Question bank — Inclined planes — with and without friction

False. Only the perpendicular slice of gravity presses into the surface, so $N=mg\cos \theta$, which is less than $mg$ for any real tilt ($\theta >0$).

False. $a=g\sin \theta$ has no mass in it — the driving force $mg\sin \theta$ and the inertia $m$ both scale with mass and cancel, so all blocks share the same acceleration.

False. Steeper tilt means gravity leans along the slope more and into it less, so $N=mg\cos \theta$ decreases as $\theta$ grows, reaching $0$ at a vertical wall.

False. Friction opposes the actual or impending relative motion. Sliding down → friction up; being pushed up → friction points down; held static on a gentle slope → friction points up to hold it.

True (marginally). It is the exact angle where the slope pull just equals maximum static friction, $mg\sin \theta ={\mu }_{s}mg\cos \theta$ — the last instant of rest before sliding begins.

False. The slope (sliding) share uses sine: at $\theta =0$ there should be zero sliding force, and only $\sin 0=0$ delivers that. "Slope gets Sine."

False. Kinetic friction only exists while sliding and only opposes motion — it can never be the cause of new motion, only a resistance to existing motion.

False. $\tan {\theta }_c={\mu }_s$ contains no mass — the repose angle depends only on the surfaces' stickiness, not on how heavy the object is.

False. Mirroring just flips "left" and "right"; $N=mg\cos \theta$ and $mg\sin \theta$ are unchanged because they depend on the tilt magnitude $\theta$, not on which way the hill faces. Only your labelling of positive-along-slope flips with the mirror.

Sine and cosine are swapped. The normal force balances the perpendicular gravity slice, so $N=mg\cos \theta$. The limit test: at $\theta =0$ we need $N=mg$, which needs $\cos$ (since $\cos 0=1$).

Wrong comparison. Sliding is decided along the slope: it slides when the along-slope pull $mg\sin \theta$ beats static friction ${\mu }_{s}mg\cos \theta$, i.e. when $\tan \theta >{\mu }_s$. Comparing $mg$ to $N$ mixes different directions.

The friction sign is wrong. Moving up, friction opposes the upward motion, so it points down the slope and must be subtracted: $F-mg\sin \theta -{\mu }_{k}mg\cos \theta =ma$. Going up, both gravity and friction fight you.

The perpendicular direction balances, but the along-slope direction does not — $mg\sin \theta$ is unopposed, so the block accelerates. Balancing one axis is not equilibrium.

A negative result here means the block never started moving ($\tan \theta <{\mu }_k$), not that kinetic friction reverses it. Kinetic friction can only slow, never drive.

The un-rotated axes give a real, solvable pair of equations — but $N\ne mg$. Along horizontal $x$: $N\sin \theta =m{a}_x$; along vertical $y$: $N\cos \theta -mg=m{a}_y$. Because the block also slides, both $a_x$ and $a_y$ are nonzero, so $N$ appears in two coupled equations and you must solve them simultaneously. Rotating the axes makes $N$ sit on one axis alone, collapsing this to the clean $N=mg\cos \theta$ — that convenience is the whole point.

So the normal force lies purely on one axis and the acceleration purely on the other — then only gravity needs resolving, instead of resolving three forces at once.

Because "vertical & horizontal" and "normal & slope" are both pairs of lines at $90^{\circ }$ to each other. Rotating the first pair by $\theta$ lands it exactly on the second, carrying the angle unchanged — see the geometry figure above. So the angle between $mg$ and $N$ equals $\theta$.

The driving force $mg\sin \theta$ is proportional to $m$, and Newton's law divides by that same $m$ (inertia). The two factors of $m$ cancel — exactly like free fall.

At slipping, $\tan {\theta }_c={\mu }_s$ — mass, length, and $g$ all cancel. Measuring the one angle where sliding starts hands you the friction coefficient directly.

Kinetic friction subtracts a term: $a=g(\sin \theta -{\mu }_k\cos \theta )$ versus $g\sin \theta$. As long as the block slides, that $-{\mu }_{k}g\cos \theta$ removes speed-up.

Because friction is proportional to $N$, and $N=mg\cos \theta$ shrinks as the slope steepens. Less "crush" into the surface means less grip.

It gives an acceleration $g\sin \theta$ that is a scaled-down $g$, letting you study gravity's effects in slow motion instead of full free fall.

The sliding force $mg\sin 0=0$ (nothing pushes it along), and $N=mg\cos 0=mg$ — the surface carries the full weight. This recovers ordinary flat-ground physics.

$N=mg\cos 90^{\circ }=0$ (nothing presses into the wall) and the along-slope pull is $mg\sin 90^{\circ }=mg$ — pure free fall straight down.

Since $\cos (-\theta )=\cos \theta$, the normal $N=mg\cos \theta$ is identical — the surface still crushes the same. But $\sin (-\theta )=-\sin \theta$, so the along-slope pull flips direction: the block tends to slide the opposite way. The magnitudes match a mirrored hill; only the sign of the slide-component reverses.

Only as big as needed to balance the pull: $f_s=mg\sin \theta$, not its maximum ${\mu }_{s}mg\cos \theta$. Static friction is a responder, matching the demand up to a ceiling.

Still static, at its maximum value ${\mu }_{s}mg\cos \theta$. Kinetic friction only appears the instant after sliding begins.

Only if ${\mu }_k$ is large enough that $\tan \theta <{\mu }_k$ (so kinetic friction exceeds the pull). Otherwise it keeps sliding, because once moving, only the (usually smaller) ${\mu }_k$ acts.

The slope block gets $g\sin 30^{\circ }=g/2$; the free-falling block gets full $g$. The incline "steals" a factor of $\sin \theta$ from gravity — see Work-Energy Theorem on Inclines for the energy view.