Motion on inclined planes: resolving weight

8 minAS Level4.4.3

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can resolve weight on an inclined plane, handle friction on a slope, and find acceleration up or down a plane.

Key to the symbols

Symbol	Meaning
$\theta$	the angle between the slope and the horizontal
$m$	mass of the particle, in kg
$g$	$10\ \text{m s}^{-2}$
$R$	normal reaction, at $90°$ to the slope, in N
$F$	friction, along the slope (in $F = ma$ , $F$ instead means the resultant force)
$\mu$	coefficient of friction; when sliding or on the point of sliding (limiting), friction $= \mu R$

Why slopes are different

On a horizontal surface, weight acts straight down and the normal reaction pushes straight up. They balance (if no vertical acceleration). But on a slope, weight doesn't line up with the surface — it points straight down while the plane is tilted.

The trick: resolve weight into two components — one perpendicular to the plane, one parallel to it. Then apply F = ma along the plane.

If the plane is inclined at angle $\theta$ to the horizontal:

Component perpendicular to the plane: $mg \cos \theta$
Component parallel to the plane (down the slope): $mg \sin \theta$

The normal reaction $R$ balances the perpendicular component (no acceleration off the plane): $R = mg \cos \theta$ .

The parallel component $mg \sin \theta$ pulls the particle down the slope.

Definition

Weight components on an inclined plane

For a particle of mass $m$ on a plane inclined at $\theta$ to the horizontal:

Perpendicular to plane: $mg \cos \theta$
Parallel to plane (down the slope): $mg \sin \theta$
Normal reaction: $R = mg \cos \theta$ (if no other perpendicular forces)

Where the trig comes from

Draw the weight vector vertically down. The plane is at angle $\theta$ to horizontal, so the angle between the weight vector and the perpendicular to the plane is also $\theta$ . (The normal is 90° to the plane; the vertical is 90° to the horizontal; the difference is $\theta$ .)

The component along the perpendicular is adjacent to $\theta$ , so it's $mg \cos \theta$ . The component along the plane is opposite to $\theta$ , so it's $mg \sin \theta$ .

Check the extremes:

$\theta = 0$ (horizontal): $\cos 0 = 1$ , $\sin 0 = 0$ . All weight is perpendicular, none parallel. Correct — no tendency to slide.
$\theta = 90^\circ$ (vertical): $\cos 90^\circ = 0$ , $\sin 90^\circ = 1$ . All weight is parallel, none perpendicular. Correct — free fall.

Worked example: Smooth inclined plane

A particle of mass $4\ \text{kg}$ is released from rest on a smooth plane inclined at $30^\circ$ to the horizontal. Find its acceleration down the plane. Take $g = 10\ \text{m s}^{-2}$ .

Have a go first?

Worked example: Rough inclined plane — moving up

A particle of mass $2\ \text{kg}$ is projected up a rough plane inclined at $30^\circ$ with initial speed $10\ \text{m s}^{-1}$ . The coefficient of friction is $\mu = 0.2$ . Find the deceleration while moving up. Take $g = 10\ \text{m s}^{-2}$ .

Have a go first?

Worked example: Rough inclined plane — moving down

The same particle ( $2\ \text{kg}$ , $\mu = 0.2$ , $\theta = 30^\circ$ ) now slides down the plane from rest. Find its acceleration.

Have a go first?

Why up and down accelerations differ on a rough plane

In the examples above:

Moving up: weight component and friction both act down the slope. They add up. Deceleration $= g\sin\theta + \mu g\cos\theta$ .
Moving down: weight component acts down, friction acts up. They oppose. Acceleration $= g\sin\theta - \mu g\cos\theta$ .

The syllabus explicitly says the acceleration up the plane differs from the acceleration down the plane. That difference is the friction term.

If $g\sin\theta \le \mu g\cos\theta$ (that is, $\tan\theta \le \mu$ ), a particle at rest stays at rest. When $g\sin\theta = \mu g\cos\theta$ exactly, it is on the point of slipping: that is limiting equilibrium, and friction has its maximum value $\mu R$ .

Check yourself

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