Exercises — Angle of friction, angle of repose — derivation

WHAT: We want the tilt at which the block just slides. WHY this tool: Angle of repose is defined by $\tan {\theta }_r={\mu }_s$ (the verge-of-slipping condition, mass cancels). ${\theta }_r=\arctan (0.5)=26.57°$ That's it — no mass, no area, just the coefficient.

WHAT: Reverse the L1.1 relation. WHY: The tilt at first slip is the angle of repose, so ${\mu }_s=\tan {\theta }_r$. ${\mu }_s=\tan 40°=0.839$

Normal: floor is flat, so $N=mg=3\times 10=30\text{N}$. Friction (at verge): $f={\mu }_{s}N=0.4\times 30=12\text{N}$. Total reaction (Pythagoras — the two are perpendicular): $R=\sqrt{N^2+f^2}=\sqrt{30^2+12^2}=\sqrt{1044}\approx 32.31\text{N}$

WHAT axes: resolve along and perpendicular to the incline (motion, if any, is along it). $N=mg\cos \theta =40\times \cos 15°=38.64\text{N}$ $mg\sin \theta =40\times \sin 15°=10.35\text{N (down-slope)}$ WHY friction points DOWN now: we are pushing the block up, so impending motion is up-slope; friction opposes it, pointing down-slope at magnitude $f={\mu }_{s}N$. $f=0.5\times 38.64=19.32\text{N}$ Verge-of-sliding-up balance (up forces = down forces): $P=mg\sin \theta +f=10.35+19.32=29.67\text{N}$

On a flat floor "normal" and "vertical" are the same line, so: $\lambda =\arctan ({\mu }_s)=\arctan (0.4)=21.80°$ Check with L1.3 numbers: $\tan \lambda =f/N=12/30=0.4$. ✓

Repose angle rises with ${\mu }_s$. The plastic coin, having the smaller ${\mu }_s$, has the smaller repose angle, so it slips first: ${\theta }_r^{\text{plastic}}=\arctan (0.3)=16.70°$ The steel coin holds until $\arctan (0.6)=30.96°$.

Start from $R=\sqrt{N^2+f^2}$ and substitute $f={\mu }_{s}N$: $R=\sqrt{N^2+({\mu }_{s}N{)}^2}=\sqrt{N^2(1+{\mu }_s^2)}=N\sqrt{1+{\mu }_s^2}$ On a flat floor $N=mg$, so $R=mg\sqrt{1+{\mu }_s^2}$. For $m=3,{\mu }_s=0.4$: $R=30\sqrt{1+0.16}=30\sqrt{1.16}=32.31\text{N}$ — matches L1.3. ✓ Reading it: $R$ is always at least $N$ (the $\sqrt{1+{\mu }_s^2}\ge 1$), and it grows as friction grows.

At repose the block is in equilibrium, so $\vec{R}+m\vec{g}=0$, i.e. $\vec{R}=-m\vec{g}$. Gravity is vertical (down), so $R$ must be vertical (up). Geometry: $R$ sits at angle $\lambda$ from the (tilted) normal; the vertical sits at angle ${\theta }_r$ from that same normal. Equilibrium forces $R$ onto the vertical, which is only possible when $\lambda ={\theta }_r$. This is the visual proof that the two angles are equal — see the figure: the plum $R$ arrow and the orange weight arrow are the same line.

(a) ${\theta }_r=\arctan (0)=0°$. On a frictionless plane any tilt slides the block — the "steepest safe angle" collapses to flat. (b) As ${\mu }_s\to \infty$, $\arctan ({\mu }_s)\to 90°$. With infinite grip the block clings even on a vertical wall; the repose angle approaches (but never reaches) $90°$. Key insight: $\arctan$ maps $[0,\infty )$ onto $[0°,90°)$ — so a real repose angle can never hit or exceed $90°$. A ramp steeper than that isn't a ramp, it's a ceiling.

A parked car on a banked surface is exactly a block on an incline. It slides down when the bank tilt exceeds the repose angle: ${\theta }_{\max }=\arctan ({\mu }_s)=\arctan (0.7)=34.99°$ Connection: in banking, this same $\arctan \mu$ sets the friction contribution to the safe-speed range. A bank steeper than ${\theta }_r$ would let a stationary car slip — which is why real banks stay modest and rely on speed for the rest of the centripetal force.

WHAT: the block is about to slide down, so friction points up the slope at max value. Resolve the horizontal push $P$ onto the incline axes:

• along incline (up-slope component): $P\cos \theta$
• perpendicular (into surface): $P\sin \theta$

Perpendicular equilibrium: $N=mg\cos \theta +P\sin \theta$ Along-incline equilibrium (verge of sliding down: down-pull = up-push + friction): $mg\sin \theta =P\cos \theta +{\mu }_{s}N$ Substitute $N$: $mg\sin \theta =P\cos \theta +{\mu }_s(mg\cos \theta +P\sin \theta )$ $mg\sin \theta -{\mu }_{s}mg\cos \theta =P(\cos \theta +{\mu }_s\sin \theta )$ $P=\frac{mg(\sin \theta -{\mu }_s\cos \theta )}{\cos \theta +{\mu }_s\sin \theta }$ Numbers ($mg=20,\theta =20°,{\mu }_s=0.25$): $P=\frac{20(0.342-0.25\times 0.940)}{0.940+0.25\times 0.342}=\frac{20(0.342-0.235)}{0.940+0.0855}=\frac{2.145}{1.0255}=2.09\text{N}$

Insight: each contact has its own repose angle, and repose angle depends only on that contact's ${\mu }_s$ (masses cancel — this is why the extra weight of B on A doesn't change A's own repose angle relative to the plane).

• B-on-A slips when tilt reaches $\arctan (0.20)=11.31°$.
• A-on-plane slips when tilt reaches $\arctan (0.35)=19.29°$.

The smaller angle wins the race: B slides off A first, at $\theta =\arctan (0.20)=11.31°$ Why B's weight doesn't help A grip more usefully: yes, B's weight increases $N$ under A, but it increases A's driving force $mg\sin \theta$ by the same factor — so A's threshold stays at its own $\arctan (0.35)$. The two thresholds are independent; you just compare $\mu$'s.

Use the L4.2 result and its mirror. With $mg=50,\theta =15°,{\mu }_s=0.30$: $\sin 15°=0.2588,\cos 15°=0.9659$.

Minimum $P$ — friction points UP (about to slide down): $P_{\min }=\frac{mg(\sin \theta -{\mu }_s\cos \theta )}{\cos \theta +{\mu }_s\sin \theta }=\frac{50(0.2588-0.30\times 0.9659)}{0.9659+0.30\times 0.2588}$ $=\frac{50(0.2588-0.2898)}{0.9659+0.0776}=\frac{50(-0.0310)}{1.0435}=-1.485\text{N}$ A negative minimum means: even with $P=0$ the block does not slide down (because $15°<\arctan 0.30=16.7°$). So $P_{\min }=0$ physically — no push needed at the low end.

Maximum $P$ — friction points DOWN (about to slide up): $P_{\max }=\frac{mg(\sin \theta +{\mu }_s\cos \theta )}{\cos \theta -{\mu }_s\sin \theta }=\frac{50(0.2588+0.2898)}{0.9659-0.0776}$ $=\frac{50(0.5486)}{0.8883}=30.88\text{N}$ Stable range: $0\le P\le 30.88\text{N}$. Beyond that the block is shoved up the hill.