Exercises — Coefficients of friction — measurement, material dependence

WHAT: we want the ratio friction ÷ normal. WHY: that ratio is the definition of the coefficient — nothing else needed. ${\mu }_s=\frac{f_{s,\max }}{N}=\frac{20}{50}=0.40$ Notice: no units. Force ÷ force = pure number. That is why $\mu$ is called dimensionless.

WHAT: the "angle at which it just begins to slip" is the angle of repose. WHY that tool: on an incline, $mg\sin \theta$ drives sliding and $mg\cos \theta$ presses the surfaces; at the verge they satisfy $mg\sin {\theta }_c={\mu }_{s}mg\cos {\theta }_c$, and the mass cancels, leaving ${\mu }_s=\tan {\theta }_c=\tan 30^{\circ }\approx \mathrm{0.577.}$ A protractor beats a force meter here — no scale, no mass, just the angle.

Larger force → larger coefficient (both divide the same $N$), so ${\mu }_s>{\mu }_k$. WHY expected: at rest the microscopic bumps settle and "cold-weld" into each other's valleys, so it takes more force to break free than to keep skimming. This is the rule ==${\mu }_s\ge {\mu }_k$== — "Static Sticks, Kinetic Skips".

Step 1 — find $N$. WHY $N=mg$: the floor is horizontal and the crate has no vertical acceleration, so by vertical force balance the upward normal force must exactly cancel the downward weight — nothing else acts vertically. Hence: $N=mg=6\times 9.8=58.8\text{N}.$ Step 2 — starting force = verge of slip. WHY ${\mu }_s=F_{\max }/N$: the $21\text{N}$ needed to just start it is the maximum static friction $f_{s,\max }$ (any less and it wouldn't have moved). Feeding that straight into the definition $\mu =f/N$ gives: ${\mu }_s=\frac{f_{s,\max }}{N}=\frac{21}{58.8}\approx \mathrm{0.357.}$ Step 3 — constant speed means $a=0$. WHY the reading equals $f_k$: with zero acceleration, Newton's Second Law says the net horizontal force is zero, so the applied $15\text{N}$ exactly balances kinetic friction — no leftover force to confuse the reading. Then ${\mu }_k=f_k/N$: ${\mu }_k=\frac{15}{58.8}\approx \mathrm{0.255.}$ Sanity: ${\mu }_s>{\mu }_k$. ✓

Step 1 — the ceiling. WHY $f_{s,\max }={\mu }_{s}N$ is the ceiling: the largest static friction the surface can supply is fixed by the definition rearranged, $f_{s,\max }={\mu }_{s}N$ — this is the most grip available before it lets go. With $N=mg=2\times 9.8=19.6\text{N}$: $f_{s,\max }={\mu }_{s}N=0.5\times 19.6=9.8\text{N}.$ Step 2 — compare push to ceiling: your push is $6\text{N}<9.8\text{N}$, so the demand stays below the ceiling and the book stays put. Step 3 — actual friction. WHY it equals the push: since the book is not moving, its horizontal acceleration is zero, so friction must exactly cancel whatever you apply — no more, no less. Friction adjusts itself to $6\text{N}$ (not $9.8\text{N}$!). This is the L1 trap in action.

WHY ${\mu }_k$ not ${\mu }_s$: the block is already moving, so kinetic friction is in play. Constant velocity → $a=0$, so along the slope the driving pull equals kinetic friction: $mg\sin \theta ={\mu }_{k}mg\cos \theta .$ WHY the mass cancels: the exact same factor $mg$ multiplies both sides — the driving term $mg\sin \theta$ and the friction term ${\mu }_{k}mg\cos \theta$. Dividing both sides by $mg$ removes it entirely, so mass never matters: ${\mu }_k=\tan \theta =\tan 18^{\circ }\approx \mathrm{0.325.}$ Same tangent trick — mass cancels again.

Step 1 — decompose weight (see figure: the slate $mg$ arrow splits into a red along-slope part $mg\sin \theta$ pulling the block down the ramp, and a mint into-slope part $mg\cos \theta$ pressing it against the surface; the lavender arrow is friction $f$ holding it up-slope).

• Driving (down-slope) part: $mg\sin \theta =3\times 9.8\times \sin 20^{\circ }\approx 10.05\text{N}$.
• Pressing part: $N=mg\cos \theta =3\times 9.8\times \cos 20^{\circ }\approx 27.63\text{N}$.

Step 2 — friction ceiling: $f_{s,\max }={\mu }_{s}N=0.6\times 27.63\approx 16.58\text{N}.$ Step 3 — compare. The pull trying to slide it ($10.05\text{N}$) is less than the ceiling ($16.58\text{N}$), so it does not slide. (b) Since it's static and stationary, friction is exactly what cancels the down-slope pull: $f_s=mg\sin \theta \approx 10.05\text{N}(\text{up the slope}).$ Not $16.58\text{N}$ — that ceiling is unused.

Alternative one-line test: compare $\tan \theta =\tan 20^{\circ }\approx 0.364$ to ${\mu }_s=0.6$. Since $\tan \theta <{\mu }_s$, no slip. Faster, and mass never appears.

Angle of repose → static: ${\mu }_s=\tan 25^{\circ }\approx 0.466$. Constant-speed slide → kinetic: ${\mu }_k=\tan 21^{\circ }\approx 0.384$. Consistency: ${\theta }_{\text{slide}}<{\theta }_c$ means ${\mu }_k<{\mu }_s$ ✓ — you can hold a shallower tilt once it's already gliding, exactly the "sticks vs skips" story.

Forecast: unchanged, $25^{\circ }$. Mass never entered ${\mu }_s=\tan {\theta }_c$. Verify: with mass $2m$, the verge-of-slip balance is $2mg\sin {\theta }_c={\mu }_s(2mg\cos {\theta }_c).$ The whole factor $2mg$ divides out: $\tan {\theta }_c={\mu }_s.$ Same equation, same angle. That's why a protractor alone measures ${\mu }_s$ — no scale required.

WHY Newton's Second Law now: it is accelerating, so we need $\sum F=ma$ along the slope, not a balance. See Newton's Second Law. Read the figure: the block sits on the $35^{\circ }$ ramp; the red arrow $mg\sin \theta$ pulls it down-slope, the lavender friction arrow $f_k={\mu }_{k}mg\cos \theta$ points up-slope (opposing motion), and the mint arrow $mg\cos \theta$ sets the normal press. The net of the red minus the lavender is what accelerates it. Along-slope forces (down positive):

• Driving: $mg\sin \theta =3\times 9.8\times \sin 35^{\circ }\approx 16.86\text{N}$ (down).
• Kinetic friction opposes motion, so it points up the slope: $f_k={\mu }_{k}N={\mu }_{k}mg\cos \theta =0.45\times 3\times 9.8\times \cos 35^{\circ }\approx 10.83\text{N}$.

Newton's Second Law: $ma=mg\sin \theta -{\mu }_{k}mg\cos \theta .$ Divide by $m$ (mass cancels again!): $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(\sin 35^{\circ }-0.45\cos 35^{\circ })\approx 2.01{\text{m/s}}^2\text{(down slope)}.$ Positive → it really does accelerate down, consistent with "it slides".

Step 1 — slip test with static coefficient: compare $\tan \theta$ to ${\mu }_s$. $\tan 35^{\circ }\approx 0.700,{\mu }_s=\mathrm{0.55.}$ Step 2 — interpret: $\tan \theta >{\mu }_s$ means the down-slope pull exceeds the static ceiling, so the block breaks free and starts sliding from rest. Step 3 — then switch coefficients: once moving, kinetic friction (${\mu }_k$) takes over, giving the acceleration found in L4.1. This is the full chain: test with ${\mu }_s$ → if it slips, compute motion with ${\mu }_k$.

Step 1 — get ${\mu }_s$ from Method A: ${\mu }_s=\tan 30^{\circ }\approx 0.577$. Step 2 — predict Method B's reading: on flat ground $N=mg=5\times 9.8=49\text{N}$, and at the verge of slipping $F_{\max }={\mu }_{s}N=0.577\times 49\approx 28.3\text{N}.$ If the spring scale reads about $28\text{N}$, the two independent methods agree — a good internal check that $\mu$ is a genuine property of the surface pair, not of the apparatus.

(a) ${\theta }_c=0^{\circ }$: ${\mu }_s=\tan 0^{\circ }=0$. A flat table that lets the block slip at zero tilt means the surfaces are perfectly slippery — no grip at all. Sensible. (b) ${\theta }_c\to 90^{\circ }$: $\tan {\theta }_c\to \infty$. A vertical wall the block clings to would need infinite grip. In reality ${\theta }_c$ never reaches $90^{\circ }$ for ordinary materials, so ${\mu }_s$ stays finite — the formula warns you that near-vertical repose demands enormous friction. Sensible as a limit. (c) Frictionless (${\mu }_s=0$): solving $\tan {\theta }_c=0$ gives ${\theta }_c=0^{\circ }$ — the block slides at the slightest tilt. Matches intuition: with zero friction there's no stable resting angle above flat.

The microscopic truth: surfaces only touch at tiny asperity tips. The real contact area $A_{\text{real}}$ is far smaller than the apparent face area, and it is set by how hard the surfaces are squeezed: $A_{\text{real}}\propto N$. Harder press → tips flatten → more true contact. Consequence: friction $\propto A_{\text{real}}\propto N$, so $f=\mu N$ with $\mu$ independent of the apparent face. Standing the brick on its end concentrates the same weight on a smaller apparent area, but the real contact area (and hence friction) is unchanged. Incline test: the brick would begin to slip at the same ${\theta }_c$ whether on its face or its end — the tangent, and thus ${\mu }_s$, doesn't budge. This is exactly the parent note's "Why area doesn't appear" callout, made testable.

Dry: ${\theta }_c=\arctan 0.6\approx 31.0^{\circ }$. Oiled: ${\theta }_c=\arctan 0.1\approx 5.71^{\circ }$. The oiled surface slips at a much shallower tilt — far less grip. Physical meaning: a thin oil film keeps the asperity tips from cold-welding and interlocking; they ride over a fluid layer instead of gouging into each other. The number $\mu$ drops because the interface changed — reinforcing that $\mu$ describes the pair plus whatever is between them, not the metal alone. See Lubrication & Tribology.

Static $\mu$ from starting force ::: ${\mu }_s=F_{\max }/(mg)$ on flat ground. One-line slip test on an incline ::: compare $\tan \theta$ with ${\mu }_s$; slips if $\tan \theta >{\mu }_s$. After it starts sliding, which coefficient sets the acceleration ::: ${\mu }_k$, via $a=g(\sin \theta -{\mu }_k\cos \theta )$. Does doubling the mass change the repose angle ::: No — $mg$ cancels in ${\mu }_s=\tan {\theta }_c$. Is friction on a resting incline always ${\mu }_{s}N$ ::: No, it equals $mg\sin \theta$ until the verge of slipping.