M1.3 Dynamics: Newton's laws, connected particles and friction
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Mechanics 1 (AS), specification sections 4.1, 4.2 and 4.4. The suvat formulae are on M1.2; friction on a particle at rest and limiting equilibrium are on the statics page (M1.5); jerks in strings are on the momentum page (M1.4).
What this page covers
6 points
- Newton's three laws for a particle, with forces given as magnitudes or in the form \(a\mathbf{i} + b\mathbf{j}\)
- Friction on a moving particle, \(F = \mu R\), including when a pulling force acts at an angle
- Motion up and down smooth and rough inclined planes
- Lifts, and vehicles towing trailers or caravans (tension and thrust in a tow bar)
- Two particles joined by a light inextensible string over a smooth pulley or peg, and the force the string exerts on the pulley
- Forces that change from one constant value to another: a string breaking, a particle hitting the ground, a pulling force removed
In one minute
6 points
- One method solves every dynamics question in M1: draw the forces on each particle, resolve along the direction of motion and write resultant force = mass × acceleration, \(F = ma\) (in vector form \(\mathbf{F} = m\mathbf{a}\)).
- Weight \(= mg\), with \(g = 9.8\) m s⁻². Give final answers to 2 or 3 significant figures.
- Friction on a moving particle is \(F = \mu R\) and acts against the motion. Find \(R\) first by resolving perpendicular to the motion.
- On a slope at angle \(\alpha\): weight component \(mg\sin\alpha\) along the slope, and \(R = mg\cos\alpha\) if nothing else pushes into the slope.
- Connected particles: a light inextensible string has the same tension throughout (over a smooth pulley too) and both particles have the same size of acceleration. Write one equation per particle.
- When a force changes (string breaks, particle lands) start a new stage: new forces, new acceleration, and the speed at the change links the stages.
Contents
13 sections
- How do Newton's laws turn forces into motion?
- How do you use F = ma when the forces are vectors?
- How does friction act on a moving particle?
- How do you handle motion on an inclined plane?
- How do you model lifts and towed vehicles?
- How do you solve connected particles over a pulley?
- What happens when a force changes during the motion?
- Exam tips for M1.3
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do Newton's laws turn forces into motion?
| Law | In words | How you use it in M1 |
|---|---|---|
| First | a particle stays at rest or moves with constant velocity unless a resultant force acts on it | constant velocity ⇒ forces balance in every direction |
| Second | the resultant force on a particle equals its mass × its acceleration, in the direction of the resultant | \(F = ma\) along the direction of motion; \(0 = m \times 0\) perpendicular to it |
| Third | if A exerts a force on B, then B exerts an equal and opposite force on A | the floor of a lift and the passenger push on each other equally; a tow bar pulls the car back as hard as it pulls the trailer forward |
The forces you will meet. Weight \(mg\), vertically down (mass in kg, weight in N). Normal reaction \(R\), perpendicular to a surface, pushing away from it. Friction \(F\), along a rough surface, opposing the motion. Tension \(T\) in a string, pulling away from the particle; a rod or tow bar can also push, a thrust. Resistance and driving force for vehicles, usually given as constants.
Newton's second law and weight
\[F = ma \qquad\qquad W = mg\]\(F\) = resultant force (N); \(m\) = mass (kg); \(a\) = acceleration (m s⁻²); \(g = 9.8\) m s⁻². Not in the formulae list: learn both. “Resultant” means you add the forces along your chosen direction, with forces the other way counted negative, before you multiply out \(ma\).
A crate of mass 40 kg is pushed from rest across a horizontal floor by a horizontal force of 150 N. The total resistance to motion is constant and equal to 62 N. The crate is modelled as a particle. Find (a) the acceleration of the crate, (b) the time it takes to reach a speed of 5.5 m s⁻¹, (c) the distance it has moved by then.
(a) Resultant force in the direction of motion, then \(F = ma\): \(150 - 62 = 40a \Rightarrow a = 88 \div 40 =\) 2.2 m s⁻²
(b) Constant force, so constant acceleration: \(5.5 = 0 + 2.2t \Rightarrow\) \(t = 2.5\) s
(c) \(s = \frac12(u + v)t = \frac12 \times 5.5 \times 2.5 =\) 6.88 m (6.875 m)
Writing \(150 = 40a\). The 150 N is only one of the forces; \(F\) in \(F = ma\) is the resultant. Writing the subtraction out in full shows every force you have used.
How do you use F = ma when the forces are vectors?
Add the forces component by component, \((a\mathbf{i} + b\mathbf{j}) + (c\mathbf{i} + d\mathbf{j}) = (a + c)\mathbf{i} + (b + d)\mathbf{j}\); then \(\mathbf{F} = m\mathbf{a}\) gives the acceleration vector by dividing each component by the mass. Magnitude: \(|p\mathbf{i} + q\mathbf{j}| = \sqrt{p^2 + q^2}\). Direction: \(\tan\theta = q/p\), then check the quadrant on a sketch. “Parallel to \(u\mathbf{i} + v\mathbf{j}\)” means the components are in the ratio \(u : v\). With constant \(\mathbf{a}\), \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) and \(\mathbf{r} = \mathbf{u}t + \frac12\mathbf{a}t^2\) work just like the scalar forms.
Two forces \(\mathbf{F}_1 = (4\mathbf{i} - 2\mathbf{j})\) N and \(\mathbf{F}_2 = (p\mathbf{i} + 5\mathbf{j})\) N act on a particle of mass 2 kg. No other forces act. The acceleration of the particle is parallel to \((2\mathbf{i} + \mathbf{j})\). (a) Find the value of \(p\). (b) Find the magnitude of the acceleration and the angle it makes with \(\mathbf{i}\). (c) The particle starts from rest. Find its velocity after 4 s.
(a) \(\mathbf{R} = (4 + p)\mathbf{i} + 3\mathbf{j}\) and \((4 + p) : 3 = 2 : 1 \Rightarrow 4 + p = 6 \Rightarrow\) \(p = 2\)
(b) \(\mathbf{a} = (6\mathbf{i} + 3\mathbf{j}) \div 2 = (3\mathbf{i} + 1.5\mathbf{j})\) m s⁻². \(|\mathbf{a}| = \sqrt{3^2 + 1.5^2} =\) 3.35 m s⁻² (3.3541…); \(\tan\theta = 1.5 \div 3 \Rightarrow\) \(\theta = 26.6^\circ\) above \(\mathbf{i}\)
(c) \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = 4(3\mathbf{i} + 1.5\mathbf{j}) =\) \((12\mathbf{i} + 6\mathbf{j})\) m s⁻¹
\(\tan^{-1}\) of a ratio of components only gives an angle between −90° and 90°; a force such as \(-3\mathbf{i} + 6\mathbf{j}\) points up and to the left, at 116.6° to \(\mathbf{i}\), not −63.4°.
How does friction act on a moving particle?
For a moving particle the friction always takes its full value \(\mu R\) (the inequality \(F \le \mu R\) matters only for a particle at rest, on the statics page). The key point is that \(R\) is not always \(mg\): find it by resolving perpendicular to the motion, where the acceleration is zero. A pull \(P\) at angle \(\theta\) above the horizontal lifts the particle a little, so \(R = mg - P\sin\theta\); a push at \(\theta\) below the horizontal presses it down, so \(R = mg + P\sin\theta\).
A sledge of mass 25 kg is pulled along rough horizontal snow by a rope inclined at 30° above the horizontal. The tension in the rope is 80 N and the coefficient of friction between the sledge and the snow is 0.15. The sledge is modelled as a particle. Find the acceleration of the sledge.
Resolve vertically (no vertical acceleration): \(R + 80\sin 30^\circ = 25g \Rightarrow R = 245 - 40 = 205\) N.
Friction: \(F = \mu R = 0.15 \times 205 = 30.75\) N.
Resolve horizontally: \(80\cos 30^\circ - 30.75 = 25a \Rightarrow a = 38.53 \div 25 =\) 1.54 m s⁻²
Using \(R = 25g = 245\) N when the rope is at an angle. That gives \(F = 36.75\) N and the wrong acceleration. Resolve perpendicular to the motion every time.
How do you handle motion on an inclined plane?
Resolve along the line of greatest slope and perpendicular to it. For a plane at angle \(\alpha\) the weight has components \(mg\sin\alpha\) down the slope and \(mg\cos\alpha\) into the slope. Perpendicular: \(R = mg\cos\alpha\) (add or subtract the perpendicular component of any other applied force). Along the slope: \(F = ma\), with friction \(\mu R\) opposing the motion: down the slope while the particle moves up, up the slope while it moves down. If \(\tan\alpha = \frac34\) you can use \(\sin\alpha = 0.6\) and \(\cos\alpha = 0.8\) exactly.
A particle of mass 2 kg is projected with speed 8 m s⁻¹ up a line of greatest slope of a rough plane inclined at \(\alpha\) to the horizontal, where \(\tan\alpha = \frac34\). The coefficient of friction is 0.25. (a) Find the distance the particle moves up the plane before coming to rest. (b) The particle then slides back down. Find its speed when it returns to its starting point.
Perpendicular to the plane: \(R = 2g\cos\alpha = 2 \times 9.8 \times 0.8 = 15.68\) N, so \(F = 0.25 \times 15.68 = 3.92\) N.
(a) Moving up, weight component and friction both act down the plane: \(-(2g\sin\alpha + 3.92) = 2a \Rightarrow a = -(11.76 + 3.92) \div 2 = -7.84\) m s⁻². Then \(0 = 8^2 - 2 \times 7.84 \times s \Rightarrow\) \(s = 4.08\) m (4.0816…)
(b) Moving down, friction now acts up the plane: \(2g\sin\alpha - 3.92 = 2a \Rightarrow a = (11.76 - 3.92) \div 2 = 3.92\) m s⁻² down the plane. \(v^2 = 0 + 2 \times 3.92 \times 4.0816\ldots = 32 \Rightarrow\) \(v = 5.66\) m s⁻¹ \((4\sqrt2)\)
The particle does slide back because the weight component, 11.76 N, is bigger than the greatest friction available, 3.92 N. Deciding this in general is a statics question (M1.5).
Keeping the friction in the same direction for the return journey. Friction opposes the current motion: when the velocity reverses, redraw the diagram and reverse the friction.
How do you model lifts and towed vehicles?
Lifts. Choose upwards as positive and use the direction of the acceleration, not the velocity: a lift slowing down on its way up has a downward acceleration. For a load of mass \(m\) on the floor, \(R - mg = ma\); the floor and the load push on each other with equal force (third law). For the cable, treat lift + contents as one particle: \(T - (M + m)g = (M + m)a\).
A lift of mass 600 kg carries a crate of mass 80 kg on its floor. The lift is raised by a vertical cable. (a) The lift accelerates upwards at 0.5 m s⁻². Find the tension in the cable and the force of the floor on the crate. (b) Later the lift is moving upwards and slowing down at 1.2 m s⁻². Find the force of the floor on the crate.
(a) Lift and crate together, upwards positive: \(T - 680g = 680 \times 0.5 \Rightarrow T = 680 \times 10.3 = 7004 =\) 7000 N (3 s.f.). The crate alone: \(R - 80g = 80 \times 0.5 \Rightarrow\) \(R = 824\) N
(b) Slowing down while moving up means \(a = -1.2\) m s⁻²: \(R - 80g = 80 \times (-1.2) \Rightarrow R = 80 \times 8.6 =\) 688 N
Check: the cable tension in this stage is \(680 \times 8.6 = 5848\) N, less than the total weight 6664 N, as a downward acceleration needs.
Towed vehicles. The car and trailer move together, so they have the same acceleration; the tow bar is modelled as light, inextensible and parallel to the road. Take the whole system first (the tow-bar force is internal and cancels), then the trailer alone for the tension, and check with the car alone. When the car brakes, the trailer pushes on the car: the tow bar is in thrust (compression). A rope could not do this.
A car of mass 1100 kg tows a caravan of mass 700 kg along a straight horizontal road by a light rigid tow bar. The resistances to motion are constant: 300 N on the car and 450 N on the caravan. (a) The engine produces a driving force of 2100 N. Find the acceleration and the tension in the tow bar. (b) The driver switches off the engine and brakes, and the car and caravan decelerate at 1.5 m s⁻². Find the force in the tow bar, stating whether it is a tension or a thrust, and the braking force on the car.
(a) Whole system: \(2100 - 300 - 450 = 1800a \Rightarrow\) \(a = 0.75\) m s⁻². Caravan alone: \(T - 450 = 700 \times 0.75 \Rightarrow\) \(T = 975\) N (check, car: \(2100 - 300 - 975 = 825 = 1100 \times 0.75\) ✓).
(b) Caravan alone, with the tow-bar force \(S\) pushing backwards on it: \(S + 450 = 700 \times 1.5 \Rightarrow\) \(S = 600\) N, a thrust. Car alone, where the thrust pushes the car forwards: \(B + 300 - 600 = 1100 \times 1.5 \Rightarrow\) \(B = 1950\) N
If you assume a tension and your answer comes out negative, the tow bar is in thrust. Say so in words: “the force in the tow bar is a thrust of 600 N”.
How do you solve connected particles over a pulley?
- Light string: same tension along its whole length. Inextensible: both particles move with the same speed and the same size of acceleration. Smooth pulley or peg: the tension is the same on both sides.
- Draw a separate force diagram for each particle and write \(F = ma\) for each in its own direction of motion. Adding the two equations removes \(T\) and gives \(a\); substitute back for \(T\).
- You may not treat the two particles as one body unless they move along the same straight line: the string turns a corner at the pulley.
- Force on the pulley: the two parts of the string each pull on it with tension \(T\). Their resultant bisects the angle between the strings, with magnitude \(2T\cos(\theta/2)\), where \(\theta\) is the angle between the two parts of the string.
A particle A of mass 3 kg lies on a rough horizontal table. The coefficient of friction between A and the table is 0.2. A is attached to a light inextensible string that passes over a small smooth pulley fixed at the edge of the table; the other end carries a particle B of mass 2 kg, which hangs freely. The string between A and the pulley is horizontal. The system is released from rest. Find (a) the acceleration of the particles, (b) the tension in the string, (c) the magnitude and direction of the force exerted by the string on the pulley.
Friction on A (resolving vertically for A, \(R = 3g\)): \(F = 0.2 \times 3g = 5.88\) N.
One equation per particle: B (down): \(2g - T = 2a\); A (towards the pulley): \(T - 5.88 = 3a\).
(a) Add: \(2g - 5.88 = 5a \Rightarrow a = 0.28g =\) 2.74 m s⁻² (2.744)
(b) \(T = 2(g - a) = 1.44g =\) 14.1 N (14.112)
(c) Two tensions at 90° to each other: \(\sqrt{T^2 + T^2} = T\sqrt2 =\) 20.0 N, at 45° below the horizontal, towards the table side
Taking the tension as the weight of the hanging particle (here 19.6 N). That is true only if B is not accelerating. While B accelerates downwards, the tension must be less than its weight.
What happens when a force changes during the motion?
In M1 a force does not vary smoothly; it switches from one constant value to another. Each stage has constant acceleration, so the suvat formulae apply within each stage, and the link between stages is the velocity at the instant of change.
- Hanging particle hits the ground: it stops; the string goes slack, so \(T = 0\) for the other particle, which carries on with its speed at that instant, acted on only by its weight, friction and the normal reaction.
- String breaks: \(T = 0\) from that instant for both particles. A pulling force removed: recompute \(R\) and the friction too.
- A sketch of the velocity–time graph keeps the stages straight: straight segments, with a change of gradient at each change of force.
A particle A of mass 2 kg rests on a rough plane inclined at 30° to the horizontal. The coefficient of friction between A and the plane is 0.25. A is attached to a light inextensible string that runs parallel to a line of greatest slope, over a small smooth pulley fixed at the top of the plane, to a particle B of mass 3 kg. B hangs freely 1.2 m above the ground. The system is released from rest with the string taut, and A does not reach the pulley. (a) Find the acceleration of the system and the tension in the string. (b) Find the magnitude of the force exerted by the string on the pulley. (c) Find the speed of B as it hits the ground. (d) B does not rebound. Find the further distance A moves up the plane before it first comes to rest.
Normal reaction and friction on A: \(R = 2g\cos 30^\circ = 16.97\) N, \(F = 0.25R = 4.244\) N.
(a) B: \(3g - T = 3a\); A: \(T - 2g\sin 30^\circ - 4.2435 = 2a\). Adding: \(29.4 - 9.8 - 4.2435 = 5a \Rightarrow\) \(a = 3.07\) m s⁻² (3.0713); \(T = 3(g - a) =\) 20.2 N
(b) The two parts of the string make 60° with each other (one along the plane at 30° above the horizontal, one vertical): \(2T\cos 30^\circ = T\sqrt3 =\) 35.0 N (34.96)
(c) B falls 1.2 m from rest with \(a = 3.0713\ldots\): \(v^2 = 2 \times 3.0713 \times 1.2 = 7.371 \Rightarrow\) \(v = 2.71\) m s⁻¹
(d) The string is slack, so for A only its weight component and friction act along the plane, both down it: \(-(g\sin 30^\circ + 0.25g\cos 30^\circ) = a \Rightarrow a = -7.022\) m s⁻². Then \(0 = 7.371 - 2 \times 7.022 \times s \Rightarrow\) \(s = 0.525\) m
Use the unrounded \(v^2 = 7.3711\) here; rounding \(v\) to 2.71 first gives 0.523 m, which shows how early rounding drifts. The mass of A cancels in (d): once the string is slack, A's motion does not depend on B at all.
Write “string taut”, “string slack” before you start, and list \(u, v, a, s, t\) for each stage. The reasoning at the change of stage, such as stating that the tension becomes zero, is part of the answer, so write it down.
Exam tips for M1.3
Draw a separate force diagram for each particle, with every force labelled and the direction of the acceleration marked. A missing force, or one pointing the wrong way, spoils a whole equation and everything that follows from it.
Perpendicular to the motion gives \(R\) (acceleration zero); along the motion gives \(F = ma\). Write each equation as “resolving parallel to the plane: …” so that each one stands on its own. Keep at least four significant figures in intermediate values and give final answers to 2 or 3, since \(g\) is only given to 2.
When a value is given (“show that the acceleration is \(0.28g\)”), every step must appear, with the forces substituted before simplifying. Then use the given value in the following parts, even if your own answer differed.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The normal reaction always equals \(mg\). | Slopes and angled pulls change the perpendicular balance. | Resolve perpendicular to the motion: \(R = mg\cos\alpha\) on a slope, \(R = mg - P\sin\theta\) for an upward pull. |
| Friction acts in a fixed direction on a given slope. | Friction opposes the relative motion. | Down the plane while moving up; up the plane while moving down. |
| The tension equals the weight of the hanging particle. | That would make the resultant force on it zero. | While it accelerates down, \(T \lt mg\); find \(T\) from \(mg - T = ma\). |
| A lift going up always has \(R \gt mg\). | The sign depends on the acceleration, not the velocity. | Slowing on the way up: acceleration downwards, so \(R \lt mg\). |
| After a string breaks, the particle stops at once. | No force can change the velocity instantly. | It keeps its speed at that instant and starts a new stage with new forces. |
| The force on a pulley is \(2T\) whatever the set-up. | The two tensions add as vectors. | \(2T\) only when both strings are parallel; otherwise \(2T\cos(\theta/2)\). |
How this topic connects
M1.2 Kinematics and M1.1 Vectors
\(F = ma\) supplies the constant acceleration that the suvat formulae need, and the \(\mathbf{i}, \mathbf{j}\) methods carry straight over to forces.
M1.4 Momentum and M1.5 Statics
When a slack string becomes taut again, the sudden jerk is a momentum problem. A particle at rest on a rough surface, where friction can take any value up to \(\mu R\), belongs to statics.
M2 and IAL Physics Unit 1
The same pulley and slope systems are solved with work and energy in M2. Free-body diagrams, lifts, towing and Newton's third law appear in Physics too, with \(g = 9.81\) N kg⁻¹ instead of 9.8 m s⁻².
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(F = ma\) · \(\mathbf{F} = m\mathbf{a}\) | Resultant force and acceleration, scalar or vector | No — learn it |
| \(W = mg\) | Weight, \(g = 9.8\) m s⁻² | No — learn it |
| \(F = \mu R\) | Friction on a moving particle | No — learn it |
| \(R = mg\cos\alpha\); \(mg\sin\alpha\) along the slope | Particle on a plane at angle \(\alpha\) | No — derive it |
| \(2T\cos(\theta/2)\) | Force of a string on a smooth pulley, \(\theta\) = angle between the two parts of the string | No — derive it |
| \(v = u + at\), \(s = ut + \frac12at^2\), \(v^2 = u^2 + 2as\), … | Each stage of constant acceleration | No — learn them |
Check your understanding
1. The only force on a particle of mass 3 kg is \((6\mathbf{i} - 9\mathbf{j})\) N. Find the magnitude of its acceleration.
\(\mathbf{a} = (2\mathbf{i} - 3\mathbf{j})\) m s⁻², so \(|\mathbf{a}| = \sqrt{13} =\) 3.61 m s⁻².
2. A box slides across a rough horizontal floor with no other horizontal force acting. The coefficient of friction is 0.4. Find its deceleration.
\(R = mg\), so friction \(= 0.4mg\) and the deceleration is \(0.4g =\) 3.92 m s⁻². The mass cancels.
3. A particle slides down a smooth plane inclined at 20° to the horizontal. Find its acceleration.
\(a = g\sin 20^\circ =\) 3.35 m s⁻² down the plane.
4. Particles of mass 4 kg and 6 kg hang at the ends of a light inextensible string over a smooth fixed pulley and are released from rest. Find the acceleration and the tension.
\(6g - T = 6a\) and \(T - 4g = 4a\), so \(a = 2g/10 =\) 1.96 m s⁻² and \(T = 4(g + a) =\) 47.0 N.
5. Why can the tension be taken as the same on both sides of a smooth pulley?
The string is light (no mass, so no resultant force is needed to accelerate it) and the pulley is smooth (no friction between string and pulley), so nothing changes the tension as the string passes over.
Specification checklist
- (M1 4.1) Apply Newton's laws of motion to a particle, using weight = \(mg\) with \(g = 9.8\) m s⁻².
- (M1 4.1) Solve problems with forces given as magnitudes or in the form \(a\mathbf{i} + b\mathbf{j}\), using \(\mathbf{F} = m\mathbf{a}\).
- (M1 4.2) Find the acceleration and tensions for two particles connected by a light inextensible string, on a line, hanging, or over a smooth fixed pulley or peg, and the force the string exerts on the pulley.
- (M1 4.2) Deal with a force that changes from one constant value to another, for example when a string breaks or a particle hits the ground.
- (M1 4.2) Solve problems about motion up or down a smooth or rough inclined plane, including forces acting at an angle.
- (M1 4.4) Use \(F = \mu R\) for a particle moving on a rough surface, with friction opposing the motion.
- (M1 4.1, 4.2) Model a lift and its contents, and a vehicle towing a trailer, including tension or thrust in a tow bar.