Push a heavy box and at first it refuses to budge, then suddenly it breaks free and slides more easily. You have just felt the difference between static friction (which holds things still, adjusting itself to your push) and kinetic friction (which resists sliding at a fixed value). Both come from the same source, rough surfaces gripping each other, and both obey the simple relationship $F = \mu N$.
A crate sits on the floor. You push it gently with $30\ \text{N}$ and it does not move; you push harder with $80\ \text{N}$ and it still does not move; at $120\ \text{N}$ it finally slides.
Before reading: while the crate is not moving, what is the friction force at each push, $30\ \text{N}$, $80\ \text{N}$, or always the maximum? Write your prediction and one sentence of reasoning.
Friction that holds still, and friction that resists sliding
Friction comes in two flavours: one that quietly holds a stationary object in place by matching your push, and one that resists an object that is already moving.
Static friction acts on an object that is not moving. It is a self-adjusting force: it grows to exactly cancel whatever push you apply, so the object stays still, but only up to a maximum, $F_{s,max} = \mu_s N$. Push harder than that maximum and the object breaks free.
Kinetic friction acts on an object that is already sliding. It is roughly constant, $F_k = \mu_k N$, and points opposite to the motion. Because $\mu_k$ is usually a little less than $\mu_s$, friction drops the moment sliding begins, which is why a box lurches forward as it breaks free.
Static friction holds a stationary object still by matching the applied force, up to a maximum $F_{s,max} = \mu_s N$. Kinetic friction acts on a sliding object and is roughly constant, $F_k = \mu_k N$. Because $\mu_k < \mu_s$, friction drops once sliding begins, so it takes more force to start than to keep an object moving.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: it usually takes a larger force to start an object sliding than to keep it sliding.
Friction depends on the normal force and the coefficient
We just saw the difference between static and kinetic friction. That raises a question: how large is the friction force in a real problem? This card answers it → use the NESA formula-sheet relationship $F = \mu N$, then feed the friction into the net force.
The size of friction is set by how hard the surfaces are pressed together (the normal force) and how rough they are (the coefficient).
The friction force is given on the NESA formula sheet by:
$$F = \mu N$$
where $N$ is the normal force (on a flat surface, $N = mg$) and $\mu$ is the coefficient of friction for the two surfaces. Use $\mu_k$ for an object that is sliding. Notice that friction does not depend on the contact area, only on $N$ and the surfaces.
A $10\ \text{kg}$ crate is pushed across a level floor with a horizontal force of $50\ \text{N}$. The coefficient of kinetic friction between crate and floor is $\mu_k = 0.30$. Take $g = 9.8\ \text{m s}^{-2}$. Find the friction force and the crate's acceleration.
Friction force: $F = \mu N$, with $N = mg$ on a flat surface. To get acceleration, subtract friction from the applied force, then use $F_{net} = ma$. For the example, $N = 98\ \text{N}$, $F_k = 0.30 \times 98 = 29.4\ \text{N}$, and $a = (50 - 29.4)/10 = 2.1\ \text{m s}^{-2}$.
Add the highlighted method to your notes before the check below.
Quick check: A $5.0\ \text{kg}$ box slides on a floor where $\mu_k = 0.40$. Using $g = 9.8\ \text{m s}^{-2}$, the kinetic friction force is closest to…
Pattern A, Calculate then reason
A $20\ \text{kg}$ box sits on a floor with $\mu_s = 0.50$ and $\mu_k = 0.35$. A person pushes it horizontally with $80\ \text{N}$. Take $g = 9.8\ \text{m s}^{-2}$. In your book: (1) calculate the maximum static friction $\mu_s N$; (2) decide whether the box moves, comparing the push to the maximum static friction; (3) if it does move, calculate the kinetic friction and the resulting acceleration.
Two truths, one lie: which statement comparing static and kinetic friction is the lie?
Dot point: Compare static and kinetic friction
The microscopic story, and why $\mu_s > \mu_k$
We just learned to calculate friction with $F = \mu N$. That raises a question: why is the maximum static friction larger than the kinetic friction in the first place? This card answers it → microscopic bonds form between stationary surfaces and have to be broken before sliding.
Surfaces that look smooth are jagged up close; when they sit still, tiny contact points settle and bond, and those bonds take extra force to break.
No surface is perfectly smooth. Microscopic bumps interlock, and where they touch, weak bonds form. When an object is stationary, these bonds have time to settle, so a larger force is needed to break them all at once, this is why maximum static friction is large. Once the object slides, the surfaces skip across the bumps without time to settle into firm contact, so kinetic friction is smaller and steadier. Because $\mu_s > \mu_k$, the friction force suddenly decreases at the instant of breaking free, and the object jerks forward.
Friction arises from microscopic roughness and bonding between surfaces. Stationary surfaces bond more firmly, giving a larger maximum static friction ($\mu_s$); a sliding object skips over the bumps, giving a smaller kinetic friction ($\mu_k$). This is why $\mu_s > \mu_k$ and starting is harder than sliding.
Pause, copy the highlighted reasoning into your book before the check below.
Last check: Decide if the statement is true or false.
Doubling the contact area of a sliding box on a floor doubles the kinetic friction force.
Fill the gaps. Friction is calculated as $F = \mu$ , where $N$ is the . The coefficient $\mu$ has no .
Dot point: Solve problems using friction: F = μN
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $15\ \text{kg}$ sled is pulled along level ground by a horizontal force of $60\ \text{N}$. The coefficient of kinetic friction is $\mu_k = 0.25$. Using $g = 9.8\ \text{m s}^{-2}$, calculate the friction force and the sled's acceleration, showing each step.
1 mark: $N = mg = 147\ \text{N}$ · 1 mark: $F_k = \mu_k N = 0.25 \times 147 = 36.75\ \text{N}$ · 1 mark: net force $= 60 - 36.75 = 23.25\ \text{N}$ · 1 mark: $a = 23.25/15 \approx 1.6\ \text{m s}^{-2}$
ApplyBand 5(4 marks) 2. A $25\ \text{kg}$ crate rests on a horizontal floor with a coefficient of kinetic friction of $0.30$. Taking $g = 9.8\ \text{m s}^{-2}$, calculate the friction force when it slides, and the horizontal force needed to slide it at constant velocity.
1 mark: $N = mg = 245\ \text{N}$ · 1 mark: applies $F = \mu N$ · 1 mark: friction $= 73.5\ \text{N}$ · 1 mark: constant velocity means the applied force equals the friction, $73.5\ \text{N}$
Dot point: Solve problems using friction: F = μN
AnalyseBand 6(3 marks) 3. Explain why a heavy box is harder to start moving than to keep moving once it is sliding, referring to static and kinetic friction.
1 mark: static friction has a larger maximum than kinetic friction · 1 mark: the applied force must exceed maximum static friction to start motion · 1 mark: once sliding, only the smaller kinetic friction must be overcome
Dot point: Compare static and kinetic friction
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): On level ground the normal force equals the weight: $N = mg = 15 \times 9.8 = 147\ \text{N}$. The kinetic friction force is $F_k = \mu_k N = 0.25 \times 147 = 36.75\ \text{N}$, opposing the motion. The net force along the ground is $F_{net} = 60 - 36.75 = 23.25\ \text{N}$. Applying Newton's second law, $a = \dfrac{F_{net}}{m} = \dfrac{23.25}{15} \approx 1.6\ \text{m s}^{-2}$, directed in the direction of the pull.
Q2 (4 marks): On a horizontal surface with no vertical applied force, the normal force equals the weight: $$N = mg = 25 \times 9.8 = 245\ \text{N}$$ The kinetic friction is then $$F = \mu N = 0.30 \times 245 = 73.5\ \text{N}$$ acting opposite to the motion. To slide the crate at constant velocity the acceleration is zero, so by Newton’s first law the net force must be zero and the applied horizontal force must exactly balance the friction: $73.5\ \text{N}$. Any larger force would accelerate the crate; any smaller one would let it slow down.
Q3 (3 marks): While the box is stationary, static friction acts, and it adjusts itself to match whatever horizontal force is applied, so the box does not move at all until the applied force exceeds the maximum static friction. For most surface pairs the coefficient of static friction is larger than the coefficient of kinetic friction, so that maximum is the largest friction force involved. The instant the box begins to slide, the friction drops to the smaller kinetic value and stays roughly constant regardless of speed. The applied force that was just enough to break the box free is now more than enough to keep it going, which is why the box tends to lurch forward at the moment it starts and why less effort is needed thereafter.
You can now compare static and kinetic friction and calculate the friction force with $F = \mu N$. Next lesson brings together everything, friction, tension, weight, to solve real-world Newton's-law problems.
↩ Back to focus-area hubAt the start you predicted the friction at each push on the stationary crate. The answer: static friction matches the push, $30\ \text{N}$ then $80\ \text{N}$, staying exactly equal to the applied force until it reaches its maximum, then the crate slides. If you said friction was always at its maximum, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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