Worked examples — Friction — static (maximum), kinetic, rolling

This page is the stress test for the friction topic. Before we grind through examples, we list every kind of situation the topic can throw at you, then guarantee that each one gets solved at least once.

Everything here rests on three facts from the parent note. Let us restate them in plain words so no symbol is used unearned:

Recall The symbols we will lean on

• $m$ (mass) is the amount of "stuff" in the object, measured in kilograms (kg). Its weight — the pull of gravity — is $mg$, where $g$ is gravity's strength.
• $N$ (the normal force) is how hard the ground pushes back, perpendicular to the surface. Units: newtons (N).
• Static friction $f_s$ is self-adjusting: it supplies exactly what is needed to prevent sliding, but never more than a ceiling $f_s^{max}={\mu }_{s}N$. So it is an inequality $f_s\le {\mu }_{s}N$.
• Kinetic friction is fixed once sliding starts: $f_k={\mu }_{k}N$, and ${\mu }_k<{\mu }_s$.
• Rolling friction $f_r={\mu }_{r}N$ resists a wheel that rolls without slipping, with ${\mu }_r\ll {\mu }_k$.

The symbol $\mu$ ("mu") is just a number — bigger $\mu$ means stickier surfaces. $g=10{\text{m/s}}^2$ throughout unless stated.

---

The scenario matrix

Every friction problem is one (or a blend) of these case classes. Each row is a distinct trap. The last column names the example that covers it.

#	Case class	The deciding question	Degenerate / limit to watch	Covered by
A	Push below the ceiling	Is applied force $<f_s^{max}$?	Zero push ⇒ zero friction	Ex 1
B	Push above the ceiling	Once it slips, use ${\mu }_k$	Push exactly $=f_s^{max}$ (verge)	Ex 2
C	Incline, block stays	Is $\tan \theta <{\mu }_s$?	$\theta =0$ (flat) ⇒ no driving force	Ex 3
D	Incline, block slides	$\theta >{\theta }_r$: net force down-slope	$\theta =90^{\circ }$ ⇒ free fall	Ex 4
E	Angled force up (lift)	Push at an angle changes $N$	Pull up hard ⇒ $N\to 0$	Ex 5
E'	Angled force down (press)	Downward angle raises $N$	More press ⇒ more grip	Ex 6
F	Friction as the driver	Sign of the "tendency to slide" flips	Max grip = ${\mu }_{s}N$	Ex 7
G	Two-body / connected	Friction on the shared motion	Zero friction ⇒ Atwood limit	Ex 8
R	Rolling resistance / grip→roll	${\mu }_r=a/R$, tiny vs ${\mu }_k$	Grip limit before roll begins	Ex 9
H	Real-world word + exam twist	Find $\mu$ from stopping distance	Wet road ⇒ $\mu$ drops	Ex 10

Notice the pattern: every problem starts by comparing a driving tendency against a ceiling ${\mu }_{s}N$. Get that comparison right and the rest is Newton's second law.

---

Example 1 — Case A: push below the ceiling (and the zero-push corner)

Step 1 — Find the ceiling. $f_s^{max}={\mu }_{s}N={\mu }_{s}mg=0.5\times 10\times 10=50\text{N}$ Why this step? Every decision compares the driving force to this ceiling. It is the referee.

Step 2 — Case (a), zero push. With no push there is no tendency to slide, so static friction supplies nothing: $f_s=0\text{N}$. Why this step? This is the degenerate corner of Case A. Friction is reactive — no cause, no effect. Beginners wrongly write $50\text{N}$ here.

Step 3 — Case (b), 30 N push. With positive = push direction, compare: $30\text{N}<50\text{N}$. The crate does not move, so by Newton's Second Law with $a=0$ the friction must exactly cancel the push: $f_s=30\text{N}$ Why this step? Static friction is self-adjusting; it copies the push, not the ceiling.

---

Example 2 — Case B: push above the ceiling (and the verge corner)

Step 1 — Recall the ceiling: $f_s^{max}=50\text{N}$ (from Ex 1). Why this step? Same referee.

Step 2 — Case (a), the verge. $50\text{N}=f_s^{max}$. The crate is on the brink: friction can just barely hold, so $a=0$ still. Why this step? The inequality $f_s\le {\mu }_{s}N$ includes equality. One hair more and it breaks free.

Step 3 — Case (b), it slides. $65>50$, so it moves. Switch to kinetic friction — the bonds are now snapping continuously: $f_k={\mu }_{k}mg=0.3\times 100=30\text{N}$ Why this step? Once sliding, the kinetic law governs, and it is smaller. Keeping ${\mu }_s$ here is a classic error.

Step 4 — Newton's second law (positive = push direction, friction negative): $a=\frac{F-f_k}{m}=\frac{65-30}{10}=3.5{\text{m/s}}^2$ Why this step? Net force / mass gives acceleration once friction is known.

---

Example 3 — Case C: incline, block stays put

Look at the figure. Gravity $mg$ points straight down. Positive direction = down the slope. We split gravity into two perpendicular pieces along the ramp's own directions (the orange and teal arrows): one down the slope (the driving piece, $+$) and one into the slope (which sets $N$).

Step 1 — Resolve gravity. From the geometry in the figure: $\text{driving}=mg\sin \theta ,N=mg\cos \theta$ Why this step? Friction reacts along the surface, so we need the force component along the surface (that is what it must fight) and the one perpendicular (that sets the ceiling).

Step 2 — Compare slope to the friction condition. The block stays if driving $\le$ ceiling: $mg\sin \theta \le {\mu }_{s}mg\cos \theta ⟺\tan \theta \le {\mu }_s$ Why this step? Dividing out $mg\cos \theta$ turns the force comparison into a clean angle comparison — this is why the incline shortcut $\tan {\theta }_r={\mu }_s$ exists.

Step 3 — Plug numbers. $\tan 20^{\circ }\approx 0.364<0.5$. It stays. Why this step? The angle-only test is the fastest possible check.

Step 4 — Actual friction value. Since it is static and $a=0$, friction (pointing up-slope, $-$) balances the driving piece: $f_s=mg\sin \theta =4\times 10\times \sin 20^{\circ }\approx 13.7\text{N}$ Why this step? Not ${\mu }_{s}N$! Static friction supplies only what is needed (Case A logic, now on a ramp).

---

Example 4 — Case D: incline, block slides (and the $90^{\circ }$ limit)

Positive direction = down the slope, as agreed.

Step 1 — Slip test. $\tan 40^{\circ }\approx 0.839>{\mu }_s=0.5$. Driving beats the ceiling ⇒ it slides. Why this step? Same referee as Ex 3, but now the inequality flips.

Step 2 — Once sliding, kinetic friction acts up the slope (opposing the downward slide, so it is negative — see the plum arrow in the figure): $f_k={\mu }_{k}mg\cos \theta =0.3\times 40\times \cos 40^{\circ }\approx 9.19\text{N}$ Why this step? Kinetic law applies during motion; friction opposes the actual sliding direction.

Step 3 — Newton's second law along the slope (driving $+$, friction $-$): $a=g\sin \theta -{\mu }_{k}g\cos \theta =10\sin 40^{\circ }-0.3\times 10\cos 40^{\circ }\approx 6.43-2.30=4.13{\text{m/s}}^2$ Why this step? Net down-slope force over mass. Friction is subtracted because it points up-slope.

Step 4 — The $\theta \to 90^{\circ }$ limit. Then $\cos \theta \to 0$, so $N\to 0$ and $f_k\to 0$; $a\to g\sin 90^{\circ }=g$. Free fall. Why this step? A vertical "ramp" presses nothing against the block, so friction vanishes — the sanity check that the model behaves at the boundary.

---

Example 5 — Case E: pull UP at an angle (the $N\to 0$ corner)

Positive direction = horizontal drag direction. The rope tension splits into a horizontal piece $T\cos \theta$ (drags it, $+$) and a vertical piece $T\sin \theta$ (lifts it, easing the ground's load) — see the teal and orange arrows.

Step 1 — Vertical balance sets $N$. The ground no longer carries the full weight: $N=mg-T\sin \theta$ Why this step? An upward pull lightens the crate, so $N$ shrinks — and $N$ controls the whole friction budget.

Step 2 — (a) Ceiling with $T=40$. $N=100-40\sin 30^{\circ }=100-20=80\text{N},f_s^{max}=0.5\times 80=40\text{N}$ Why this step? The lift lowers the ceiling from 50 N (Ex 1) to 40 N.

Step 3 — Horizontal drive vs ceiling. $T\cos 30^{\circ }=40\times 0.866=34.6\text{N}<40\text{N}\Rightarrow \text{stays.}{f}_s=34.6\text{N}$ Why this step? Compare the actual pull to the reduced ceiling; static friction matches the pull.

Step 4 — (b) The $N=0$ corner. Set $N=0$: $mg=T\sin \theta \Rightarrow T=\frac{mg}{\sin 30^{\circ }}=\frac{100}{0.5}=200\text{N}$ Why this step? Beyond this the crate lifts off the ground entirely — friction dies because there is nothing to press against.

---

Example 6 — Case E': push DOWN at an angle (mirror case — more grip)

Positive direction = horizontal shove direction. Now the vertical piece $P\sin \theta$ points down, adding to the ground's load — the mirror image of Ex 5 (see the orange arrow pressing into the floor).

Step 1 — Vertical balance sets $N$. The downward push increases the load: $N=mg+P\sin \theta$ Why this step? Pressing down means the ground pushes back harder, so $N$ grows — and a bigger $N$ raises the friction ceiling.

Step 2 — (a) Ceiling with $P=40$. $N=100+40\sin 30^{\circ }=100+20=120\text{N},f_s^{max}=0.5\times 120=60\text{N}$ Why this step? The press raises the ceiling from 50 N (Ex 1) to 60 N.

Step 3 — Horizontal drive vs ceiling. $P\cos 30^{\circ }=40\times 0.866=34.6\text{N}<60\text{N}\Rightarrow \text{stays.}{f}_s=34.6\text{N}$ Why this step? Compare the actual shove to the raised ceiling; static friction matches the shove.

Step 4 — Mirror comparison. Pull-up ratio was $34.6/40=0.865$; push-down ratio is $34.6/60=0.577$. Same horizontal drive, but the downward version is much further from slipping. Why this step? This is the whole point of the mirror case: angling down buys grip, angling up loses it.

---

Example 7 — Case F: friction as the DRIVER (walking / accelerating car)

Positive direction = the car's forward motion. Here friction is not the villain but the hero. The tyre's contact patch tends to slip backward against the road (the driven wheel pushes road backward), so static friction pushes the tyre — and the car — forward (orange arrow in the figure). This is the free-body insight the parent note flagged.

Step 1 — Max grip is the ceiling. The forward force cannot exceed static friction's maximum: $F_{\max }={\mu }_{s}N={\mu }_{s}mg=0.8\times 1200\times 10=9600\text{N}$ Why this step? Static friction still obeys $f_s\le {\mu }_{s}N$; the ceiling caps how hard the road can shove the car.

Step 2 — Newton's second law. $a_{\max }=\frac{F_{\max }}{m}=\frac{9600}{1200}=8{\text{m/s}}^2$ Why this step? At the grip limit, this is the largest acceleration possible. Push the engine harder and the tyre exceeds ${\mu }_{s}N$, breaks into kinetic (spinning) friction, and grip drops.

Step 3 — Degenerate note. On ice, ${\mu }_s\to 0.1$, so $a_{\max }\to 1{\text{m/s}}^2$ — the wheels spin almost at once. Why this step? Shows the model scaling with surface, matching everyday experience.

---

Example 8 — Case G: connected bodies (with the frictionless limit)

Positive direction = the direction the system tends to move: $A$ toward the pulley, $B$ downward.

Step 1 — Does it move? Compare the drive to the STATIC ceiling. Block $B$'s weight $m_{B}g$ pulls the cord, and that tension is the only thing trying to drag $A$ across the table. Before assuming any motion we must ask: can static friction on $A$ hold that pull? Its maximum is $f_s^{max}={\mu }_s{m}_{A}g=0.25\times 3\times 10=7.5\text{N},m_{B}g=20\text{N}.$ Since $20\text{N}>7.5\text{N}$, the drive beats the static ceiling ⇒ the system moves. Why this step? A two-body system is still governed by the same referee: nothing moves until the driving pull exceeds $f_s^{max}={\mu }_s{m}_{A}g$. We use ${\mu }_s$ (not ${\mu }_k$) here because we are testing whether it starts — kinetic friction only exists once sliding has already begun, so it would be circular to use it in the "does it move?" test.

Step 2 — Now that it slides, use kinetic friction on $A$ (opposing its slide toward the pulley, so it is negative): $f_k={\mu }_k{m}_{A}g=0.2\times 3\times 10=6\text{N}$ Why this step? Motion is confirmed, so the kinetic law now applies; friction fights the actual motion of $A$.

Step 3 — Newton's second law for the whole system. The falling weight drives ($+$), friction resists ($-$); the two blocks share one acceleration and the internal tension cancels: $a=\frac{m_{B}g-f_k}{m_A+m_B}=\frac{20-6}{5}=\frac{14}{5}=2.8{\text{m/s}}^2$ Why this step? Treating the pair as one mass removes the unknown tension in one stroke.

Step 4 — Tension (isolate block $B$: $m_{B}g-T=m_{B}a$): $T=m_B(g-a)=2(10-2.8)=14.4\text{N}$ Why this step? Isolate one body to extract the internal force.

Step 5 — Frictionless limit. Set ${\mu }_k=0$: $a=20/5=4{\text{m/s}}^2$, $T=2(10-4)=12\text{N}$ — the classic Atwood-on-a-table result. Why this step? Confirms our formula reduces correctly when friction is switched off.

---

Example 9 — Case R: rolling resistance and the grip→roll boundary

Positive direction = the cart's forward motion.

Step 1 — Rolling coefficient from the geometry. The parent note derived ${\mu }_r=a/R$ (normal force shifts ahead by $a$, making a backward torque $Na$ that balances $f_{r}R$): ${\mu }_r=\frac{a}{R}=\frac{0.0015}{0.30}=0.005$ Why this step? This turns a tiny deformation length into the pure number that scales resistance.

Step 2 — Rolling-resistance force. $f_r={\mu }_{r}N={\mu }_{r}mg=0.005\times 50\times 10=2.5\text{N}$ Why this step? $N=mg$ on flat ground; multiply by ${\mu }_r$ to get the gentle backward drag that a rolling wheel feels.

Step 3 — (b) The grip ceiling (roll→slip boundary). As long as the drive stays below the static grip limit at the contact patch, the wheel rolls; exceed it and the patch slips: $f_{grip}^{max}={\mu }_{s}N=0.7\times 50\times 10=350\text{N}$ Why this step? Rolling-without-slipping is held together by static friction at the contact point — the same ${\mu }_{s}N$ ceiling. Cross it and the tyre skids into kinetic friction.

Step 4 — Compare the two regimes. $\frac{f_{grip}^{max}}{f_r}=\frac{350}{2.5}=140$. Why this step? It shows in numbers why wheels are wonderful: the everyday drag ($2.5\text{N}$) is $140\times$ smaller than the grip you can call on before slipping.

---

Example 10 — Case H: real-world word problem + exam twist

Positive direction = the car's forward motion; friction (backward) is negative and decelerates it.

Step 1 — Deceleration from friction. With brakes locked, kinetic friction is the whole force: $ma={\mu }_{k}mg\Rightarrow a={\mu }_{k}g$ Why this step? Mass cancels — stopping depends only on ${\mu }_k$ and $g$, a beautiful energy-friendly fact.

Step 2 — Kinematics link. Using $v^2=u^2-2ad$ with final $v=0$: $0=u^2-2({\mu }_{k}g)d\Rightarrow {\mu }_k=\frac{u^2}{2gd}$ Why this step? Connects the measurable distance to the unknown ${\mu }_k$.

Step 3 — Plug in (a). ${\mu }_k=\frac{20^2}{2\times 10\times 25}=\frac{400}{500}=0.8$ Why this step? Direct substitution; a plausibly grippy dry road.

Step 4 — Twist (b). Wet: ${\mu }_k^{\prime }=0.4$. Since $d=\frac{u^2}{2{\mu }_{k}g}$, halving ${\mu }_k$ doubles $d$: $d^{\prime }=\frac{20^2}{2\times 0.4\times 10}=\frac{400}{8}=50\text{m}$ Why this step? $d\propto 1/{\mu }_k$ at fixed speed — the exam twist rewards spotting the proportionality, not re-deriving.

---

Recap — the one habit that solves them all