Drawing a useful free-body diagram
You will be able to: Draw and label only the forces acting on one chosen object.
Which arrows belong on the object?
A book rests on a desk. Earth pulls it down. The desk pushes it up. Two interactions explain why it stays at rest; no extra “rest force” is needed.
A useful starting point: Choosing a system →
Words and symbols before equations
- Force
- An interaction: name what exerts it and what receives it.
- Weight mg
- Earth’s gravitational pull, in N. Here g=10 N/kg.
- Normal force N
- A surface’s perpendicular push; normal means perpendicular, not ordinary.
- Tension T
- A pull along a taut rope.
What this picture assumes
Horizontal net-force investigation: right is positive. Resistance is a prescribed force, not a static-friction solver. Vertical support balances weight; g=10 m/s². Velocity is not specified.
Connect the picture to the physics
Replace the object with a dot or simple box, choose axes, then draw each force ON it from that dot. Label the source and type. Longer arrows represent larger forces when a scale is specified.
Contact forces require contact. Gravity can act without contact. Forces from separate interactions combine by vector addition: add x-components together and y-components together.
Do not draw velocity or acceleration as forces. Draw them separately if needed. Do not count a force and its components as three independent forces. The weight arrow stays vertically down even on a ramp.
A worked example, step by step
A 2 kg book is pulled 8 N right across a level desk while friction acts 3 N left. There is no vertical acceleration. Construct its force diagram and find the net force.
- System: book. External sources: Earth, desk, and pulling hand or rope.
- Weight is 2×10 = 20 N down. Vertical balance requires normal force 20 N up.
- Draw applied force 8 N right and friction 3 N left. Horizontal net force is 8−3 = 5 N right.
- The 20 N forces cancel vertically. A velocity arrow would not be included in this sum.
The normal force equals mg only when the vertical force balance and geometry require it.
Should “net force” be a fifth force arrow added to the four real forces?
Compare with an explanation
No. Net force is their vector sum, not another interaction.
Predict. Change one thing. Explain.
Keep mass at 2 kg and resistance at 3 N. Change the applied force through 3 N. Compare horizontal arrow lengths and the net-force readout. The vertical pair stays balanced.
Net horizontal force 5 N; acceleration 2.5 m/s² (right positive). Force arrows share a scale of 4.75 drawing units/N.
Horizontal net-force investigation: right is positive. Resistance is a prescribed force, not a static-friction solver. Vertical support balances weight; g=10 m/s². Velocity is not specified.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 3 kg box on a level surface is pulled 12 N right and experiences 5 N friction left, with no vertical acceleration. (a) Name its four forces, (b) calculate weight, (c) calculate normal force, and (d) give the net force.
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Compare with the answer and four-point rubric
- 1 point: Gravity, surface normal, applied pull, kinetic friction.
- 1 point: Weight 30 N down.
- 1 point: Normal 30 N up because vertical acceleration is zero.
- 1 point: Net force 7 N right; opposing vertical forces cancel.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does normal mean?
Perpendicular to the surface.
RECALL 2Is velocity a force?
No. Motion is not an additional interaction.
RECALL 3Why label the force’s source?
It prevents inventing or double-counting forces.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Drawing a useful free-body diagram
- ΣF_x = sum of signed horizontal force components.
- ΣF_y = sum of signed vertical force components.
- 1 N = 1 kg·m/s².
Remember: The normal force equals mg only when the vertical force balance and geometry require it.
Conditions: Horizontal net-force investigation: right is positive. Resistance is a prescribed force, not a static-friction solver. Vertical support balances weight; g=10 m/s². Velocity is not specified.
Refresh Kid · Unit 2 · Objectives 2.2.A, 2.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.2, objectives 2.2.A, 2.2.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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