A ramp tilts the whole problem. The trick to every inclined-plane question is to tilt your axes too: line one axis up the slope and the other perpendicular to it. Do that and weight splits cleanly into a part that pushes the object down the slope and a part that presses it into the surface, and the normal force is no longer just $mg$.
A block sits on a ramp. You gradually tilt the ramp from flat up towards vertical.
Before reading: what happens to the normal force (the push of the surface on the block) as the ramp gets steeper, does it increase, stay the same, or decrease? Write your prediction and one sentence of reasoning.
Gravity, normal force, friction, and the net force they leave
On a ramp an object feels the same three forces as on flat ground, gravity, the surface push, and friction, but now they point in different directions, so we draw them carefully.
For an object on an inclined plane, draw these forces in the free-body diagram:
On an inclined plane an object feels weight $W = mg$ (straight down), the normal force $N$ (perpendicular to the surface), and friction $f$ (along the slope, opposing motion). The net force is their vector sum; on a frictionless slope it points straight down the incline.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: the normal force on an object on a slope points perpendicular to the slope surface, not straight up.
Only the perpendicular slice of weight presses into the surface
We just saw the four forces on a slope. That raises a question: how big is the normal force when the ground is tilted? This card answers it → resolve the weight perpendicular to the slope; the surface supports only that part, giving $N = mg\cos\theta$.
The surface only has to hold up the part of the weight that presses straight into it, and on a slope that part shrinks as the angle grows.
Choose axes along and perpendicular to the slope. The weight $mg$ splits into a component pressing into the surface, $mg\cos\theta$, and a component along the slope, $mg\sin\theta$. Perpendicular to the slope there is no acceleration (the object stays on the ramp), so the normal force balances the perpendicular weight component:
$$N = mg\cos\theta$$
As $\theta$ increases, $\cos\theta$ decreases, so $N$ falls, at $\theta = 0$ (flat) $N = mg$, and at $\theta = 90°$ (vertical) $N = 0$.
A $12\ \text{kg}$ box rests on a ramp inclined at $30°$ to the horizontal. Using $g = 9.8\ \text{m s}^{-2}$, find the normal force the ramp exerts on the box.
On an incline the normal force balances only the perpendicular component of weight: $N = mg\cos\theta$. This is less than $mg$, and it falls as the slope gets steeper. For the example, $N = 12 \times 9.8 \times \cos 30° \approx 102\ \text{N}$.
Add the highlighted formula and method to your notes before the check below.
Quick check: A $10\ \text{kg}$ block rests on a slope of $60°$. Using $g = 9.8\ \text{m s}^{-2}$, the normal force is closest to…
Pattern A, Calculate then reason
A $20\ \text{kg}$ crate rests on a ramp at $25°$ to the horizontal. Take $g = 9.8\ \text{m s}^{-2}$. In your book: (1) draw a free-body diagram with tilted axes, labelling weight, normal force and friction; (2) calculate the normal force $N = mg\cos\theta$; (3) explain in one sentence what happens to $N$ if the ramp angle is increased to $40°$.
Two truths, one lie: which statement about an object on an incline is the lie?
Dot point: Determine the normal force on objects on inclined planes
Choose axes along and perpendicular to the slope, not horizontal and vertical
We just learned that $N = mg\cos\theta$. That raises a question: why resolve along the slope instead of using ordinary horizontal and vertical axes? This card answers it → tilted axes line up with the motion and the normal force, so two of the three forces lie cleanly on the axes.
By rotating the axes to match the ramp, the normal force and the sliding motion each fall along a single axis, leaving only the weight to split.
If you used horizontal and vertical axes, both the normal force and the motion would point at awkward angles, and you would have to resolve two forces. By tilting the axes to align with the slope, the normal force lies entirely along the perpendicular axis and the motion (and friction) lie entirely along the slope axis. Only the weight needs resolving, into $mg\cos\theta$ (perpendicular, balanced by $N$) and $mg\sin\theta$ (along the slope, the part that drives sliding). This is the standard set-up for every incline problem that follows.
Always tilt the axes to lie along and perpendicular to the slope. Then the normal force and the motion each sit on one axis, and only the weight must be resolved: $mg\cos\theta$ perpendicular (balanced by $N$) and $mg\sin\theta$ along the slope (the driving component).
Pause, copy the highlighted set-up into your book before the check below.
Last check: Decide if the statement is true or false.
On a steeper inclined plane the normal force on a resting block increases.
Fill the gaps. On a slope of angle $\theta$, the normal force is $N = mg$ $\theta$. As $\theta$ increases, $N$ . At $\theta = 0$ the normal force equals .
Dot point: Determine the normal force on objects on inclined planes
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.
UnderstandBand 5(4 marks) 1. An $8.0\ \text{kg}$ block rests on a ramp inclined at $35°$ to the horizontal. Using $g = 9.8\ \text{m s}^{-2}$, calculate the normal force, and explain why it is less than the block's weight.
1 mark: identifies $N = mg\cos\theta$ · 1 mark: substitutes $8.0 \times 9.8 \times \cos 35°$ · 1 mark: $N \approx 64\ \text{N}$ · 1 mark: explains only the perpendicular component of weight ($mg\cos\theta$) presses into the surface
ApplyBand 5(4 marks) 2. A $20\ \text{kg}$ crate rests on a frictionless ramp inclined at $30^\circ$. Taking $g = 9.8\ \text{m s}^{-2}$, draw or describe the free-body diagram and calculate the normal force.
1 mark: three forces identified, weight vertically down and normal perpendicular to the surface (no friction) · 1 mark: axes tilted along and perpendicular to the slope · 1 mark: uses $N = mg\cos\theta$ · 1 mark: $N = 170\ \text{N}$
Dot point: Determine the normal force on objects on inclined planes
UnderstandBand 5(3 marks) 3. Explain why the axes are conventionally tilted to lie along and perpendicular to an inclined plane rather than kept horizontal and vertical.
1 mark: the acceleration is along the slope, so one axis matches it · 1 mark: normal force and friction then lie entirely along the axes and need no resolving · 1 mark: only the weight must be resolved, so there is one resolution instead of several
Dot point: Construct free-body diagrams to show the gravitational, normal, frictional and net force acting on an object on an inclined 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 (4 marks): Resolving the weight perpendicular to the slope, the normal force balances the perpendicular component: $N = mg\cos\theta = 8.0 \times 9.8 \times \cos 35° = 78.4 \times 0.819 \approx 64\ \text{N}$. This is less than the block's full weight ($mg = 78.4\ \text{N}$) because on a slope only the component of weight perpendicular to the surface, $mg\cos\theta$, presses into the ramp; the remaining component $mg\sin\theta$ acts along the slope and is not supported by the normal force.
Q2 (4 marks): On a frictionless ramp only two forces act on the crate: its weight $mg$ acting vertically downward, and the normal force $N$ acting perpendicular to the ramp surface. Choosing axes tilted so that one runs along the slope and the other perpendicular to it means the normal force lies entirely along one axis and only the weight needs resolving. Perpendicular to the slope there is no acceleration, so the forces balance in that direction: $$N = mg\cos\theta = 20 \times 9.8 \times \cos 30^\circ = 196 \times 0.866 = 170\ \text{N}$$ This is less than the $196\ \text{N}$ weight, because on a slope only the component of the weight perpendicular to the surface presses the crate against it.
Q3 (3 marks): An object on an incline accelerates along the slope, never vertically, so choosing one axis along the slope means the whole acceleration lies on a single axis and the perpendicular direction has zero acceleration. That immediately gives one balance equation and one $F = ma$ equation instead of two coupled ones. The choice also aligns the axes with most of the forces: the normal force is perpendicular to the surface by definition and friction acts along it, so both lie entirely on an axis and need no resolving at all. Only the weight, which is vertical, has to be resolved, into $mg\sin\theta$ along the slope and $mg\cos\theta$ perpendicular to it. Keeping horizontal and vertical axes would instead force the normal force, the friction and the acceleration all to be resolved.
You can now draw a complete free-body diagram on a slope and calculate the normal force $N = mg\cos\theta$. Next lesson resolves the weight fully and derives the incline acceleration formula.
↩ Back to focus-area hubAt the start you predicted what happens to the normal force as a ramp gets steeper. The answer: it decreases, because $N = mg\cos\theta$ and $\cos\theta$ shrinks as $\theta$ grows, at vertical ($90°$) the normal force is zero. If you predicted it stays at $mg$, 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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