Motion on a Vertical or Inclined Plane (9709 Mechanics 4.4) — Mathematics

Introduction

Many real-world mechanics situations involve objects accelerating vertically (under gravity alone) or along a slope (under gravity, normal reaction, friction, and possibly an applied force). The 9709 syllabus requires you to apply Newton's second law ( $F = ma$ ) to these situations, correctly resolving forces along and perpendicular to the plane, and handling the subtlety that friction reverses direction when the particle changes from moving up to moving down. This subtlety is a frequent source of exam marks — and errors.

Core Concept

Vertical Motion

For a particle moving vertically with no surface, the only force is weight $W = mg$ acting downward. Taking upward as positive:

F_{\text{net}} = -mg \implies a = -g

This gives the familiar constant acceleration $g \approx 9.8\ \text{m s}^{-2}$ downward.

Motion on an Inclined Plane

When a particle of mass $m$ rests on a plane inclined at angle $\theta$ to the horizontal, forces are most conveniently resolved along the plane and perpendicular to the plane.

Perpendicular to the plane (no acceleration in this direction):

N = mg\cos\theta

Along the plane (taking up the plane as positive):

Component of weight down the plane: $mg\sin\theta$
Friction force $F$ : magnitude $\mu N = \mu mg\cos\theta$ , direction opposing motion

The Key Asymmetry: Rough Plane

This is the most tested idea in this subtopic. Because friction always opposes motion:

Direction of Motion	Friction Acts	Net force down the plane	Resulting acceleration
Up the plane	Down the plane	$mg\sin\theta + \mu mg\cos\theta$	$a_{\text{up}} = g(\sin\theta + \mu\cos\theta)$ (deceleration)
Down the plane	Up the plane	$mg\sin\theta - \mu mg\cos\theta$	$a_{\text{down}} = g(\sin\theta - \mu\cos\theta)$ (acceleration downward, provided $\tan\theta > \mu$ )

Because $\mu > 0$ , we always have $a_{\text{up}} > a_{\text{down}}$ in magnitude — the particle decelerates faster going up than it accelerates going down.

Note: If $\tan\theta \leq \mu$ , the plane is rough enough that the particle will not slide back down after coming to rest.

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