Newton's three laws of motion are the foundation of classical mechanics. They explain why objects start moving, keep moving, speed up, slow down, and interact with one another. In Grade 11 you extend these laws to two-dimensional situations including inclined planes, friction, and the universal law of gravitation.
2.1 Forces and Free-Body Diagrams
Before applying Newton's laws, you must identify ALL the forces acting on an object. A FREE-BODY DIAGRAM (FBD) is a drawing that shows the object isolated from its surroundings, with all the forces acting ON it represented as arrows. The arrow length is proportional to the force's magnitude; the arrowhead shows the direction; the arrow's tail or tip is placed at the object. A good FBD is the first step in solving ANY mechanics problem.
Definition
Normal force
The normal force, N, is the force exerted by a surface on an object in contact with it.
Definition
Frictional force
Frictional force is the force that opposes the motion of an object in contact with a surface and it acts parallel to the surface the object is in contact with.
Rules for drawing a free-body diagram
- Represent the object as a dot or a simple box.
- Draw ALL forces as arrows pointing away from the object (or toward it for attractive forces like weight).
- Label each force clearly: W or Fg (weight), N (normal), f or Ff (friction), T (tension), FA (applied force).
- Make arrow lengths roughly proportional to magnitudes.
- Do NOT include internal forces or forces the object exerts on other objects.
- Include a reference axis (x and y) on or beside the FBD.
Watch Out
COMMON MISTAKE — The normal force is NOT always equal to the weight. The normal force equals the weight only when the surface is horizontal AND there is no vertical component of applied force. On an inclined plane N = mg cosθ, not mg. Always derive N from Newton's second law in the perpendicular direction.
Worked Example
A 5 kg book rests on a horizontal table. A horizontal applied force of 20 N pushes the book. Kinetic friction acts on the book. Draw the free-body diagram and list all forces with their directions.
Given
- m = 5 kg
- Applied force FA = 20 N horizontal (right)
- Surface is horizontal, kinetic friction is present
Find
Free-body diagram and list of forces
Solution
- 1Weight W = mg = 5 × 9,8 = 49 N, direction: downward.
- 2Normal force N, direction: upward (perpendicular to surface). Since the surface is horizontal and no vertical applied force: N = W = 49 N.
- 3Applied force FA = 20 N, direction: to the right (horizontal).
- 4Kinetic friction fk, direction: to the LEFT (opposing motion; book moves right).
- 5FBD shows: W down, N up, FA right, fk left — all arrows originating from the block.
Practice Question
A 10 kg crate rests on a rough horizontal floor. A rope pulls it at 30° above the horizontal with a force of 60 N. Draw the free-body diagram and write down all four forces acting on the crate, including their directions.
(6 marks)
2.2 Newton's First and Second Laws
Newton's First Law is the LAW OF INERTIA. Inertia is the tendency of objects to resist changes in their state of motion. A heavy truck is hard to start from rest AND hard to stop once moving — it has a LOT of inertia. A tennis ball is easy to start and stop — it has very little inertia. If the net force on an object is ZERO, the object either stays at rest or keeps moving in a straight line at constant speed. We call this condition EQUILIBRIUM.
Definition
Newton's first law
An object continues in a state of rest or uniform motion (motion with a constant velocity) unless it is acted on by an unbalanced (net or resultant) force.
Newton's Second Law links the resultant force to the resulting acceleration. The acceleration is in the SAME DIRECTION as the net force. If you double the net force, the acceleration doubles. If you double the mass, the acceleration halves. This law allows us to calculate the acceleration of any object if we know all the forces on it.
Definition
Newton's second law
If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. F_net = ma
Formula
Newton's second law
Fnet = net (resultant) force (N), m = mass (kg), a = acceleration (m·s⁻²)
SI unit: N
Exam Tip
EXAM TIP — The step-by-step method for Newton's 2nd Law problems: (1) Draw a free-body diagram. (2) Choose a positive direction. (3) Write Fnet = Σ(all forces) in that direction. (4) Substitute Fnet = ma. (5) Solve for the unknown. Always state the direction of acceleration in your answer.
Worked Example
A 20 kg box rests on a smooth (frictionless) horizontal surface. A horizontal force of 80 N is applied to the right, and another horizontal force of 20 N acts to the left. (a) Draw the free-body diagram. (b) Calculate the net force. (c) Calculate the acceleration.
Given
- m = 20 kg
- FA = 80 N right
- FB = 20 N left
- Surface is smooth (frictionless, horizontal)
Find
(a) FBD (b) Fnet (c) acceleration a
Solution
- 1(a) FBD: W down, N up (equal, vertical equilibrium), FA = 80 N right, FB = 20 N left.
- 2(b) Taking rightward as positive: Fnet = 80 + (−20) = +60 N (to the right).
- 3(c) Fnet = ma → a = Fnet/m = 60/20 = 3 m·s⁻²
- 4Direction: 3 m·s⁻² to the right.
Worked Example
A 1 200 kg car accelerates from rest to 20 m·s⁻¹ in 8 s on a level road. The frictional force on the car is 800 N. Calculate the driving force (engine force) that the car's wheels exert.
Given
- m = 1200 kg
- u = 0 m·s⁻¹
- v = 20 m·s⁻¹
- t = 8 s
- fk = 800 N (opposing motion)
Find
Engine force FE
Solution
- 1Find acceleration: a = (v − u)/t = (20 − 0)/8 = 2,5 m·s⁻²
- 2Apply Newton's 2nd Law (taking forward as positive):
- 3Fnet = ma → FE − fk = ma
- 4FE − 800 = 1200 × 2,5
- 5FE = 3000 + 800 = 3800 N
Practice Question
A 3 kg toy car on a horizontal surface is pushed forward by a 15 N force. Friction acts on it with a magnitude of 3 N. (a) Calculate the net force. (b) Calculate the acceleration. (c) If friction increases to 15 N, what happens to the toy car's acceleration? Explain using Newton's First Law.
(7 marks)
2.3 Newton's Third Law and Action-Reaction Pairs
Imagine pressing your hand against a wall. You push the wall. Can you feel the wall pushing back on your hand? That 'push back' is Newton's Third Law in action. Forces always come in pairs: if object A pushes or pulls object B, then object B ALWAYS pushes or pulls object A with an equal force in the OPPOSITE direction. These are called an ACTION-REACTION PAIR, or a Newton's Third Law pair.
Definition
Newton's third law
If body A exerts a force on body B, then body B exerts a force of equal magnitude on body A, but in the opposite direction.
Properties of Newton's Third Law pairs
- Equal in magnitude — both forces have exactly the same size.
- Opposite in direction — they point in exactly opposite directions (180° apart).
- Same type of force — both forces are the same type (both gravitational, both normal, both friction, etc.).
- Act on DIFFERENT objects — one force acts on body A; the other acts on body B.
- They act simultaneously — the action and reaction occur at the same instant.
- They NEVER cancel each other — because they act on different objects, they cannot be added on the same free-body diagram.
Watch Out
COMMON MISTAKE — Newton's Third Law pairs act on DIFFERENT objects. They do NOT cancel. If you are drawing the free-body diagram of object A, you include the force that B exerts on A — you do NOT include the force A exerts on B (that acts on B, not A). Students often include both forces of a pair on the same FBD, which is wrong.
Identifying Newton's Third Law Pairs
| Property | Action force (on B) | Reaction force (on A) |
|---|---|---|
| Book on table | Book pushes table downward (normal force) | Table pushes book upward (normal force) |
| Rocket and gas | Rocket pushes exhaust gas backward | Exhaust gas pushes rocket forward |
| Foot and ground | Foot pushes ground backward | Ground pushes foot forward (friction) |
| Earth and Moon | Earth pulls Moon toward Earth (gravity) | Moon pulls Earth toward Moon (gravity) |
Worked Example
A 60 kg skater (A) pushes a 90 kg skater (B) with a force of 270 N to the right. (a) State the reaction force on skater A. (b) Calculate the acceleration of skater B. (c) Calculate the acceleration of skater A. Assume frictionless ice.
Given
- mA = 60 kg
- mB = 90 kg
- Force A exerts on B: 270 N to the right
Find
(a) Reaction force (b) aB (c) aA
Solution
- 1(a) By Newton's Third Law: B exerts 270 N to the LEFT on A.
- 2(b) aB = Fnet/mB = 270/90 = 3 m·s⁻² to the right.
- 3(c) aA = Fnet/mA = 270/60 = 4,5 m·s⁻² to the left.
Practice Question
A tennis ball of mass 0,06 kg is struck by a racket. The racket exerts a force of 300 N on the ball for 0,004 s. (a) State the force the ball exerts on the racket (Newton's Third Law). (b) Calculate the acceleration of the ball. (c) Why does the racket barely accelerate, even though the ball exerts an equal force on it?
(7 marks)
2.4 Friction Forces
Friction is a force that opposes relative motion between surfaces in contact. Think of it as 'microscopic roughness': even surfaces that look smooth are covered with tiny peaks and valleys that interlock and resist sliding. There are two types: STATIC friction (before the object moves) and KINETIC friction (while the object is sliding). Static friction is always greater than kinetic friction — that is why it is harder to START an object sliding than to KEEP it sliding.
Formula
Maximum static friction
fs(max) = maximum static friction force (N), μs = coefficient of static friction (no units), N = normal force (N)
SI unit: N
Formula
Kinetic friction
fk = kinetic friction force (N), μk = coefficient of kinetic friction (no units), N = normal force (N)
SI unit: N
Static Friction vs Kinetic Friction
| Property | Static friction (fs) | Kinetic friction (fk) |
|---|---|---|
| When it acts | Before the object starts sliding | While the object is sliding |
| Magnitude | Varies from 0 to a maximum of μsN | Constant = μkN (for a given surface) |
| Coefficient | μs (coefficient of static friction) | μk (coefficient of kinetic friction) |
| Relative size | μs > μk always | μk < μs always |
| Formula | fs ≤ μsN (the object moves when applied force > μsN) | fk = μkN (constant once sliding) |
Watch Out
COMMON MISTAKE — The normal force N is NOT always equal to mg. On a horizontal surface with no vertical applied force: N = mg. On an inclined plane: N = mg cosθ. If an applied force has a vertical component pulling upward: N = mg − Fy. Always calculate N from Newton's 2nd Law in the perpendicular (y) direction before calculating friction.
Worked Example
A 10 kg wooden crate rests on a rough horizontal floor. μs = 0,5 and μk = 0,35. (a) Calculate the maximum static friction force. (b) Calculate the kinetic friction force. (c) A horizontal force of 45 N is applied. Does the crate move? (d) If the force is increased to 55 N, calculate the acceleration of the crate.
Given
- m = 10 kg
- μs = 0,5
- μk = 0,35
- Surface horizontal
- g = 9,8 m·s⁻²
Find
(a) fs(max) (b) fk (c) Does it move at FA = 45 N? (d) a at FA = 55 N
Solution
- 1Normal force: N = mg = 10 × 9,8 = 98 N
- 2(a) fs(max) = μsN = 0,5 × 98 = 49 N
- 3(b) fk = μkN = 0,35 × 98 = 34,3 N
- 4(c) Applied force (45 N) < fs(max) (49 N) → static friction holds → the crate does NOT move.
- 5(d) Applied force (55 N) > fs(max) (49 N) → crate slides. Now use kinetic friction:
- 6Fnet = FA − fk = 55 − 34,3 = 20,7 N
- 7a = Fnet/m = 20,7/10 = 2,07 m·s⁻²
Practice Question
A 15 kg box is on a horizontal surface. μs = 0,4 and μk = 0,3. (a) Calculate N. (b) Calculate the maximum static friction. (c) A force of 70 N is applied horizontally. Does the box move? (d) If the box is already moving, calculate the net force and acceleration.
(8 marks)
2.5 Inclined Plane Problems
On an inclined plane (a ramp at angle θ), the weight W = mg still acts STRAIGHT DOWN. But the surface is not horizontal — so we rotate our coordinate system to align with the surface. The x-axis now runs ALONG the slope and the y-axis runs PERPENDICULAR (normal) to the slope. Then we resolve the weight into two components: one ALONG the slope (causing the object to slide) and one PERPENDICULAR to the slope (pressing the object into the surface).
Formula
Weight component along slope
W∥ = component of weight along the incline (N), m = mass (kg), g = 9,8 m·s⁻², θ = angle of incline
SI unit: N
Formula
Weight component perpendicular to slope
W⊥ = component of weight perpendicular to the incline (N), m = mass (kg), g = 9,8 m·s⁻², θ = angle of incline
SI unit: N
Note
On an inclined plane: Normal force N = W⊥ = mg cosθ (because there is no acceleration perpendicular to the slope). Therefore friction on an incline: fk = μk × mg cosθ.
Worked Example
A 5 kg block slides down a smooth (frictionless) incline angled at 25° to the horizontal. Calculate (a) the component of weight along the slope, (b) the normal force, and (c) the acceleration of the block down the slope.
Given
- m = 5 kg
- θ = 25°
- Smooth surface (frictionless)
- g = 9,8 m·s⁻²
Find
(a) W∥ (b) N (c) a
Solution
- 1(a) W∥ = mg sinθ = 5 × 9,8 × sin25° = 49 × 0,4226 = 20,7 N (down the slope)
- 2(b) N = mg cosθ = 5 × 9,8 × cos25° = 49 × 0,9063 = 44,4 N
- 3(c) Fnet along slope = W∥ − 0 (no friction) = 20,7 N (down slope)
- 4a = Fnet/m = 20,7/5 = 4,14 m·s⁻² down the slope
Worked Example
An 8 kg crate on a 30° incline has μk = 0,2. It is given an initial push up the slope. Calculate the acceleration of the crate as it slides back DOWN the slope after the push.
Given
- m = 8 kg
- θ = 30°
- μk = 0,2
- g = 9,8 m·s⁻²
- Object slides downward
Find
Acceleration a down the slope
Solution
- 1Normal force: N = mg cosθ = 8 × 9,8 × cos30° = 78,4 × 0,866 = 67,9 N
- 2Kinetic friction: fk = μkN = 0,2 × 67,9 = 13,6 N (acts UP the slope, opposing downward motion)
- 3Weight component along slope: W∥ = mg sinθ = 8 × 9,8 × sin30° = 78,4 × 0,5 = 39,2 N (down slope)
- 4Fnet (down slope) = W∥ − fk = 39,2 − 13,6 = 25,6 N
- 5a = Fnet/m = 25,6/8 = 3,2 m·s⁻² down the slope
Exam Tip
EXAM TIP — When a block slides DOWN an incline, friction acts UP the slope (opposing motion). When a block slides UP an incline, friction also acts DOWN the slope (opposing motion). The direction of friction always opposes the direction of motion, NOT the direction of any applied force.
Practice Question
A 12 kg suitcase is on a ramp at 20° to the horizontal. The coefficient of static friction is 0,5. (a) Calculate W∥ (component of weight along the ramp). (b) Calculate the maximum static friction force. (c) Will the suitcase slide? (d) If μk = 0,3 and the suitcase is sliding, calculate its acceleration.
(10 marks)
2.6 Newton's Law of Universal Gravitation
The same force that makes an apple fall from a tree also keeps the Moon in orbit around the Earth and the Earth in orbit around the Sun. Newton realised this and formulated the LAW OF UNIVERSAL GRAVITATION: every object in the universe attracts every other object with a gravitational force. This force depends on how massive the two objects are and how far apart they are.
Definition
Newton's law of universal gravitation
F = Gm₁m₂/d²
Formula
Newton's law of universal gravitation
F = gravitational force (N), G = universal gravitational constant = 6,67 × 10⁻¹¹ N·m²·kg⁻², m₁ and m₂ = masses (kg), d = distance between their centres (m)
SI unit: N
Note
G = 6,67 × 10⁻¹¹ N·m²·kg⁻² is the universal gravitational constant. It is a very small number, which is why we only notice gravitational attraction when at least one object is very massive (like a planet). Two people standing next to each other exert a gravitational force of only about 10⁻⁷ N on each other — completely undetectable.
INVERSE SQUARE LAW: Notice that d² appears in the denominator. If you double the distance between two objects, the gravitational force becomes 1/4 as strong (not half). If you triple the distance, the force becomes 1/9 as strong. This is called an inverse square relationship. Gravity weakens rapidly with distance.
Worked Example
The Earth has a mass of 5,97 × 10²⁴ kg and the Moon has a mass of 7,35 × 10²² kg. The distance between their centres is 3,84 × 10⁸ m. Calculate the gravitational force between the Earth and the Moon.
Given
- mE = 5,97 × 10²⁴ kg
- mM = 7,35 × 10²² kg
- d = 3,84 × 10⁸ m
- G = 6,67 × 10⁻¹¹ N·m²·kg⁻²
Find
Gravitational force F
Solution
- 1F = Gm₁m₂/d²
- 2F = (6,67 × 10⁻¹¹)(5,97 × 10²⁴)(7,35 × 10²²) / (3,84 × 10⁸)²
- 3Numerator: 6,67 × 10⁻¹¹ × 5,97 × 10²⁴ × 7,35 × 10²² = 6,67 × 5,97 × 7,35 × 10⁻¹¹⁺²⁴⁺²² = 292,5 × 10³⁵ = 2,925 × 10³⁷
- 4Denominator: (3,84 × 10⁸)² = 14,75 × 10¹⁶ = 1,475 × 10¹⁷
- 5F = 2,925 × 10³⁷ / 1,475 × 10¹⁷ = 1,98 × 10²⁰ N
Worked Example
Two identical spheres each of mass 5 kg are 0,2 m apart (centre-to-centre). (a) Calculate the gravitational force between them. (b) If the distance is doubled to 0,4 m, what is the new force?
Given
- m₁ = m₂ = 5 kg
- d₁ = 0,2 m
- d₂ = 0,4 m
- G = 6,67 × 10⁻¹¹ N·m²·kg⁻²
Find
(a) F at d = 0,2 m (b) F at d = 0,4 m
Solution
- 1(a) F = G × 5 × 5 / (0,2)² = 6,67 × 10⁻¹¹ × 25 / 0,04 = 166,75 × 10⁻¹¹ / 0,04 = 4169 × 10⁻¹¹ = 4,17 × 10⁻⁸ N
- 2(b) Distance doubled → force reduced by factor of 4 (inverse square law):
- 3F₂ = F₁/4 = 4,17 × 10⁻⁸ / 4 = 1,04 × 10⁻⁸ N
- 4OR directly: F = 6,67 × 10⁻¹¹ × 25 / (0,4)² = 6,67 × 10⁻¹¹ × 25 / 0,16 = 1,04 × 10⁻⁸ N ✓
Watch Out
COMMON MISTAKE — Always use the distance between the CENTRES of mass, not the distance between the surfaces. For a 1 m diameter ball and a 2 m diameter ball touching each other, d = 0,5 + 1,0 = 1,5 m (sum of radii), not 0.
Practice Question
A satellite of mass 800 kg orbits the Earth at a height of 400 km above the surface. The Earth's mass is 5,97 × 10²⁴ kg and its radius is 6,38 × 10⁶ m. (a) Calculate the distance d from the satellite to the Earth's centre. (b) Calculate the gravitational force on the satellite. (c) By Newton's Third Law, what is the gravitational force on the Earth due to the satellite?
(9 marks)