram

Four forces show up on student diagrams again and again, and none of them is real in this context. Learning to spot them is faster than learning any formula.

1. The “force of motion”

A ball rolling across a floor has no forward arrow. It is moving because nothing has stopped it yet — motion needs no cause, only changes in motion do. Velocity is not a force and never earns an arrow.

A ball coasting to the right, already moving, nothing pushing it Wrong N W f F(motion) Nothing is touching it from behind. Right N W f v Velocity is not a force. It never gets an arrow.

The phantom “force of motion” is the most common wrong arrow in introductory mechanics.

2. Centrifugal force

Swing a conker on a string and the string pulls the conker inward, toward your hand. There is no outward arrow on the conker. The outward feeling belongs to your hand, which the conker really does pull.

3. The third-law partner

Newton’s third law pairs always act on different bodies, so they can never appear on the same diagram. The book pushes down on the table; the table pushes up on the book. Only the second one belongs on the book’s diagram.

4. The net force itself

ΣF and ma are results, not inputs. Drawing an extra arrow labelled “ma” double-counts forces you have already drawn, and the equation collapses.

Common Forces and Which Way They Point

Most diagrams are built from the same short list of forces. This table is the one to memorise — direction first, magnitude second.

Force Symbol Direction of the arrow Magnitude
WeightWAlways vertically down, whatever the surface is doingmg, with g ≈ 9.81 m/s²
Normal forceNPerpendicular to the contact surface, pushing away from itWhatever balances the perpendicular direction — not automatically mg
FrictionfAlong the surface, opposing sliding or the tendency to slideKinetic: μkN. Static: anything up to μsN
TensionTAlong the rope, away from the body — ropes pull, never pushUsually an unknown you solve for
Applied forceFWhichever way the push or pull actsGiven in the problem
DragFdDirectly opposite the velocity through the fluidGrows with speed
Spring forceFsOpposite the stretch or compression, back toward natural lengthkx

Notice how many entries describe direction by a rule rather than a picture. That is deliberate — the rules survive when the geometry gets strange, and a good grasp of the different types of forces in physics is what lets you populate a diagram quickly.

Free Body Diagrams on an Inclined Plane

On an inclined plane, tilt your axes so x runs along the slope and y runs perpendicular to it, then resolve the weight into mg sin θ down the slope and mg cos θ into the surface. This one choice removes almost all the algebra.

Why does it help so much? Because on a slope, N and f already lie along the tilted axes. Leave the axes horizontal and you must split three forces instead of one.

Tilt the axes, then split the weight θ x W = mg mg sin θ mg cos θ N (y) f Along the slope (x): mg sin θ – f = ma Into the slope (y): N = mg cos θ N is smaller than mg — the slope carries only part of it. With x along the slope, only the weight needs splitting — N and f already lie on an axis.

Tilting the axes on an incline turns a three-force resolution into a one-force resolution.

The perpendicular equation is where the insight hides. Since the block does not accelerate into the ramp, N = mg cos θ — smaller than the weight, and smaller still as the slope steepens.

That is why friction fades on a steep ramp: friction depends on N, and N is shrinking exactly when gravity’s pull along the slope is growing. Our full guide to inclined plane physics works through the sliding condition in detail.

Real-World Examples of Free Body Diagrams

These diagrams are not a classroom ritual. Every one of the situations below is solved this way in professional practice.

A lift accelerating upward

Standing in a lift, you feel heavier as it starts to rise. Your diagram has just two arrows: N up from the floor, W down. Because you accelerate upward, N must exceed W — and N is exactly what a bathroom scale reads.

A climber on a rope

A hanging climber has weight down and tension up, and while they hang still the two are equal. The moment they are lowered with acceleration, tension drops below weight — which is why a controlled lower feels gentler than a sudden stop.

A plane in level flight

Four arrows: lift up, weight down, thrust forward, drag back. Cruising at constant speed and height means both pairs cancel exactly, and the whole of aerodynamics starts from that balance.

A crate on a lorry that brakes

Here friction is the only horizontal force on the crate, and it must point backward to slow the crate with the lorry. If the required friction exceeds μsN, the crate slides forward — the calculation that sets load-securing rules.

A parked car on a hill

Weight down, normal force out of the slope, friction up the slope, and nothing accelerating. Engineers size handbrakes by asking whether friction alone can supply mg sin θ.

Common Misconceptions About Free Body Diagrams

“The normal force always equals the weight”

It equals the weight only in the narrow case of a flat surface with nothing else acting vertically. Tilt the surface, add a rope pulled at an angle, or accelerate vertically, and N changes immediately.

This is the single most expensive mistake in force problems, because friction depends on N. Get N wrong and every friction value downstream is wrong.

“Static friction equals μsN”

That product is the maximum static friction, not its actual value. Static friction is whatever it needs to be to prevent sliding, right up until it cannot manage any more.

A book resting on a gentle slope needs only a few newtons of friction, even if the surface could supply thirty. Read more in our guide to what friction is and how it works.

“If it’s moving, something must be pushing it”

Constant velocity means zero acceleration, which means the forces balance exactly. A car at a steady 70 mph has thrust and drag in perfect opposition — it is a case of equilibrium, not of a winning force.

“Two cables each carry half the weight”

Only if both cables hang vertically. Angle them and each carries more than half, because only the vertical component of each tension does the lifting — a point Worked Problem 4 below makes concrete.

How Free Body Diagrams Fit With Newton’s Laws

Each of Newton’s three laws of motion maps onto a different part of the diagram, which is why the technique feels natural once it clicks.

  • First law — tells you what a balanced diagram means: at rest or at constant velocity, every arrow cancels.
  • Second law — turns an unbalanced diagram into numbers through ΣF = ma.
  • Third law — tells you which arrows are forbidden, since a reaction force lives on the other body’s diagram.

The diagram is also where vector skills earn their keep. Forces are vectors, so direction carries as much information as size, and confidence with vector and scalar quantities is what makes resolving components feel routine.

One more connection worth holding on to: ropes and pulleys are just tension arrows with a geometric constraint attached. Our guide to tension force covers the ideal-rope assumptions that let you carry one symbol T across two diagrams.

Worked Problems

Problem 1
A 2.0 kg book rests on a level table. Draw the free body diagram and find the normal force. Take g = 9.81 m/s².
Show Solution
Solution: Step 1: Two arrows only — W down, N up. The book is at rest, so a = 0 and ΣFy = 0. Step 2: ΣFy = N − W = 0, so N = W = mg. Step 3: N = 2.0 kg × 9.81 m/s² = 19.62 N. Answer: N = 19.6 N (3 s.f.), directed vertically upward.
Problem 2
A 65 kg person stands in a lift accelerating upward at 1.2 m/s². Find the normal force from the floor.
Show Solution
Solution: Step 1: Arrows are N up and W down. Take up as positive; the acceleration is upward, so ay = +1.2 m/s². Step 2: ΣFy = N − mg = ma, so N = m(g + a). Step 3: N = 65 kg × (9.81 + 1.2) m/s² = 65 × 11.01 = 715.65 N. Sanity check: at rest N would be 637.65 N, so the person genuinely feels heavier. Answer: N = 716 N (3 s.f.).
Problem 3
A 12 kg crate on a level floor is pushed horizontally with 45 N. The coefficient of kinetic friction is 0.25. Find the acceleration.
Show Solution
Solution: Step 1: Vertical direction first — no vertical acceleration, so N = mg = 12 × 9.81 = 117.72 N. Step 2: Kinetic friction f = μkN = 0.25 × 117.72 = 29.43 N, opposing the push. Step 3: ΣFx = 45 − 29.43 = 15.57 N, so a = 15.57 / 12 = 1.2975 m/s². Answer: a = 1.30 m/s² (3 s.f.), in the direction of the push.
Problem 4
An 8.0 kg sign hangs from two cables, each making 30° with the horizontal. Find the tension in each cable.
Show Solution
Solution: Step 1: Three arrows — W down, and two tensions T pulling up and outward at 30° above horizontal. By symmetry both tensions are equal. Step 2: Horizontal components cancel. Vertically: 2T sin 30° − W = 0. Step 3: W = 8.0 × 9.81 = 78.48 N, so T = 78.48 / (2 × 0.500) = 78.48 N. Note the result: each cable carries the full weight, not half — because at 30° only half of each tension acts vertically. Answer: T = 78.5 N (3 s.f.) in each cable.
Problem 5
A 5.0 kg block slides down a frictionless 30° incline. Find its acceleration and the normal force.
Show Solution
Solution: Step 1: Tilt the axes — x down the slope, y perpendicular. Resolve W into mg sin θ along x and mg cos θ along −y. Step 2: Along x: ΣFx = mg sin θ = ma, so a = g sin θ = 9.81 × 0.500 = 4.905 m/s². Step 3: Along y: N − mg cos θ = 0, so N = 5.0 × 9.81 × cos 30° = 49.05 × 0.8660 = 42.48 N. Note that a does not depend on mass — every object slides at the same rate on a frictionless slope. Answer: a = 4.91 m/s² down the slope; N = 42.5 N (3 s.f.).
Problem 6
A 3.0 kg block sits on a 20° incline with μs = 0.45. Does it slide? If not, what is the friction force acting on it?
Show Solution
Solution: Step 1: Find the pull along the slope: mg sin θ = 3.0 × 9.81 × sin 20° = 29.43 × 0.3420 = 10.066 N. Step 2: Find the maximum static friction available: μsN = μsmg cos θ = 0.45 × 29.43 × 0.9397 = 12.445 N. Step 3: Since 10.07 N < 12.44 N, the block stays put. Static friction supplies only what is needed to hold it, not its maximum. Equivalent check: the block slides only when tan θ > μs, and tan 20° = 0.364 < 0.45. Answer: It does not slide. Friction = 10.1 N (3 s.f.) up the slope — not 12.4 N.
Problem 7
A 4.0 kg block on a frictionless table is joined by a light cord over a frictionless pulley to a 2.0 kg block hanging freely. Find the acceleration and the cord tension.
Show Solution
Solution: Step 1: Draw two diagrams. Block 1 (on the table): T forward, N up, W down. Block 2 (hanging): T up, m2g down. The cord fixes both accelerations to the same magnitude a. Step 2: Block 1 horizontally: T = m1a. Block 2 vertically: m2g − T = m2a. Step 3: Add the two equations to eliminate T: m2g = (m1 + m2)a, so a = (2.0 × 9.81) / 6.0 = 19.62 / 6.0 = 3.27 m/s². Step 4: Back-substitute: T = m1a = 4.0 × 3.27 = 13.08 N. Check with block 2: m2g − T = 19.62 − 13.08 = 6.54 N, and m2a = 2.0 × 3.27 = 6.54 N. Consistent. Answer: a = 3.27 m/s²; T = 13.1 N (3 s.f.). Note T is less than the hanging weight of 19.6 N.
Problem 8
A 20 kg sled is pulled by a 60 N force at 30° above the horizontal across ground with μk = 0.15. Find the normal force and the acceleration.
Show Solution
Solution: Step 1: Resolve the pull. Fx = 60 cos 30° = 51.96 N; Fy = 60 sin 30° = 30.00 N upward. Step 2: Vertically there is no acceleration, so N + Fy − mg = 0, giving N = mg − Fy = 196.2 − 30.0 = 166.2 N. Step 3: Friction f = μkN = 0.15 × 166.2 = 24.93 N. Then ΣFx = 51.96 − 24.93 = 27.03 N and a = 27.03 / 20 = 1.3516 m/s². The lesson: the upward component of the pull lightens the sled, so N is 166 N rather than 196 N — pulling at an angle beats pushing down at one. Answer: N = 166 N; a = 1.35 m/s² (3 s.f.).

Frequently Asked Questions

What is a free body diagram in simple terms?
A free body diagram is a sketch of one object on its own, with an arrow drawn for every force pushing or pulling on it. You remove everything around the object — floors, ropes, other blocks — and replace each removed contact with the force it was applying. The arrows then add up as vectors to give the net force.
What forces should be included in a free body diagram?
Include only external forces acting on the chosen body: weight, plus one contact force for each thing physically touching it. A rough surface contributes two, a normal force and friction. Exclude any force the body exerts on something else, any net force or ma term, and any invented force such as a forward “force of motion” or centrifugal force.
Do you include acceleration in a free body diagram?
No — acceleration is not a force and does not get an arrow among the forces. It is the result the diagram helps you calculate through ΣF = ma. Many textbooks allow you to note the acceleration direction beside the diagram, clearly separated and often in a different colour, so it is never mistaken for a force.
Why is the normal force not always equal to the weight?
The normal force equals the weight only when the surface is horizontal, nothing else acts vertically, and there is no vertical acceleration. On a slope it becomes mg cos θ. In an accelerating lift it becomes m(g ± a). If a rope pulls partly upward, the normal force drops by that upward component.
How do you draw a free body diagram for two connected objects?
Draw a separate diagram for each object, never one combined picture. Use the same symbol T for the tension in a light cord, since it is the same throughout an ideal rope over a frictionless pulley. Each diagram gives one equation, and the cord supplies a constraint linking the accelerations, so two equations solve for two unknowns.
What is the difference between a free body diagram and a force diagram?
The terms are used almost interchangeably, but a free body diagram is stricter. It isolates exactly one body, usually shrinks it to a point, and shows only external forces acting on that body. A general force diagram may keep the original picture and show forces on several objects at once, which makes it easier to mix up which forces belong where.

Free body diagrams reward practice more than memorisation. Draw one for the next five problems you meet — even the easy ones — and the harder ones stop looking hard. For a rigorous parallel treatment, the OpenStax University Physics section on drawing free-body diagrams works through coupled blocks in detail, and MIT OpenCourseWare’s 8.01 Week 2 materials include filmed worked examples on stacked blocks and pulley systems.

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