Free-Body Diagrams: Drawing Forces the Right Way

Quick answer: A free-body diagram shows every force acting ON one object as a labelled arrow drawn from a single point, with the arrow's length matching the force's size and its direction matching the push or pull. Draw only real forces - weight (mg, always straight down), the normal contact force (perpendicular to the surface), tension, friction and air resistance - never a "force of motion," and never the forces the object exerts on other things.

The first time a Cambridge past paper told me to "draw a free-body diagram," I drew a neat box with six arrows - including one pointing forwards, proudly labelled "movement." Instant lost mark. My teacher's margin note: "there's no engine on a falling apple." A free-body diagram is not a picture of what the object is doing; it is a map of what is being done to it. Once that lands, the arrows almost draw themselves.

What a free-body diagram actually is

• One object, on its own, mentally "freed" from everything around it.
• Represent it as a dot or a small box.
• Every force acting on it is one arrow drawn from that dot.
• Arrow length shows the force's size, its direction the push or pull; label each with its cause.

What does NOT go on it: velocity, acceleration, the resultant, or any force the object exerts on something else. Those reaction forces belong on the other object's diagram - that is Newton's third law.

AP Physics is strict: object as a dot, each force its own labelled vector, and any components drawn on a separate diagram. Writing "ma" as an arrow costs you - ma is the result of the forces, not one of them.

The forces you are allowed to draw

• Weight (W = mg): always straight down, toward the Earth's centre. On a 30 degree slope it still points vertically down, not into the slope. AP and A-Level use g = 9.8 N/kg; GCSE and IGCSE usually allow g = 10 N/kg - check your data sheet.
• Normal contact force (N or R): perpendicular to the surface of contact - straight up on flat ground, but leaning on a slope. AQA wants the exact words "normal contact force," not just "reaction."
• Tension (T): along a rope or cable, always pulling away from the object.
• Friction (f): along the surface, opposing motion or the tendency to move; at right angles to the normal.
• Air resistance / drag: opposite to the motion through the air, growing with speed. On a falling object it points up.

If a force has no source - no rope, surface, field or hand touching it - it does not go on the diagram.

Resultant force is the payoff

Once the arrows are drawn, the resultant (net) force is just their vector sum. Balanced forces give a resultant of zero - constant velocity or rest (Newton's first law). Unbalanced forces give a non-zero resultant, and the object accelerates in that direction, following F = ma (Newton's second law). The diagram is the setup; the resultant is where the marks are cashed in.

Worked example 1: a box on a slope

A box of mass 4 kg sits on a slope at 30 degrees to the horizontal. Friction acts up the slope with a size of 12 N. Take g = 10 N/kg. Find the acceleration.

1. Isolate and draw. One box. Three arrows from its centre: weight straight down, normal perpendicular to the slope, friction up along the slope.
2. Find the weight. W = mg = 4 x 10 = 40 N, straight down.
3. Choose smart axes. Point them along the slope and perpendicular to it - the slope trick that keeps the maths clean.
4. Resolve the weight. Along the slope (down-slope): W sin 30 = 40 x 0.5 = 20 N. Perpendicular: W cos 30 = 40 x 0.866 = 34.6 N.
5. Perpendicular direction. Nothing accelerates into the slope, so the normal force balances the perpendicular part of the weight: N = 34.6 N.
6. Along the slope. Down-slope pull is 20 N, friction up is 12 N. Resultant = 20 - 12 = 8 N down the slope.
7. Apply F = ma. a = F/m = 8 / 4 = 2 m/s^2, directed down the slope.

Notice that the normal force (34.6 N) is not equal to the weight (40 N). On a slope it never is - and that is the single mark most people drop.

Worked example 2: a lift accelerating

You have a mass of 60 kg and stand on bathroom scales inside a lift. The lift accelerates upward at 2 m/s^2. What do the scales read? Take g = 10 N/kg.

1. Isolate yourself. Two forces only: your weight down, and the normal force R pushing up from the scales. The scales read R.
2. Find the weight. W = mg = 60 x 10 = 600 N, down.
3. Set up F = ma. Take up as positive: R - W = ma.
4. Solve. R = m(g + a) = 60 x (10 + 2) = 720 N. The scales read 720 N - you feel heavier.
5. Flip it. If the lift accelerated downward at 2 m/s^2 instead: R = m(g - a) = 60 x (10 - 2) = 480 N - same weight, lighter reading. In free fall (a = 10 down) R = 0, which is weightlessness.

Your actual weight never changed - it stayed 600 N. The reading is the contact force, which is exactly what the free-body diagram tracks.

If any step felt fuzzy, paste the exact question into our step-by-step explainer and ask it to add the forces one at a time - watching them appear in order is what made resolving on a slope click for me.

The mistakes that cost marks

• The phantom driving force. A ball thrown up, a car coasting in neutral, a puck sliding on ice - none gets a forward arrow. Once the hand or engine stops touching it, that push is gone.
• Making the normal equal the weight out of habit. True only on flat ground with nothing else acting vertically - not on a slope, not in an accelerating lift.
• Arrows that do not touch the object, or are unlabelled. Examiners want each arrow starting at the body with a word saying what it is; an unlabelled arrow often scores nothing.
• Drawing the Newton's third-law partner. The ground pushing up on the box goes on the box's diagram; the box pushing down on the ground goes on the ground's, not the box's.
• Confusing mass and weight. Mass is in kilograms; weight is a force in newtons (W = mg).

Test yourself

1. A skydiver falls at constant terminal velocity. Draw the forces. What is the resultant?
2. A 2 kg book rests on a flat table. Give the size of the normal contact force. (g = 10 N/kg)
3. A 5 kg box is pulled along flat ground by a horizontal rope with tension 30 N; friction is 18 N. Find the acceleration.

Quick answers:

1. Two arrows only - weight down and air resistance (drag) up, equal in length. Resultant = 0, because at constant velocity the forces are balanced.
2. N = weight = mg = 2 x 10 = 20 N, upward.
3. Resultant = 30 - 18 = 12 N; a = F/m = 12 / 5 = 2.4 m/s^2 in the direction of the pull.

Want more of these, marked instantly? Generate a set with our physics quiz maker, or check your own working on any force problem with the math solver.

FAQ

What is the difference between a free-body diagram and a force diagram?

For GCSE and IGCSE they mean the same thing: the forces on one object shown as arrows. "Free-body" just stresses that you have mentally removed the object from its surroundings and drawn only what acts on it. AP always calls it an FBD.

Do the arrows have to be drawn to scale?

Not exactly to scale, but relative lengths should make sense - a bigger force gets a longer arrow, and balanced forces get equal-length arrows. Examiners do notice when a clearly larger force is drawn shorter than a smaller one.

Should I include the resultant force on the diagram?

No - a free-body diagram shows the individual forces only. Work out the resultant separately, as a calculation or a second diagram. On AP papers, adding the net force or "ma" to the FBD can lose you the point.

Why does a moving object not need a forward force?

Newton's first law: an object keeps moving at constant velocity when there is no net force. Motion does not need a force to continue - only a change in motion (speeding up, slowing, turning) does. So a coasting object has no forward arrow, just the forces actually touching or pulling it.

In short: isolate one object, draw every real force as a labelled arrow from a single point, keep the weight vertical and the normal perpendicular to the surface, and leave off anything that is not a genuine push or pull. Do that and the resultant - and the marks - follow. Stuck on a particular diagram? Ask the explainer to build it force by force.