The shape of the lesson
By the end of the lesson, learners can
- Zero resultant, two equations: In the lab, set the hanging weight to 40 N and the string angles to \(30^\circ\) and \(50^\circ\).
- Smooth slopes: A contact is "smooth" when we model it as having no friction: the only contact force is the normal reaction, perpendicular to the surface.
- Finding an unknown force: When the size and direction of a force are both unknown, resolve to find its two components, then rebuild it with Pythagoras and inverse tan.
- Diagram: every force, weight \(mg\) with \(g = 10\).
- Directions: horizontal/vertical for strings; parallel/perpendicular for slopes.
- Equations: sum of components \(= 0\) in each direction.
- Solve: substitute, or square-and-add and divide for a force of unknown size and direction.
- Check: steeper strings carry more tension; answers to 3 s.f., angles to 1 d.p.
One picture every idea returns to
A particle is in equilibrium when the vector sum of the forces on it is zero. Equivalently, the sum of the components in ANY direction is zero, so resolving in two perpendicular directions gives two equations for two unknowns.
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, Everything cancels.. A particle in equilibrium has zero resultant force, so the components in any direction add up to zero. 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: Zero resultant, two equations | In the lab, set the hanging weight to 40 N and the string angles to \(30^\circ\) and \(50^\circ\). Project tab 2 of the lab (Forces lab, equilibrium of three forces) and let a student drive it while the class predicts. Model the worked example A weight on two strings on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Smooth slopes | A contact is "smooth" when we model it as having no friction: the only contact force is the normal reaction, perpendicular to the surface. Model the worked example Held by a horizontal force on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Finding an unknown force | When the size and direction of a force are both unknown, resolve to find its two components, then rebuild it with Pythagoras and inverse tan. Model the worked example Magnitude and direction on the board, one line at a time. Your turn: A particle of mass 3 kg hangs on a light string. A horizontal force of 12 N holds it in equilibrium with the string inclined to the vertical. Find the tension and the angle of the string to the vertical. | Connect-Extend-Challenge |
| 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, 26 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 Finding an unknown force, 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.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
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 |
|---|---|
| Keep at least 4 significant figures in the middle of the working. | Rounding \(T_2\) to 35 before finding \(T_1\) gives 26.0 instead of 26.1 and can cost the accuracy mark. |
| The normal reaction is NOT always \(mg\cos\theta\). | Any other force with a component perpendicular to the plane changes it. Always write the full perpendicular equation. |
| Dividing the two equations gives \(\tan\theta\) only if you put the right one on top: \(\dfrac{P\sin\theta}{P\cos\theta} = \tan\theta\). | Check which component used sin before you divide. |
Equilibrium of a particle: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
A particle of mass 5 kg hangs on a light string. A horizontal force of 20 N holds it in equilibrium. Find the tension in the string and the angle the string makes with the vertical.
Forces of 7 N and 24 N act on a particle at right angles to each other. A third force keeps the particle in equilibrium. Find its magnitude and the angle it makes with the 7 N force.
A particle of weight 30 N hangs in equilibrium from two light strings inclined at \(40^\circ\) and \(60^\circ\) to the horizontal. Find the tension in each string.
A particle of mass 4 kg rests on a smooth plane inclined at \(30^\circ\) to the horizontal. It is held in equilibrium by a force of 25 N acting at an angle \(\alpha\) above a line of greatest slope, up the plane. Find \(\alpha\) and the normal reaction.
A particle of mass 3 kg rests on a smooth plane inclined at \(30^\circ\) to the horizontal, held by a horizontal force of \(P\) N.
(a) Find \(P\) and the normal reaction, giving exact answers.
(b) State one limitation of modelling the plane as smooth.
A lamp of mass 2.4 kg hangs from a point on two light wires fixed to two walls. One wire makes \(15^\circ\) with the horizontal and the other makes \(25^\circ\) with the horizontal. Model the lamp as a particle and find the tension in each wire.
Answer key and marking notes
- \(T\cos\alpha = 50\), \(T\sin\alpha = 20\).
- \(T = \sqrt{50^2 + 20^2} = 53.9\) N.
- \(\alpha = \tan^{-1}\dfrac{20}{50} = 21.8^\circ\).
- The third force is equal and opposite to the resultant of the other two.
- Magnitude \(= \sqrt{7^2 + 24^2} = 25\) N.
- The resultant makes \(\tan^{-1}\dfrac{24}{7} = 73.7^\circ\) with the 7 N force, so the third force makes \(180^\circ - 73.7^\circ = 106.3^\circ\).
- Horizontally: \(T_1\cos 40^\circ = T_2\cos 60^\circ\), so \(T_2 = 1.532\,T_1\).
- Vertically: \(T_1\sin 40^\circ + T_2\sin 60^\circ = 30\).
- \(0.6428\,T_1 + 1.3268\,T_1 = 30\), so \(T_1 = 15.2\) N.
- \(T_2 = 1.532 \times 15.23 = 23.3\) N.
- Parallel: \(25\cos\alpha = 40\sin 30^\circ = 20\), so \(\cos\alpha = 0.8\) and \(\alpha = 36.9^\circ\).
- Then \(\sin\alpha = 0.6\).
- Perpendicular: \(R = 40\cos 30^\circ - 25 \times 0.6 = 34.64 - 15\).
- \(R = 19.6\) N.
- (a) Parallel: \(P\cos 30^\circ = 30\sin 30^\circ\), so \(P = 30\tan 30^\circ = 10\sqrt{3}\).
- Perpendicular: \(R = 30\cos 30^\circ + P\sin 30^\circ = 15\sqrt{3} + 5\sqrt{3} = 20\sqrt{3}\) N.
- (Check: horizontally and vertically, \(R\cos 30^\circ = 30\) gives \(R = 20\sqrt{3}\) too.)
- (b) A real plane has friction, so a smaller force could hold the particle; the smooth model ignores this.
- Weight \(= 24\) N.
- Horizontally: \(T_1\cos 15^\circ = T_2\cos 25^\circ\), so \(T_1 = 0.9383\,T_2\).
- Vertically: \(T_1\sin 15^\circ + T_2\sin 25^\circ = 24\).
- \(0.2428\,T_2 + 0.4226\,T_2 = 24\), so \(T_2 = 36.1\) N.
- \(T_1 = 0.9383 \times 36.07 = 33.8\) N.
- Both tensions are larger than the weight because the wires are shallow.
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 particle of weight 40 N hangs from a light string. A horizontal force of \(P\) N holds it in equilibrium with the string at \(30^\circ\) to the vertical, as shown. Find \(P\).
- A23.1
- B20
- C34.6
- D69.3
A particle is in equilibrium under the action of several forces. Which statement MUST be true?
- AThe sum of the components of the forces in any direction is zero
- BThe particle is at rest
- CThe normal reaction on the particle equals its weight
- DAll the forces acting on the particle have the same magnitude
A particle of mass 3 kg rests in equilibrium on a smooth plane inclined at \(20^\circ\) to the horizontal, held by a force of \(P\) N acting up the plane along a line of greatest slope. Find \(P\) and the normal reaction \(R\).
- A\(P = 10.3\), \(R = 28.2\)
- B\(P = 28.2\), \(R = 10.3\)
- C\(P = 10.3\), \(R = 30\)
- D\(P = 1.03\), \(R = 2.82\)
Three forces act on a particle in equilibrium. Two of them are 9 N due east and 12 N due north. Find the third force.
- A15 N at \(53.1^\circ\) south of west
- B15 N at \(53.1^\circ\) north of east
- C15 N at \(36.9^\circ\) south of west
- D21 N at \(53.1^\circ\) south of west
A particle of weight 50 N hangs in equilibrium from two light strings attached to a horizontal ceiling. The strings make angles of \(35^\circ\) and \(55^\circ\) with the horizontal, as shown. Find the tensions \(T_1\) and \(T_2\).
- A\(T_1 = 28.7\) N, \(T_2 = 41.0\) N
- B\(T_1 = 41.0\) N, \(T_2 = 28.7\) N
- C\(T_1 = T_2 = 35.9\) N
- D\(T_1 = 43.6\) N, \(T_2 = 30.5\) N
A particle of mass 4 kg rests on a smooth plane inclined at \(35^\circ\) to the horizontal. It is held in equilibrium by a horizontal force of \(P\) N, as shown. Find \(P\) and the normal reaction \(R\).
- A\(P = 28.0\), \(R = 48.8\)
- B\(P = 28.0\), \(R = 32.8\)
- C\(P = 22.9\), \(R = 32.8\)
- D\(P = 57.1\), \(R = 69.7\)
A particle of weight 25 N is in equilibrium under its weight, a horizontal force of 10 N, and a force of \(P\) N acting at \(\theta^\circ\) to the upward vertical. Find \(P\) and \(\theta\).
- A\(P = 26.9\), \(\theta = 21.8\)
- B\(P = 26.9\), \(\theta = 68.2\)
- C\(P = 22.9\), \(\theta = 23.6\)
- D\(P = 35\), \(\theta = 21.8\)
A particle of mass 3 kg rests on a smooth plane inclined at \(25^\circ\) to the horizontal. It is held in equilibrium by a force of \(P\) N acting at \(15^\circ\) above a line of greatest slope, up the plane, as shown. Find \(P\) and the normal reaction \(R\).
- A\(P = 13.1\), \(R = 23.8\)
- B\(P = 13.1\), \(R = 27.2\)
- C\(P = 13.1\), \(R = 30.6\)
- D\(P = 12.2\), \(R = 24.0\)
Quiz key
Vertically: \(T\cos 30^\circ = 40\), so \(T = 46.19\) N. Horizontally: \(P = T\sin 30^\circ = 23.1\). (In one step, \(P = 40\tan 30^\circ\).) The tempting 20 is \(40\sin 30^\circ\), which treats the weight as if it were the tension; the tension is bigger than the weight because it also balances \(P\). 34.6 is \(40\cos 30^\circ\), and 69.3 is \(40 \div \tan 30^\circ\), using the angle from the horizontal.
Equilibrium means the resultant force is zero, so the components in every direction add to zero (that is why resolving in two directions gives two equations). The tempting "at rest" is not forced: a particle moving with constant velocity is also in equilibrium. The normal reaction equals the weight only in special cases (horizontal surface, no other vertical forces), and the forces need not be equal in size.
Weight \(= 30\) N. Parallel to the plane: \(P = 30\sin 20^\circ = 10.3\). Perpendicular: \(R = 30\cos 20^\circ = 28.2\). The tempting swap (\(P = 28.2\)) uses cos for the component down the slope; a check is that a gentle slope should need only a small force. \(R = 30\) takes the reaction as the full weight, which is only true on level ground, and 1.03 and 2.82 forget to multiply the mass by \(g\).
The resultant of the first two is \(\sqrt{9^2 + 12^2} = 15\) N at \(\tan^{-1}\dfrac{12}{9} = 53.1^\circ\) north of east. The third force balances it: 15 N at \(53.1^\circ\) south of west. The tempting "north of east" answer is the resultant itself; adding it would double the resultant, not cancel it. \(36.9^\circ\) is measured from the wrong axis, and 21 N adds the sizes.
From the horizontal equation, \(T_2 = T_1\dfrac{\cos 35^\circ}{\cos 55^\circ} = 1.428T_1\). Then \(T_1(\sin 35^\circ + 1.428\sin 55^\circ) = 50\) gives \(T_1 = 28.7\) N and \(T_2 = 41.0\) N. The steeper string carries more tension. The tempting swap comes from resolving the wrong way (cos vertically, sin horizontally). Equal tensions of 35.9 N ignore the horizontal equation, and 43.6 N, 30.5 N give each string half the weight.
Along the plane: \(P\cos 35^\circ = 40\sin 35^\circ\), so \(P = 40\tan 35^\circ = 28.0\). Perpendicular: \(R = 40\cos 35^\circ + P\sin 35^\circ = 32.77 + 16.06 = 48.8\). The tempting \(R = 32.8\) uses \(mg\cos\alpha\) only and misses the horizontal force pushing the particle into the plane. 22.9 is the weight component alone (the push is not parallel to the plane), and 57.1, 69.7 use the angle the wrong way round.
Horizontally: \(P\sin\theta = 10\). Vertically: \(P\cos\theta = 25\). So \(P = \sqrt{10^2 + 25^2} = 26.9\) and \(\tan\theta = \dfrac{10}{25}\), \(\theta = 21.8\). The tempting \(68.2\) divides the equations upside down, giving the angle to the horizontal instead of the vertical. \(P = 22.9\) uses \(\sqrt{25^2 - 10^2}\) as if \(P\) were a side rather than the hypotenuse, and 35 adds the sizes.
Along the plane: \(P\cos 15^\circ = 30\sin 25^\circ = 12.68\), so \(P = 13.1\). Perpendicular: \(R + P\sin 15^\circ = 30\cos 25^\circ\), so \(R = 27.19 - 3.40 = 23.8\). The tempting \(R = 27.2\) assumes \(R = mg\cos\alpha\) and ignores the lifting component of \(P\). \(R = 30.6\) adds that component instead of subtracting it, and \(P = 12.2\) multiplies by \(\cos 15^\circ\) instead of dividing.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.