Exercises — Inclined planes — with and without friction

WHAT we choose: On a frictionless incline the only along-slope force is $mg\sin \theta$. By Newton's Second Law, $ma=mg\sin \theta$, so the mass cancels: $a=g\sin \theta$ That is answer (b). WHY not $\cos$? Look at the master figure — $\cos$ points into the slope (green), which is the direction with no motion; it cannot cause sliding. WHY not $\tan$? $\tan$ never appears in a force balance here — it only shows up later as a ratio of two forces (angle of repose). Compute: $a=9.8\times \sin 30°=9.8\times 0.5=4.9{\text{m/s}}^2$.

WHAT: Perpendicular direction has no acceleration, so $N$ balances only the crush component (green arrow). $N=mg\cos \theta =4\times 9.8\times \cos 30°=4\times 9.8\times 0.8660=33.9\text{N}$ WHY less than $mg=39.2\text{N}$? Because the slope "carries" part of the weight along itself; only the perpendicular part presses in. $N<mg$ for any tilt.

WHY find $N$ first: friction needs $N$. $N=mg\cos \theta =2\times 9.8\times \cos 30°=16.97\text{N}$ WHAT next: the block moves down, so kinetic friction (orange) points up the slope, opposing motion. Along-slope Newton's law: $ma=mg\sin \theta -{\mu }_{k}mg\cos \theta$ Divide by $m$ (it cancels): $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(0.5-0.20\times 0.8660)=9.8(0.5-0.1732)=3.20{\text{m/s}}^2$ Sanity: less than the frictionless $4.9{\text{m/s}}^2$, as friction should slow it. ✓

WHY this works: at the angle of repose the down-slope pull just equals maximum static friction: $mg\sin {\theta }_c={\mu }_{s}mg\cos {\theta }_c\Rightarrow {\mu }_s=\tan {\theta }_c$ Mass, $g$, and coin size all cancel — a pure-angle measurement. ${\mu }_s=\tan 26.6°=0.50$

WHAT: We already found $a=3.20{\text{m/s}}^2$ (constant). Use the kinematic relation $v^2=u^2+2as$ with $u=0$: $v=\sqrt{2as}=\sqrt{2\times 3.20\times 5}=\sqrt{32.0}=5.66\text{m/s}$ WHY kinematics here: acceleration is constant (all forces are constant along the slope), so the standard constant-$a$ equations apply. This is the same idea as Work-Energy Theorem on Inclines would give — energy in, kinetic energy out.

WHAT to test first: compare the driving pull with the maximum static friction.

• Down-slope pull: $mg\sin \theta =3\times 9.8\times \sin 25°=12.43\text{N}$.
• Max static friction: ${\mu }_{s}mg\cos \theta =0.60\times 3\times 9.8\times \cos 25°=0.60\times 26.64=15.99\text{N}$. Since $12.43<15.99$, the pull cannot overcome friction ⇒ it does NOT slide. Equivalent quick test: $\tan 25°=0.466<{\mu }_s=0.60$ ✓ (below the angle of repose). WHAT the friction actually is: friction is NOT at its maximum when static. It only supplies what's needed to keep balance. So the actual friction (pointing UP the slope) equals the down-slope pull: $f_s=mg\sin \theta =12.43\text{N}(\text{not }15.99\text{N})$

WHAT changes: the block moves UP, so kinetic friction (orange) flips to point DOWN the slope — friction always opposes the actual motion. Now BOTH gravity's pull and friction point down the slope, both fighting the upward motion. $ma=-mg\sin \theta -{\mu }_{k}mg\cos \theta$ The magnitude of deceleration: $a=g(\sin \theta +{\mu }_k\cos \theta )=9.8(\sin 20°+0.30\cos 20°)$ $=9.8(0.3420+0.30\times 0.9397)=9.8(0.3420+0.2819)=9.8\times 0.6239=6.11{\text{m/s}}^2$ WHY a plus sign here (vs. minus in Problem 2.1): direction of friction flipped. Going down, friction subtracts from the pull; going up, friction adds to gravity because both now point down-slope. Compare the orange arrow's two possible directions in the master figure.

WHAT the forces are (block moving up): applied $F$ up; gravity component $mg\sin \theta$ down; kinetic friction ${\mu }_{k}mg\cos \theta$ down (opposes upward motion). Newton's law along slope, taking up-slope as positive: $ma=F-mg\sin \theta -{\mu }_{k}mg\cos \theta$ Compute the pieces:

• $mg\sin \theta =5\times 9.8\times 0.5=24.5\text{N}$
• $mg\cos \theta =5\times 9.8\times 0.8660=42.44\text{N}$, so friction $=0.25\times 42.44=10.61\text{N}$ $ma=60-24.5-10.61=24.89\text{N}$ $a=\frac{24.89}{5}=4.98{\text{m/s}}^2\text{up the slope}$ WHY both subtractions: going up, the block fights BOTH gravity (naturally downhill) AND friction (resisting the uphill motion). Both are "negative" relative to the push.

WHAT decides sliding-back: at the top the block is momentarily at rest. Static friction now tries to hold it. It slides back only if the down-slope pull beats maximum static friction. The problem gives only ${\mu }_k$; assume ${\mu }_s\approx {\mu }_k=0.25$ for the check.

• Down-slope pull: $mg\sin 30°=24.5\text{N}$
• Max static friction: ${\mu }_{s}mg\cos 30°=0.25\times 42.44=10.61\text{N}$ Since $24.5>10.61$, yes it slides back down. (Equivalently $\tan 30°=0.577>0.25$.) Slide-back acceleration: now moving down, friction flips to point UP: $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(0.5-0.25\times 0.8660)=9.8(0.5-0.2165)=2.78{\text{m/s}}^2\text{down}$ WHY smaller than the up-deceleration: going down, friction helps against gravity (subtracts), so the net down-slope force is smaller than the net force fighting the upward trip.

WHAT to invert: we know the sliding-down formula and want ${\mu }_k$. Start from $a=g(\sin \theta -{\mu }_k\cos \theta )$ Solve for ${\mu }_k$: ${\mu }_k=\frac{g\sin \theta -a}{g\cos \theta }=\frac{9.8\times 0.6018-3.0}{9.8\times 0.7986}=\frac{5.898-3.0}{7.826}=\frac{2.898}{7.826}=0.370$ WHY this is mastery: we reversed the usual direction — from a measured motion back to a surface property. This is exactly how experimentalists measure ${\mu }_k$.

WHAT the energy accounting is: over a slope distance $d$, gravity does positive work $mg\sin \theta \cdot d$ (component along motion), friction does negative work $-{\mu }_{k}mg\cos \theta \cdot d$. The net work equals the gain in kinetic energy (starts from rest): $\frac{1}{2}m{v}^2=mg\sin \theta d-{\mu }_{k}mg\cos \theta d$ Mass cancels: $v^2=2gd(\sin \theta -{\mu }_k\cos \theta )$ Compute: $\sin 45°=\cos 45°=0.7071$. $v^2=2\times 9.8\times 3\times (0.7071-0.40\times 0.7071)=58.8\times (0.7071-0.2828)=58.8\times 0.4243=24.94$ $v=\sqrt{24.94}=4.99\text{m/s}$ Cross-check with kinematics: $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8\times 0.4243=4.158{\text{m/s}}^2$; then $v=\sqrt{2ad}=\sqrt{2\times 4.158\times 3}=\sqrt{24.95}=4.99\text{m/s}$ ✓. Both routes agree — energy and force are two views of the same physics.

WHAT we compute: the net down-slope force once moving is $F_{\text{net}}=mg\sin \theta -{\mu }_{k}mg\cos \theta =mg\cos \theta (\tan \theta -{\mu }_k)$ At $\theta ={\theta }_c$ we have $\tan {\theta }_c={\mu }_s={\mu }_k$, so $\tan {\theta }_c-{\mu }_k=0\Rightarrow F_{\text{net}}=0$. Beyond ${\theta }_c$: $\tan \theta$ increases (it's a rising function on $0$ to $90°$), so $\tan \theta -{\mu }_k>0\Rightarrow F_{\text{net}}>0$: the block accelerates. Interpretation: the angle of repose is a knife-edge. Right at it, the block is in neutral balance (constant velocity if nudged, with ${\mu }_k={\mu }_s$); any steeper and it genuinely accelerates. This is WHY ${\theta }_c$ is the sharp threshold between "stays put" and "runs away." Notice the factor $mg\cos \theta$ is always positive for $\theta <90°$, so the sign of $F_{\text{net}}$ is decided entirely by $(\tan \theta -{\mu }_k)$.

Moving DOWN ::: friction points UP → $a=g(\sin \theta -{\mu }_k\cos \theta )$. Moving UP ::: friction points DOWN → deceleration $=g(\sin \theta +{\mu }_k\cos \theta )$. At rest, not slipping ::: friction = whatever balances the pull, up to max ${\mu }_{s}mg\cos \theta$.