Rods on pegs, rough walls and limiting equilibrium

Smooth pegs and rails

A rod resting on a smooth peg or rail receives a normal reaction perpendicular to the rod. If the rod is at θ to the horizontal, this force has components N sin θ horizontally and N cos θ vertically. Take moments about the end on the ground to find N first.

Rough walls

A rough wall gives a normal reaction S (horizontal) and friction μS (vertical, usually upwards to stop the top sliding down). If the ladder is on the point of slipping at both ends, friction is limiting at both: F_ground = μ_1 R and F_wall = μ_2 S.

Exam technique

• Draw every force with a label before writing equations.
• Choose a moments point that removes two unknowns.
• Pearson's examiners (June 2025) note that most candidates resolved for one part of a rigid body question, and that attempts at taking moments about another point often left out a force.
• Trigonometric identities such as sec² θ = 1 + tan² θ may be needed to reach a given form.

Worked example

A uniform rod AB of length 4 m and weight W rests with A on rough horizontal ground and touches a smooth peg at C, where AC = 3 m. The rod makes 30° with the horizontal and is about to slip. Find μ.

1. M(A): N × 3 = W × 2 cos 30°, so N = (2√3/6)W = 0.577W.
2. Horizontally: F = N sin 30° = 0.289W.
3. Vertically: R + N cos 30° = W, so R = W - 0.5W = 0.5W.
4. μ = F/R = 0.289W/0.5W = 0.577 (= 1/√3).

Answer: μ = 0.577

Check yourself

1. In the worked example take W = 60 N. Find the reaction at the peg, to 3 significant figures.
   Show answer

   34.6 N. N = 60 × 2 cos 30°/3 = 34.6 N.

2. With W = 60 N, find the friction at A, to 3 significant figures.
   Show answer

   17.3 N. F = N sin 30° = 17.3 N.

3. With W = 60 N, find the normal reaction at A.
   Show answer

   30 N. R = 60 - N cos 30° = 60 - 30 = 30.0 N.

4. A uniform ladder rests in a vertical plane perpendicular to a rough vertical wall (μ = 0.4), with its foot on rough horizontal ground (μ = 0.5). It is about to slip at both ends. Find its angle with the horizontal, in degrees to 1 decimal place.
   Show answer

   38.7 °. Horizontally S = 0.5R; vertically R + 0.4S = W. Moments about the foot give tan θ = (1 - μ_1μ_2)/(2μ_1) = 0.8/1 = 0.8, θ = 38.7°.

5. For the same ladder, the weight is 100 N. Find the normal reaction at the wall, to 3 significant figures.
   Show answer

   41.7 N. R(1 + 0.5 × 0.4) = 100, so R = 83.3 N and S = 0.5R = 41.7 N.

6. A rod rests on a smooth peg. In which direction does the peg push on the rod?
   • Vertically upwards
   • Horizontally
   • Perpendicular to the rod
   • Along the rod
   Show answer

   Perpendicular to the rod. A smooth contact gives only a normal reaction, which is perpendicular to the rod's surface.

7. A ladder against a rough wall on rough ground is in limiting equilibrium at both ends. Which pair of equations holds?
   • F_ground = μ_1 R and F_wall = μ_2 S
   • F_ground = μ_1 S and F_wall = μ_2 R
   • F_ground = F_wall
   • R = S
   Show answer

   F_ground = μ_1 R and F_wall = μ_2 S. Limiting friction at each contact equals μ times the normal reaction at that contact.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy
• Pearson Edexcel IAL Mechanics M2 (WME02) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.