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tan θ = μ is one of the shortest results in mechanics, and one of the most useful. It gives a way to measure friction with a protractor, explains why the mass of the block drops out, and leads to the angle of friction, which solves a classic minimum-force question in two lines. This post derives both angles and works three examples.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "Static and Kinetic friction, laws of friction, rolling friction", and the NEET (UG) 2026 syllabus has the same line. The derivation is in section 4.9.1 of the NCERT Class 11 Physics chapter, Laws of Motion (2026-27 reprint). The CBSE 2026-27 physics curriculum includes a practical that finds the coefficient of friction between a block and a horizontal surface. Check the current documents.
Where does tan θ = μs come from?
Put a block on a plane inclined at θ. NCERT resolves the forces: weight mg down, normal force N, and static friction fs up the slope. In equilibrium, mg sin θ = fs and mg cos θ = N.
As θ increases, static friction adjusts upward until it reaches its maximum, μsN, at θmax. Dividing the two equations there gives
tan θmax = μs, so θmax = tan⁻¹ μs.
Mass cancels, so θmax depends only on μs. NCERT's example: if θmax = 15°, μs = tan 15° = 0.27. This θmax is the angle of repose.
What is the angle of friction?
The surface pushes on the block with one contact force. Its component perpendicular to the surface is N; its component along the surface is friction f. The angle of friction λ is the angle between the total contact force and the normal, so tan λ = f/N. At limiting friction, f = μsN, so tan λ = μs.
The angle of repose and the angle of friction are therefore equal in value, but defined differently: one is a tilt of the surface, the other is a direction of a force. The size of the full contact force is √(N² + f²) = N√(1 + μs²) at the limit.
Worked example 1: measuring μs with a protractor
Sample numbers for a classroom test. A wooden block on a board starts to slide when the board is tilted to 31°. Then μs = tan 31° ≈ 0.60. Repeat three times and average the angle before taking the tangent; a single reading is sensitive to how gently you tilt.
Worked example 2: the least force to drag a block
A 10 kg block rests on a floor with μs = 0.75. Take g = 10 m/s², so mg = 100 N.
Pulling horizontally. You need F = μsmg = 0.75 × 100 = 75 N.
Pulling at angle α above the horizontal. Lifting reduces N to mg − F sin α. At the point of sliding, F cos α = μs(mg − F sin α), so F = μsmg / (cos α + μs sin α).
This is smallest when tan α = μs, that is when α equals the angle of friction: λ = tan⁻¹ 0.75 ≈ 36.9°. Then sin λ = 0.6 and cos λ = 0.8.
F = 75 / (0.8 + 0.75 × 0.6) = 75 / 1.25 = 60 N. The shortcut F_min = mg sin λ = 100 × 0.6 = 60 N agrees. Pulling at the angle of friction saves 15 N.
Worked example 3: a slope steeper than the angle of repose
The same surface is tilted to 40°, beyond 36.9°, and the block slides. With μk = 0.6, a = g(sin θ − μk cos θ) = 10(0.643 − 0.6 × 0.766) ≈ 1.83 m/s². Once sliding starts, use μk, not μs.
Which mistakes cost marks?
Writing sin θ = μ or cos θ = μ. It is the ratio of the two components, so tan.
Using μk for the angle of repose. The block starts to slide when static friction hits its maximum, so it is μs.
Thinking a heavier block needs a steeper angle. Mass cancels.
Assuming friction is always μsN below θmax. It equals mg sin θ, only what is needed.
Pulling horizontally in a minimum-force question. Check whether the angle is free to choose.
Our post on friction and stacked-block problems covers the "only what is needed" rule in more detail.
A practice routine
Derive tan θmax = μs from a free-body diagram, without notes.
Find F_min and its angle for μs = 0.5 and m = 5 kg. (You should get about 22.4 N at 26.6°.)
Measure μs for a book on a tilted notebook cover at home, three trials.
Solve one incline problem above and one below the angle of repose, and say which friction applies.
If you want to see the contact force tilt as the slope rises, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
This week, redo Example 2 with μs = 1 and check that the best pulling angle is 45°.
