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VCE Physics (Year 12) · Motion and Forces · 20m read

Newton's Laws and Free-Body Diagrams

Ever wondered why drawing a simple picture before solving a force problem can mean the difference between full marks and zero? A free-body diagram is your superpower—it shows every force acting on an object so you can apply Newton's three laws (force causes acceleration, forces come in pairs, and objects keep doing what they're doing unless pushed). From pushing cars up hills to understanding why seatbelts matter, these tools let you predict motion in the real world and crack every VCE Physics force question.

Learning objective: Apply Newton's three laws and free-body diagrams to solve single-body, connected-body and inclined-plane force problems.

By the end of this lesson you can:

  • I draw a complete, labelled FBD for any object in a force scenario.
  • I apply $\sum F_x = ma_x$ and $\sum F_y = ma_y$ independently.
  • I correctly identify Newton's third-law pairs and explain why they act on different objects.
  • I decompose weight on inclined planes using $mg\sin\theta$ and $mg\cos\theta$.
  • I solve connected-body problems by analysing each body separately and using equal accelerations.

Free-body diagrams are a tug-of-war scoreboard. Draw every force acting on one object; the net force decides which way it accelerates.

Newton's three laws — precise statements

### Newton's First Law (Law of Inertia)

An object remains at rest or in uniform motion (constant velocity) unless acted upon by a non-zero net force.

> *Implication:* An object moving at constant velocity has $\sum F = 0$. This does NOT mean no forces act — it means the forces balance. A car at constant speed on a level road has balanced engine force and friction.

Inertia is an object's resistance to changes in its motion. It is measured by mass $m$ (in kg). Greater mass → greater inertia → harder to accelerate.

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### Newton's Second Law (Law of Acceleration)

$$\sum \vec{F} = m\vec{a}$$

The net (resultant) force equals mass times acceleration. Both $\sum F$ and $a$ are vectors in the same direction. Apply component-wise:

$$\sum F_x = ma_x \qquad \sum F_y = ma_y$$

This law applies to the net force — you must identify all forces, then sum them with correct signs. A 10 N applied force with 6 N friction gives $\sum F = 4$ N, not 10 N.

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### Newton's Third Law (Law of Interaction Pairs)

For every force body A exerts on body B, body B exerts an equal magnitude, opposite direction force on body A.

$$\vec{F}_{AB} = -\vec{F}_{BA}$$

Critical point: The two forces in a Newton's 3rd law pair:
- Are always equal in magnitude and opposite in direction
- Always act on different objects (one on A, one on B)
- Are always the same type of force

Common mistake: Students think normal force and weight are a Newton's 3rd pair because they're equal and opposite for a stationary object. They are NOT — they act on the *same* object. The real N3L pair for weight (Earth pulls book down) is: book pulls Earth up with the same force.

Forces you need to know and how to draw them

Every object in a VCAA problem experiences a subset of these forces. The FBD contains only forces on the object of interest, not forces the object exerts on others.

| Force | Symbol | Direction | Magnitude |
|-------|--------|-----------|-----------|
| Weight (gravity) | $W$ or $F_g$ | Vertically downward | $mg$ |
| Normal | $N$ | Perpendicular to surface, away from it | Depends on situation |
| Friction | $f$ | Along surface, opposing relative motion | $f \leq \mu N$ |
| Tension | $T$ | Along string/rope, away from object | Set up equations |
| Applied force | $F_{app}$ | As specified | As specified |
| Air resistance | $F_{drag}$ | Opposing velocity | Depends on speed |

### Drawing the FBD — step by step

  • Isolate the object — draw it as a dot or box
  • Identify every surface contact — each contact gives a normal force and potentially friction
  • Identify weight — always act downward from the centre of mass
  • Identify any strings — tension acts along the string, away from the object
  • Label magnitudes — use symbols like $N$, $T$, $mg$
  • Indicate direction — arrows must be proportional in length to magnitude

### Axes on inclined planes

Tilt the coordinate system so $x$ runs along the slope and $y$ runs perpendicular to it:

$$W_x = mg\sin\theta \quad \text{(component along slope, down the incline)}$$
$$W_y = mg\cos\theta \quad \text{(component into the surface)}$$

The normal force balances $W_y$ (if there's no vertical acceleration perpendicular to the surface):
$$N = mg\cos\theta$$

Newton's Laws Formula Sheet

  • Newton 2 (net): \sum \vec{F} = m\vec{a} (Net force — always sum first)
  • Weight: W = mg ($g = 9.8 \ \text{m/s}^2$ on Earth (VCAA standard))
  • Normal (flat): N = mg (Horizontal surface, no vertical acceleration)
  • Weight — along incline: W_\parallel = mg\sin\theta (Down the slope, with $\theta$ from horizontal)
  • Weight — into surface: W_\perp = mg\cos\theta (Perpendicular to slope)
  • Normal on incline: N = mg\cos\theta (No perpendicular acceleration)
  • Connected bodies: a = \dfrac{F_{net}}{m_1 + m_2} (Apply to system as a whole)

Force problem protocol

  • 1. Draw FBD: Every force labelled. Do this before touching equations.
  • 2. Choose axes: Horizontal+vertical OR along+perpendicular to slope.
  • 3. Write ΣF equations: $\sum F_x = ma_x$ and $\sum F_y = ma_y$ separately.
  • 4. Substitute and solve: Plug in known values, solve for the unknown.
  • 5. N3 pairs: different objects: If both forces are on the same body, it's N2 not N3.
  • 6. Connected bodies: Find system acceleration first ($a = F/M_{total}$), then internal tension.

Worked example (Easy)

A 5.0 kg box is pulled along a frictionless horizontal surface by a horizontal force of $30 \ \text{N}$. Find the acceleration and the normal force.

  1. Draw the FBD. Forces: Weight $W = mg = 5.0 \times 9.8 = 49 \ \text{N}$ downward; Normal $N$ upward; Applied force $F = 30 \ \text{N}$ to the right. No friction (frictionless).
  2. Vertical equation ($y$-direction). No vertical acceleration, so $\sum F_y = 0$: N - mg = 0 \implies N = mg = 49 \ \text{N}
  3. Horizontal equation ($x$-direction). Only the applied force acts horizontally: \sum F_x = ma_x \implies 30 = 5.0 \times a \implies a = 6.0 \ \text{m/s}^2

Answer: Acceleration $a = 6.0 \ \text{m/s}^2$ (to the right); Normal force $N = 49 \ \text{N}$ (upward).

Worked example (Medium)

A 4.0 kg block on a horizontal surface is pushed by a $25 \ \text{N}$ horizontal force. Friction is $9.0 \ \text{N}$. Find the acceleration.

  1. FBD forces. $W = 39.2 \ \text{N}$ down, $N = 39.2 \ \text{N}$ up, $F_{app} = 25 \ \text{N}$ right, $f = 9.0 \ \text{N}$ left.
  2. Net horizontal force. Friction opposes applied force: \sum F_x = F_{app} - f = 25 - 9.0 = 16 \ \text{N}
  3. Apply Newton 2. a = \frac{\sum F_x}{m} = \frac{16}{4.0} = 4.0 \ \text{m/s}^2

Answer: Acceleration $a = 4.0 \ \text{m/s}^2$ (in the direction of the applied force).

Worked example (Hard)

Two blocks are connected by a light string over a frictionless pulley. Block A ($m_A = 6.0 \ \text{kg}$) sits on a frictionless incline at $30°$ to the horizontal. Block B ($m_B = 4.0 \ \text{kg}$) hangs vertically. Find the acceleration of the system and the tension in the string.

  1. Set up the system. Block A tends to slide down the incline; Block B pulls it. Let $a$ be positive down the incline / down for B.
  2. Forces on Block A (along incline). Component of weight down slope = $m_A g \sin 30°$; tension $T$ up slope. m_A g \sin 30° - T = m_A a \implies 6.0(9.8)(0.5) - T = 6.0a \implies 29.4 - T = 6a \quad\text{...(1)}
  3. Forces on Block B (vertically). Weight down, tension up. m_B g - T = m_B a \implies 4.0(9.8) - T = 4.0a \implies 39.2 - T = 4a \quad\text{...(2)}
  4. Solve simultaneously. Subtract (1) from (2): $39.2 - 29.4 = 4a - 6a + (T - T)$. Wait — let me add equations eliminating $T$: \text{(2)}: \ T = 39.2 - 4a \ \text{; substitute into (1):} \ 29.4 - (39.2 - 4a) = 6a
  5. Solve for a. 29.4 - 39.2 + 4a = 6a \implies -9.8 = 2a \implies a = -4.9 \ \text{m/s}^2
  6. Interpret sign and find T. Negative $a$ means Block B actually pulls upward on A — Block A moves UP the slope. Magnitude $a = 4.9 \ \text{m/s}^2$. T = 39.2 - 4.0(-4.9) = 39.2 + 19.6 = 58.8 \ \text{N} \approx 59 \ \text{N}

Answer: $a = 4.9 \ \text{m/s}^2$ (Block A accelerates up the slope, Block B accelerates upward — because $m_B g > m_A g \sin 30°$); $T \approx 59 \ \text{N}$.

Common mistakes

  • Claiming normal force and weight are a Newton's 3rd law pair Newton's 3rd law pairs act on different objects. Normal and weight both act on the same object (e.g., the book). They are equal in a static situation because of Newton's 2nd law ($\sum F = 0$, not because they are a reaction pair). Fix: Find N3L pairs by asking: "What two objects are interacting?" The pair for weight (Earth pulls book down) is: book pulls Earth upward. The pair for normal (table pushes book up) is: book pushes table downward.
  • Using total weight instead of component along incline On an inclined plane, not all of $mg$ acts along the slope. Only $mg\sin\theta$ acts along the slope. Using $mg$ instead gives an overestimate of the acceleration. Fix: Always decompose weight using the tilt angle: $W_{\parallel} = mg\sin\theta$ along slope, $W_{\perp} = mg\cos\theta$ into surface.
  • Forgetting that friction doesn't have a direction until you know motion direction Friction opposes relative motion (or tendency of motion). If the problem doesn't specify direction of movement, you must infer which way the object would tend to move without friction, then draw friction opposing that. Fix: Determine the direction of motion (or tendency) first, then draw friction opposing it. If static, check whether applied force exceeds maximum static friction.

Newton's laws in extreme conditions — where they break down and what replaces them

Newton's three laws are extraordinarily successful for everyday physics — cars, planes, satellites, and sport. But they have limits.

At very high speeds ($v \gtrsim 0.1c$), Newton's second law $F = ma$ must be replaced by the relativistic form:

$$F = \frac{dp}{dt} = \frac{d(\gamma mv)}{dt}$$

where $\gamma = 1/\sqrt{1 - v^2/c^2}$. At low speeds, $\gamma \approx 1$ and this reduces to $F = ma$. At $v = 0.99c$, $\gamma \approx 7.1$, so the effective "inertia" is over seven times the rest mass. No finite force can accelerate an object to the speed of light — $\gamma \to \infty$ as $v \to c$.

At very small scales (atoms, electrons, photons), Newtonian mechanics is replaced by quantum mechanics. Particles do not have definite positions and momenta simultaneously (Heisenberg uncertainty principle: $\Delta x \cdot \Delta p \geq \hbar/2$). Instead of $F = ma$, the Schrödinger equation governs the evolution of a probability wavefunction.

In strong gravitational fields (near black holes, neutron stars), Newton's law of gravitation $F = GMm/r^2$ is replaced by Einstein's general relativity, which describes gravity as the curvature of spacetime rather than a force.

Despite these limitations, Newton's laws remain the foundation of engineering. Bridges, buildings, vehicles, and spacecraft are all designed using $F = ma$ and free-body diagrams — the same tools you learn at VCE level. Understanding where these laws apply (and where they don't) is the hallmark of a mature physicist.

Practice questions

  1. A $3.0 \ \text{kg}$ book rests on a table. The Newton's third law reaction partner to the weight force (Earth pulling the book down with $29.4 \ \text{N}$) is:

    1. The table pushing the book up with $29.4 \ \text{N}$
    2. The book pulling Earth upward with $29.4 \ \text{N}$
    3. The book pushing the table down with $29.4 \ \text{N}$
    4. Gravity at the centre of the Earth

    (1 marks)

  2. A $5.0 \ \text{kg}$ block sits on a frictionless surface. A horizontal force of $20 \ \text{N}$ is applied. The acceleration is:

    1. $4.0 \ \text{m/s}^2$
    2. $2.0 \ \text{m/s}^2$
    3. $0.25 \ \text{m/s}^2$
    4. $100 \ \text{m/s}^2$

    (2 marks)

  3. A block of mass $m$ sits on a smooth incline at angle $\theta$. What is the acceleration of the block along the slope?

    1. $g$
    2. $g\cos\theta$
    3. $g\sin\theta$
    4. $g\tan\theta$

    (3 marks)

  4. Two blocks ($m_1 = 2.0 \ \text{kg}$ and $m_2 = 3.0 \ \text{kg}$) are connected by a string on a frictionless horizontal surface. A $10 \ \text{N}$ force pulls $m_1$. Find (a) the acceleration and (b) the tension in the string between them.

    (3 marks)

  5. A $4.0 \ \text{kg}$ object slides down a $40°$ frictionless incline. Calculate (a) the acceleration and (b) the normal force.

    (4 marks)

Summary

  • Newton 1: no net force → constant velocity (or rest). Net force → acceleration.
  • Newton 2: $\sum \vec{F} = m\vec{a}$. Apply component-wise: $\sum F_x = ma_x$, $\sum F_y = ma_y$.
  • Newton 3: every force has an equal, opposite partner on a different object.
  • FBD: draw every force on the object before writing any equations.
  • Inclined planes: tilt axes. Use $mg\sin\theta$ along slope, $mg\cos\theta$ perpendicular to slope.
  • Connected bodies: find system acceleration from net force / total mass, then find tension from one body.

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