Newton's Laws and Friction
When the Sydney Harbour Bridge opened in 1932, structural engineer Ralph Freeman calculated that its 2.5 million rivets — with a combined friction coefficient of $\mu \approx 0.45$ — carry approximately 40% of the lateral load through friction alone: roughly 52,000 kN per span. This was the first large-scale Australian structure designed to exploit friction ($f = \mu F_N$) and Newton's Second Law simultaneously in its load calculations.
Practise this lesson
Newton's Laws and Friction worksheets.
A delivery driver brakes suddenly at 60 km/h. The van stops. The packages in the back fly forward and slam into the cabin. The driver's seatbelt prevents the same. Which of Newton's three laws best explains each event?
A 1200 kg car accelerates from rest to 25 m/s in 10 s. The net force is approximately:
Know
- Newton's First Law — formal statement with inertia
- Newton's Second Law: $F_{net} = ma$
- Friction formula: $f = \mu F_N$
- Difference between static and kinetic friction ($\mu_s > \mu_k$)
Understand
- Why packages fly forward when a van brakes (N1)
- Why $F$ in $F_{net} = ma$ is the net (total) force
- Why $\mu_s > \mu_k$ — it takes more force to start sliding than to keep sliding
- Why constant velocity does not mean no forces
Can Do
- Draw an FBD for a braking scenario
- Calculate net force and acceleration with friction
- Find acceleration from $F_{net} = ma$
- Distinguish between Newton's First and Second Law applications
Core Content
A delivery van brakes sharply. The van stops — but the unsecured packages in the back fly forward and hit the cabin wall. No one pushed the packages forward. The brakes acted on the van, not on the packages, so the packages kept moving at the van's original speed until the wall stopped them.
Formal statement: An object will remain at rest or continue moving at constant velocity in a straight line unless acted upon by a net external force.
Inertia is the property responsible — it is not a force. It is the tendency to resist change. A heavier object has more inertia and requires more force to change its motion.
In the delivery van scenario: the van brakes due to friction between tyres and road. The packages have no such braking force — they continue at the van's original velocity. The net force on the packages is zero (in the horizontal direction) until they hit the cabin wall.
Newton's First Law: an object remains at rest or at constant velocity when net force is zero; inertia (resistance to change in motion) is proportional to mass and is not itself a force.
Pause — copy the highlighted law and clarification into your book before moving on.
True or false: A car travelling at constant 60 km/h on a straight highway has no forces acting on it.
We just saw that inertia keeps objects at constant velocity unless a net force acts. That raises a question: when a net force does act, how does it determine the resulting acceleration? This card answers it → Newton's Second Law gives the quantitative relationship: $F_{net} = ma$.
Acceleration is proportional to net force and inversely proportional to mass. Push harder, accelerate more. Push the same force on something heavier, accelerate less.
Critical: $F$ in this equation is the net force — the vector sum of all forces acting. If a car engine pushes with 3000 N and friction acts with 800 N backward, the net force is 2200 N forward, giving $a = 2200/m$.
A 1400 kg car has engine force 4200 N forward. Friction is 700 N backward. Find the acceleration.
Newton's Second Law: $F_{net} = ma$ where $F_{net}$ is the vector sum of ALL forces (N), $m$ is mass (kg), and $a$ is acceleration (m s$^{-2}$); 1 N = 1 kg m s$^{-2}$; direction of acceleration equals direction of net force.
Pause — copy the highlighted law and unit into your book before moving on.
A 60 kg person stands on bathroom scales in an accelerating lift. The scales read 720 N. What is the acceleration of the lift? ($g = 9.8$ m/s²)
We just saw that $F_{net} = ma$ requires knowing the net force — which includes friction. That raises a question: how do we calculate the friction force, and does it change once the object starts moving? This card answers it → static friction is variable (up to $\mu_s F_N$) while kinetic friction is constant at $\mu_k F_N$, with $\mu_s > \mu_k$.
It takes more force to start an object sliding than to keep it sliding — that's why $\mu_s > \mu_k$.
Static friction is a variable force — it exactly equals the applied force until the maximum is reached. Once sliding begins, kinetic friction ($\mu_k F_N$) takes over and is approximately constant.
The normal force $F_N$ on a horizontal surface equals $mg$. On a slope, $F_N = mg\cos\theta$ (from L03).
Maximum static friction $f_{s,max} = \mu_s F_N$ (variable up to this limit); kinetic friction $f_k = \mu_k F_N$ (constant while sliding); always $\mu_s > \mu_k$; normal force $F_N = mg$ on horizontal surfaces and $F_N = mg\cos\theta$ on slopes.
Pause — copy the highlighted formulae and the $\mu_s > \mu_k$ rule into your book before moving on.
Which statement about friction is the odd one out (incorrect)?
Activities
A 2000 kg delivery van travelling at 15 m/s brakes to rest. The braking force (friction from tyres) is 8000 N backward. Draw the FBD and calculate: (a) net force, (b) deceleration, (c) time to stop from 15 m/s.
For each scenario, identify which of Newton's laws applies (N1, N2, or N3) and explain why:
- A parcel slides forward in a braking van
- A heavier trolley accelerates less than a lighter one under the same force
- A swimmer pushes the pool wall and moves backward
- A car moves at constant speed on a highway
Pick your answer, then rate your confidence.
UnderstandBand 3(3 marks) 1. Explain using Newton's First and Second Laws why a passenger in a car that brakes suddenly feels as though they are thrown forward, while in reality they are simply continuing forward at the original speed.
ApplyBand 4(3 marks) 2. A 750 kg car accelerates from rest. The engine exerts a forward force of 3000 N and friction is 450 N opposing motion. Calculate the net force, acceleration, and velocity after 5 s.
AnalyseBand 5(4 marks) 3. A 50 kg block rests on a horizontal surface. The coefficient of static friction is 0.55 and kinetic friction is 0.40. A horizontal force is applied. (a) What is the minimum force needed to start the block sliding? (b) Once sliding, if the same force is maintained, calculate the acceleration. ($g = 9.8$ m/s²)
Show all answers
Short Answer — Model Answers
Q1 (3 marks): By Newton's First Law, the passenger's body tends to remain at its original velocity unless a net force acts on it. When the car brakes, a large braking force acts on the car via tyre-road friction — but no equivalent force acts on the passenger horizontally. The passenger continues forward at the original speed while the car decelerates. This appears as if the passenger is thrown forward, but they are actually staying still while the car's interior decelerates around them. Newton's Second Law: $F_{net} = ma$; for the car, $F_{net}$ is large and backward. For the unsecured passenger, $F_{net} \approx 0$, so $a \approx 0$.
Q2 (3 marks): $F_{net} = 3000 - 450 = 2550\text{ N}$ (forward). $a = 2550/750 = 3.4\text{ m/s}^2$ (forward). $v = u + at = 0 + 3.4 \times 5 = 17\text{ m/s}$.
Q3 (4 marks): (a) $F_N = mg = 50 \times 9.8 = 490\text{ N}$. $f_{s,max} = \mu_s F_N = 0.55 \times 490 = 269.5\text{ N}$. Minimum force to start sliding = 269.5 N (just above). (b) Once sliding: $f_k = 0.40 \times 490 = 196\text{ N}$. $F_{net} = 269.5 - 196 = 73.5\text{ N}$. $a = 73.5/50 = 1.47\text{ m/s}^2$.
Five timed questions on Newton's Laws and Friction. Beat the boss to bank a tier.
⚔ Enter the arenaThe Sydney Harbour Bridge (1932) shows that friction ($f = \mu F_N$) is an engineering tool, not just a hazard: its 2.5 million rivets with $\mu \approx 0.45$ carry 52,000 kN of lateral load per span. In the van braking scenario, Newton's First Law explains why packages fly forward (no net force acted on them), while Newton's Second Law ($F_{net} = ma$) and $f = \mu F_N$ together determine the van's deceleration rate — exactly the same two tools Freeman used on the bridge.