Friction revisited: rough surfaces, μ and the angle of friction
Week 1 · Statics prerequisites (Sec 2 friction revisited): smooth and rough surfaces, static friction and its limit, the coefficient of friction, kinetic friction, the resultant reaction and the angle of friction, equilibrium of a body on a rough horizontal plane and on a rough inclined plane · 3 lessons
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Limiting friction F = μN, kinetic friction F = μk N, the resultant reaction R′ and the angle of friction λ with tan λ = μ
Applied Mathematics Sec 3 Term 1, Statics prerequisites
GoalUse F = μN at the limit, find the friction on a sliding body, and find R′ and the angle of friction λ.
New words
coefficient of friction μ · the constant ratio Fs / N for two surfaces; a bigger μ means rougher surfaces, μ = 0 means smoothkinetic friction Fk· the friction on a body that is sliding; Fk = μk N, opposite to the motionresultant reaction R′ · the single force that is the resultant of N and Fangle of friction λ · the angle between R′ and N when the body is about to slide
Explanation
For two given surfaces the ratio of the limiting friction to the normal reaction is a constant, the coefficient of static friction: μ = Fs / N, so at the limit Fs = μN. It does not depend on how big the touching area is; it measures how rough the pair of surfaces is. Before the limit the friction is smaller than μN. Once the body slides, the friction is the kinetic friction Fk = μk N, opposite to the motion, and μk is usually a little smaller than μ. The normal reaction N and the friction F act at the same point at right angles, so they have one resultant, the resultant reaction R′ = √(N² + F²). R′ leans away from the normal by an angle whose tangent is F / N. As F grows towards its limit this angle grows, and at the limit it has its largest value, the angle of friction λ. So tan λ = Fs / N = μ. The angle λ belongs to the limiting case only: for a body not about to slide the angle of R′ with the normal is smaller than λ. At the limit R′ = N√(1 + μ²).
Worked examples
A 15 kg crate (weight 150 N) on a level floor is about to slide when pulled with a horizontal force of 60 N. N = 150 N and Fs = 60 N, so μ = 60/150 = 0.4. R′ = √(150² + 60²) = √26100 ≈ 161.6 N, and tan λ = 0.4 gives λ ≈ 21.8°.
The same crate is now sliding, and the floor has μk = 0.3. The kinetic friction is Fk = 0.3 × 150 = 45 N, opposite to the motion. The 60 N pull is bigger than this, so the crate speeds up (we do not study that here). The kinetic angle is tan⁻¹ 0.3 ≈ 16.7°, smaller than the static λ ≈ 21.8°, because μk < μ.
Measuring from the angle: for a book on a wooden shelf, λ = 25°. Then μ = tan 25° ≈ 0.466. For a 30 N book the limiting friction is 0.466 × 30 ≈ 13.99 N.
Class activity
Hook a rubber band round a school bag and pull it slowly along a desk with the band horizontal. The band stretches more and more and then the bag starts to move: the stretch just before it moves shows the limiting friction. Compare the bare desk with a sheet of paper or a cloth under the bag. Pull only a bag, never a person.
Practice
1
A 80 N block is about to move under a horizontal pull of 28 N. Find μ.
μ = 28/80 = 0.35.
2
Two surfaces have μ = 0.45. What is the limiting friction when N = 200 N?
Fs = 0.45 × 200 = 90 N.
3
At the limit N = 90 N and F = 40 N. Find R′ and λ.
