Angle of friction, angle of repose — derivation: Worked examples

This page is the drill ground for the parent derivation. We derived that friction hands the surface a tilted reaction, and that a ramp tipped to $arctan (μ_{s})$ is on the verge of slipping. Now we run that single idea through every case class it can appear in — small angles, the exact $45°$ pivot, steep angles, a zero-friction limit, a push-along-the-floor problem, and an exam twist where the answer is not simply $arctan (μ)$ .

Before any symbol appears: $μ_{s}$ (read "mew-sub-ess") is the static coefficient of friction, a plain number that measures how "grippy" a surface pair is — see Static and Kinetic Friction. $arctan (x)$ (read "arc-tangent of $x$ ") is the question "which angle has tangent equal to $x$ ?". $tan (θ)$ of an angle is the ratio (opposite side) $/$ (adjacent side) of the right triangle that angle sits in — the steeper the angle, the bigger this ratio.

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are each labelled with the cell(s) they cover.

Cell	What makes it distinct	Covered by
A. Small angle ( $θ_{r} < 45°$ , so $μ_{s} < 1$ )	gentle slope, "typical" surface	Ex 1
B. The pivot ( $θ_{r} = 45°$ , $μ_{s} = 1$ )	tangent equals exactly one	Ex 2
C. Steep angle ( $θ_{r} > 45°$ , $μ_{s} > 1$ )	very grippy (rubber, rough rock)	Ex 3
D. Degenerate: $μ_{s} = 0$	frictionless limit, angle $\to 0$	Ex 4
E. Reaction geometry (horizontal floor, find $R$ and $λ$ )	vector sum of $N$ and $f$	Ex 5
F. Below vs at vs above repose (is it moving?)	inequality reasoning, not just equality	Ex 6
G. Real-world word problem	translate a story into $arctan (μ)$	Ex 7
H. Exam twist: extra applied force	answer is NOT plain $arctan (μ)$	Ex 8

Case A — Small angle

Case B — The pivot at 45°

Case C — Steep angle

Case D — Degenerate frictionless limit

Case E — Reaction geometry on a horizontal floor

Figure — Angle of friction, angle of repose — derivation

Worked example Ex 5 (Cell E) — Total reaction of a pushed block

A $2 kg$ block sits on a horizontal floor, $μ_{s} = 0.5$ , and is pushed horizontally until it is just about to slide. Take $g = 10 m/s^{2}$ . Find (a) the angle $λ$ of the total contact reaction $R$ from the vertical, and (b) the magnitude of $R$ .

Forecast: Will $R$ be bigger or smaller than the weight $20 N$ ? Guess before computing.

Step 1. Normal balances weight: $N = m g = 2 \times 10 = 20 N$ . Why this step? Vertically the block does not accelerate, so $N = m g$ (equilibrium).

Step 2. Maximum friction: $f = μ_{s} N = 0.5 \times 20 = 10 N$ . Why this step? "Just about to slide" means friction has reached its ceiling $μ_{s} N$ .

Step 3. Angle of reaction: $λ = arctan (f / N) = arctan (0.5) \approx 26.57°$ . Why this step? Look at the figure — $N$ points straight up (adjacent to $λ$ ), $f$ points sideways (opposite to $λ$ ); tangent is opposite/adjacent $= f / N$ .

Step 4. Magnitude: $R = N^{2} + f^{2} = 2 0^{2} + 1 0^{2} = 500 \approx 22.36 N$ . Why this step? $N$ and $f$ are perpendicular, so their vector sum is the hypotenuse — Pythagoras (see Resolving Vectors into Components).

Verify: $R = N 1 + μ_{s}^{2} = 20 1.25 \approx 22.36 N$ — same number, so the shortcut and the direct sum agree. Units: newtons throughout. ✓ And $R > 20 N$ (friction adds a sideways slice), matching the forecast.

Case F — Below / at / above repose

Worked example Ex 6 (Cell F) — Does it move at three different tilts?

A ramp has $μ_{s} = 0.577$ (so $θ_{r} = 30°$ ). A block sits on it. Decide whether it stays put at (i) $θ = 20°$ , (ii) $θ = 30°$ , (iii) $θ = 40°$ .

Forecast: Guess the outcome of each before working.

Step 1. The block stays if and only if the required friction $m g sin θ$ does not exceed the ceiling $μ_{s} m g cos θ$ , i.e. $tan θ \leq μ_{s}$ . Why this step? Equilibrium demands friction supply the full down-slope pull; but friction can supply at most $μ_{s} N$ . This is an inequality, not just the equality at the verge.

Step 2 (i). $θ = 20°$ : $tan 20° \approx 0.364 < 0.577$ . Stays put (friction spare capacity). Why? Required grip is below the ceiling.

Step 3 (ii). $θ = 30°$ : $tan 30° \approx 0.577 = μ_{s}$ . On the verge — the exact repose angle. Why? Required grip exactly equals the ceiling; the tiniest extra tilt breaks it.

Step 4 (iii). $θ = 40°$ : $tan 40° \approx 0.839 > 0.577$ . Slides (impossible to balance). Why? Demand exceeds the friction ceiling; the surplus $m g (sin θ - μ_{s} cos θ)$ accelerates the block down-slope.

Verify: The three verdicts stay / verge / slide straddle $θ_{r} = 30°$ exactly as the matrix predicts. In the figure, the red down-slope arrow overtakes the green friction cap only past $30°$ . ✓

Case G — Real-world word problem

Worked example Ex 7 (Cell G) — Grain silo cone

Dry sand poured on the ground naturally forms a cone whose sides make $34°$ with the horizontal — this is literally its angle of repose. Estimate the effective $μ_{s}$ between sand grains, and hence the height of a cone with a $10 m$ base radius.

Forecast: The pile refuses to get steeper than $34°$ — what does that tell you about grain-on-grain grip?

Step 1. $μ_{s} = tan θ_{r} = tan 34° \approx 0.675$ . Why this step? A settled pile is at its steepest stable slope — that is the angle of repose, so the same $tan θ_{r} = μ_{s}$ relation applies to loose grains.

Step 2. Cone height $h = r tan θ_{r} = 10 \times tan 34° \approx 6.75 m$ . Why this step? The cone's slant makes angle $θ_{r}$ with the base; height/radius is exactly $tan θ_{r}$ (opposite/adjacent of that slope triangle).

Verify: $h / r = 6.75/10 = 0.675 = μ_{s} = tan 34°$ . ✓ Units: metres. The same $arctan (μ)$ idea that limits a ramp also caps how tall a sand pile can grow.

Case H — Exam twist (answer is NOT plain arctan μ)

Worked example Ex 8 (Cell H) — Repose with an extra downward push

A block on a rough incline ( $μ_{s} = 0.5$ ) is pressed against the slope by an extra force $P$ directed perpendicular to the surface (into it). At the verge of sliding, is the critical tilt still $arctan (0.5)$ ? Work out the condition in general and comment.

Forecast: Does adding an inward push make the block harder or easier to hold? Guess how the repose angle shifts.

Step 1. Perpendicular balance now includes $P$ : $N = m g cos θ + P$ . Why this step? The extra push presses the block into the surface, so it adds to the normal force (see Block on an Inclined Plane).

Step 2. Verge condition (down-slope pull = friction ceiling): $m g sin θ = μ_{s} N = μ_{s} (m g cos θ + P)$ . Why this step? Only gravity drives down-slope ( $P$ has no along-slope component, being perpendicular), while friction grows because $N$ grew.

Step 3. Rearrange: $tan θ = μ_{s} (1 + \frac{P}{m g cos θ}) > μ_{s}$ . Why this step? The extra $P$ term is strictly positive, so the critical $tan θ$ exceeds $μ_{s}$ — the block tolerates a steeper tilt.

Step 4 (special check). If $P = 0$ , this collapses to $tan θ = μ_{s}$ , the ordinary repose angle $arctan (0.5) \approx 26.57°$ . Why this step? A good general formula must reduce to the known case when the new ingredient vanishes.

Verify: With $P = 0$ , $arctan (0.5) = 26.57°$ recovers the standard result. ✓ With $P > 0$ , the pressing force lets the ramp go steeper — matching intuition that pushing a block into a slope helps it grip. The naive "answer is always $arctan μ$ " trap is exactly what this cell tests.

Recall Checkpoint

Below repose, is friction at its maximum value? ::: No — friction only supplies what is needed ( $m g sin θ$ ), staying below its ceiling $μ_{s} N$ until the verge. Why did $R$ come out larger than the weight in Ex 5? ::: Because $R$ is the hypotenuse of $N$ and the sideways friction $f$ ; adding a perpendicular slice always lengthens the resultant. In Ex 8, why doesn't $P$ appear in the down-slope driving term? ::: $P$ is perpendicular to the surface, so it has zero component along the slope — it only boosts $N$ .

Connections

Static and Kinetic Friction — where $μ_{s}$ comes from.
Block on an Inclined Plane — the resolution setup reused in Ex 6 and 8.
Resolving Vectors into Components — Pythagoras and $sin / cos$ splitting.
Newton's First Law (Equilibrium) — the $\sum F = 0$ backbone of every case.
Banking of Roads — the same $arctan (μ)$ recipe for safe cornering speed.