Coefficients of friction — measurement, material dependence: Worked examples

Before any example, one idea must be crystal clear, because half of all friction mistakes come from ignoring it.

Definition Static friction is a

range, not a value Two coefficients appear throughout this page, so name them once and for all:

$μ_{s}$ is the static-friction coefficient — it sets the maximum grip while the object is still stuck.
$μ_{k}$ is the kinetic-friction coefficient — it sets the (roughly constant) drag once the object is sliding.

Kinetic friction (while sliding) has a fixed size: $f_{k} = μ_{k} N$ . Static friction (while stuck) is whatever it needs to be to keep the object still — anywhere from $0$ up to a ceiling $f_{s, m a x} = μ_{s} N$ . $0 \leq f_{s} \leq μ_{s} N$ So the very first question in ANY problem is: does it slide or not? You answer that by comparing the force trying to cause motion against the ceiling $μ_{s} N$ . Only if the pushing force exceeds the ceiling does the object move — and only then may you write $f = μ_{k} N$ .

Figure 1 — the grip curve.

Figure — Coefficients of friction — measurement, material dependence

Look at the red bar: as you push harder (moving right), static friction rises to match you, tracing the diagonal — friction and push are equal, nothing moves. At the peak ( $μ_{s} N$ ) the block breaks free; friction then drops to the flat kinetic level $μ_{k} N$ . This single picture explains why $μ_{s} > μ_{k}$ shows up as a "jerk" when things start to move. We will point back to this curve in Ex 1 and Ex 2.

The scenario matrix

Every friction problem is one (or a mix) of these cells. Each example below is tagged with the cell(s) it covers.

#	Cell (case class)	What makes it tricky	Example
A	Horizontal pull, does it even move?	must compare push vs ceiling $μ_{s} N$	Ex 1
B	Horizontal pull, it moves → find $a$	use $μ_{k}$ , apply $\sum F = ma$	Ex 2
C	Angled pull (up)	pull changes $N$ → changes friction	Ex 3
D	Push down at an angle	$N$ grows, harder to move	Ex 4
E	Incline, angle of repose (verge)	$μ_{s} = tan θ_{c}$ , mass cancels	Ex 5
F	Incline, block slides → find $a$	net of gravity-pull and $μ_{k} N$	Ex 6
G	Incline, block stays put (below repose)	$f_{s}$ is less than ceiling	Ex 7
H	Degenerate / limiting: $θ = 0$ , $θ = 9 0^{\circ}$ , $μ = 0$	sanity-check the formulas	Ex 8
I	Real-world word problem	translate story → numbers	Ex 9
J	Exam twist: two-surface / trick	a subtle catch examiners love	Ex 10

We rely on the Free Body Diagrams and coordinate decomposition from the parent, and on Newton's Second Law $\sum F = ma$ . Every example below opens with its own free-body diagram (FBD) — the picture of all forces on the object — because friction problems are impossible to reason about safely without one.

Ex 1 — Does it even move? (Cell A)

Worked example Below the ceiling

A $5 kg$ crate sits on a floor with $μ_{s} = 0.40$ , $μ_{k} = 0.30$ . You push horizontally with $15 N$ . Does it move? What is the friction force right now?

Forecast: Guess before reading — will it slide, and is the friction $μ_{s} N$ , $μ_{k} N$ , or something else?

FBD — Figure 2.

Four forces: weight

m g

down, normal

N

up, your push right, friction left.

Step 1 — Normal force. Floor is horizontal, no vertical pull, so vertical balance gives $N = m g = 5 \times 9.8 = 49 N$ . Why this step? Friction is always $μ N$ ; we can't touch friction until we know $N$ . See Normal Force.

Step 2 — Compute the ceiling. $f_{s, m a x} = μ_{s} N = 0.40 \times 49 = 19.6 N$ . Why this step? This is the maximum grip available — the peak of Figure 1's grip curve. If our push beats it, the crate slides.

Step 3 — Compare. Push $= 15 N < 19.6 N = f_{s, m a x}$ . So it does not move. Why this step? This is the "does it slide?" test — the heart of every static problem. On Figure 1 we are still on the rising diagonal, not yet at the peak.

Step 4 — State the actual friction. Since it's stuck and not accelerating, friction exactly cancels the push: $f_{s} = 15 N$ (NOT $19.6$ , NOT $μ_{k} N$ ). Why this step? Static friction adapts to whatever is needed, up to the ceiling. It's using only $15$ of its available $19.6 N$ — a point on the diagonal of Figure 1.

Verify: $\sum F = 15 - 15 = 0 \Rightarrow a = 0$ . ✓ Consistent with "not moving". Units: all newtons. ✓

Ex 2 — It moves, find the acceleration (Cell B)

Worked example Past the ceiling

Same crate ( $5 kg$ , $μ_{s} = 0.40$ , $μ_{k} = 0.30$ ). Now you push with $25 N$ . Find the acceleration.

Forecast: Do we use $μ_{s}$ or $μ_{k}$ in $\sum F = ma$ ?

FBD: identical to Figure 2, but now the push arrow is longer than the friction arrow — a net force remains.

Step 1 — Test for sliding. Ceiling $= μ_{s} N = 0.40 \times 49 = 19.6 N$ . Push $25 > 19.6$ , so it slides. Why this step? Only after confirming motion may we switch to kinetic friction. On Figure 1 we have pushed past the peak.

Step 2 — Kinetic friction. Once moving, $f_{k} = μ_{k} N = 0.30 \times 49 = 14.7 N$ . Why this step? While sliding, friction has the fixed value $μ_{k} N$ — the flat red level in Figure 1, below the peak.

Step 3 — Newton's second law. $\sum F = 25 - 14.7 = 10.3 N$ , so $a = \frac{10.3}{5} = 2.06 m/s^{2}$ . Why this step? Net force divided by mass gives acceleration. See Newton's Second Law.

Verify: Units $N / kg = m/s^{2}$ . ✓ Because $μ_{k} < μ_{s}$ , the friction dropped from a potential $19.6$ to $14.7$ — the "jerk into motion" (the drop) in Figure 1. ✓

Ex 3 — Angled pull (upward) (Cell C)

Worked example Pulling up changes the grip

A $10 kg$ sled, $μ_{s} = 0.50$ . You pull with a rope at $3 0^{\circ}$ above horizontal. What tension $T$ just starts it moving?

Forecast: Will you need more or less force than a purely horizontal pull? Think about what the upward component does to $N$ .

FBD — Figure 3.

The red pull

T

tilts up; its vertical part fights weight, its horizontal part fights friction.

Step 1 — Decompose the pull. Split $T$ into horizontal $T cos 3 0^{\circ}$ (drives motion) and vertical $T sin 3 0^{\circ}$ (lifts). Why this step? Different directions do different jobs; only the horizontal part fights friction, only the vertical part changes $N$ . Same decomposition idea as Inclined Plane Dynamics.

Step 2 — New normal force. Vertical balance: $N + T sin 3 0^{\circ} = m g$ , so $N = m g - T sin 3 0^{\circ}$ . Why this step? The upward pull relieves some weight, so the floor pushes back less — friction weakens.

Step 3 — Verge-of-slip condition. $T cos 3 0^{\circ} = μ_{s} N = μ_{s} (m g - T sin 3 0^{\circ})$ . Why this step? At the exact start of motion, horizontal pull equals maximum friction.

Step 4 — Solve for $T$ . $T cos 3 0^{\circ} + μ_{s} T sin 3 0^{\circ} = μ_{s} m g$ $T = \frac{μ _{s} m g}{c o s 3 0 ^{\circ} + μ _{s} s i n 3 0 ^{\circ}} = \frac{0.50 \times 10 \times 9.8}{0.8660 + 0.50 \times 0.5} = \frac{49}{1.1160} \approx 43.9 N$ Why this step? Collect the $T$ terms, then isolate.

Verify: A horizontal pull would need $μ_{s} m g = 49 N$ . We got only $43.9 N$ — less, confirming the upward tilt helps by cutting $N$ . ✓ Units: N. ✓

Ex 4 — Push down at an angle (Cell D)

Worked example Pushing down works against you

Same $10 kg$ sled, $μ_{s} = 0.50$ , but now you push at $3 0^{\circ}$ below horizontal (like shoving a lawnmower). What force $F$ just starts it?

Forecast: Compare to Ex 3 — more force or less needed now?

FBD — Figure 4.

The red push

F

tilts down; its vertical part adds to weight, so

N

grows.

Step 1 — Decompose. Horizontal $F cos 3 0^{\circ}$ drives; vertical $F sin 3 0^{\circ}$ presses down. Why this step? The downward component now adds to the squeeze.

Step 2 — Normal force. $N = m g + F sin 3 0^{\circ}$ . Why this step? Pushing down means the floor must push back harder → bigger $N$ → more friction to overcome.

Step 3 — Verge condition. $F cos 3 0^{\circ} = μ_{s} (m g + F sin 3 0^{\circ})$ . Why this step? At the exact start of motion the horizontal drive equals the maximum friction $μ_{s} N$ , and here $N = m g + F sin 3 0^{\circ}$ — so the very force $F$ that drives also inflates the friction it must beat.

Step 4 — Solve. $F (cos 3 0^{\circ} - μ_{s} sin 3 0^{\circ}) = μ_{s} m g$ $F = \frac{0.50 \times 10 \times 9.8}{0.8660 - 0.50 \times 0.5} = \frac{49}{0.6160} \approx 79.5 N$ Why this step? Now the $μ_{s} sin$ term subtracts, shrinking the denominator, blowing up $F$ .

Verify: $79.5 N > 49 N$ (horizontal) $> 43.9 N$ (Ex 3, upward). Pushing down is the worst strategy — matches intuition. ✓

Ex 5 — Angle of repose (Cell E)

Worked example The verge on an incline

A block just begins to slide when a ramp reaches $θ_{c} = 2 7^{\circ}$ . Find $μ_{s}$ .

Forecast: Do you need the block's mass?

FBD — Figure 5.

Weight $m g$ splits into $m g sin θ$ (down-slope, red) and $m g cos θ$ (into-slope); normal $N$ and friction $f$ balance them.

Step 1 — Verge-of-slip on incline. From the parent note, $μ_{s} = tan θ_{c}$ . Why this step? At the critical angle, $m g sin θ_{c} = μ_{s} m g cos θ_{c}$ , and $m g$ cancels.

Step 2 — Evaluate. $μ_{s} = tan 2 7^{\circ} \approx 0.510$ . Why this step? Once we know the general result $μ_{s} = tan θ_{c}$ , plugging in the measured critical angle $2 7^{\circ}$ turns it into an actual number — that number is the coefficient we set out to find.

Verify: Dimensionless (tan of an angle). ✓ Mass never entered — that's the beauty of the ramp method. ✓

Ex 6 — Incline, block slides, find acceleration (Cell F)

Worked example Sliding down a steep ramp

A $2 kg$ block on a $3 5^{\circ}$ incline with $μ_{k} = 0.25$ . It's already sliding down. Find its acceleration.

Forecast: Positive (speeding up) or is friction enough to slow it?

FBD: same layout as Figure 5, but now friction $f = μ_{k} N$ is at its fixed kinetic size, pointing up-slope against the downhill slide. (Recall the tilted axes: $x$ along-slope, $y$ perpendicular.)

Step 1 — Perpendicular balance. $N = m g cos 3 5^{\circ} = 2 \times 9.8 \times 0.8192 = 16.06 N$ . Why this step? No motion off the surface, so the perpendicular forces cancel.

Step 2 — Kinetic friction. $f_{k} = μ_{k} N = 0.25 \times 16.06 = 4.01 N$ , pointing up the slope (opposing downhill motion). Why this step? Kinetic friction always opposes the direction of sliding.

Step 3 — Along-slope Newton's law. Downhill gravity pull $= m g sin 3 5^{\circ} = 2 \times 9.8 \times 0.5736 = 11.24 N$ . $\sum F = m g sin 3 5^{\circ} - f_{k} = 11.24 - 4.01 = 7.23 N$ $a = \frac{7.23}{2} = 3.62 m/s^{2} (down the slope)$ Why this step? Net driving force divided by mass.

Verify: Since $tan 3 5^{\circ} = 0.70 > μ_{k} = 0.25$ , gravity beats friction, so acceleration is positive (down-slope). ✓ Units N/kg = m/s². ✓

Ex 7 — Incline, block stays put (Cell G)

Worked example Below the repose angle

A $3 kg$ block on a $1 5^{\circ}$ incline with $μ_{s} = 0.40$ . Does it slide? What is the actual friction force?

Forecast: Is the friction here equal to $μ_{s} N$ ?

FBD: same layout as Figure 5, but the block is at rest — static friction $f_{s}$ (up-slope) is whatever is needed, not necessarily its ceiling. (Same tilted axes as Ex 6.)

Step 1 — Compute the ceiling. $N = m g cos 1 5^{\circ} = 3 \times 9.8 \times 0.9659 = 28.40 N$ ; $f_{s, m a x} = μ_{s} N = 0.40 \times 28.40 = 11.36 N$ . Why this step? We need the maximum grip the surface can offer so we can later test whether the downhill pull exceeds it — the "does it slide?" test on an incline.

Step 2 — Downhill pull. $m g sin 1 5^{\circ} = 3 \times 9.8 \times 0.2588 = 7.61 N$ . Why this step? This along-slope component of gravity is the only force trying to slide the block; we must know it to compare against the ceiling from Step 1.

Step 3 — Compare. Downhill pull $7.61 < 11.36$ ceiling → block stays put. Equivalently, $tan 1 5^{\circ} = 0.268 < μ_{s} = 0.40$ . Why this step? The repose test: it slides only if $tan θ > μ_{s}$ .

Step 4 — Actual friction. Since static and still, friction exactly balances the downhill pull: $f_{s} = 7.61 N$ (up the slope), NOT $11.36$ . Why this step? Static friction supplies only what's needed — echoing Ex 1.

Verify: $\sum F_{along} = 7.61 - 7.61 = 0$ . ✓ It's using $7.61$ of its $11.36 N$ budget. ✓

Ex 8 — Degenerate & limiting cases (Cell H)

Worked example Sanity-check the extremes

Test the incline formulas at every boundary: $θ = 0$ , $θ \to 9 0^{\circ}$ , and $μ = 0$ .

Case (i) — Flat floor, $θ = 0$ . The repose formula $μ_{s} = tan θ_{c}$ is a formula for the critical tilt angle, not for $μ_{s}$ itself — the material coefficient $μ_{s}$ is a fixed property of the surface pair and does not become $0$ just because the ramp is flat. At $θ = 0$ there is simply no down-slope component of gravity ( $m g sin 0 = 0$ ), so nothing drives sliding and the repose test is never triggered. The formula only applies at the verge of slipping; at $θ = 0$ we are nowhere near that verge, so reading off $tan 0 = 0$ is meaningless here, not a claim that $μ_{s} = 0$ . Also note $N = m g cos 0 = m g$ , the full weight — the familiar flat-ground case. ✓

Case (ii) — Vertical wall, $θ \to 9 0^{\circ}$ . $tan 9 0^{\circ} \to \infty$ . Interpretation: you'd need an infinite $μ_{s}$ to hold a block by an incline this steep — i.e. it always slides. Physically, at $9 0^{\circ}$ the normal force $N = m g cos 9 0^{\circ} = 0$ , so there is no squeeze, hence no friction at all ( $f_{s, m a x} = μ_{s} N = 0$ ) — nothing holds the block and it free-falls. ✓

Case (iii) — Frictionless, $μ = 0$ . Put $μ_{k} = 0$ into the Ex 6 result $a = g (sin θ - μ_{k} cos θ)$ , giving $a = g sin θ$ . At $θ = 3 0^{\circ}$ : $a = 9.8 \times 0.5 = 4.9 m/s^{2}$ . This is the pure-gravity slide — the fastest an object can accelerate down that angle. ✓

Verify: All three limits agree with $N = m g cos θ$ and the repose logic; $θ = 0$ gives $N = m g$ , $θ = 9 0^{\circ}$ gives $N = 0$ , $μ = 0$ gives $a = g sin θ$ . No formula breaks. ✓

Ex 9 — Real-world word problem (Cell I)

Worked example The braking car

A car ( $1200 kg$ ) skids to a stop on dry asphalt, $μ_{k} = 0.80$ . Its wheels are locked (pure sliding). What is its deceleration, and how far does it travel from $20 m/s$ ?

Forecast: Does the car's mass affect the stopping distance?

FBD: flat-ground layout like Figure 2 — weight $m g$ down, normal $N$ up, and the only horizontal force is kinetic friction $f_{k}$ pointing backward.

Step 1 — Normal force. Flat road: $N = m g = 1200 \times 9.8 = 11760 N$ . Why this step? Friction is $μ_{k} N$ , so we need $N$ first; on flat ground it equals the full weight.

Step 2 — Friction = the only horizontal force. $f_{k} = μ_{k} N = 0.80 \times 11760 = 9408 N$ , backward. Why this step? Locked wheels mean pure sliding, so kinetic friction (not static) is what stops the car.

Step 3 — Deceleration. $a = \frac{f _{k}}{m} = \frac{9408}{1200} = 7.84 m/s^{2}$ . Notice $a = μ_{k} g = 0.80 \times 9.8 = 7.84$ — mass cancels! Why this step? $f_{k} = μ_{k} m g$ and $a = f_{k} / m$ , so $m$ divides out.

Step 4 — Stopping distance. Kinematics $v^{2} = v_{0}^{2} - 2 a d \Rightarrow 0 = 2 0^{2} - 2 (7.84) d$ , so $d = \frac{400}{15.68} = 25.5 m$ . Why this step? Constant deceleration lets us use $v^{2} = v_{0}^{2} - 2 a d$ .

Verify: Mass never appears in $d$ — a heavy truck and light car with the same $μ_{k}$ stop in the same distance. ✓ Units: m. ✓

Ex 10 — Exam-style twist (Cell J)

Worked example The "which surface breaks first" catch

A block rests on a table; the table rests on the floor. Table+block are pushed as one. Block–table $μ_{s} = 0.30$ , table–floor $μ_{s} = 0.50$ . If you push the table horizontally, and block mass $= 2 kg$ , what is the maximum acceleration of the system before the block slips on the table?

Forecast: Which surface limits the acceleration — the grippier or the slipperier one at the top?

FBD — Figure 6.

Isolate the block: weight $m g$ down, normal from table up, and the only horizontal force is the red static friction from the table — this alone must accelerate the block.

Step 1 — Isolate the block. The only horizontal force on the block is friction from the table beneath it. That friction must supply all the block's acceleration. Why this step? Draw the block's own free body diagram — see Free Body Diagrams.

Step 2 — Max friction the top surface can give. $f_{s, m a x} = μ_{s, top} m g = 0.30 \times 2 \times 9.8 = 5.88 N$ . Why this step? This is the biggest push-forward the table can give the block through friction alone.

Step 3 — Max acceleration. $a_{m a x} = \frac{f _{s, m a x}}{m} = \frac{5.88}{2} = 2.94 m/s^{2}$ . Note $a_{m a x} = μ_{s, top} g = 0.30 \times 9.8$ . Why this step? Beyond this, friction can't keep the block accelerating with the table → it slips back.

Verify: $a_{m a x} = μ_{s} g$ , independent of block mass. ✓ The top ( $0.30$ ) surface limits it, not the floor ( $0.50$ ) — the exam trap is thinking the bigger $μ$ matters. It doesn't for this question. ✓

Recall Self-test: name the cell

"You push a box and it doesn't budge; find the friction." ::: Cell A — friction equals your push, below $μ_{s} N$ . "A rope pulls a crate at $2 0^{\circ}$ up; find starting tension." ::: Cell C — $T = μ_{s} m g / (cos θ + μ_{s} sin θ)$ . "Block slides down a $4 0^{\circ}$ ramp; find $a$ ." ::: Cell F — $a = g (sin θ - μ_{k} cos θ)$ . "A car skids; find stopping distance." ::: Cell I — $a = μ_{k} g$ , then kinematics. "Ramp tilted to $θ_{c}$ where it just slips." ::: Cell E — $μ_{s} = tan θ_{c}$ .

Connections

Newton's Second Law — every "find $a$ " example is $\sum F = ma$ .
Static vs Kinetic Friction — the "does it move?" branch chooses between them.
Inclined Plane Dynamics — Ex 5–8 use the along/perpendicular split.
Normal Force — Ex 3–4 hinge on how angled forces change $N$ .
Free Body Diagrams — the tool behind Ex 10's isolation.
Lubrication & Tribology — why $μ_{k}$ differs across the surface tables.

Worked examples — Coefficients of friction — measurement, material dependence