Newton's Three Laws of Motion — Simply Explained

Newton's Laws of Motion are the backbone of mechanics in Physical Science. They appear in NSC and IEB Paper 1 every year — in questions about cars braking, blocks on inclined planes, objects connected by ropes, projectiles, and more. If you can apply all three laws systematically to any diagram and set of forces, you will score well in mechanics.

Understanding the laws is not enough. You need to be able to use them in calculations, and that requires a specific technique. This guide covers the laws, the technique, and the exam mistakes that cost marks.

The Free-Body Diagram: Draw First, Calculate Second

Before any calculation, draw a free-body diagram (FBD). This is a simple sketch showing a single object as a dot, with labelled arrows representing every force acting on it. The length and direction of each arrow represents the magnitude and direction of that force.

Every mechanics problem you encounter should start with an FBD. Students who skip this step make substantially more errors, especially in multi-object or inclined-plane problems.

Forces you might include (depending on the problem):

• Weight: $W = mg$, always pointing straight down
• Normal force: $N$, perpendicular to the contact surface
• Friction: $f$, along the surface, opposing motion (or opposing potential motion)
• Applied force: $F$, in the direction of application
• Tension: $T$, along a string or rope, away from the object
• Air resistance: opposing velocity

Once you have the FBD, you can work with Newton's Laws.

Newton's First Law: The Law of Inertia

Statement: An object remains at rest, or moves at constant velocity in a straight line, unless acted upon by a net external force.

What This Actually Means

Objects resist changes to their state of motion. An object at rest wants to stay at rest. An object moving at constant velocity wants to keep doing exactly that — constant speed, straight line.

This resistance to change is called inertia. Importantly, Newton's First Law is a special case of the Second Law where the net force equals zero.

First Law in Exam Questions

The First Law appears in two forms:

Conceptual questions — "Explain why a passenger lurches forward when a car brakes suddenly." Answer: The passenger's body tends to continue moving forward (inertia) — the car decelerates under them but no net force has been applied to the passenger's upper body quickly enough to decelerate it at the same rate.

Calculation questions — When an object is in equilibrium (at rest or constant velocity), the net force is zero. This lets you set up equations where forces balance.

Worked Example 1: Object in Equilibrium

A 5 kg box is pushed along a horizontal surface at constant velocity by a 20 N horizontal force. Calculate the friction force acting on the box.

Since velocity is constant, $F_{\text{net}} = 0$ (Newton's First Law).

$F_{\text{applied}} - f = 0$

$20 - f = 0$

$f = \mathbf{20\ \text{N}}$

The friction force equals the applied force in magnitude, opposing the direction of motion.

Newton's Second Law: $F_{\text{net}} = ma$

Statement: The net force acting on an object is equal to the product of the object's mass and its acceleration. The direction of the acceleration is the same as the direction of the net force.

$F_{\text{net}} = ma$

The Four-Step Method

1. Draw the FBD — label every force acting on the object
2. Choose a positive direction — explicitly state which direction you are taking as positive (usually the direction of acceleration)
3. Write the net force equation — add up all forces, using $+$ for forces in your chosen positive direction and $-$ for forces opposing it
4. Substitute and solve

This method works for every $F = ma$ problem at matric level. The same steps, every time.

Worked Example 2: Single Object, Horizontal Surface

A 10 kg crate is pushed along a horizontal floor by a force of 50 N. The friction force is 18 N. Calculate the acceleration of the crate.

FBD: Weight down, Normal up, Applied force right (positive), Friction left (negative)

Taking right as positive:

$F_{\text{net}} = F_{\text{applied}} - f = 50 - 18 = 32\ \text{N}$

$F_{\text{net}} = ma$

$32 = 10 \times a$

$a = \mathbf{3.2\ \text{m·s}^{-2}} \text{ (to the right)}$

Worked Example 3: Inclined Plane

A 4 kg block slides down a frictionless inclined plane at $30^\circ$ to the horizontal. Calculate its acceleration.

On an inclined plane, weight ($mg$) must be resolved into components parallel and perpendicular to the surface:

• Parallel to slope (down the slope): $mg\sin 30^\circ$
• Perpendicular to slope: $mg\cos 30^\circ$

Taking down the slope as positive:

$F_{\text{net}} = mg\sin 30^\circ = 4 \times 10 \times 0.5 = \mathbf{20\ \text{N}}$

$F_{\text{net}} = ma$

$20 = 4 \times a$

$a = \mathbf{5\ \text{m·s}^{-2}} \text{ (down the slope)}$

Note: The normal force and the perpendicular component of weight cancel — they do not affect the acceleration along the slope.

Worked Example 4: Two Objects Connected by a String

A 3 kg block (A) and a 5 kg block (B) are connected by a light inextensible string. Block B hangs off the edge of a frictionless table. Block A is on the table. Find the acceleration of the system and the tension in the string.

For the system as a whole (taking the direction of motion as positive):

The only unbalanced force is the weight of block B:

$F_{\text{net}} = m_B \times g = 5 \times 10 = 50\ \text{N}$

Total mass $= 3 + 5 = 8\ \text{kg}$

$50 = 8a$

$a = \mathbf{6.25\ \text{m·s}^{-2}}$

For block A only (to find tension):

$T = m_A \times a = 3 \times 6.25 = \mathbf{18.75\ \text{N}}$

Check using block B:

$m_B g - T = m_B a$

$50 - T = 5 \times 6.25 = 31.25$

$T = 50 - 31.25 = 18.75\ \text{N}\ ✓$

Common Second Law Mistakes

Using total weight instead of net force. The equation is $F_{\text{net}} = ma$, not $F_{\text{any}} = ma$. You must sum all forces in your chosen direction first.

Mixing up mass and weight. Weight $= mg$ (in newtons). Mass $=$ kg. Never write $W$ in place of $m$ in $F = ma$.

Wrong sign for friction or tension. Forces opposing the positive direction get a minus sign. If you take right as positive and friction acts left, it is $-f$ in the equation.

Treating a multi-object system incorrectly. For the whole system, internal forces (like string tension between the objects) cancel out. For an individual object, tension is an external force that must be included.

Newton's Third Law: Action–Reaction

Statement: When object A exerts a force on object B, object B simultaneously exerts a force of equal magnitude and opposite direction on object A.

The Most Important Clarification

Newton's Third Law pairs are always:

• Equal in magnitude
• Opposite in direction
• Acting on different objects
• The same type of force

This last point is often missed. If Earth exerts a gravitational force on you downward, you exert a gravitational force on Earth upward — not a normal force, not a contact force, but a gravitational force. Action-reaction pairs are always the same interaction type.

Why Don't Action-Reaction Forces Cancel?

Students frequently ask: if every action has an equal and opposite reaction, why does anything ever accelerate?

The answer: Newton's Second Law applies to one object at a time. Action-reaction forces act on different objects, so they cannot cancel. Only forces acting on the same object can cancel each other in the net force calculation.

Example: You push a wall with 100 N. The wall pushes you back with 100 N. These forces act on different objects (you vs the wall). To find your acceleration, you only consider forces acting on you — including the 100 N from the wall.

Worked Example 5: Identifying Action-Reaction Pairs

A horse pulls a cart forward. Identify the Newton's Third Law pair for the force the horse exerts on the cart.

Force: Horse exerts a forward force on the cart.

Reaction pair: The cart exerts a backward force of equal magnitude on the horse.

(Note: These forces cannot cancel — they act on different objects. Whether the system accelerates depends on the net force on each object individually.)

Third Law in the Exam

Third Law questions most commonly appear as:

• Identify the action-reaction pair for a named force
• Explain why two equal forces do not mean zero acceleration
• Questions about rockets, swimming, jets, or collisions

The key phrase to use in written answers: "...of equal magnitude and opposite direction, acting on a different object."

Exam Strategy: Multi-Object Problems

The majority of high-mark mechanics questions involve two or more objects. Use this approach:

1. Draw a separate FBD for each object
2. Write separate $F_{\text{net}} = ma$ equations for each object
3. Identify shared quantities: the objects connected by a string have the same acceleration (if the string is inextensible) and the tension in the string is the same at both ends
4. You now have a system of simultaneous equations — solve for the unknowns

This approach never fails. The only variable is how many objects and how many forces there are.

Practice Problems

1. A 2 kg block sits on a frictionless horizontal surface. A 12 N horizontal force is applied. Find the acceleration.
2. A 6 kg block is on a rough surface ($f = 15\ \text{N}$). A 39 N force pushes it horizontally. Find the acceleration.
3. Block A (8 kg) sits on a frictionless table connected via a string to Block B (2 kg) hanging off the edge. Find the system acceleration and string tension.
4. A 5 kg block slides down a $40^\circ$ frictionless incline. Find its acceleration. (Use $g = 10\ \text{m·s}^{-2}$)
5. A car of mass 1200 kg accelerates from rest to $20\ \text{m·s}^{-1}$ in 8 seconds. Find the net force acting on it.

Answers: 1) $6\ \text{m·s}^{-2}$, 2) $4\ \text{m·s}^{-2}$, 3) $a = 2\ \text{m·s}^{-2}$, $T = 16\ \text{N}$, 4) $6.43\ \text{m·s}^{-2}$, 5) $3000\ \text{N}$

Need Help With Newton's Laws?

Mechanics is one of those topics where working through problems with someone — having your FBDs checked and your reasoning corrected in real time — produces faster progress than working alone. If inclined planes, two-body problems, or friction are giving you trouble, book a session.

I cover all of Grade 11 and Grade 12 Physical Science, including mechanics, electricity, waves, and chemistry, for both NSC and IEB students.