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HKDSE Physics | Newton’s Laws and Free-Body Diagrams | F = ma, Apparent Weight and Pulleys
Students can recite F = ma, then miss a force on the diagram, confuse mass with weight, and treat an action-reaction pair as if it cancelled. This lesson does one thing only: draw every force first, then write the equation.
Objectives — Students can: 1) draw a complete free-body diagram and label where each force comes from; 2) use F = ma with friction to find acceleration; 3) separate mass from weight and compute apparent weight in a lift; 4) link both sides of a pulley with one shared acceleration.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Vectors | Adding and resolving forces | Add in the same direction, subtract in the opposite |
| Kinematics | u, v, a and s | v equals u plus at |
| Units | One newton is one kilogram metre per second squared | One N accelerates one kg at one m/s squared |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-10 min | Three laws | Inertia, F = ma, action and reaction | One everyday example each |
| 10-22 min | Free-body diagram | Weight, normal, friction, tension | All four forces on every diagram |
| 22-34 min | Basic F = ma | 3 kg pushed by 12 N | a is 4.0 m/s squared |
| 34-44 min | Adding friction | Subtract 4 N | a is 2.67 m/s squared |
| 44-56 min | Apparent weight and pulleys | Lift and Atwood machine | Apparent weight 708.6 N, a is 1.962 |
| 56-60 min | Exit check | Three questions collected and marked | Done independently |
03 Teaching script (read it out)
Scene 1: Draw every force and the equation appears
A free-body diagram holds only forces acting on one body: weight mg downwards, the normal force perpendicular to the surface, friction opposing relative motion, and tension along the string away from the body.
The three laws divide the work. The first says zero net force means constant velocity, not stillness. The second, F net equals ma, does the calculations. The third pair acts on different bodies, so it never cancels inside one diagram.
The three laws divide the work. The first says zero net force means constant velocity, not stillness. The second, F net equals ma, does the calculations. The third pair acts on different bodies, so it never cancels inside one diagram.
Ask the student: a book rests on a table. Are the book weight and the table push an action-reaction pair? No, both act on the book.
Scene 2: F = ma with friction: find the net force first
A 3 kg body pushed by 12 N with no friction accelerates at 12 divided by 3, which is 4.0 metres per second squared. With 4 N of friction the net force is 8 N, so a is 8 divided by 3, which is 2.67.
Three fixed steps: choose a positive direction, write the net force adding and subtracting by direction, then divide by mass. A net force of zero means the body is at rest or moving at constant velocity, and the student should be able to say that.
Three fixed steps: choose a positive direction, write the net force adding and subtracting by direction, then divide by mass. A net force of zero means the body is at rest or moving at constant velocity, and the student should be able to say that.
Ask the student to compute: if friction rises to 12 N, what is a? Zero.
Scene 3: Apparent weight and pulleys: the two examined settings
Apparent weight in a lift is the normal force. For a 60 kg person with the lift accelerating up at 2 metres per second squared, N minus mg equals ma, so N is 60 times (9.81 plus 2), which is 708.6 N, more than the standing weight of 589 N. Accelerating down reverses the sign, and in free fall N is zero.
For an Atwood machine with 3 kg and 2 kg the accelerations are equal in size and opposite, so (m two minus m one) g equals (m one plus m two) a and a is 1.962. Substituting back into the lighter side gives T equal to 23.5 N, between 19.6 and 29.4 N.
For an Atwood machine with 3 kg and 2 kg the accelerations are equal in size and opposite, so (m two minus m one) g equals (m one plus m two) a and a is 1.962. Substituting back into the lighter side gives T equal to 23.5 N, between 19.6 and 29.4 N.
Reminder: the commonest pulley error is using different accelerations; one string means one a.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculate | 3 kg pushed by 12 N, no friction | a equals F over m | 4.0 metres per second squared |
| Calculate | The same with 4 N of friction | (F minus f) over m | 2.67 metres per second squared |
| Calculate | 60 kg lift accelerating up at 2 | N equals m(g plus a) | 708.6 N |
| Calculate | Pulley with 3 kg and 2 kg | (m two minus m one) g over the total mass | a is 1.962 |
Sanity check (teach this): Acceleration is in metres per second squared, so an answer in newtons means the mass was never divided out. The frictionless value must exceed the frictional one (4.0 above 2.67). A lift accelerating upwards shows more than body weight (709 above 589). Pulley tension must sit between the two weights (19.6 to 29.4 N).
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Acceleration of 3 kg under 12 N with no friction | 4.0 metres per second squared | a equals F over m |
| The same with 4 N of friction | 2.67 metres per second squared | (F minus f) over m |
| Apparent weight of 60 kg accelerating up at 2 | 708.6 N | N equals m(g plus a) |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Treating weight as mass | Writes 60 kg as a force | Kilograms are mass; multiply by 9.81 for newtons. |
| Leaving the normal force off the diagram | Draws only weight | Any contact surface produces a normal force. |
| Cancelling an action-reaction pair | Says the book pushes the table so it cannot fall | The pair acts on two bodies, so it never cancels. |
| Calling any downward lift weightless | Sets N to zero whenever the lift moves down | Look at the acceleration, not the direction. |
07 Homework (paper, section and question numbers)
- Secondary Physics chapter index — All chapters
- Prep guide master index — Other chapters in this subject
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Acceleration of 3 kg under 12 N | 4.0 metres per second squared | All three right: move to momentum |
| Acceleration with 4 N of friction | 2.67 metres per second squared | One wrong: reteach net force |
| Apparent weight of 60 kg at 2 upwards | 708.6 N | Two wrong: restart scene three |
09 Tutor FAQ
What is the difference between mass and weight?
Mass is the amount of matter in kilograms and does not change; weight is the gravitational force in newtons and depends on g.
Why do action and reaction never cancel?
They act on two different bodies, so only one of them belongs on any single free-body diagram.
Does a lift moving down always mean weightlessness?
No. With downward acceleration below g the apparent weight is smaller but not zero, and only free fall gives zero.
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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.
