Weight always points straight down, but on a slope that single force does two jobs: one part presses the object into the ramp, the other drags it down the slope. Split the weight into these two components and Newton's second law along the slope hands you a single, powerful formula for the acceleration, $a = g(\sin\theta - \mu\cos\theta)$, that works for any angle and any roughness.
Two blocks, one $2\ \text{kg}$ and one $8\ \text{kg}$, slide down the same frictionless ramp from rest.
Before reading: do they accelerate at the same rate, or does the heavier block accelerate faster? Write your prediction and one sentence explaining why.
One down for sliding, one into the surface
A single downward weight, seen with tilted axes, becomes two useful pieces: one that slides the object down the ramp and one that presses it into the ramp.
With axes along and perpendicular to the slope, the weight $mg$ (straight down) resolves into:
$$W_\parallel = mg\sin\theta \quad (\text{down the slope}) \qquad W_\perp = mg\cos\theta \quad (\text{into the slope})$$
The parallel component $mg\sin\theta$ is the only thing trying to slide the object; the perpendicular component $mg\cos\theta$ is balanced by the normal force. Notice the geometry: because the slope angle $\theta$ appears between the weight and the perpendicular axis, the down-slope part takes the $\sin$ and the into-surface part takes the $\cos$.
On a slope of angle $\theta$, resolve weight into $mg\sin\theta$ (parallel, down the slope, the driving component) and $mg\cos\theta$ (perpendicular, into the surface, balanced by the normal force $N$). The down-slope component uses $\sin$, the into-surface component uses $\cos$.
Pause, copy the highlighted resolution into your book before moving on.
Did you get this? Which component of weight drives an object to slide down a slope of angle $\theta$?
Newton's second law along the slope, step by step
We just saw how weight splits into two components. That raises a question: what acceleration does a block actually have as it slides down a rough slope? This card answers it → apply $F_{net} = ma$ along the slope and watch the mass cancel.
Adding up the forces along the slope and dividing by mass gives a formula where the mass disappears entirely, the acceleration depends only on the angle and the roughness.
Along the slope, two forces act: the driving component $mg\sin\theta$ (down) and friction $\mu mg\cos\theta$ (up, opposing motion, using $N = mg\cos\theta$). Apply Newton's second law along the slope and the mass cancels, leaving the master incline formula.
Derive an expression for the acceleration of a block of mass $m$ sliding down a rough slope of angle $\theta$ with coefficient of kinetic friction $\mu$.
Newton's second law along the slope gives $mg\sin\theta - \mu mg\cos\theta = ma$. Dividing by $m$, the mass cancels: $a = g(\sin\theta - \mu\cos\theta)$. On a frictionless slope ($\mu = 0$) this reduces to $a = g\sin\theta$, acceleration is independent of mass.
Add the highlighted derivation and result to your notes before the check below.
Quick check: True or false: on a frictionless slope, a $2\ \text{kg}$ block and an $8\ \text{kg}$ block accelerate at the same rate.
Change the angle, the mass and the friction, and watch which forces change
Use the simulation. With $\mu = 0.30$, the block holds at $16^\circ$ and slides at $17^\circ$. Which statement explains where the change happens?
Use the simulation. While the block is sliding, double its mass. True or false: the acceleration doubles, because the down-slope force doubles.
Pattern A, Calculate then reason
A block slides down a slope inclined at $30°$ with coefficient of kinetic friction $\mu = 0.20$. Take $g = 9.8\ \text{m s}^{-2}$. In your book: (1) calculate the acceleration using $a = g(\sin\theta - \mu\cos\theta)$; (2) recalculate the acceleration if the slope were frictionless; (3) explain in one sentence why the block's mass does not appear in your answer.
Two truths, one lie: which statement about resolving weight on an incline is the lie?
Dot point: Resolve the weight force into perpendicular and parallel components on inclined planes using mg cosθ, mg sinθ
Angle, friction, and the point where sliding begins
We just derived $a = g(\sin\theta - \mu\cos\theta)$. That raises a question: what does this formula predict as we change the angle or the roughness? This card answers it → it shows when a block stays still, when it slides, and how fast.
The formula is a compact summary: it tells you whether an object will slide at all, and if so, how its acceleration grows with the steepness of the slope.
Reading the formula $a = g(\sin\theta - \mu\cos\theta)$:
If the calculation gives a negative value, it means friction is strong enough to hold the block still (use static friction instead), a moving object does not spontaneously accelerate up the slope.
The formula $a = g(\sin\theta - \mu\cos\theta)$ shows acceleration rises with a steeper slope and falls with more friction. The block is on the verge of sliding when $\tan\theta = \mu$ (acceleration zero); a negative result means friction holds it stationary.
Pause, copy the highlighted interpretation into your book before the check below.
Last check: Decide if the statement is true or false.
Increasing the slope angle increases the down-slope acceleration of a sliding block.
Fill the gaps. Along the slope, $ma = mg\sin\theta - \mu N$ with $N = mg$ $\theta$. Cancelling $m$ gives $a = g(\sin\theta - \mu$ $\theta)$, which shows the acceleration does not depend on the .
Dot point: Derive a = g(sinθ − μcosθ) to describe the acceleration of an object on an inclined plane with friction using Newton's second law parallel to the plane
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
AnalyseBand 6(5 marks) 1. A block slides down a rough inclined plane of angle $\theta$ with coefficient of kinetic friction $\mu$. Derive, using Newton's second law along the slope, the expression $a = g(\sin\theta - \mu\cos\theta)$, clearly showing how the mass cancels.
1 mark: resolves weight as $mg\sin\theta$ (along) and $mg\cos\theta$ (perpendicular) · 1 mark: states $N = mg\cos\theta$ and friction $= \mu mg\cos\theta$ · 1 mark: applies $F_{net} = mg\sin\theta - \mu mg\cos\theta = ma$ · 1 mark: divides by $m$ to cancel mass · 1 mark: states result $a = g(\sin\theta - \mu\cos\theta)$
AnalyseBand 6(5 marks) 2. Starting from Newton’s second law applied parallel to the plane, derive $a = g(\sin\theta - \mu\cos\theta)$ for an object sliding down a rough incline. State clearly what each step assumes.
1 mark: resolves the weight into $mg\sin\theta$ along and $mg\cos\theta$ perpendicular · 1 mark: perpendicular balance gives $N = mg\cos\theta$ · 1 mark: writes $ma = mg\sin\theta - \mu N$ · 1 mark: substitutes $N$ · 1 mark: cancels $m$ and notes the acceleration is mass-independent
Dot point: Derive a = g(sinθ − μcosθ) to describe the acceleration of an object on an inclined plane with friction using Newton's second law parallel to the plane
EvaluateBand 6(3 marks) 3. Use $a = g(\sin\theta - \mu\cos\theta)$ to explain what happens physically when $\tan\theta = \mu$, and what it means if the formula returns a negative value.
1 mark: $a = 0$ when $\tan\theta = \mu$ · 1 mark: this is the angle at which the object slides at constant velocity, or is on the point of sliding · 1 mark: a negative result means gravity cannot overcome friction, so a stationary object stays at rest rather than accelerating up the slope
Dot point: Derive a = g(sinθ − μcosθ) to describe the acceleration of an object on an inclined plane with friction using Newton's second law parallel to the plane
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (5 marks): Choosing axes along and perpendicular to the slope, the weight resolves into $mg\sin\theta$ down the slope and $mg\cos\theta$ into the surface. Perpendicular to the slope there is no acceleration, so the normal force is $N = mg\cos\theta$, and the kinetic friction is $f = \mu N = \mu mg\cos\theta$, acting up the slope. Applying Newton's second law along the slope (down-slope positive): $F_{net} = mg\sin\theta - \mu mg\cos\theta = ma$. Dividing both sides by $m$, the mass cancels, giving $a = g(\sin\theta - \mu\cos\theta)$. The acceleration therefore depends only on the angle and the coefficient of friction, not on the mass.
Q2 (5 marks): Resolve the weight into a component $mg\sin\theta$ acting down the slope and $mg\cos\theta$ acting perpendicular to it. Perpendicular to the plane there is no acceleration, so those forces balance and $$N = mg\cos\theta$$ Parallel to the plane, the object accelerates down the slope under $mg\sin\theta$, opposed by kinetic friction $f = \mu N$ acting up the slope. Newton’s second law parallel to the plane gives $$ma = mg\sin\theta - \mu N$$ Substituting the normal force, $$ma = mg\sin\theta - \mu mg\cos\theta$$ Every term carries a factor $m$, so dividing through gives $$a = g(\sin\theta - \mu\cos\theta)$$ The assumptions are that the object is already sliding, so kinetic rather than static friction applies, that $\mu$ is constant, and that air resistance is negligible. The mass cancelling means two objects of different mass on the same surface slide down at the same rate.
Q3 (3 marks): Setting $a = 0$ gives $\sin\theta = \mu\cos\theta$, that is $\tan\theta = \mu$. At this angle the component of the weight down the slope exactly equals the maximum friction, so an object already moving slides down at constant velocity and one at rest is on the verge of slipping. This angle is a direct experimental route to $\mu$. If the formula returns a negative value, $\theta$ is below that critical angle and the friction available exceeds the driving component of the weight. It does not mean the object accelerates up the slope, which would be physically absurd: it means the assumption that the object is sliding was wrong. A stationary object simply stays at rest, held by static friction, and the equation no longer applies.
You can now resolve weight on a slope and derive $a = g(\sin\theta - \mu\cos\theta)$. Next lesson puts this formula to work on a full set of inclined-plane problems and a hands-on practical.
↩ Back to focus-area hubAt the start you predicted whether a heavier block accelerates faster down a frictionless slope. The answer: they accelerate equally, because $a = g\sin\theta$ has no mass in it, the mass cancels in the derivation. If you predicted the heavier block wins, revisit Card 2.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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