Friction — static (maximum), kinetic, rolling: Question bank

0. Symbol recap — read this first

Before any trap, here is every symbol used on this page, defined in plain words and anchored to the figure that follows.

Definition The cast of symbols

$N$ = normal force: the perpendicular ("normal" means at 90°) push the surface gives back to the object. Look at the red arrow in Figure 1 pointing straight out of the floor.
$m g$ = weight: the object's mass $m$ times gravity $g$ , pulling straight down (black down-arrow, Figure 1). On flat ground the surface must supply $N = m g$ to stop the object sinking.
$f_{s}$ = static friction: the along-surface force acting while the object is not yet sliding. It self-adjusts.
$f_{s}^{ma x}$ = maximum static friction: the ceiling $f_{s}$ can reach just before sliding, $f_{s}^{ma x} = μ_{s} N$ . Past this the bonds snap and motion begins.
$f_{k}$ = kinetic friction: the along-surface force while the object is sliding, $f_{k} = μ_{k} N$ , roughly constant.
$μ_{s}, μ_{k}, μ_{r}$ = coefficients of static, kinetic, rolling friction: pure numbers (no units) capturing surface stickiness.
$θ$ = the tilt angle of an incline; $θ_{r}$ = the angle of repose, the tilt at which sliding just begins.
$R$ = wheel radius; $a$ = how far the normal force shifts forward on a rolling wheel.

Figure — Friction — static (maximum), kinetic, rolling

1. Two must-see pictures

Figure 2 is the incline decomposition every incline trap on this page rests on. Figure 3 shows why static-vs-kinetic is a jump, not a slide.

3. Spot the error

"The box is pushed with 30 N and

f_{s}^{ma x} = 50

N, so friction is 50 N."

Error: it isn't sliding, so static friction supplies only the 30 N needed to balance the push, not the ceiling value

f_{s}^{ma x}

. Friction here is 30 N.

"It's sliding, so I'll use

f_{k} = μ_{s} N

Error: sliding means the kinetic law applies,

f_{k} = μ_{k} N

, and

μ_{k} < μ_{s}

. Using

μ_{s}

overestimates the friction of a moving object.

"On the incline the driving force is

m g

, so it slides when

μ_{s} m g < m g

Error: only the component along the plane,

m g sin θ

, drives sliding, and friction opposes it as

μ_{s} m g cos θ

. The correct condition is

tan θ > μ_{s}

"The normal force always equals

m g

Error:

N = m g

only for a horizontal surface with no vertical push. On an incline

N = m g cos θ

; a downward push adds to

N

, an upward pull reduces it.

"Angle of repose depends on the block's mass, since heavier blocks grip more."

Error: mass cancels —

m g sin θ_{r} = μ_{s} m g cos θ_{r}

gives

tan θ_{r} = μ_{s}

, mass-free. Repose angle depends only on the surfaces.

"To keep a box moving at constant velocity you must keep pushing harder than

f_{s}^{ma x}

Error: once moving you only fight kinetic friction

μ_{k} N

, which is less than the ceiling

f_{s}^{ma x}

. Push equal to

f_{k}

gives constant velocity (

a = 0

"Static friction points backward on the driving tyre, so friction slows the car down."

Error: the tyre surface tends to slip backward relative to the road, so static friction on it points forward (opposite the slip, per our sign convention) — it is the accelerating force.

5. Edge cases

What is the friction on a block in free-fall alongside a vertical wall it isn't pressed against?

Zero. With no normal force pressing block to wall (

N = 0

f = μ N = 0

— friction needs contact pressure to exist.

A block sits on an incline steep enough that

tan θ = μ_{s}

exactly. What happens?

It is on the exact verge: driving force equals the friction ceiling. It neither accelerates nor has spare grip — the boundary between staying and sliding.

You push a resting box with a force exactly equal to

f_{s}^{ma x}

. Does it move?

At the exact ceiling it is on the threshold — friction can just barely match, so acceleration is zero, but any infinitesimal extra push starts sliding.

A block slides, then you stop pushing entirely. What friction acts?

While still moving it feels kinetic friction

μ_{k} N

opposing its motion, decelerating it; once it stops, friction drops to whatever (zero here) is needed to keep it stopped.

Two identical boxes, one stacked on the other, on the floor. Is the floor friction ceiling on the bottom box doubled?

Yes — the bottom box now carries both weights, so

N

doubles and

f_{s}^{ma x} = μ_{s} N

doubles. Friction scaled through

N

, not area.

On a perfectly smooth (

μ = 0

) incline, what is the block's acceleration?

g sin θ

down the slope. With no friction the only along-plane force is the weight component

m g sin θ

, giving

a = g sin θ

Can static friction ever exceed

μ_{s} N

momentarily?

No.

μ_{s} N

is the physical breaking point of the bonds; demand more and the bonds fail, so the object begins to slide and kinetic friction takes over.

A wheel rolls without slipping. Which friction acts at the contact point, static or kinetic?

Static — "without slipping" means the contact point is instantaneously at rest relative to the ground, so there is no sliding for kinetic friction to resist.

Recall Self-check

Did every "false" answer name the specific wrong assumption (ceiling vs actual, area vs $N$ , speed independence)? ::: If you only wrote "false" without the mechanism, revisit — the reasoning is the whole point. From Figure 2, can you re-derive $tan θ_{r} = μ_{s}$ by balancing $m g sin θ$ against $μ_{s} m g cos θ$ ? ::: If not, reread Section 1 and Inclined Plane Problems.

Question bank — Friction — static (maximum), kinetic, rolling

0. Symbol recap — read this first

1. Two must-see pictures

2. True or false — justify

3. Spot the error

4. Why questions

5. Edge cases