Worked examples — Inclined planes — with and without friction

Before we start, one reminder of the whole toolkit, so no symbol appears unexplained:

The scenario matrix

Every incline problem lives in one of these cells. The last column names the example that covers it.

Ex 1 — The two degenerate limits (Cells A & B)

Step 1 — Flat case, $\theta =0^{\circ }$. $mg\sin 0^{\circ }=5(9.8)(0)=0\text{N}N=mg\cos 0^{\circ }=5(9.8)(1)=49\text{N}$ Why this step? We test the formula at its edge. $\sin 0=0$ means no along-slope pull — nothing tries to slide the block, exactly what "flat" should mean. $\cos 0=1$ means the surface carries the entire weight, $N=mg=49\text{N}$.

Step 2 — Vertical case, $\theta =90^{\circ }$. $mg\sin 90^{\circ }=5(9.8)(1)=49\text{N}N=mg\cos 90^{\circ }=5(9.8)(0)=0\text{N}$ Why this step? At $90^{\circ }$ the "slope" is a vertical wall. The whole weight $49\text{N}$ becomes along-slope pull (the block is in free fall), and $N=0$ — a vertical wall presses on nothing resting against it. Look at the right figure: the blue normal arrow has shrunk to zero.

Ex 2 — Frictionless slide, mass cancels (Cell C)

Step 1 — Use $a=g\sin \theta$. $a=9.8\sin 37^{\circ }=9.8(0.6018)=5.90{\text{m/s}}^2\text{ down the slope}$ Why this step? Frictionless means nothing opposes the along-slope pull, so Newton's 2nd law along the slope is simply $mg\sin \theta =ma$; the $m$'s divide out.

Step 2 — Double the mass. $m$ never appears in $a=g\sin \theta$, so $a=5.90{\text{m/s}}^2$ again. Why this step? This is the "beautiful" cancellation from the parent — driving force $\propto m$, inertia $\propto m$, so mass drops out. A feather and a boulder tie.

Ex 3 — Rough but stays put; find the REAL friction (Cell D)

Step 1 — Slide test with $\tan \theta$ vs ${\mu }_s$. $\tan 20^{\circ }=0.364\le {\mu }_s=0.60$ Why this step? The block slips only when the along-slope pull beats max friction, i.e. $\tan \theta >{\mu }_s$. Here $0.364\le 0.60$, so it stays at rest.

Step 2 — Actual friction = whatever balances the pull. Because it is static and not moving, along-slope forces cancel: $f_s=mg\sin \theta =3(9.8)\sin 20^{\circ }=3(9.8)(0.342)=10.05\text{N (up the slope)}$ Why this step? Static friction is not automatically ${\mu }_{s}N$. It only supplies as much as needed to hold the block — here just enough to cancel the $10.05\text{N}$ pull. The maximum available is ${\mu }_{s}N=0.60\times 3(9.8)\cos 20^{\circ }=16.58\text{N}$, comfortably more.

Ex 4 — Rough, actually slides down (Cell E)

Step 1 — Confirm motion is possible. $\tan 40^{\circ }=0.839>{\mu }_k=0.25$, so once moving it keeps sliding (the pull beats kinetic friction). Why this step? If $\tan \theta <{\mu }_k$, the friction term would swamp the pull and the formula would give a nonsensical "negative" acceleration meaning "it doesn't slide." We check first.

Step 2 — Apply the sliding-down formula. $a=g(\sin \theta -{\mu }_k\cos \theta )=9.8(\sin 40^{\circ }-0.25\cos 40^{\circ })$ $=9.8(0.6428-0.25\times 0.766)=9.8(0.6428-0.1915)=9.8(0.4513)=4.42{\text{m/s}}^2$ Why this step? Kinetic friction acts up the slope (opposing the downhill motion), so it subtracts ${\mu }_{k}g\cos \theta$ from the pull.

Ex 5 — Reverse: find ${\mu }_s$ from the slip angle (Cell F)

Step 1 — Use the angle of repose. ${\mu }_s=\tan {\theta }_c=\tan 31^{\circ }=0.601$ Why this step? At the verge of slipping, $mg\sin \theta ={\mu }_{s}mg\cos \theta$. Mass and $g$ cancel, leaving ${\mu }_s=\tan {\theta }_c$ — a stickiness number read straight off a protractor. This is the elegant measurement trick from the parent, linked to Friction — Static and Kinetic.

Ex 6 — Pushed UP the slope: both forces fight you (Cell G)

Step 1 — List every along-slope force with its sign. Up = positive. Gravity's slope-pull is down; friction opposes the upward motion, so it too points down: $F-mg\sin \theta -{\mu }_{k}mg\cos \theta =ma$ Why this step? Going up, both nature (gravity) and friction point downhill — see the figure's two down-arrows against your single up-arrow.

Step 2 — Plug numbers. $mg\sin 25^{\circ }=4(9.8)(0.4226)=16.57\text{N}$ ${\mu }_{k}mg\cos 25^{\circ }=0.30\times 4(9.8)(0.9063)=10.66\text{N}$ $a=\frac{40-16.57-10.66}{4}=\frac{12.77}{4}=3.19{\text{m/s}}^2\text{ up the slope}$ Why this step? Net up-slope force $=F$ minus the two opponents, then divide by mass (Newton's Second Law).

Ex 7 — Real-world word problem: kinematics on a ramp (Cell H)

Step 1 — Acceleration down the slope. $a=g\sin 30^{\circ }=9.8(0.5)=4.9{\text{m/s}}^2$ Why this step? Frictionless, so $a=g\sin \theta$; mass is irrelevant.

Step 2 — Final speed via $v^2=u^2+2as$. Starts from rest ($u=0$), travels $s=3\text{m}$: $v=\sqrt{2as}=\sqrt{2(4.9)(3)}=\sqrt{29.4}=5.42\text{m/s}$ Why this step? We know distance and acceleration but not time yet, so the time-free kinematic equation is the natural choice.

Step 3 — Time via $v=u+at$. $t=\frac{v}{a}=\frac{5.42}{4.9}=1.11\text{s}$ Why this step? Now that $v$ is known, this is the quickest route to time.

Ex 8 — Exam twist: minimum force to STOP a block sliding (Cell I)

Step 1 — Set up the verge-of-sliding-down balance. On the brink of sliding down, both $F$ and maximum static friction point up, gravity's pull points down: $F+{\mu }_{s}mg\cos \theta =mg\sin \theta$ Why this step? We want the smallest $F$; friction is a free ally here, so we let it work at full strength ${\mu }_{s}N$ up the slope. Anything less than this $F$ and the block slips down.

Step 2 — Solve for $F$. $mg\sin 35^{\circ }=2(9.8)(0.5736)=11.24\text{N}$ ${\mu }_{s}mg\cos 35^{\circ }=0.30\times 2(9.8)(0.8192)=4.82\text{N}$ $F=11.24-4.82=6.42\text{N}$ Why this step? The applied force only needs to make up the shortfall between gravity's pull and what friction alone can hold.

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