Resolution of forces and inclined planes
Master resolution of forces and inclined planes for Year 12 Specialist Mathematics in Queensland (QCAA). You split a force into two perpendicular parts, then apply them to a particle resting or sliding on a smooth slope.
You will learn to resolve the weight into mg sin theta down the slope and mg cos theta across it, find the normal reaction and the acceleration, and work out the force needed for equilibrium — a core application of Newton's second law in modelling motion.
Theory
Resolution of forces and inclined planes in Year 12 Specialist Mathematics (QCAA, Queensland) is about splitting a force into two perpendicular parts. For a particle on a smooth slope, the weight \(mg\) resolves into \(mg\sin\theta\) down the line of greatest slope and \(mg\cos\theta\) perpendicular to it. This page shows how to find the normal reaction, the acceleration \(a=g\sin\theta\), and the force needed for equilibrium.
To resolve a force means to replace it with two perpendicular components that together have the same effect. A force of magnitude \(F\) acting at an angle \(\theta\) to a chosen direction has a component \(F\cos\theta\) along that direction and \(F\sin\theta\) at right angles to it.
On an inclined plane the natural directions are along the line of greatest slope and perpendicular to the surface. The particle's weight \(mg\) acts vertically down, so relative to a slope of angle \(\theta\) it resolves into \(mg\sin\theta\) down the slope and \(mg\cos\theta\) into the plane.
A smooth plane is frictionless, so the surface can only push outwards: the normal reaction \(N\) is perpendicular to the plane and balances the perpendicular part of the weight, giving \(N=mg\cos\theta\). (This course excludes rough planes, pulleys and connected bodies.)
Along the slope, Newton's second law \(F=ma\) governs the motion. In equilibrium the forces along the slope cancel; in non-equilibrium the resultant along the slope produces an acceleration. A particle released on a smooth incline has only \(mg\sin\theta\) acting along it, so \(a=g\sin\theta\) — independent of the mass.
For a particle of mass \(m\) on a smooth plane inclined at \(\theta\), resolve the weight along and perpendicular to the slope:
The normal reaction balances the perpendicular component, and a particle released from rest accelerates down the slope:
To hold the particle at rest, apply a force \(F\) up the slope or a horizontal force \(P\):
Solving an inclined-plane problem
- Set axes along the line of greatest slope and perpendicular to the surface, and draw every force from the particle.
- Resolve the weight: \(mg\sin\theta\) down the slope and \(mg\cos\theta\) perpendicular to it.
- Balance perpendicular: the surface gives no motion into it, so \(N=mg\cos\theta\) (plus any perpendicular part of an applied force).
- Apply \(F=ma\) along the slope: add the along-slope forces; set the resultant to \(0\) for equilibrium, or to \(ma\) to find the acceleration.
First the weight, then its along-slope component \(mg\sin\theta\):
| \(mg\) | \(=\) | \(6 \times 9.8\) |
| \(=\) | \(58.8\) | |
| \(mg\sin\theta\) | \(=\) | \(58.8 \times \sin 30^\circ\) |
| \(=\) | \(58.8 \times \dfrac{1}{2}\) | |
| \(=\) | \(29.4\text{ N}\) |
Perpendicular to a smooth plane the reaction balances \(mg\cos\theta\):
| \(N\) | \(=\) | \(58.8 \times \cos 30^\circ\) |
| \(=\) | \(58.8 \times \dfrac{\sqrt{3}}{2}\) | |
| \(=\) | \(29.4\sqrt{3}\) | |
| \(\approx\) | \(50.9\text{ N}\) |
Down the slope \(29.4\text{ N}\); normal reaction \(N\approx 50.9\text{ N}\).
Along a smooth slope the only force is \(mg\sin\theta\), so \(ma=mg\sin\theta\) and the mass cancels:
| \(ma\) | \(=\) | \(mg\sin\theta\) |
| \(a\) | \(=\) | \(g\sin\theta\) |
| \(=\) | \(9.8 \times \sin 40^\circ\) | |
| \(=\) | \(9.8 \times 0.6428\ldots\) | |
| \(\approx\) | \(6.3\text{ m/s}^2\) |
The acceleration is \(6.3\ \text{m/s}^2\) down the slope.
Resolve along the slope. The horizontal force contributes \(P\cos\theta\) up the slope; for equilibrium it balances \(mg\sin\theta\):
| \(P\cos\theta\) | \(=\) | \(mg\sin\theta\) |
| \(P\) | \(=\) | \(mg\tan\theta\) |
| \(=\) | \(49 \times \tan 30^\circ\) | |
| \(=\) | \(49 \times \dfrac{1}{\sqrt{3}}\) | |
| \(=\) | \(\dfrac{49\sqrt{3}}{3}\text{ N}\) |
The horizontal force is \(P=\dfrac{49\sqrt{3}}{3}\text{ N}\ (\approx 28.3\text{ N})\).
Find the resultant along the slope: the push acts up, the weight component \(mg\sin\theta\) acts down:
| \(mg\sin\theta\) | \(=\) | \(5 \times 9.8 \times \dfrac{1}{2}\) |
| \(=\) | \(24.5\text{ N}\) | |
| \(\text{net}\) | \(=\) | \(40 - 24.5\) |
| \(=\) | \(15.5\text{ N up the slope}\) |
Apply \(F=ma\) along the slope:
| \(a\) | \(=\) | \(\dfrac{\text{net}}{m}\) |
| \(=\) | \(\dfrac{15.5}{5}\) | |
| \(=\) | \(3.1\text{ m/s}^2\) |
The acceleration is \(3.1\ \text{m/s}^2\) up the slope.
Common pitfalls
Frequently asked questions
Why does the weight split into mg sin theta and mg cos theta?
The weight \(mg\) points straight down. Measured relative to a slope of angle \(\theta\), the part along the slope is \(mg\sin\theta\) and the part perpendicular to it is \(mg\cos\theta\). Together these two perpendicular components add back to the full weight.
What is the normal reaction on a smooth inclined plane?
The surface pushes out perpendicular to itself and balances the perpendicular part of the weight, so \(N=mg\cos\theta\). It is not equal to the full weight \(mg\) unless the plane is horizontal.
How fast does a particle accelerate down a smooth slope?
Only \(mg\sin\theta\) acts along a smooth slope, so \(ma=mg\sin\theta\) gives \(a=g\sin\theta\). With \(g=9.8\), a \(30^\circ\) slope gives \(a=4.9\ \text{m/s}^2\).
Does the mass change the acceleration down a smooth incline?
No. In \(a=g\sin\theta\) the mass has cancelled, so a heavy block and a light block released together on the same smooth slope accelerate at exactly the same rate.
What force keeps a particle in equilibrium on a smooth incline?
A force applied up the line of greatest slope must equal \(mg\sin\theta\). If instead the holding force is horizontal, it must equal \(mg\tan\theta\).
How do I resolve a force at an angle into components?
A force \(F\) at angle \(\theta\) to a chosen direction has a component \(F\cos\theta\) along that direction and \(F\sin\theta\) perpendicular to it. Pick the two directions that make the problem simplest — on a slope, along and perpendicular to the surface.