The shape of the lesson
By the end of the lesson, learners can
- Two parts of one contact force: When two rough surfaces touch, the contact force has a component perpendicular to the surface (the normal reaction \(R\)) and a component along it (friction \(F\)).
- Limiting equilibrium on the level: In limiting equilibrium you can replace \(F\) by \(\mu R\).
- Rough slopes: On a slope, resolve parallel and perpendicular to the plane.
- Diagram: weight, normal reaction, friction and any applied forces.
- Perpendicular to the surface: find \(R\), including components of applied forces.
- Direction of friction: opposite to the way the particle moves or is about to move.
- Limiting? If moving or about to slip, \(F = \mu R\); otherwise \(F \le \mu R\).
- Parallel to the surface: resolve and solve.
One picture every idea returns to
A rough contact gives a normal reaction \(R\) and a friction force \(F\), with \(F \le \mu R\). While the particle stays at rest, \(F\) is only as large as needed. In limiting equilibrium ("about to slip") \(F = \mu R\), acting against the direction the particle is about to move.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Only as much as it needs.. Friction is lazy: it supplies just enough force to stop slipping, up to a maximum of \(\mu R\). Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Two parts of one contact force | When two rough surfaces touch, the contact force has a component perpendicular to the surface (the normal reaction \(R\)) and a component along it (friction \(F\)). Model the worked example Does it move? on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Limiting equilibrium on the level | In limiting equilibrium you can replace \(F\) by \(\mu R\). Model the worked example Pulled at an angle on the board, one line at a time. Your turn: A block of mass 6 kg on rough horizontal ground is pushed by a force of 40 N acting at \(30^\circ\) BELOW the horizontal. The block is about to move. Find \(\mu\). | Pairs |
| 21 to 30 min | Teach and check: Rough slopes | On a slope, resolve parallel and perpendicular to the plane. Model the worked example Will it slide? on the board, one line at a time. Your turn: A particle rests on a rough plane inclined at angle \(\alpha\), with no other forces acting, and is about to slip. Show that \(\mu = \tan\alpha\), and find \(\mu\) when \(\alpha = 25^\circ\). | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 24 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Rough slopes, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Never write \(F = \mu R\) automatically. | It is only true when the particle is moving or about to slip. A box at rest with a small push has friction equal to the push. |
| Using \(R = mg\) when the applied force has a vertical component. | For the 5 kg block this gives \(\mu = 27.19 \div 50 = 0.544\), a very common wrong answer. |
| Friction changes direction between "about to slip down" and "about to slip up". | Using the same sign for both gives one answer twice instead of a range. |
Friction and limiting equilibrium: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
A crate of mass 12 kg rests on rough horizontal ground, \(\mu = 0.35\). A horizontal force of 30 N acts on it. Determine whether the crate moves, and state the friction force.
A particle of mass 0.8 kg is about to slip down a rough plane inclined at \(28^\circ\) to the horizontal. No other force acts. Find \(\mu\).
A block of mass 6 kg rests on rough horizontal ground. A force of 40 N pushes it at \(30^\circ\) below the horizontal, and the block is on the point of moving. Find the coefficient of friction.
A particle of mass 5 kg rests on a rough plane inclined at \(30^\circ\), \(\mu = 0.4\). A force of \(P\) N acts on it, parallel to a line of greatest slope and up the plane. Find the set of values of \(P\) for which the particle stays at rest.
A block of mass 4 kg on rough horizontal ground is about to move under a horizontal force of 18 N. Find \(\mu\), the magnitude of the total contact force, and the angle the contact force makes with the vertical.
A sledge of mass 15 kg rests on a slope inclined at \(10^\circ\) to the horizontal. The coefficient of friction between sledge and snow is 0.1. A rope pulls the sledge with tension \(T\) N at \(20^\circ\) above the line of greatest slope, up the slope. The sledge is about to move up the slope. Find \(T\).
Answer key and marking notes
- \(R = 120\) N, so maximum friction \(= 0.35 \times 120 = 42\) N.
- \(30 < 42\), so the crate stays at rest.
- Friction \(= 30\) N, opposing the applied force.
- Limiting with no other forces: \(mg\sin 28^\circ = \mu mg\cos 28^\circ\).
- \(\mu = \tan 28^\circ = 0.532\). (The mass does not matter.)
- Vertically: \(R = 60 + 40\sin 30^\circ = 80\) N.
- Horizontally: \(F = 40\cos 30^\circ = 34.64\) N.
- \(\mu = \dfrac{34.64}{80} = 0.433\).
- \(R = 50\cos 30^\circ = 43.30\) N, so \(\mu R = 17.32\) N. Weight component down the plane \(= 50\sin 30^\circ = 25\) N.
- About to slip down (friction up): \(P = 25 - 17.32 = 7.68\).
- About to slip up (friction down): \(P = 25 + 17.32 = 42.3\).
- \(7.68 \le P \le 42.3\).
- \(R = 40\) N and \(F = 18\) N, so \(\mu = 0.45\).
- Contact force \(= \sqrt{18^2 + 40^2} = 43.9\) N.
- Angle with the vertical \(= \tan^{-1}\dfrac{18}{40} = 24.2^\circ\) (in limiting equilibrium this angle is \(\tan^{-1}\mu\)).
- Perpendicular: \(R = 150\cos 10^\circ - T\sin 20^\circ\).
- Parallel (friction down the slope): \(T\cos 20^\circ = 150\sin 10^\circ + 0.1R\).
- \(T\cos 20^\circ = 150\sin 10^\circ + 15\cos 10^\circ - 0.1T\sin 20^\circ\).
- \(T(\cos 20^\circ + 0.1\sin 20^\circ) = 26.05 + 14.77\), so \(0.9739T = 40.82\).
- \(T = 41.9\) N (3 s.f.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A box of mass 10 kg rests on rough horizontal ground. The coefficient of friction is 0.35. A horizontal force of 25 N is applied to the box. What happens, and what is the friction force?
- AIt stays at rest; friction is 25 N
- BIt moves; friction is 35 N
- CIt stays at rest; friction is 35 N
- DIt moves; friction is 25 N
For a particle on a rough surface with normal reaction \(R\) and coefficient of friction \(\mu\), when is the friction force exactly equal to \(\mu R\)?
- AOnly when the particle is sliding or on the point of slipping
- BAlways, whenever the surface is rough
- COnly when the particle is at rest
- DOnly when the surface is horizontal
A block of mass 6 kg rests on rough horizontal ground and is on the point of slipping when a horizontal force of 21 N is applied. Find the magnitude of the total contact force from the ground on the block.
- A63.6 N
- B81 N
- C60 N
- D56.2 N
A particle of mass 2 kg rests on a rough plane inclined at \(25^\circ\) to the horizontal. The coefficient of friction is 0.6. Find the friction force acting on the particle.
- A8.45 N up the plane
- B10.9 N up the plane
- C8.45 N down the plane
- D18.1 N up the plane
A block of mass 4 kg rests on rough horizontal ground. A force of 20 N acting at \(30^\circ\) above the horizontal is applied, as shown, and the block is on the point of slipping. Find the coefficient of friction.
- A0.577
- B0.433
- C0.346
- D0.667
A particle of mass 5 kg is on a rough plane inclined at \(30^\circ\) to the horizontal, with coefficient of friction 0.3. A force of \(P\) N acting up the plane along a line of greatest slope keeps it in equilibrium, as shown. Find the range of possible values of \(P\).
- A\(12.0 \le P \le 38.0\)
- B\(10.0 \le P \le 40.0\)
- C\(12.0 \le P \le 25.0\)
- D\(35.8 \le P \le 50.8\)
A particle of mass 3 kg is on a rough plane inclined at \(20^\circ\) to the horizontal, with coefficient of friction 0.4. A horizontal force of \(P\) N pushes it towards the plane, as shown, and the particle is on the point of sliding up the plane. Find \(P\).
- A26.8
- B22.9
- C20.0
- D10.9
A particle rests on a rough plane. The angle of the plane is slowly increased, and the particle starts to slide when the angle reaches \(31^\circ\). Find the coefficient of friction between the particle and the plane.
- A0.601
- B0.515
- C0.857
- D1.66
Quiz key
\(R = 100\) N, so the greatest possible friction is \(0.35 \times 100 = 35\) N. The push of 25 N is less than this, so the box stays at rest and friction is only as big as needed: 25 N. The tempting "stays at rest, 35 N" uses \(F = \mu R\) automatically; 35 N of friction against a 25 N push would give a resultant of 10 N and the box would accelerate backwards, which is impossible.
Friction can take any value up to \(\mu R\). It reaches \(\mu R\) only in limiting equilibrium (about to slip) or when the particle is sliding. The tempting "always" is the most common error in friction questions: a particle at rest under a small push has friction equal to the push, less than \(\mu R\). The surface need not be horizontal; the rule is the same on a slope.
Vertically \(R = 60\) N; horizontally friction \(F = 21\) N (it balances the push). The total contact force is \(\sqrt{60^2 + 21^2} = 63.6\) N. The tempting 60 N is only the normal reaction, which is one component of the contact force. 81 N adds perpendicular components as if parallel, and 56.2 N subtracts the squares.
The weight component down the plane is \(20\sin 25^\circ = 8.45\) N. The greatest friction is \(0.6 \times 20\cos 25^\circ = 10.9\) N, which is more than 8.45 N, so the particle stays at rest and friction is 8.45 N up the plane. The tempting 10.9 N assumes limiting equilibrium; it would leave a resultant of 2.4 N UP the plane. Friction acts up the plane because the particle tends to slide down.
Vertically: \(R + 20\sin 30^\circ = 40\), so \(R = 30\) N. Horizontally: \(F = 20\cos 30^\circ = 17.32\) N. Limiting, so \(\mu = \dfrac{17.32}{30} = 0.577\). The tempting 0.433 uses \(R = 40\), ignoring the lift from the pull. 0.346 adds the vertical component to the weight (as if the force pushed down), and 0.667 uses the whole 20 N as the friction.
\(R = 50\cos 30^\circ = 43.30\) N, so the greatest friction is \(0.3 \times 43.30 = 12.99\) N, and the weight component down the plane is \(50\sin 30^\circ = 25\) N. Least \(P = 25 - 12.99 = 12.0\); greatest \(P = 25 + 12.99 = 38.0\). The tempting \(10.0 \le P \le 40.0\) uses friction \(\mu mg = 15\) N, forgetting that \(R = mg\cos 30^\circ\) on a slope. \(12.0 \le P \le 25.0\) misses that friction reverses direction, and \(35.8 \le P \le 50.8\) swaps sin and cos.
Perpendicular: \(R = 30\cos 20^\circ + P\sin 20^\circ\). Along the plane: \(P\cos 20^\circ = 30\sin 20^\circ + 0.4R\). Substituting, \(P(\cos 20^\circ - 0.4\sin 20^\circ) = 30\sin 20^\circ + 12\cos 20^\circ\), so \(P = \dfrac{21.54}{0.8029} = 26.8\). The tempting 22.9 uses \(R = 30\cos 20^\circ\), missing the push into the plane. 20.0 subtracts \(P\sin 20^\circ\) in \(R\) (as if \(P\) pulled away from the plane), and 10.9 ignores friction altogether.
In limiting equilibrium, \(mg\sin 31^\circ = \mu mg\cos 31^\circ\). The mass cancels, so \(\mu = \tan 31^\circ = 0.601\). The tempting 0.515 is \(\sin 31^\circ\), which comes from setting friction equal to \(\mu mg\) instead of \(\mu mg\cos\alpha\). 0.857 is \(\cos 31^\circ\), and 1.66 is \(\dfrac{1}{\tan 31^\circ}\), the ratio inverted.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.