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A rod on rough ground against a smooth wall: the least extra horizontal force that makes it slip, and the coefficient of friction at limiting equilibrium
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 3
GoalFind the friction and the least μ for a uniform rod resting on rough ground against a smooth wall at the angle θ, and the extra horizontal force that makes it slip.
New words
tan theta · height divided by distance from the wallextra horizontal force · a push or pull on the foot that changes the friction needed
Explanation
A uniform rod of length L and weight W rests on rough ground and a smooth wall, at the angle θ to the ground. Put the wall on the right; up and to the right are positive and anticlockwise moments are positive. Moments about the foot: S × (L sin θ) = W × (L/2) cos θ, so S = W/(2 tan θ). Then F = S and N = W. The limiting value F = μN gives μ = 1/(2 tan θ). Notice that S does not depend on the friction. A steeper rod (larger tan θ) needs less friction. With an extra horizontal force X at the foot, pointing away from the wall, the friction needed becomes S + X, so the largest X is μN − S.
Worked examples
A rod of weight 90 N makes the angle θ with the ground, where tan θ = 3, and μ = 0.4. S = 90/6 = 15 N, N = 90 N and F = 15 N. The most friction possible is 0.4 × 90 = 36 N, so an extra horizontal force X pointing away from the wall can be at most 36 − 15 = 21 N.
A rod rests with tan θ = 5/2. Then μ ≥ 1/(2 × 5/2) = 1/5.
Class activity
Lean a pencil on a book at different angles on a rough mat. Which angle makes it slip? Link what you see to μ = 1/(2 tan θ).
Practice
1
A rod of weight 60 N has tan θ = 2 and μ = 0.5. What is the largest extra horizontal force X, pointing away from the wall, that it can take?
S = 60/4 = 15 N. The most friction is 0.5 × 60 = 30 N, so X = 30 − 15 = 15 N.
2
The same rod, but now X points towards the wall. What is the largest X before it slips?
S stays 15 N. The friction then acts away from the wall: X − 15 = 30, so X = 45 N.
3
Find the least μ if tan θ = 2.5.
μ = 1/(2 × 2.5) = 1/5 = 0.2.
4
A rod rests at θ = 60° to the ground. Find the least μ.
tan 60° = √3, so μ = 1/(2√3) ≈ 0.289.
5
Does S depend on μ?
No: the moments about the foot do not involve the friction.