Friction — One Unified Model for Equilibrium and Motion on a Rough Plane

A parcel rests on a rough plane inclined at $15°$ to the horizontal, held in place by nothing but gravity, the normal reaction, and friction — no other applied force. Which of these correctly describes the friction force $F$ acting on it?

• $F$ acts up the plane, and $F = mg\sin15°$, whatever the actual value of $\mu$ happens to be (provided $\mu$ is large enough for equilibrium to hold at all)

  Correct. Resolving along the plane with the parcel stationary (resultant zero) forces $F$ (up-slope, opposing the tendency to slide down) to exactly balance the down-slope weight component — this is true regardless of $\mu$'s specific value, as long as $F \le \mu R$ is actually satisfiable, because F ≤ μR only says friction COULD supply up to that much, not that it always does.

• B$F$ acts down the plane, opposing the parcel's tendency to be pushed up by the normal reaction

  The normal reaction doesn't push a particle 'up the plane' — it acts perpendicular to the surface, not along it, so it creates no along-plane tendency to oppose. Gravity is the only force with an along-plane component here, and it pulls down-slope, so any tendency to slide (and any friction opposing it) is about that, not about the normal reaction.

• C$F = \mu R$ necessarily, since friction is present whenever there's contact with a rough surface

  Presence of friction doesn't mean it's at its LIMITING value — that only holds if the particle is stated to be on the point of sliding or actually moving. Here the parcel is simply described as resting in equilibrium, so all that's guaranteed is $F \le \mu R$; the actual value of F is fixed by the resolving equation, at $mg\sin15°$, independently of what the ceiling $\mu R$ happens to be (provided it's above that value).

• D$F = 0$, since the parcel isn't accelerating

  Not accelerating means the resultant is zero, not that friction itself is zero — here gravity's down-slope component is genuinely nonzero ($mg\sin15° \ne 0$), so something has to balance it, and with no other applied force, that something is friction.

Traps tested: Frictional force direction reversed on a rough plane · Limiting friction equation applied without a limiting condition · Static particle assumed to have zero friction

A box rests on a rough plane inclined at $25°$ to the horizontal, with no other applied forces, and is stated to be on the point of sliding (limiting equilibrium). In which direction does the friction force act, and why?

• Up the plane, because the box's weight component tends to pull it down the slope, and friction opposes that tendency

  Correct. With gravity the only force having an along-plane component, and gravity always pulling down-slope, the box's only possible tendency to slide is downward — so limiting friction, opposing that tendency, acts up the plane. This is exactly the direction call a real WME01 examiner report records candidates getting wrong: "the most common error was having the frictional force acting down rather than up the plane" (Oct 2022, Q3).

• BDown the plane, because that's the direction the box would move in if friction disappeared

  This IS the direction the box would move without friction — which is precisely why friction has to act the OPPOSITE way to stop it: up the plane, not down. This choice describes the tendency correctly and then draws friction pointing the same way as that tendency instead of against it — the exact reversal the Oct 2022 report documents.

• CPerpendicular to the plane, since limiting friction always acts opposite to the normal reaction

  Friction acts ALONG the surface at every stage — limiting or not, moving or not — never perpendicular to it. The perpendicular direction belongs entirely to the normal reaction; conflating the two makes it impossible to resolve them as separate, independent quantities.

• DAlong the plane, but its direction can't be determined without knowing $\mu$

  Friction's DIRECTION is fixed entirely by the geometry (which way gravity's along-plane component points, here down-slope) — it never depends on the actual numerical value of $\mu$. $\mu$ only affects the SIZE of the limiting friction force, $\mu R$, not which way that force points.

Traps tested: Frictional force direction reversed on a rough plane · Friction direction confused with normal reaction direction · Overclaims uncertainty

A sled slides down a rough plane, and its acceleration is calculated in part (a) of a question. In part (b), a rope now pulls the sled up the plane with an additional force while it continues to slide down, and a student reuses part (a)'s acceleration value as the answer to part (b) without further working. What has actually gone wrong?

• The net force along the plane has changed because a new force is now included in the system — the acceleration has to be recalculated from a fresh equation of motion, not carried over

  Correct. This is exactly the failure a real WME01 examiner report records on a two-part incline question with a changed force condition: "a substantial number did not recognise that the system was subject to a different acceleration" (Jun 2024, Q6). The sled continuing to move in the same direction across both parts says nothing about whether the FORCES acting on it — and therefore its acceleration — have stayed the same.

• BNothing has gone wrong — since the sled is still sliding down the plane in both parts, the friction and gravity terms are unchanged, so the acceleration must be the same

  This is precisely the documented error, restated as if it were correct reasoning. The DIRECTION of the sled's motion staying the same does not mean the FORCES acting on it have — a new applied force genuinely changes the along-plane resultant, and therefore the acceleration, even while the sled keeps moving the same way.

• CFriction must have reversed direction once the new rope force was added, so the whole equation needs rebuilding with F acting down the plane instead

  Friction's direction is decided by the sled's ACTUAL motion, not by which new forces have been added — and the sled is still moving down the plane in part (b), so friction is still up the plane, exactly as in part (a). This answer correctly spots that something needs rebuilding, but rebuilds the wrong quantity: it's the NET FORCE (from the new applied force) that changes, not friction's direction.

• DThe sled's mass must be different in part (b), since a different scenario should mean a different mass, not a different acceleration

  Nothing in the scenario suggests the mass has changed between parts — only that an additional force has been introduced. Reaching for a changed mass here confuses which quantity in the equation of motion the question has actually altered.

Traps tested: Changed force condition assumed not to change the acceleration · Friction direction changed unnecessarily when a new force is added · Unrelated quantity blamed for the real change

A crate is on the point of sliding down a rough plane inclined at $35°$ to the horizontal, held in equilibrium only by friction (no other applied forces). What is the coefficient of friction $\mu$, and which method finds it fastest?

• $\mu = \tan35° \approx 0.700$ (3 s.f.), found by resolving perpendicular ($R=mg\cos35°$) and parallel ($F=mg\sin35°$) then dividing the two equations directly — eliminating R (and m, and g) in one step rather than computing R as a decimal first

  Correct. At limiting equilibrium, $F=\mu R$, so $\mu(mg\cos35°) = mg\sin35°$, and dividing both sides by $mg\cos35°$ gives $\mu = \tan35°$ directly — the mass and $g$ cancel completely, and no decimal value of $R$ ever needs computing. This is the elimination technique a real WME01 examiner report verifies as a genuinely used, credited method on this exact scenario family (Jan 2022, Q5).

• B$\mu = mg\sin35°$, since limiting friction equals the weight component down the plane

  This reports a FORCE (units of newtons) as if it were the coefficient of friction, which is dimensionless. $F=mg\sin35°$ is correct as far as it goes, but $\mu$ comes from $F = \mu R$, i.e. $\mu = F/R$ — R has been dropped from the equation entirely here, not just computed differently.

• C$\mu = \sin35°$, since $R \approx mg$ for a plane close to horizontal

  $35°$ is not close to horizontal, and even where an approximation like this might tempt someone, $R = mg\cos35°$ is the correctly resolved value — using $mg$ in place of $R$ mixes up the FULL weight with the weight's PERPENDICULAR component, which are only equal when the plane is exactly horizontal.

• D$\mu$ cannot be found without knowing the crate's actual mass

  Mass cancels completely out of a limiting-equilibrium-under-gravity-alone equation, as the correct working shows — $\mu = \tan35°$ needs only the angle, nothing about how heavy the crate actually is.

Traps tested: Normal reaction omitted from the limiting friction equation · Normal reaction conflated with the full weight · Overclaims uncertainty