Learn9709Paper 4 · MechanicsForces and equilibriumInclined planes

Forces and equilibrium · Subtopic 4 of 4

Equilibrium on an Inclined Plane

Tilt the axes to match the slope. Weight splits into $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into it, and friction points whichever way stops the slide.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Resolve weight parallel and perpendicular to an inclined plane.
Find the force needed to hold a particle on a smooth slope.
Decide which way friction acts on a rough slope.
Find the range of forces that keep a particle in equilibrium on a rough slope.

Pause and think

A book lies on a tilted desk. As you slowly raise one end, at some angle it starts to slide. What two things change as the angle increases?

Section 2 of 14

Core idea

Start hereWill the box slide down the ramp?

A box sits on a ramp. Tilt the ramp a little more and at some angle it starts to slide. Which angle? It depends only on $\mu$ , not on the box's weight. The key is splitting weight into "down the slope" and "into the slope".

The big idea

On a slope at angle $\alpha$ : weight components $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into it. With no other perpendicular forces, $R=mg\cos\alpha$ . Friction acts along the slope, opposing the (tendency to) motion.

1. Splitting the weight

Forces on a block on a 30° slope (about to slide down)

Split the weight into two parts:

mg\sin\alpha

pulling down the slope and

mg\cos\alpha

pushing into it. Friction acts up the slope, opposing the motion (or the tendency to move).

Why the angle between weight and the normal is

\alpha

$\text{slope tilted }\alpha\text{ from horizontal}$
Given.
$\text{normal to slope tilted }\alpha\text{ from vertical}$
Rotating a figure rotates its perpendicular by the same angle.
$\text{into slope}=mg\cos\alpha,\quad\text{down slope}=mg\sin\alpha$
Resolve the vertical weight.

Worked exampleAbout to slide on $30^\circ$ : find $\mu$

A block is on the point of sliding down a $30^\circ$ slope. Find $\mu$ .

Goal: $\mu$ .
Strategy: Limiting friction up the slope balances the down-slope weight.

$R=mg\cos30^\circ$
Perpendicular.
$\mu R=mg\sin30^\circ$
Along.
$\mu=\tan30^\circ=0.577$
Divide; $mg$ cancels.

Answer

\mu=0.577

✓ CheckMass cancelled: any block of that material slips at

30^\circ

✓.

Worked exampleRange of forces that hold a block

A 4 kg block on a $30^\circ$ slope, $\mu=0.2$ . Find the least and greatest force $P$ (up the slope) that keep it at rest.

Goal: Two values.
Strategy: Friction can act either way.

$R=40\cos30^\circ=34.64,\ F_{\max}=6.93$
Limit.
$\text{about to slide down: }P+6.93=20\Rightarrow P=13.1$
Friction helps $P$ .
$\text{about to slide up: }P=20+6.93=26.9$
Friction opposes $P$ .

Answer

13.1\le P\le26.9

✓ CheckWithout friction,

P=20

exactly; friction widens the range symmetrically ✓.

Solving any forces problem

Drawing diagram…

The same routine solves equilibrium, friction, slopes, connected particles and lifts.

Pause and think

Can a block rest on a smooth slope with no other force?

Summary

$mg\sin\alpha$ along, $mg\cos\alpha$ into.
Friction direction depends on the tendency.

Inclined plane: A slope.

HistoryStevin's wreath of spheres. In 1586 the Flemish engineer Simon Stevin imagined a loop of equally spaced balls draped over a triangular prism. If the chain moved, it would move forever, which is impossible. So the few balls on the steep side must balance the many on the gentle side, which gives the rule for forces on a slope. He was so pleased he put the picture on his title page with the motto "Wonder en is gheen wonder" ("A wonder, yet no wonder").

Fun factThe angle of repose. Pour sand, salt or gravel into a pile and the sides settle at a fixed steepest angle, called the angle of repose. At that angle a grain on the surface is in limiting equilibrium, so

\tan\alpha=\mu

. Dry sand settles at about

34^\circ

; that is why sand dunes and salt heaps have such similar slopes.

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