Free-body diagrams, explained with examples
A free-body diagram shows every force acting on one object. It is the first step in almost every mechanics problem, and AP Physics free-response questions often award points for the diagram itself.
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What a free-body diagram is
A free-body diagram (FBD) isolates a single object and shows all the forces acting on it, as arrows starting from the object. The object is usually drawn as a dot or a simple box. Each arrow points in the direction of the force, and its length should roughly show the force's relative size. Forces the object exerts on other things do not belong on its diagram.
The forces you will meet
- Gravitational force (weight), Fg = mg: always straight down, towards the centre of the Earth.
- Normal force, FN: from a surface, perpendicular to the surface, pushing away from it.
- Friction, f: parallel to the surface, opposing sliding (or the tendency to slide).
- Tension, T: along a rope or string, pulling away from the object.
- Spring force: along the spring, towards its natural length.
- Applied force: a push or pull from a person or machine.
- Air resistance or drag: opposite to the velocity.
Every force must have a source: something touching the object, or gravity. If you cannot name what is exerting a force, it is not there.
Forces that do not exist
Two invented forces cost students many points. There is no “force of motion” that keeps a moving object going; by Newton's first law, no force is needed to keep moving at constant velocity. And there is no separate “centripetal force” arrow; centripetal force is the net inward force provided by the real forces already on the diagram. Similarly, “ma” is not a force and never goes on a free-body diagram.
A method that always works
- Choose the object and draw it as a dot.
- Draw weight first, straight down.
- Look for every point of contact: each surface gives a normal force and possibly friction; each rope gives tension.
- Add any applied forces and drag.
- Label each arrow clearly, and make lengths sensible: balanced forces should be equal lengths.
- Choose axes, ideally with one axis along the direction of acceleration.
- Write Newton's second law, ΣF = ma, separately for each axis.
Worked examples
Example 1: a box pulled across a floor
A 10 kg box is pulled to the right by a 60 N horizontal rope across a floor with a 20 N friction force. Draw the FBD and find the acceleration. (g = 9.8 m/s²)
- Forces: weight 98 N down, normal 98 N up, tension 60 N right, friction 20 N left.
- Vertical: FN − mg = 0. Horizontal: 60 − 20 = 10a.
- a = 4.0 m/s² to the right.
Example 2: a hanging mass
A 3.0 kg lamp hangs at rest from a single cord. Find the tension.
- Forces: tension up, weight down. At rest, so they balance.
- T = mg = 29.4 N.
Example 3: a lift
A 70 kg person stands in a lift accelerating upwards at 2.0 m/s². Find the normal force from the floor.
- Forces: normal up, weight down. Net force upwards because the acceleration is upwards.
- FN − 686 = 70 × 2.0, so FN = 826 N. A scale would read more than the person's weight.
Inclined planes
On a slope, tilt your axes so that x runs along the slope and y is perpendicular to it. Then the normal force and friction lie along the axes, and only weight needs to be split into components:
- along the slope (downhill): mg sinθ
- perpendicular to the slope (into it): mg cosθ
A quick check: when θ = 0 (a flat surface), the along-slope part is zero and the perpendicular part is the full weight, as it should be.
Example 4: a block on a frictionless slope
A 5.0 kg block slides down a frictionless 30° slope. Find its acceleration and the normal force.
- Along the slope: mg sin 30° = ma, so a = 9.8 × 0.5 = 4.9 m/s².
- Perpendicular: FN = mg cos 30° = 5.0 × 9.8 × 0.866 ≈ 42 N, less than the weight.
Example 5: a block held still by friction
A 2.0 kg block rests on a 20° slope. Find the friction force.
- At rest, friction balances the downhill component of weight.
- f = mg sin 20° = 2.0 × 9.8 × 0.342 ≈ 6.7 N up the slope.
Connected objects
For two objects joined by a string, such as a block on a table connected over a pulley to a hanging mass, draw a separate FBD for each object. The tension is the same throughout a light string over a frictionless pulley. Write ΣF = ma for each object and solve the two equations together. Alternatively, treat both objects as one system to find the acceleration quickly, then use one object's FBD to find the tension.
Example 6: a table and a hanging mass
A 4.0 kg block on a frictionless table is connected over a pulley to a hanging 1.0 kg mass. Find the acceleration and the tension.
- System: the only unbalanced force is the hanging weight, 9.8 N, acting on 5.0 kg in total. a = 9.8/5.0 = 1.96 m/s².
- Block on the table: T = 4.0 × 1.96 ≈ 7.8 N, less than the hanging weight, as it must be for the mass to accelerate downwards.
Common mistakes
- Adding a “force of motion” or “ma” arrow.
- Drawing forces the object exerts on others.
- Assuming the normal force always equals mg; on slopes and in lifts it does not.
- Drawing weight perpendicular to a slope instead of straight down.
- Making balanced forces visibly different lengths.
Practice questions
Try each one before opening the answer.
Q1. A book rests on a table. What forces belong on its FBD?
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Q2. A skydiver falls at terminal velocity. Compare the forces.
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Q3. Find the normal force on a 10 kg block on a 40° slope.
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Q4. A 2 kg ball is thrown upwards. Ignoring air resistance, what forces act on it while it rises?
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Frequently asked questions
Are free-body diagrams on the AP Physics exam?
Yes. AP Physics 1 and C free-response questions often ask you to draw an FBD and then use it, and points are given for correct, labelled forces with sensible relative lengths.
Should I draw components on the FBD?
Draw the actual forces on the diagram. If you need components, show them separately or as dashed arrows, so it is clear which arrows are real forces.
Key takeaways
- Show only forces acting on the chosen object.
- Every force needs a source: contact or gravity.
- No force of motion, no centripetal arrow, no ma arrow.
- Tilt axes on slopes: mg sinθ along, mg cosθ into the slope.
- Write ΣF = ma for each axis, and for each object.