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DSE Physics | Forces and Newton Laws Tutor Prep Guide | Free Body Diagrams, Friction and Lift Readings
Students can recite F equals ma, yet freeze in front of a box on a slope or a scale inside a lift. The formula is not the problem; they never list the forces first. This lesson drills one habit: diagram first, algebra second.
Objectives — Students can: (1) draw a complete free body diagram and fix a positive direction; (2) find acceleration from the resultant force, friction included; (3) separate weight from apparent weight and compute a lift scale reading; (4) explain why an action-reaction pair cannot cancel.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Vector addition | 6 N forward against 3.9 N back | 2.1 N in the larger direction |
| Units | What is 1 N in base units | 1 kg m s-2 |
| Graph slope | What the slope of a velocity-time graph means | acceleration |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Free body diagram | only forces on that one body | draw the 2.0 kg block |
| 8-22 min | Resultant and acceleration | sum F equals ma, direction first | 3.0 m s-2 |
| 22-36 min | Friction | f equals mu N, find N first | 1.04 m s-2 |
| 36-48 min | Apparent weight | R equals m times g plus a | 709 N |
| 48-56 min | Third law | a pair acts on two bodies | write one pair |
| 56-60 min | Exit check | three questions collected | independent work |
03 Teaching script (read it out)
Scene 1: First move: diagram before algebra
A 2.0 kg block on a smooth table is pushed with 6.0 N: a = F over m = 6.0 over 2.0 = 3.0 m s-2. The diagram carries only forces: weight 19.62 N down, normal force the same up, 6.0 N to the right.
Ask the student: why does the normal force equal the weight here? (There is no vertical acceleration, so the vertical resultant is zero.) Student task: change the push to 9.0 N and recompute (4.5 m s-2).
Ask the student: why does the normal force equal the weight here? (There is no vertical acceleration, so the vertical resultant is zero.) Student task: change the push to 9.0 N and recompute (4.5 m s-2).
Scene 2: Friction: N first, then f, then the resultant
Same block on a rough table with mu equal to 0.20: f = mu N = 0.20 times 2.0 times 9.81 = 3.924 N, resultant = 6.0 minus 3.924 = 2.076 N, so a = 2.076 over 2.0 = 1.038 m s-2, about 1.04 m s-2.
The order never changes: find N, find the friction limit, write the resultant, divide by mass. Ask the student: is friction always opposite to motion? (Kinetic friction is, static friction need not be.) Student task: repeat with mu equal to 0.40 and decide whether the block moves (f would need 7.85 N, more than 6.0 N, so it stays put).
The order never changes: find N, find the friction limit, write the resultant, divide by mass. Ask the student: is friction always opposite to motion? (Kinetic friction is, static friction need not be.) Student task: repeat with mu equal to 0.40 and decide whether the block moves (f would need 7.85 N, more than 6.0 N, so it stays put).
Scene 3: Apparent weight: the scale reads the normal force
A 60 kg student stands on a scale in a lift accelerating upward at 2.0 m s-2: R minus mg equals ma, so R = m(g + a) = 60 times (9.81 plus 2.0) = 708.6 N, which is 1.20 times the true weight of 588.6 N.
Accelerating downward gives m(g minus a), and in free fall R is zero. Ask the student: what does the scale read while the lift rises at constant speed? (The weight, because a is zero.) Student task: compute the reading when a is 2.0 m s-2 downward.
Accelerating downward gives m(g minus a), and in free fall R is zero. Ask the student: what does the scale read while the lift rises at constant speed? (The weight, because a is zero.) Student task: compute the reading when a is 2.0 m s-2 downward.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Find acceleration | mass and one or more horizontal forces | diagram, resultant, divide by mass | 3.0 m s-2 |
| With friction | mu and the normal force | N then f then resultant | 1.04 m s-2 |
| Lift reading | direction and size of a | R equals m times g plus or minus a | 709 N |
| Third law pair | one force given | same size, opposite, two bodies | book on table, table on book |
Sanity check (teach this): The acceleration must point the same way as the resultant; a negative answer means the assumed direction was wrong, not that acceleration can be negative in a different sense. The friction coefficient carries no units and the friction force never exceeds mu N. Here R = 708.6 N is larger than the weight 588.6 N because the resultant points up; a value below the weight means the sign of a was flipped. Keep g consistent at 9.81 or 10 throughout, and match the significant figures of the question.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Acceleration of 2.0 kg under 6.0 N on a smooth table | 3.0 m s-2 | a equals F over m |
| Same block with mu equal to 0.20 | 1.04 m s-2 | subtract f equals mu mg |
| Scale reading for 60 kg in a lift at 2.0 m s-2 upward | 709 N | R equals m times g plus a |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Confusing weight with mass | answers m equals 6.0 over 9.81 kg | 6.0 N is a force, and the mass is already given as 2.0 kg. |
| Leaving out the normal force | writes only F minus mg equals ma | For horizontal motion the vertical resultant is zero, so N and mg must both appear. |
| Multiplying mu by the applied force | answers 0.20 times 6.0 equals 1.2 N | mu multiplies the normal force, not the push. |
| Saying the student got heavier | writes 709 N with no explanation | The scale reads a normal force; the weight is unchanged, the resultant points up so the reading rises. |
07 Homework (paper, section and question numbers)
- Secondary Physics chapter index — mechanics set, questions 2 and 5
- Prep guide home — other chapters in the subject
- DSE Physics practice papers — forces set, questions 1 to 6
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Acceleration of 2.0 kg under 6.0 N | 3.0 m s-2 | three correct means move on to friction |
| Acceleration with mu equal to 0.20 | 1.04 m s-2 | one wrong means reteach N and f |
| Direction of the scale reading when the lift speeds up | greater than the weight | two or more wrong means back to act one |
09 Tutor FAQ
Why do action and reaction not cancel?
They act on two different bodies, and only the resultant on one body sets its acceleration.
What is the difference between static and kinetic friction?
Static friction adjusts up to mu N, while kinetic friction is fixed and usually smaller once sliding starts.
Why is the normal force on a slope not mg?
Weight splits into a component along the slope and one perpendicular to it, and N equals mg cos theta.
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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.
