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Why a ladder slips on a smooth floor

A leaning ladder stays put only when friction at the floor balances the outward push at the base. On a smooth floor there is too little friction, so the balance breaks and the ladder slides.

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A ladder resting against a wall is in equilibrium only if all the forces on it balance and there is no net turning effect. Three main forces act: its weight pulling down at its centre, the wall pushing back near the top, and the floor pushing up and providing friction at the base.

The wall pushes the bottom of the ladder outward, and only friction at the floor stops the base from sliding away. If the floor is smooth, like polished tiles, friction is too small to hold the base, so the sideways forces no longer balance and the ladder slips.

This is why a ladder is safer on a rough surface, set at a sensible angle, and safest with someone standing on the bottom rung to add a downward force that increases friction.

In SPM you should analyse a ladder using balanced forces and the principle of moments about a point.

Common misconceptions

  • A ladder slips because it is too heavy -> A heavier ladder can even press down harder; slipping is about too little friction at the base, not weight alone.
  • The wall being smooth is what causes slipping -> The floor's friction is what usually holds the base; a smooth floor is the common cause.
  • If forces balance, the ladder is safe regardless of angle -> Turning effects (moments) must also balance, and a poor angle can still cause it to slip or topple.

Force and Motion II

The physics behind it

A leaning ladder is a problem in static equilibrium: the forces must cancel and the turning effects, or moments, must also cancel. A moment is a force multiplied by its perpendicular distance from a pivot, measured in newton metres.

Take a uniform ladder of weight W = 200 N leaning at 60° to the floor against a smooth wall, and take moments about its foot. The wall's push acts at the top and the weight acts at the centre, giving the wall reaction F = W ÷ (2 × 1.73) = 200 N ÷ 3.46 = 57.8 N. For the base not to slide, the floor must supply a friction force equal to this, so f = 57.8 N. The floor also carries the full weight, so the normal reaction N = 200 N. Friction cannot exceed μN, so the smallest coefficient of friction that will hold the ladder is μ = f ÷ N = 57.8 N ÷ 200 N = 0.29.

If the floor offers a coefficient below 0.29, the friction available is too small, the horizontal forces no longer balance, and the base slides out.

An everyday example at home

Picture a long broom stood at a slant against the kitchen wall just after the tiles have been mopped. On dry tiles the broom stays put for days, but on the wet, slippery surface it slowly slides at the foot and clatters down.

Nothing about the broom changed; only the friction under its base fell.

The wall still pushes the top of the broom outward, and with too little grip at the floor there is nothing to balance that push. The same thing happens to a bamboo laundry pole propped in a corner, or a bicycle leaned against a wall on smooth, damp cement after rain.

You can feel the physics if you try to stand a ruler at a shallow angle against a book on a polished table: set it too flat and it skates away, because a smaller angle needs a larger friction force at the base, exactly as the moment calculation shows. Roughening the floor, or standing something heavy on the foot, restores the grip and keeps it in place.

How this comes up in SPM

In Paper 2 this sits in the Force and Motion II chapter and is examined with command words such as explain, describe, state and determine. You might be asked to state the conditions for equilibrium, to explain why a smooth floor lets a ladder slip, or to determine an unknown force using the principle of moments.

A clear force diagram showing weight, the wall reaction, the floor's normal reaction and friction is usually expected.

The idea draws on several neighbours in the same chapter: the resolution of forces into horizontal and vertical components, the concept of a resultant force, and above all the principle of moments, which states that for a body in equilibrium the sum of clockwise moments about any point equals the sum of anticlockwise moments. It also links to the position of the centre of gravity, since the weight acts there.

Practise setting up both the force balance and the moment balance together, because a ladder question usually needs both to reach the answer.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026

Frequently asked questions

How is this examined in SPM?
It can appear in Paper 1 and Paper 2. We do not predict questions.
Does making the ladder heavier stop it slipping?
Not by itself. A heavier ladder presses down harder, raising both the normal reaction and the friction available, but it also increases the sideways push from the wall in the same proportion. The required coefficient of friction stays about the same, so a smooth floor still lets it slip.
Why is a steeper ladder less likely to slide?
As the angle to the floor increases, the wall reaction and therefore the friction needed both fall, because the weight's turning effect about the foot shrinks. A steeper ladder needs a smaller coefficient of friction, so it is safer on a slippery floor than a shallow one.
Why does standing on the bottom rung help?
Your weight on the foot increases the normal reaction from the floor, and since maximum friction is μN, more normal force means more grip available at the base. That extra friction helps balance the wall's outward push and stops the base sliding.

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