Explanation

A uniform ladder leans on a smooth wall and rough ground, so this rigid body is in equilibrium. Ground normal N equals the weight, wall reaction S equals friction F at the foot, and it stands while F ≤ μN. μ = 0 is smooth ground, which cannot hold a smooth wall.

The ladder is a rigid body in equilibrium, so the forces resolve and the moments about any point are zero. There is no acceleration to find.

The wall is smooth, so the reaction S there is horizontal. The ground is rough, so it supplies an upward normal N and a frictionF toward the wall. The weight mg acts at the midpoint.

• Vertical: N = mg
• Horizontal: F = S
• Moments about the foot: S × ℓ sin θ = mg × (ℓ/2) cos θ

The length cancels, so S = mg / (2 tan θ) andF = S. Limiting friction is μN = μmg, so the ladder stands while μ ≥ 1 / (2 tan θ).

If μ is smaller than that least value, friction cannot supplyF and the ladder slips. If μ = 0 the ground is smooth and the ladder cannot rest.

Worked example

A uniform ladder of mass 10 kg and length 4 m rests with one end on rough horizontal ground and the other against a smooth vertical wall, at 60° to the ground. Take g = 9.81 m/s². Find the reactions and the least coefficient of friction at the ground for which the ladder can rest.

1. Vertical: N = 10 × 9.81 = 98.10 N
2. Moments about the foot: S × 4 sin 60° = 98.10 × 2 cos 60°
3. 4 × 0.8660 S = 98.10 × 1, so 3.464 S = 98.10
4. S = 28.32 N, and F = S = 28.32 N
5. Least μ = F/N = 28.32 / 98.10 = 0.289

The coefficient of friction at the ground is 0.40. Show that the ladder can rest.

0.40 > 0.289, so the ground can supply the 28.32 N required, and the ladder is in equilibrium.

If instead μ = 0.20, then 0.20 < 0.289, so friction cannot reach 28.32 N and the ladder slips. The required F does not change — only whether the ground can supply it.