HookThe rocket that landed itself on a pillar of fire
On the evening of 21 December 2015, a SpaceX Falcon 9 did something no orbital-class rocket had done before: having launched eleven satellites toward orbit, its \(40\)-metre first stage turned around, fell back through the atmosphere, relit a single engine and settled upright on a concrete pad at Cape Canaveral. Every phase of that landing is Newton's laws made visible. The engine hurls exhaust gas violently downward and the gas pushes the rocket up with an equal and opposite force — Newton's third law. Whether the rocket accelerates up, hovers, or slows on the way down is decided by the resultant of that thrust and its weight — Newton's second law. And in the vacuum between burns, it coasts in a straight line, needing no force to keep moving — Newton's first law.
This section is those laws turned into a method. You start with the idea of a force and Newton's first law of equilibrium, move to Newton's second law \(F=ma\) and the resolving of forces that act at an angle, pin down weight and the value of \(g\), use Newton's third law to handle equilibrium, connected particles and smooth pulleys, add forces into a single resultant for dynamics in a plane, and finish with the \(F\le\mu R\) model of friction on rough surfaces. The habit that carries all of it is the same one that lands rockets: draw the forces, choose a direction, and apply \(F=ma\) along it.
ModelForce and Newton's first law — motion needs no cause, change does
A force is a push or a pull — anything that can change a body's motion — and it is a vector measured in newtons, with a size and a direction. The forces in this course are a short list: weight (gravity, always straight down), the normal reaction (a surface pushing perpendicular to itself), tension (a string or rod pulling along its length), thrust or the driving force of an engine, friction (opposing sliding along a surface) and air resistance or drag. Every problem starts by drawing these as arrows on the body — the force diagram — and almost no marks are safe without one.
Newton's first law states that a body stays at rest, or keeps moving in a straight line at constant velocity, unless a resultant force acts on it. The revolutionary half is the second clause: motion needs no force to sustain it, only to change it. A puck on frictionless ice would glide forever; it slows on a real rink only because friction is a resultant force. This tendency to keep doing what it is already doing is a body's inertia, and it is why a car passenger is thrown forward in a sudden stop — the car decelerates, but the passenger continues at the old speed until the seatbelt supplies a force. When the resultant force is zero the body is in equilibrium: the forces balance, and it is either still or moving steadily.
MechanismNewton's second law — \(F=ma\) and resolving forces
When the resultant force is not zero, Newton's second law says the body accelerates in the direction of that resultant, with \(F=ma\): resultant force equals mass times acceleration. Force and acceleration are vectors pointing the same way; mass is the scalar that says how much force a given acceleration costs. The equation is used one direction at a time — you pick a direction, add up the force components along it, and set the total equal to \(ma\) in that direction.
That is why resolving matters. A force acting at an angle contributes to motion only through its component along the direction of interest. A force \(F\) at angle \(\theta\) to the horizontal has a horizontal component \(F\cos\theta\) and a vertical component \(F\sin\theta\), and you use whichever axis you are applying \(F=ma\) along. The Edexcel course keeps this to two dimensions, so every problem reduces to two perpendicular equations — typically one along the motion and one across it — solved together.
A child pulls a \(20\ \text{kg}\) sledge across smooth ice by a rope inclined at \(30^{\circ}\) above the horizontal, with a tension of \(50\ \text{N}\). Find the sledge's acceleration.
Resolve horizontally, the direction of motion. The only horizontal force is the component of the tension, \(50\cos 30^{\circ}=50\times 0.866=43.3\ \text{N}\). The ice is smooth, so there is no friction to oppose it. Apply \(F=ma\) horizontally: \(43.3=20a\), so \(a=\dfrac{43.3}{20}=2.17\ \text{m s}^{-2}\). Only the horizontal component of the pull drives the sledge; the vertical component \(50\sin 30^{\circ}=25\ \text{N}\) merely lightens the load on the ice — reducing the normal reaction — and does no driving. Using the full \(50\ \text{N}\) as if it were horizontal is the classic error, and it inflates the acceleration to \(2.5\ \text{m s}^{-2}\).
ModelWeight, \(g\), and apparent weight in a lift
Weight is the force of gravity on a mass, \(W=mg\), directed vertically downward, with \(g\approx 9.8\ \text{m s}^{-2}\) at the Earth's surface. A body falling freely under gravity alone therefore has, by \(F=ma\), an acceleration of exactly \(g\): \(mg=ma\Rightarrow a=g\), independent of the mass, which is why — without air resistance — a hammer and a feather fall together.
The subtle point examiners probe is that the force a surface feels from a body, the normal reaction, need not equal the body's weight. The two are equal only when there is no vertical acceleration and no other vertical force. Put a person on bathroom scales in a lift and the scales read the normal reaction, which is their apparent weight: when the lift accelerates upward the floor must both support and accelerate them, so the reaction exceeds \(mg\); when it accelerates downward the reaction is less. That is the lightness you feel as a lift starts down, and it is a direct reading of \(F=ma\) applied vertically.
A person of mass \(70\ \text{kg}\) stands on scales in a lift that accelerates upward at \(2\ \text{m s}^{-2}\). Find the reading — the normal reaction \(R\). Take \(g=9.8\ \text{m s}^{-2}\).
Draw the two vertical forces on the person: the reaction \(R\) up and the weight \(mg\) down. The person accelerates up at \(2\ \text{m s}^{-2}\), so apply \(F=ma\) upward: \(R-mg=ma\). Hence \(R=m(g+a)=70(9.8+2)=70\times 11.8=826\ \text{N}\). Their true weight is \(mg=70\times 9.8=686\ \text{N}\), so the scales over-read by \(140\ \text{N}\) — about \(14\ \text{kg}\) of apparent extra mass — purely because of the acceleration. If instead the lift accelerated downward at \(2\ \text{m s}^{-2}\), the same equation with \(a=-2\) gives \(R=70\times 7.8=546\ \text{N}\); and in genuine free fall (\(a=-g\)) the reaction would be zero — weightlessness.
MechanismNewton's third law, connected particles and smooth pulleys
Newton's third law states that if body A exerts a force on body B, then B exerts an equal and opposite force on A. The forces in the pair are the same size, opposite in direction, and — crucially — act on different bodies, which is why they never cancel each other out. The Falcon 9 pushes exhaust gas down; the gas pushes the rocket up. You push down on the floor; the floor pushes up on you, and that push is the normal reaction.
This unlocks connected particles. Two masses joined by a light, inextensible string over a smooth pulley share two facts: because the string does not stretch, both move with the same acceleration; and because the string is light and the pulley smooth, the tension is the same throughout. You then apply \(F=ma\) to each mass separately — the heavier one accelerating down, the lighter one up — and solve the pair of equations together. For a particle in equilibrium the same law says the forces sum to zero, which in two dimensions means the components balance in each of two perpendicular directions.
Masses of \(3\ \text{kg}\) and \(5\ \text{kg}\) hang from the ends of a light inextensible string passing over a smooth fixed pulley, and are released from rest. Find the acceleration of the system and the tension in the string. Take \(g=9.8\ \text{m s}^{-2}\).
The \(5\ \text{kg}\) mass falls and the \(3\ \text{kg}\) mass rises, both with acceleration \(a\), and the tension \(T\) is the same on both sides. Apply \(F=ma\) to each. For the \(5\ \text{kg}\) mass, taking down as positive: \(5g-T=5a\). For the \(3\ \text{kg}\) mass, taking up as positive: \(T-3g=3a\). Add the two equations to eliminate \(T\): \(5g-3g=8a\), so \(a=\dfrac{2g}{8}=\dfrac{2\times 9.8}{8}=2.45\ \text{m s}^{-2}\). Substitute back: \(T=3(g+a)=3(9.8+2.45)=36.75\ \text{N}\). Check with the other equation: \(5(g-a)=5(9.8-2.45)=36.75\ \text{N}\) — the two agree, which is the sign that the working is sound. Writing an \(F=ma\) equation for each mass, then adding to cancel the tension, is the reliable method for every pulley problem.
MechanismResultant forces and dynamics in a plane
Forces add as vectors, tip to tail, and the single arrow that could replace them all is the resultant. The practical method is to work in components: add the \(\mathbf{i}\) parts and the \(\mathbf{j}\) parts separately, giving a resultant \((X\mathbf{i}+Y\mathbf{j})\), whose magnitude is \(\sqrt{X^2+Y^2}\) by Pythagoras and whose direction is \(\tan^{-1}\!\left(\dfrac{Y}{X}\right)\) from the horizontal. A resultant of zero means equilibrium; a non-zero resultant means acceleration.
Once you have the resultant, dynamics in a plane is just \(F=ma\) read as a vector equation, \(\mathbf{F}=m\mathbf{a}\). The acceleration points along the resultant force and has magnitude \(\dfrac{|\mathbf{F}|}{m}\); equivalently you divide each component of the resultant by the mass to get the components of the acceleration. This is the same content as resolving, seen the other way round: instead of splitting one awkward force into components, you gather several forces into one.
Two forces \((5\mathbf{i}+2\mathbf{j})\ \text{N}\) and \((-\mathbf{i}+3\mathbf{j})\ \text{N}\) act on a particle of mass \(2\ \text{kg}\). Find the resultant force, its magnitude, and the particle's acceleration.
Add the components: the resultant is \((5-1)\mathbf{i}+(2+3)\mathbf{j}=(4\mathbf{i}+5\mathbf{j})\ \text{N}\). Its magnitude is \(\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}\approx 6.40\ \text{N}\). The acceleration follows from \(\mathbf{a}=\dfrac{\mathbf{F}}{m}\): divide each component by \(2\ \text{kg}\) to get \(\mathbf{a}=(2\mathbf{i}+2.5\mathbf{j})\ \text{m s}^{-2}\), whose magnitude is \(\sqrt{2^2+2.5^2}=\sqrt{10.25}\approx 3.20\ \text{m s}^{-2}\) — exactly half the force magnitude, as \(F=ma\) with \(m=2\) demands. Keeping everything in \(\mathbf{i}\), \(\mathbf{j}\) form means the direction is carried through automatically, with no separate angle to track.
CaseFriction and the \(F\le\mu R\) model
A rough surface resists sliding with a frictional force, and the model for it is an inequality, not an equation: \(F\le\mu R\), where \(R\) is the normal reaction and \(\mu\) is the coefficient of friction for the two surfaces. Friction is self-adjusting. While a body is not sliding, friction takes exactly whatever value is needed to hold it still, up to a maximum of \(\mu R\). At the instant it is on the point of slipping — limiting equilibrium — friction reaches that maximum, \(F=\mu R\), and once the body is actually moving you take friction as \(\mu R\) opposing the motion.
That 'less than or equal to' is the whole subtlety. To decide whether a body moves, compare the force trying to move it with the maximum friction \(\mu R\) available: if the driving force exceeds \(\mu R\), the body slides and you find the acceleration from the surplus; if not, it stays put and friction simply matches the driving force. On a slope you must resolve the weight into a component along the surface, which tries to slide the body, and a component perpendicular to it, which sets the normal reaction and hence the friction available.
A \(10\ \text{kg}\) crate rests on a rough plane inclined at \(20^{\circ}\) to the horizontal; the coefficient of friction is \(\mu=0.3\). Determine whether it slides, and if so, its acceleration. Take \(g=9.8\ \text{m s}^{-2}\).
Resolve the weight \(mg=98\ \text{N}\) into components along and perpendicular to the slope. The component pulling the crate down the slope is \(mg\sin 20^{\circ}=98\times 0.342=33.5\ \text{N}\). Perpendicular to the slope the crate is in balance, so the normal reaction is \(R=mg\cos 20^{\circ}=98\times 0.940=92.1\ \text{N}\). The maximum friction available is \(\mu R=0.3\times 92.1=27.6\ \text{N}\). Compare them: the driving component \(33.5\ \text{N}\) exceeds the maximum friction \(27.6\ \text{N}\), so the crate slides. The resultant force down the slope is the surplus, \(33.5-27.6=5.9\ \text{N}\), so by \(F=ma\) the acceleration is \(a=\dfrac{5.9}{10}=0.59\ \text{m s}^{-2}\) down the plane. Had \(\mu\) been \(0.4\), the maximum friction \(36.8\ \text{N}\) would have exceeded the \(33.5\ \text{N}\) driving force, and the crate would have stayed still with friction holding at \(33.5\ \text{N}\), not \(36.8\).
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Nearly every mark in this section is earned by the same opening move: draw a clear force diagram with every force labelled — weight, normal reaction, tension, friction, applied forces — and then choose two perpendicular directions to resolve along. State \(F=ma\) in the direction of the acceleration and the equilibrium condition (resultant zero) in the perpendicular direction; writing these as separate equations is where the method marks live.
Use the modelling words precisely, because they set up your equations: smooth means take friction as zero, light means the string is massless so tension is uniform, and inextensible means connected particles share one acceleration. For pulley problems, write an \(F=ma\) equation for each mass and add them to eliminate the tension. For friction, never assume the body moves — compare the driving force with the maximum friction \(\mu R\) first, and only use \(F=\mu R\) once you have shown the body is sliding or in limiting equilibrium. Quote the value of \(g\) you are told to use, put weight (in newtons) on the diagram rather than mass, and check a pulley tension by substituting it back into both equations — agreement is your proof the working is sound.