Distinguish a vector quantity from a scalar quantity and identify examples of each.
Resolve a force at an angle into perpendicular components, including weight on an inclined surface.
Determine the resultant and the equilibrant of two or more concurrent forces.
Draw a fully-labelled free-body diagram for an object, including two objects connected by a string over a pulley.
State and apply Newton's first, second and third laws of motion to single and connected objects.
Apply Newton's Law of Universal Gravitation to calculate the force between two masses and the gravitational acceleration at a given distance.
📖 KEY DEFINITIONVector: a quantity with both magnitude and direction (force, weight, displacement, velocity, momentum). Scalar: a quantity with magnitude only (mass, distance, speed, energy, time). Newton's First Law: an object remains at rest or moves at constant velocity in a straight line unless a non-zero resultant force acts on it. Newton's Second Law: the resultant force acting on an object equals the product of its mass and the acceleration it produces, in the direction of the resultant force. Newton's Third Law: when object A exerts a force on object B, object B exerts a force of equal magnitude but opposite direction back on object A. Newton's Law of Universal Gravitation: every object with mass attracts every other object with mass, with a force directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centres.
Core content · vectors, components and resultants
To resolve a force into components on a chosen set of x- and y-axes, project it onto each axis using the angle it makes with that axis - sketch the triangle every time rather than memorising a fixed "cos goes here" rule, since which ratio applies flips depending on which axis the angle is measured from.
To find the resultant of several forces, resolve every force into x- and y-components first, add the components on each axis separately, then recombine the two totals with Pythagoras (magnitude) and inverse tangent (direction).
The equilibrant is the single force that would hold the object in equilibrium - equal in magnitude to the resultant of the other forces, but opposite in direction.
Component on a slope
Formula
Effect
Parallel to the slope
Fg∥ = mg sinθ
pulls the object down the incline
Perpendicular to the slope
Fg⊥ = mg cosθ
presses the object into the surface
Normal force
FN
reaction to the perpendicular component - NOT automatically equal to mg
Core content · Newton's three laws and the free-body diagram
Law
In words
Formula-sheet form
1st (inertia)
an object keeps its state of rest or constant velocity unless a resultant force acts on it
ΣF = 0 → a = 0
2nd
acceleration is directly proportional to the resultant force and inversely proportional to mass
Fnet = ma
3rd
every force has an equal, opposite reaction force acting on the OTHER object
FA on B = −FB on A
The forces that typically appear in a free-body diagram: weight (Fg = mg, always straight down), normal force (FN, perpendicular to the contact surface), applied force or tension (FA or FT), and friction (Ff, parallel to the surface, opposing motion or the tendency to move).
Static friction resists an object starting to move and only grows as large as it needs to, up to a maximum: fs(max) = μsN. Below that maximum the object stays put. Kinetic friction acts once the object is already sliding and is treated as constant: fk = μkN. μs is normally larger than μk for the same pair of surfaces.
Connected objects (a string over a pulley, for example): draw a separate free-body diagram per object. The string applies the same tension to both ends it connects, and the two objects share the same magnitude of acceleration because the string does not stretch. The identical method - one equation per object, then combine - also applies to two blocks pushed together in contact or to systems with more than one pulley.
⚡ MUST MEMORISE
Fnet = ma Fg = mg fs(max) = μsN and fk = μkN F = Gm1m2⁄d2 g = GM⁄d2, with G = 6,67×10−11 N·m2·kg−2 (given on the data sheet, never derived)
Core content · Newton's Law of Universal Gravitation
Every object with mass attracts every other object with mass; the force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centres: F = Gm1m2⁄d2.
d is measured centre-to-centre, in metres. G is a universal constant, never calculated from other quantities.
This force is always attractive, and by Newton's third law the two masses pull on each other with equal magnitude but opposite direction.
The gravitational acceleration produced by a body of mass M at a distance d from its centre is g = GM⁄d2 - the object's own mass cancels out completely, which is why g is the same for a marble and a boulder at the same distance from a planet's centre.
⭐ EXAM FAVOURITE
A combined "forces on a slope" question: show whether an object slides using the maximum static friction test, then calculate its acceleration once moving using kinetic friction, often extended into a connected two-mass system linked over a pulley. This exact combination is close to guaranteed on Paper 1 every year.
🔥 FREQUENTLY TESTED
Resolving weight into Fg∥ and Fg⊥ on a slope, deciding whether static or kinetic friction applies, and setting up one Fnet = ma equation per object for a connected system.
⚠️ COMMON MISTAKE
FN is not automatically equal to mg - the moment there is a vertical applied force or the object sits on a slope, FN must come from the perpendicular equilibrium condition. A Newton's third law pair can never appear together in one object's free-body diagram, since the two forces act on two different objects. Learners often mix up sinθ and cosθ on a slope - always sketch the angle rather than guessing.
💡 EASY MARK
"Is force X a vector or a scalar?" and "state Newton's first/second/third law in words" are 1-2 mark gifts that need no calculation at all.
Worked example 1 · Friction and Newton's second law on a slope
Q: A 45 kg crate rests on a ramp inclined at 25° to the horizontal. The coefficient of static friction between crate and ramp is 0,42; the coefficient of kinetic friction is 0,35. (a) Show whether the crate slides down under gravity alone. (b) Calculate its acceleration once it is sliding.
Fg∥ = mg sinθ = (45)(9,8)(sin25°) = 186,4 N
FN = Fg⊥ = mg cosθ = (45)(9,8)(cos25°) = 399,7 N
Maximum static friction = μsN = (0,42)(399,7) = 167,9 N
Since 186,4 N > 167,9 N, the driving component exceeds the maximum available static friction, so the crate slides.
Once sliding: fk = μkN = (0,35)(399,7) = 139,9 N (acting up the slope)
Fnet = Fg∥ − fk = 186,4 − 139,9 = 46,5 N down the slope
Fnet = ma → a = 46,5 ÷ 45 = 1,03 m·s−2 down the slope
∴ The crate slides, accelerating at 1,03 m·s−2 down the incline
Worked example 2 · Connected objects over a pulley
Q: Block A (8 kg) sits on a rough horizontal tabletop (μk = 0,25) and is connected by a light inextensible string, over a frictionless pulley at the table's edge, to a hanging Block B (5 kg). Calculate the acceleration of the system and the tension in the string.
Taking the direction each block actually moves as positive:
Block B (vertical): mBg − T = mBa → (5)(9,8) − T = 5a → 49 − T = 5a
Block A (horizontal): T − fk = mAa, where fk = μkmAg = (0,25)(8)(9,8) = 19,6 N
So T − 19,6 = 8a
Adding the two equations to eliminate T: 49 − 19,6 = 13a → 29,4 = 13a → a = 2,26 m·s−2
Substituting back: T = 8(2,26) + 19,6 = 37,7 N
∴ The system accelerates at 2,26 m·s−2, with a string tension of 37,7 N
Worked example 3 · Newton's Law of Universal Gravitation
Q: Calculate the gravitational force of attraction between the Earth (mass 5,97×1024 kg) and a 900 kg satellite orbiting at a distance of 7,0×106 m from the Earth's centre.
F = Gm1m2⁄d2 = (6,67×10−11)(5,97×1024)(900) ÷ (7,0×106)2
= (3,98×1014)(900) ÷ (4,9×1013)
= 3,58×1017 ÷ 4,9×1013
∴ F ≈ 7,31×103 N
Worked example 4 · Resultant of two forces at angles
Q: Two ropes pull on a tyre in a tug-of-war rig. Rope 1 exerts 60 N at 30° above the horizontal, pulling up and to the right. Rope 2 exerts 45 N at 50° above the horizontal on the other side, pulling up and to the left. Calculate the magnitude and direction of the resultant force.
Taking right as positive x and up as positive y:
F1x = 60cos30° = 51,96 N, F1y = 60sin30° = 30,00 N
F2x = −45cos50° = −28,93 N, F2y = 45sin50° = 34,47 N
ΣFx = 51,96 − 28,93 = 23,03 N ΣFy = 30,00 + 34,47 = 64,47 N
R = √(23,032 + 64,472) = √4686,8 ≈ 68,5 N
θ = tan−1(64,47 ÷ 23,03) ≈ 70,4° above the horizontal, toward the side where ΣFx is positive
∴ The resultant is 68,5 N at 70,4° above the horizontal
Nothing moves without a reason, and nothing changes its motion without a push - that is the whole of mechanics in one line.- on Newton's laws
Why it matters: almost every Mechanics answer starts with a free-body diagram, and so do the marks. Learners lose them by writing F N = mg out of habit on a slope, or by putting a Newton's third law pair into one object's diagram. Draw the forces first, calculate second.
🧠 ACTIVE RECALL
Which of the following is a vector quantity?(2)
A crate slides across a rough horizontal floor at a constant velocity. What can you conclude about the resultant force acting on it?(2)
A 6 kg block rests on a frictionless horizontal table, with no other vertical force applied. The normal force on the block is closest to:(2)
If the distance between two masses is doubled, the gravitational force between them:(2)
A book rests on a table. The Newton's third law reaction pair to the table pushing up on the book is:(2)
A small horizontal force is applied to a heavy crate, too small to move it. The static friction acting on the crate is:(2)
A 12 kg block rests on a frictionless slope inclined at 18° to the horizontal. Calculate the component of the block's weight acting parallel to the slope.(4)
A 6 kg trolley on a frictionless table is connected by a string over a pulley to a hanging 4 kg mass. Set up the two equations of motion and calculate the acceleration of the system and the tension in the string.(5)
Calculate the gravitational force between an 80 kg person standing on the Earth's surface (mass 5,97×1024 kg, radius 6,38×106 m) and the Earth itself.(3)
Explain this for a Grade 8: what do Newton's second and third laws actually mean in everyday life?
ONE-MINUTE SUMMARY
Draw the free-body diagram first, every time. Resolve weight into Fg∥ and Fg⊥ on a slope, check static friction against its maximum before assuming motion, then apply Fnet = ma per object. A Newton's third law pair never appears twice in the same diagram - it acts on the other object. FN is only equal to mg on a flat surface with no other vertical force. Gravitation follows the same inverse-square shape as every other field you will meet this year: F = Gm1m2⁄d2.
IN THE EXAM
Mechanics as a whole (this chapter at 18 marks, plus momentum, projectile motion and work, energy and power) is the highest-weighted content group on Paper 1, carrying 65 of its 150 marks, and it appears in full in every sitting.
Expect a structured question combining a slope, friction, and a connected two-mass system, The ten Paper 1 MCQs (1.1 to 1.10, 2 marks each) are spread across the whole paper's topic range, so expect vectors, friction or Newton's laws there too; those marks already sit inside this topic's 18 and are not added on top.
Marks are earned for a correctly labelled free-body diagram and a clearly substituted formula line, not just a final numeric answer.
🔒 MOMENTUM AND IMPULSE
🔒 VERTICAL PROJECTILE MOTION
🔒 WORK, ENERGY AND POWER
🔒 DOPPLER EFFECT
🔒 ELECTROSTATICS
🔒 ELECTRIC CIRCUITS
🔒
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