Exercises — Friction — static (maximum), kinetic, rolling

WHAT: No horizontal push means no tendency to slide. WHY: Static friction is self-adjusting — it only supplies what is needed to prevent sliding. Needed $=0$, so it supplies $0$. Answer: $f_s=0\text{N}$. Friction is not ${\mu }_{s}N$ here; that is only the ceiling, not the actual value.

Step 1 — find $N$. $N=mg=4\times 10=40\text{N}$. Why: the block does not move vertically, so the upward normal force must exactly balance the downward weight; nothing else pushes vertically, hence $N=mg$. Step 2 — apply the ceiling formula. $f_s^{max}={\mu }_{s}N=0.6\times 40=24\text{N}$. Why: the most static friction can supply is the ceiling ${\mu }_{s}N$, by definition of $f_s^{max}$. Answer: $24\text{N}$.

Step 1 — ceiling. $N=mg=100\text{N}$, so $f_s^{max}=0.5\times 100=50\text{N}$. Why: we must know the highest force static friction can supply before we can decide whether the box breaks free. Step 2 — compare. Push $=30\text{N}<50\text{N}$ ⇒ does not slide. Why: if the push is below the ceiling, static friction can still match it, so no sliding occurs. Step 3 — actual friction. Static and not moving ⇒ net force zero ⇒ friction exactly cancels the push: $f_s=30\text{N}$. Why: $a=0$ means (Newton's 2nd law) the horizontal forces sum to zero, so friction must equal the push. Step 4 — acceleration. $a=0$. Why: the box does not slide and nothing else moves it, so its velocity is unchanged, i.e. $a=0$.

Step 1 — ceiling. Still $f_s^{max}=50\text{N}$. Why: mass, ${\mu }_s$ and $g$ are unchanged, so the static ceiling ${\mu }_{s}mg$ is the same as before. Step 2 — compare. $70>50$ ⇒ box slides. Why: the push now exceeds the maximum static friction, so friction can no longer hold the box and it breaks free. Step 3 — switch to kinetic. Once moving the kinetic law applies: $f_k={\mu }_{k}N=0.3\times 100=30\text{N}$. Why: sliding has begun, so the sliding-bond model ($f_k={\mu }_{k}N$, smaller) replaces the static one. Step 4 — Newton's 2nd law. Net $=70-30=40\text{N}$, so $a=\frac{40}{10}=4{\text{m/s}}^2$. Why: with friction now known and constant, the leftover (unbalanced) force divided by mass gives the acceleration.

WHY this tool: at the tilt where sliding just begins, the down-slope pull equals the maximum friction — see the figure. The red arrow is the down-slope pull $mg\sin \theta$; the green arrow is the maximum static friction $f_s^{max}={\mu }_{s}mg\cos \theta$ pointing up the slope. At the verge of slipping these are equal and opposite. One-line reminder of why $\tan {\theta }_r={\mu }_s$. Setting red $=$ green: $mg\sin {\theta }_r={\mu }_{s}mg\cos {\theta }_r$. Cancel $mg$ and divide by $\cos {\theta }_r$: $\frac{\sin {\theta }_r}{\cos {\theta }_r}={\mu }_s$, i.e. $\tan {\theta }_r={\mu }_s$. Step 1: ${\mu }_s=\tan {\theta }_r=\tan 37^{\circ }=0.75$. Why: the tilt at first slip is exactly the angle of repose, so the derived condition $\tan {\theta }_r={\mu }_s$ applies directly. Answer: ${\mu }_s=0.75$.

Step 1 — resolve gravity. Split $mg$ into a part along the slope (drives sliding) and a part into the slope (pressed by the surface). See the figure's decomposition. Why: gravity points straight down but the block can only slide along the slope, so we separate the part that drives motion from the part the surface must resist.

• Along slope: $mg\sin \theta =2\times 10\times 0.5=10\text{N}$.
• Into slope: $mg\cos \theta =2\times 10\times 0.866=17.32\text{N}$. Step 2 — normal force. $N=mg\cos \theta =17.32\text{N}$. Why: the block does not move perpendicular to the slope, so $N$ must balance only the into-slope part of gravity. Step 3 — ceiling. $f_s^{max}={\mu }_{s}N=0.7\times 17.32=12.12\text{N}$. Why: this is the most friction the surface can supply, needed before we can judge whether the block holds. Step 4 — compare driving vs ceiling. Driving $=10\text{N}<12.12\text{N}$ ⇒ does not slide. Why: the down-slope pull is below the friction ceiling, so static friction can fully cancel it. Step 5 — actual friction. Static, no acceleration ⇒ friction balances the driving pull: $f_s=10\text{N}$ (up the slope). Why: $a=0$ ⇒ forces along the slope sum to zero, so friction equals the $10\text{N}$ pull, not the $12.12\text{N}$ ceiling. Answer: stays put; $f_s=10\text{N}$.

Key insight: both the driving force $mg\sin \theta$ and the ceiling ${\mu }_{s}mg\cos \theta$ scale with $m$. Their ratio is $\tan \theta /{\mu }_s$, independent of mass. Check: driving $=4\times 10\times 0.5=20\text{N}$; ceiling $=0.7\times 4\times 10\times 0.866=24.25\text{N}$. Still $20<24.25$ ⇒ no slide. Why unchanged: sliding depends on whether $\tan \theta >{\mu }_s$, which has no $m$ in it. Since $\tan 30^{\circ }=0.577<0.7$, no mass can make it slide on this slope.

In the figure the blue arrow is the push $P$ up the slope, the white arrow is gravity's down-slope pull $mg\sin \theta$, and the two red arrows show the two possible directions of friction — up the slope when $P$ is small (block tends to slide down) and down the slope when $P$ is large (block tends to slide up). Step 0 — sanity check. $\tan 30^{\circ }=0.577>{\mu }_s=0.4$, so without help it slides down. Good — friction alone can't hold it. Why: if $\tan \theta >{\mu }_s$ the down-slope pull already beats the friction ceiling, so a push is genuinely needed. Setup. $N=mg\cos \theta =5\times 10\times 0.866=43.3\text{N}$, so $f_s^{max}={\mu }_{s}N=0.4\times 43.3=17.32\text{N}$; gravity's down-slope pull is $mg\sin \theta =5\times 10\times 0.5=25\text{N}$. Why: we collect the fixed quantities (ceiling and gravity pull) before testing the two extreme cases. Case A — smallest $P$ (block about to slide down). Friction points up at full strength to help $P$ hold gravity: $P_{min}+f_s^{max}=mg\sin \theta ⟹P_{min}=25-17.32=7.68\text{N}.$ Why: if $P$ were any smaller, even maximal up-slope friction could not stop the block sliding down. Case B — largest $P$ (block about to slide up). Friction flips to point down at full strength, resisting $P$: $P_{max}=mg\sin \theta +f_s^{max}=25+17.32=42.32\text{N}.$ Why: if $P$ were any larger, even maximal down-slope friction could not stop the block sliding up. Answer: the block stays static for $7.68\text{N}\le P\le 42.32\text{N}$.

Part (a) — acceleration via Newton's 2nd law.

• Driving down: $mg\sin \theta =25\text{N}$.
• Kinetic friction (opposes motion, so up the slope): $f_k={\mu }_{k}mg\cos \theta =0.3\times 43.3=12.99\text{N}$.
• Net along slope: $25-12.99=12.01\text{N}$.
• $a=\frac{12.01}{5}=2.40{\text{m/s}}^2$ down the slope.

Part (b) — speed via the Work-Energy Theorem. Why this tool: work–energy links net force over a distance directly to the change in kinetic energy, so we get the final speed without ever needing the time. Why it reduces to (net force)×d: the net force here is constant (all three forces are constant) and points along the straight-line motion, so its work is simply force × distance with no angle factor. The block starts from rest, so the change in kinetic energy is just the final $\frac{1}{2}m{v}^2$: $\frac{1}{2}m{v}^2=(\text{net force})\times d=12.01\times 4=48.05\text{J}$ $v=\sqrt{\frac{2\times 48.05}{5}}=\sqrt{19.22}=4.38\text{m/s}.$ Answer: $a\approx 2.40{\text{m/s}}^2$, $v\approx 4.38\text{m/s}$.

Recall from the glossary that $f_r={\mu }_{r}N$ is the small resisting force on a rolling wheel, and (from the parent note) ${\mu }_r=a/R$, where $a$ is how far ahead the normal force shifts because the wheel and ground deform slightly. Step 1 — use ${\mu }_r=a/R$. Why: the shifted normal force makes a backward torque $Na$; balancing it against $f_{r}R$ gives $f_r=Na/R$, i.e. ${\mu }_r=a/R$. ${\mu }_r=\frac{a}{R}=\frac{0.5\times 10^{-3}\text{m}}{0.25\text{m}}=\mathrm{0.002.}$ Step 2 — compare. $\frac{{\mu }_k}{{\mu }_r}=\frac{0.3}{0.002}=150$. Why: dividing the two coefficients shows how many times harder sliding resists than rolling. Answer: ${\mu }_r=0.002$ — about 150 times smaller than kinetic friction. This is why rolling a load beats dragging it. See also Rolling without Slipping.

Direction: to accelerate forward the foot pushes the ground backward; by static friction the ground pushes the foot forward. Friction is the forward driving force here. Why: friction opposes the relative sliding at the contact, and the foot tends to slide backward, so friction points forward. Step 1 — required force. $F=ma=60\times 1.5=90\text{N}$ (this is static friction). Why: the only horizontal force on the person is this friction, so Newton's 2nd law fixes its size at $ma$. Step 2 — required ${\mu }_s$. This must fit under the ceiling ${\mu }_{s}N={\mu }_{s}mg={\mu }_s\times 600$. Why: static friction cannot exceed ${\mu }_{s}N$, so the surface must be sticky enough to supply the needed $90\text{N}$. ${\mu }_s\times 600\ge 90⟹{\mu }_s\ge \mathrm{0.15.}$ Answer: friction supplies $90\text{N}$ forward; any ${\mu }_s\ge 0.15$ works.

Actual static friction on an un-pushed box? ::: Zero — it supplies only what is needed. On an incline, what carries the normal force? ::: Only the perpendicular part, $N=mg\cos \theta$. Why is the incline slide-or-not answer mass-independent? ::: Both driving and ceiling scale with $m$; slide iff $\tan \theta >{\mu }_s$. After sliding starts, which coefficient governs friction? ::: The kinetic one, ${\mu }_k$ (smaller than ${\mu }_s$). Why can static friction have two limits in the pushed-incline problem? ::: Friction can point up or down depending on which way the block tends to slide. Why is rolling friction so tiny? ::: ${\mu }_r=a/R$ with the deformation offset $a\ll R$.