The shape of the lesson
By the end of the lesson, learners can
- Drawing the forces: Before any calculation, draw a force diagram: the particle as a dot, every force as an arrow starting from it, each labelled.
- Resolving into components: A force is a vector: it has a size and a direction.
- Finding the resultant: The resultant is the single force with the same effect as all the forces together.
- Draw: a force diagram with every force labelled, weight as \(mg\) with \(g = 10\).
- Choose directions: horizontal/vertical, or along/perpendicular to a slope.
- Resolve: next to the angle is \(\cos\), across from it is \(\sin\).
- Add with signs: \(X\) and \(Y\), negative for left/down.
- Rebuild: \(R = \sqrt{X^2+Y^2}\), \(\tan\alpha = Y/X\), and say which way it points.
One picture every idea returns to
A force \(F\) at angle \(\theta\) to a direction has component \(F\cos\theta\) along that direction and \(F\sin\theta\) perpendicular to it. To find a resultant, add all the components in two perpendicular directions to get \(X\) and \(Y\); then \(R = \sqrt{X^2 + Y^2}\) and \(\tan\alpha = \dfrac{Y}{X}\).
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, Split it, then add.. Every Paper 4 question starts the same way: draw the forces, split each one into two perpendicular components, and add the components. 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: Drawing the forces | Before any calculation, draw a force diagram: the particle as a dot, every force as an arrow starting from it, each labelled. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Resolving into components | A force is a vector: it has a size and a direction. Model the worked example Horizontal and vertical components on the board, one line at a time. Your turn: A horizontal force of 15 N acts on a particle on a plane inclined at \(30^\circ\). Find its components parallel to and perpendicular to the plane. | Pairs |
| 21 to 30 min | Teach and check: Finding the resultant | The resultant is the single force with the same effect as all the forces together. Model the worked example Two perpendicular forces on the board, one line at a time. | Frayer Model |
| 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, 23 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 the resultant, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 |
|---|---|
| Weight and normal reaction are NOT a third-law pair: they act on the same body and are different kinds of force. | A third-law pair always acts on two different bodies and is the same type of force. |
| Do not learn "horizontal is cos". | The rule is: the component next to the angle you are given uses cos. If the angle is measured from the vertical, the vertical component is the cos one. |
| Keep the signs. | A force pointing left or down gives a NEGATIVE component; adding its size instead is the most common way to lose every mark in a resultant question. |
Forces, components and resultants: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
A force of 40 N acts at \(25^\circ\) above the horizontal. Find its horizontal and vertical components.
Forces of 9 N horizontally to the right and 12 N vertically downwards act on a particle. Find the magnitude and direction of the resultant.
Three coplanar forces act at a point: 25 N along the positive \(x\)-axis, 18 N along the positive \(y\)-axis and 30 N at \(30^\circ\) below the negative \(x\)-axis. Find the magnitude of the resultant and the angle it makes with the positive \(x\)-axis.
Forces of \(P\) N and 20 N act at a point, and the angle between their directions is \(120^\circ\). The resultant is perpendicular to the \(P\) N force. Find \(P\) and the magnitude of the resultant, giving the resultant in exact form.
A particle of weight 60 N is on a plane inclined at \(35^\circ\) to the horizontal. A force of 25 N acts on it at \(20^\circ\) above a line of greatest slope, pointing up the plane. Find the total component of these two forces (a) parallel to the plane, (b) perpendicular to the plane. State the direction of each.
A barge is pulled along a straight canal by two ropes. One rope has tension 300 N and makes \(25^\circ\) with the direction of the canal; the other has tension \(T\) N and makes \(40^\circ\) with the canal, on the other side. The resultant of the two tensions acts along the canal. Find \(T\) and the magnitude of the resultant.
Answer key and marking notes
- Horizontal: \(40\cos 25^\circ = 36.3\) N.
- Vertical: \(40\sin 25^\circ = 16.9\) N.
- \(R = \sqrt{9^2 + 12^2} = 15\) N.
- \(\tan\alpha = \dfrac{12}{9}\), so \(\alpha = 53.1^\circ\).
- The resultant is 15 N at \(53.1^\circ\) below the horizontal, to the right.
- \(X = 25 - 30\cos 30^\circ = 25 - 25.98 = -0.981\).
- \(Y = 18 - 30\sin 30^\circ = 18 - 15 = 3\).
- \(R = \sqrt{0.981^2 + 3^2} = 3.16\) N.
- \(X < 0,\ Y > 0\): the resultant points up and to the left, at \(\tan^{-1}\dfrac{3}{0.981} = 71.9^\circ\) above the negative \(x\)-axis.
- Angle with the positive \(x\)-axis \(= 180^\circ - 71.9^\circ = 108.1^\circ\).
- Resolve along the \(P\) N force: \(P + 20\cos 120^\circ = 0\), so \(P - 10 = 0\), \(P = 10\).
- Resolve perpendicular: \(R = 20\sin 120^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\) N (17.3 N).
- (a) Up the plane: \(25\cos 20^\circ - 60\sin 35^\circ = 23.49 - 34.41 = -10.9\).
- So 10.9 N down the plane.
- (b) Away from the plane: \(25\sin 20^\circ - 60\cos 35^\circ = 8.55 - 49.15 = -40.6\).
- So 40.6 N into the plane (this is what the normal reaction must balance).
- Across the canal the components cancel: \(300\sin 25^\circ = T\sin 40^\circ\).
- \(T = \dfrac{300 \times 0.4226}{0.6428} = 197\) N (3 s.f.).
- Along the canal: \(R = 300\cos 25^\circ + 197.2\cos 40^\circ = 271.9 + 151.1\).
- \(R = 423\) 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 force of 24 N acts at \(35^\circ\) to the upward vertical, as shown. Find its horizontal component.
- A13.8 N
- B19.7 N
- C29.3 N
- D41.8 N
A box rests on the floor of a lift. Which of these is a Newton's third law pair?
- AThe box pushes down on the floor; the floor pushes up on the box
- BThe weight of the box; the normal reaction of the floor on the box
- CThe weight of the box; the box's push down on the floor
- DThe tension in the lift cable; the weight of the lift
Forces of 7 N due east and 24 N due north act on a particle. Find the magnitude and direction of their resultant.
- A25 N at \(73.7^\circ\) to the 7 N force
- B31 N at \(73.7^\circ\) to the 7 N force
- C25 N at \(16.3^\circ\) to the 7 N force
- D17 N due north
A particle of mass 6 kg rests on a plane inclined at \(40^\circ\) to the horizontal. Find the component of its weight perpendicular to the plane.
- A46.0 N
- B38.6 N
- C4.60 N
- D78.3 N
Three forces act at a point O: 12 N along the positive \(x\)-axis, 8 N at \(30^\circ\) above the negative \(x\)-axis, and 10 N along the negative \(y\)-axis, as shown. Find the magnitude and direction of the resultant.
- A7.86 N at \(49.8^\circ\) below the positive \(x\)-axis
- B7.86 N at \(40.2^\circ\) below the positive \(x\)-axis
- C14.9 N at \(70.1^\circ\) above the positive \(x\)-axis
- D19.9 N at \(17.6^\circ\) below the positive \(x\)-axis
Forces of 10 N and \(P\) N act at a point, and the angle between their directions is \(120^\circ\). Their resultant is perpendicular to the 10 N force. Find \(P\) and the magnitude of the resultant.
- A\(P = 20\), resultant 17.3 N
- B\(P = 20\), resultant 22.4 N
- C\(P = 11.5\), resultant 5.77 N
- D\(P = 5\), resultant 4.33 N
Two forces of 8 N and 5 N act at a point. Their resultant has magnitude 11 N. Find the angle between the two forces.
- A\(66.4^\circ\)
- B\(113.6^\circ\)
- C\(36.9^\circ\)
- D\(23.6^\circ\)
A particle of mass 2 kg is on a smooth plane inclined at \(30^\circ\) to the horizontal. A horizontal force of 25 N pushes it towards the plane, as shown. Find the resultant of the 25 N force and the weight in the direction up the plane.
- A11.7 N up the plane
- B31.7 N up the plane
- C2.5 N up the plane
- D4.33 N up the plane
Quiz key
The horizontal direction is across from the \(35^\circ\) angle, so the horizontal component is \(24\sin 35^\circ = 13.8\) N. The tempting 19.7 N is \(24\cos 35^\circ\), which is the VERTICAL component here because the angle is measured from the vertical. 29.3 N and 41.8 N divide by cos and sin instead of multiplying; a component can never be bigger than the force itself.
The box pushing on the floor and the floor pushing on the box are both contact forces, act on different bodies, and are always equal and opposite: a third-law pair. The tempting weight and normal reaction both act on the box and are different kinds of force; they balance only when the lift is not accelerating, which is Newton's first law, not the third. Weight is a gravitational force between the box and the Earth, so its partner is the box pulling the Earth.
\(R = \sqrt{7^2 + 24^2} = \sqrt{625} = 25\) N, and \(\tan\alpha = \dfrac{24}{7}\) gives \(\alpha = 73.7^\circ\) to the 7 N (east) force. The tempting \(16.3^\circ\) is \(\tan^{-1}\dfrac{7}{24}\), the angle to the 24 N force, not the 7 N one. 31 N adds the sizes, which only works for forces in the same direction; 17 N subtracts them.
Weight \(= 6 \times 10 = 60\) N. The angle between the vertical and the normal to the plane is \(40^\circ\), so the perpendicular component is \(60\cos 40^\circ = 46.0\) N. The tempting 38.6 N is \(60\sin 40^\circ\), the component DOWN the plane. 4.60 N forgets to multiply the mass by \(g\), and 78.3 N divides by \(\cos 40^\circ\).
\(X = 12 - 8\cos 30^\circ = 5.072\) and \(Y = 8\sin 30^\circ - 10 = -6\). \(R = \sqrt{5.072^2 + 6^2} = 7.86\) N, and \(\tan\alpha = \dfrac{6}{5.072}\) gives \(\alpha = 49.8^\circ\) below the positive \(x\)-axis. The tempting \(40.2^\circ\) is \(\tan^{-1}\dfrac{5.072}{6}\), measured from the negative \(y\)-axis instead. 14.9 N adds the 10 N to the 4 N upward component instead of subtracting; 19.9 N adds the 8 N force's horizontal component instead of subtracting it.
Along the 10 N force: \(10 + P\cos 120^\circ = 0\), so \(10 - 0.5P = 0\) and \(P = 20\). The resultant is the perpendicular component: \(20\sin 120^\circ = 17.3\) N. The tempting 22.4 N is \(\sqrt{10^2 + 20^2}\), Pythagoras on two forces that are NOT perpendicular. \(P = 11.5\) resolves with sin instead of cos, and \(P = 5\) multiplies by \(\cos 60^\circ\) instead of dividing.
\(X = 8 + 5\cos\theta\), \(Y = 5\sin\theta\), so \(R^2 = 64 + 80\cos\theta + 25 = 89 + 80\cos\theta\). Setting \(121 = 89 + 80\cos\theta\) gives \(\cos\theta = 0.4\) and \(\theta = 66.4^\circ\). The tempting \(113.6^\circ\) is the angle in the triangle of forces (\(180^\circ - 66.4^\circ\)), not the angle between the force directions. \(36.9^\circ\) loses the 2 in \(2 \times 8 \times 5\cos\theta\), and \(23.6^\circ\) takes \(\sin^{-1} 0.4\).
The horizontal force makes \(30^\circ\) with the plane, so its component up the plane is \(25\cos 30^\circ = 21.65\) N. The weight is 20 N, with component \(20\sin 30^\circ = 10\) N DOWN the plane. Resultant up the plane \(= 21.65 - 10 = 11.7\) N. The tempting 31.7 N adds the weight component instead of subtracting it. 2.5 N uses \(25\sin 30^\circ\) for the push (that is its component into the plane), and 4.33 N uses \(\cos 30^\circ\) for the weight too.
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.