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DSE Physics: forces and motion, free-body diagrams and F = ma (tutor lesson plan)
Give a student two or three forces and the answer comes out right or wrong by luck. This lesson drills one habit: draw the free-body diagram first, then write one equation.
Objectives — Students can: (1) draw a free-body diagram showing only the forces acting on that one body; (2) find the resultant, decide its direction and use F = ma to get the acceleration; (3) explain in a two-body pulley problem why the accelerations have the same magnitude but opposite directions.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Vector addition | How do you add two forces at right angles? | Square root of the sum of squares, direction from tan |
| Equations of motion | Displacement from rest under constant acceleration? | s = ut + ½at², so s = ½at² when u is zero |
| Unit conversion | How many grams in a kilogram? | 1000 grams |
02 Lesson flow (45 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-5 min | Answer one item cold | Student answers with no diagram | Wrong or unsure |
| 5-15 min | Three steps of the diagram | Isolate, mark external forces, fix a positive direction | Three forces plus an arrow |
| 15-25 min | Resultant then equation | 6 N pull, 4 N friction, 2 kg | a = 1.0 m/s² |
| 25-38 min | Connected bodies | Two masses, one string, one acceleration | T = 23.5 N |
| 38-45 min | Exit check | Mark live and correct live | Three out of three |
03 Teaching script (read it out)
Scene 1: Draw first, calculate later: isolate the body
The commonest error is putting the forces the body exerts on other things into the same diagram as the forces acting on it. A free-body diagram contains only the second kind.
Three fixed steps: circle the body in dashes; mark only external forces, weight downwards, normal force perpendicular to the surface, friction opposing the relative sliding, tension along the string towards the other end; then draw one arrow for the positive direction and keep it for every equation.
The floor, the wall and the hand in the story are never drawn, only the forces they give the body. No diagram, no calculation.
Three fixed steps: circle the body in dashes; mark only external forces, weight downwards, normal force perpendicular to the surface, friction opposing the relative sliding, tension along the string towards the other end; then draw one arrow for the positive direction and keep it for every equation.
The floor, the wall and the hand in the story are never drawn, only the forces they give the body. No diagram, no calculation.
Ask the student: if you push a desk, does the free-body diagram of the desk show both your push and the desk pushing back?
Scene 2: The resultant is not the sum: same direction adds, opposite direction subtracts
A box of 2 kg is pulled with 6 N while friction gives 4 N in the opposite direction. The resultant is 2 N and the acceleration is 1.0 m/s².
Before every equation the student must say one sentence: which way the resultant points and how big it is. If that sentence comes out, the equation almost never goes wrong; if it does not, the diagram is not finished.
From rest, in 3 seconds the box travels 4.5 m using s = ½at². Most of the marks sit after the resultant, and one wrong sign at the start loses all of them, so check the direction before substituting numbers.
Before every equation the student must say one sentence: which way the resultant points and how big it is. If that sentence comes out, the equation almost never goes wrong; if it does not, the diagram is not finished.
From rest, in 3 seconds the box travels 4.5 m using s = ½at². Most of the marks sit after the resultant, and one wrong sign at the start loses all of them, so check the direction before substituting numbers.
Ask the student: when does friction act in the same direction as the motion?
Scene 3: Connected bodies: one acceleration, and tension is not weight
With 3 kg on a smooth table joined over a smooth pulley to 2 kg hanging, a = (m₂ − m₁)g/(m₁ + m₂) = 1.96 m/s² and T = m₁(g + a) = 23.5 N.
Three points must be said out loud: the string cannot stretch, so both masses share one acceleration magnitude; the tension is the same throughout one light string; and the tension never equals either weight while the system accelerates.
Put two numbers side by side: the hanging weight is 19.6 N but the tension is only 23.5 N. Seeing those differ is what kills the instinct that tension equals weight.
Three points must be said out loud: the string cannot stretch, so both masses share one acceleration magnitude; the tension is the same throughout one light string; and the tension never equals either weight while the system accelerates.
Put two numbers side by side: the hanging weight is 19.6 N but the tension is only 23.5 N. Seeing those differ is what kills the instinct that tension equals weight.
Ask the student: why must the two accelerations have the same magnitude?
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Free-body diagram | Draw every external force on the body | Isolate, mark external forces, fix a direction | Three forces plus one arrow |
| Find the acceleration | Forces and mass given | Resultant first, divide by mass, state the unit | 2 N over 2 kg gives 1.0 m/s² |
| Find the tension | Tension in a connected system | Whole system for a, then isolate one body for T | T = 23.5 N |
| Direction of friction | Decide which way friction acts | Look at relative sliding, not absolute speed | Down the slope when sliding up |
Sanity check (teach this): Check three numbers: resultant 2 N, acceleration 1.0 m/s², tension 23.5 N. The sanity test is one line: acceleration times total mass must equal driving force minus resistance, 1.96 × 5 = 9.8 N against (m₂ − m₁)g = 9.8 N. If they differ, one force is missing from the equation.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| 2 kg box, 6 N pull, 4 N friction: find the acceleration. | 1.0 m/s² | Whether the resultant was written before the division |
| Same box from rest, displacement after 3 seconds. | 4.5 m | Whether ½at² was used rather than at² |
| 3 kg hanging, 2 kg on the table: find a and T. | a = 1.96 m/s², T = 23.5 N | Whether the whole system came first and the single body second |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Action and reaction in one diagram | The diagram shows the hand pushing and the box pushing back | One diagram, one body. You are not drawing the whole room. |
| Friction forgotten | Writing 6 over 2 to get 3.0 m/s² | There is contact and sliding, so ask where the friction went. |
| Tension treated as weight | Writing T = 19.6 N | An accelerating body never has tension equal to its weight. Two numbers, two steps. |
| Units not converted | Substituting 2000 grams into F = ma | The formula wants kilograms, so convert at the start. |
07 Homework (paper, section and question numbers)
- DSE Physics mechanics practice paper — Questions 1 to 3: draw the free-body diagram and mark the positive direction before writing any equation; answers without a diagram do not count.
- Same paper, connected bodies — Question 5: write the whole-system equation and the isolated equation, and explain why the accelerations share a magnitude, in 12 minutes.
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Which forces belong in a free-body diagram | Only external forces acting on that body | Answering all forces means reteach the first script |
| 6 N pull, 4 N friction, 2 kg: acceleration? | 1.0 m/s² | Answering 3.0 means friction was dropped |
| Can tension equal weight | Only at rest or at constant velocity | Answering always means reteach the third script |
09 Tutor FAQ
Must the three laws be memorised word for word?
No. Students need to know what each one governs: zero resultant, non-zero resultant, and a pair of forces on two bodies.
Does friction always oppose motion?
No. Walking and a car pulling away rely on forward friction; the direction follows the relative sliding.
How detailed should the diagram be?
Three forces and one direction arrow. Its job is to stop missing forces and wrong signs, not to look neat.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
