Angle of friction, angle of repose — derivation: Question bank

True or false — justify

A heavier block has a larger angle of repose

False. Both the down-slope pull

m g sin θ

and the friction limit

μ_{s} m g cos θ

carry the same

m g

, so it cancels;

θ_{r} = arctan (μ_{s})

knows nothing about mass.

A larger contact area lowers the angle of repose

False. Neither

N = m g cos θ

nor

f_{m a x} = μ_{s} N

contains area, so

θ_{r}

is area-independent — this is baked into the Coulomb friction model.

The angle of friction

λ

is measured from the surface, not from the normal

False. By definition

λ

is the tilt of

R

away from the normal; the angle it makes with the surface is

9 0^{\circ} - λ

λ

and

θ_{r}

are equal, so they are measured from the same reference line

False. They are equal in value only:

λ

is measured from the normal,

θ_{r}

is the plane's tilt from horizontal. Same number, different geometry.

On a horizontal floor with no push, the reaction

R

points straight up along the normal

True. With no applied horizontal force,

f = 0

, so

R = N

and

λ = arctan (0) = 0^{\circ}

— the reaction sits exactly on the normal.

At the angle of repose the block is in equilibrium

True. "On the verge" means still not moving: forces balance (

\sum F = 0

per Newton's First Law (Equilibrium)), with friction sitting at its maximum but not yet exceeded.

Once the ramp is tilted past

θ_{r}

the block accelerates because static friction disappears

False. Static friction is replaced by kinetic friction

μ_{k} N

; since

μ_{k} < μ_{s}

, the down-slope pull now exceeds friction and the net force accelerates the block.

μ_{s} = 0

(frictionless), the angle of repose is

0^{\circ}

True.

θ_{r} = arctan (0) = 0^{\circ}

: the block slips at the slightest tilt, because there is no friction to hold it.

The angle of friction can never exceed

9 0^{\circ}

True.

λ = arctan (μ_{s})

and

arctan

approaches but never reaches

9 0^{\circ}

even as

μ_{s} \to \infty

, so

λ < 9 0^{\circ}

always.

Doubling

μ_{s}

doubles the angle of repose

False. The relation is

θ_{r} = arctan (μ_{s})

, which is nonlinear; doubling

μ_{s}

increases

θ_{r}

by less than double (e.g.

arctan (0.5) = 26. 6^{\circ}

but

arctan (1) = 4 5^{\circ}

, not

53. 1^{\circ}

Why questions

Why do the normal forces

N

cancel when computing

tan λ

Because at the verge

f = μ_{s} N

, so

tan λ = f / N = μ_{s} N / N = μ_{s}

— the same

N

appears top and bottom and divides out, leaving a pure coefficient.

Why do we resolve the weight along and perpendicular to the incline rather than horizontal/vertical?

Because slipping happens along the incline; choosing those axes puts the driving force and the normal force on clean, separate directions so no single force needs re-splitting (see Resolving Vectors into Components).

Why is the angle of friction equal to the angle of repose?

Both derivations independently end at

tan (angle) = μ_{s}

. Geometrically, at repose gravity tilts from the normal by exactly the amount the reaction tilts from the normal, and equilibrium forces the two tilts to match.

Why must we use static (not kinetic) friction for both

λ

and

θ_{r}

Both describe the maximum grip just before motion, which is the static limit

μ_{s} N

. Kinetic friction only applies once sliding has already begun.

Why does the same

arctan (μ)

idea reappear in Banking of Roads?

Because banking asks the mirror question — at what tilt do gravity and contact forces balance for a given friction — so the critical/safe angle again comes out as an

arctan

of a friction-related ratio.

Why can two perpendicular forces

N

and

f

be combined by Pythagoras?

Because perpendicular vectors form the two legs of a right triangle, and

R

is its hypotenuse, so

R = N^{2} + f^{2}

— a direct application of the Pythagorean theorem to the force triangle.

Edge cases

What is

θ_{r}

when the surface is perfectly smooth (

μ_{s} = 0

θ_{r} = arctan (0) = 0^{\circ}

: any nonzero tilt makes it slide, since there is no friction to resist the down-slope pull.

What happens to

θ_{r}

μ_{s} \to \infty

(extremely grippy surface)?

θ_{r} = arctan (μ_{s}) \to 9 0^{\circ}

but never reaches it. Even an infinitely grippy surface cannot hold a block on a perfectly vertical wall by friction against gravity alone.

On a flat horizontal surface with no horizontal push, what are

f

λ

, and

R

f = 0

, so

λ = 0^{\circ}

and

R = N = m g

pointing straight up — the reaction coincides with the normal because nothing tries to slide the block.

If a horizontal push exactly reaches the verge on flat ground with

μ_{s} = 1

, what is

λ

λ = arctan (1) = 4 5^{\circ}

; the reaction leans

4 5^{\circ}

from vertical because friction and normal are then equal in size.

At exactly

θ = θ_{r}

, is friction at its maximum or is it exceeded?

Exactly at its maximum,

f = f_{m a x} = μ_{s} N

, and not exceeded — that equality is the definition of the repose angle.

A block sits on an incline with

θ < θ_{r}

; how big is the friction force?

Only as big as needed for balance,

f = m g sin θ

, which is less than

μ_{s} N

. Friction is not "always at max" — below repose it self-adjusts to whatever keeps equilibrium.

Recall One-line summary to lock in

Everything on this page radiates from a single equation: $tan (that angle) = μ_{s}$ . Mass and area vanish, static friction rules the verge, and the reaction's tilt from the normal equals the ramp's tilt from horizontal.

Question bank — Angle of friction, angle of repose — derivation

True or false — justify

Spot the error

Why questions

Edge cases