JEE Mains Physics · Laws of Motion

Static and Kinetic Friction

Static friction takes any value up to μs N to stop sliding; once a body slides, kinetic friction is fixed at μk N, so every friction question is first a question about the normal reaction N.

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Why this matters

Thirty PYQs, twenty-five of them multiple choice, and four from 2026: the largest page in the chapter. Fourteen are on a level surface: stopping distances, a force pulled or pushed at an angle, blocks that must move together, and the work done against friction. Eight hold a block on an incline or push it up the slope; eight let it slide down, and five of those compare the time on a rough slope with the time on a smooth one.

Concept 1 of 3: Friction on a level surface

Friction depends on how hard the surfaces are pressed together, the normal reaction N, and on the materials, through μ. It does not depend on the area of contact. Before a body slides, static friction is only as large as it needs to be. A force at an angle changes N, and so changes the friction: pulling slightly upward makes a heavy box easier to start.

Definition

Static: $f_s \le \mu_s N$ , self-adjusting. Kinetic: $f_k = \mu_k N$ once sliding. Neither depends on the area of contact.
A force F at $\theta$ above the horizontal: $N = mg - F\sin\theta$ . Below the horizontal (a push): $N = mg + F\sin\theta$ .
Least pull at $\theta$ to start the body: $F = \dfrac{\mu mg}{\cos\theta + \mu\sin\theta}$ .
Sliding to a stop: deceleration $\mu g$ , distance $\dfrac{v^{2}}{2\mu g}$ , time $\dfrac{v}{\mu g}$ .
Two stacked blocks pushed by a force on the lower one: the top block can reach at most $\mu_s g$ , so they move together only up to $F = (m + M)\mu_s g$ .
Work done against friction over a distance s: $\mu N s$ .
A bag dropped on a belt moving at v slips $\dfrac{v^{2}}{2\mu g}$ relative to the belt. A chain on a table can hang over the edge by at most the fraction $\dfrac{\mu}{1 + \mu}$ .

Friction and the least starting pull

f_s \le \mu_s N,\quad f_k = \mu_k N, \qquad F_{\min} = \frac{\mu mg}{\cos\theta + \mu\sin\theta}

Worked example

A 20 kg crate rests on a floor with

\mu_s = 0.5

. It is pulled by a rope at

37^{\circ}

above the horizontal (

\sin 37^{\circ} = 0.6

\cos 37^{\circ} = 0.8

). Find the least force that starts it, and compare with a horizontal pull. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q108Moderate

Example 1 · Laws of Motion · Static and Kinetic Friction

As shown in the figure a block of mass

10\text{ }kg

lying on a horizontal surface is pulled by a force

F

acting at an angle

30^{\circ}

, with horizontal. For

\mu_{s}= 0.25

, the block will just start to move for the value of

F

: [Given

g = 10{\text{ }ms}^{- 2}

]

N is not mg when the force is at an angle

An angled pull lifts part of the weight and an angled push adds to it. Using μmg for the friction here gives the horizontal-force answer, which is always among the options.

Static friction is only as large as it needs to be

A 10 kg block with μs = 0.5 pushed by 20 N does not move, and the friction is 20 N, not 50 N. μs N is the limit, reached only at the point of slipping.

Area of contact does not matter

Turning a brick on its side changes the area but not the friction. Both static and kinetic friction depend on the materials (through μ) and on N only.

Drill 13 more on friction on a level surface

Concept 2 of 3: Holding and pushing a block on a rough incline

Friction always opposes the motion, or the motion that would happen without it. Pushing a block up the slope, friction acts down the slope and adds to the weight's pull. Holding it from sliding down, friction acts up the slope and helps you. That flip of direction is the whole difference between the two forces.

Definition

Angle of repose: the block just starts to slide when $\tan\theta = \mu_s$ .
Least force along the slope to push it up: $F_1 = mg(\sin\theta + \mu\cos\theta)$ .
Least force along the slope to stop it sliding down: $F_2 = mg(\sin\theta - \mu\cos\theta)$ . $F_1 - F_2 = 2\mu mg\cos\theta$ .
If $\mu > \tan\theta$ the block stays put by itself; to slide it down at constant velocity, push with $mg(\mu\cos\theta - \sin\theta)$ .
A block at rest, or sliding at constant velocity: the surface's total force (N plus friction) equals mg, straight up.
On a curved surface y = f(x), a block can rest wherever the slope $dy/dx \le \mu$ .

Push up and hold

F_{\text{up}} = mg(\sin\theta + \mu\cos\theta), \qquad F_{\text{hold}} = mg(\sin\theta - \mu\cos\theta)

Worked example

A 10 kg block lies on a

37^{\circ}

incline with

\mu = 0.5

(

\sin 37^{\circ} = 0.6

). Find the least force along the slope (a) to push it up and (b) to stop it sliding down. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q100Moderate

Example 2 · Laws of Motion · Static and Kinetic Friction

Consider a block kept on an inclined plane (inclined at

45^{\circ}

) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane

(\mu)

is equal to:

Friction flips direction between push and hold

Pushing up, friction points down the slope: add μmg cos θ. Holding against sliding, it points up the slope: subtract it. Using the same sign in both gives F₁ − F₂ = 0.

The contact force is the full weight

For a block at rest or at constant velocity on an incline, N and friction together balance mg. The total contact force is mg, not mg cos θ.

μ > tan θ means it will not slide by itself

Then the force to move it down is mg(μ cos θ − sin θ). A negative answer from the 'hold' formula is the sign that the block needs a push, not a brake.

Drill 7 more on holding and pushing a block on a rough incline

Concept 3 of 3: Sliding down a rough incline

A block sliding down feels the weight's pull along the slope, mg sin θ, minus kinetic friction, μmg cos θ. From rest, the time to cover a length goes as one over the square root of the acceleration. So a block that takes twice as long on a rough slope has a quarter of the acceleration it would have on a smooth one.

Definition

Sliding down: $a = g(\sin\theta - \mu_k\cos\theta)$ . Moving up and slowing: $a = g(\sin\theta + \mu_k\cos\theta)$ , down the slope.
From rest over a length l: $t = \sqrt{2l/a}$ , $v = \sqrt{2al}$ .
Rough time = n × smooth time: $a_{\text{rough}} = a_{\text{smooth}}/n^{2}$ , so $\mu = \tan\theta\left(1 - \dfrac{1}{n^{2}}\right)$ . At $45^{\circ}$ , $\mu = 1 - \dfrac{1}{n^{2}}$ .
Constant velocity down the slope: $\mu_k = \tan\theta$ .
Any extra force (an electric force, a push) with a part perpendicular to the slope changes N; resolve it as well.

Sliding down

a = g(\sin\theta - \mu\cos\theta), \qquad \mu = \tan\theta\left(1 - \frac{1}{n^{2}}\right)

Worked example

A block slides from rest down 8 m of a

37^{\circ}

incline with

\mu_k = 0.25

(

\sin 37^{\circ} = 0.6

\cos 37^{\circ} = 0.8

). Find its acceleration, the time taken and its speed at the bottom. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q3Moderate

Example 3 · Laws of Motion · Static and Kinetic Friction

The time taken by a block of mass

m

to slide down from the highest point to the lowest point on a rough inclined plane is

50\%

more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at

45^{\circ}

with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is

\_\_\_\_

Time ratios are squared

Time goes as 1/√a. Taking 50% more time means the acceleration is smaller by a factor of 1.5² = 2.25, not 1.5. That is why the answer at 45° is 1 − 1/2.25 = 5/9.

Friction on a slope is μ mg cos θ

On an incline, N = mg cos θ. Writing the friction as μmg, the level-floor value, makes it too large and gives a wrong μ.

An extra force can change N

A horizontal electric force on a charged block has a part pressing it into the slope or lifting it off. Add that part to mg cos θ before multiplying by μ.

Drill 7 more on sliding down a rough incline

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Friction on a level surface
Friction and the least starting pull
$f_s \le \mu_s N,\quad f_k = \mu_k N, \qquad F_{\min} = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$
Holding and pushing a block on a rough incline
Push up and hold
$F_{\text{up}} = mg(\sin\theta + \mu\cos\theta), \qquad F_{\text{hold}} = mg(\sin\theta - \mu\cos\theta)$
Sliding down a rough incline
Sliding down
$a = g(\sin\theta - \mu\cos\theta), \qquad \mu = \tan\theta\left(1 - \frac{1}{n^{2}}\right)$

Watch out for (9)

N is not mg when the force is at an angle
→ Friction on a level surface
Static friction is only as large as it needs to be
→ Friction on a level surface
Area of contact does not matter
→ Friction on a level surface
Friction flips direction between push and hold
→ Holding and pushing a block on a rough incline
The contact force is the full weight
→ Holding and pushing a block on a rough incline
μ > tan θ means it will not slide by itself
→ Holding and pushing a block on a rough incline
Time ratios are squared
→ Sliding down a rough incline
Friction on a slope is μ mg cos θ
→ Sliding down a rough incline
An extra force can change N
→ Sliding down a rough incline

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