A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by

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Concept understanding — Static Friction Limit

The Static Friction Limit: Why a Heavy Box Won't Move Until You Really Push

Imagine you're trying to push a heavy wooden crate across a rough floor. You lean into it gently — nothing happens. You push a little harder — still nothing. The crate stays perfectly still, as if glued to the spot. Then, at some point, you push just a bit more, and suddenly the crate lurches forward.

That invisible "sticking point" — the exact moment the crate finally gives way — is the static friction limit.

The Intuition: Friction as a "Smart" Force

Friction between two surfaces that aren't sliding is called static friction. What makes it special is that it's self-adjusting. It doesn't have a fixed value. Instead, it automatically grows to match whatever force you apply — up to a point.

Think of it like a tug-of-war where your opponent (static friction) matches your pull exactly, but only until you exceed their maximum strength. As long as you pull less than their limit, you both stay in equilibrium and nothing moves. The moment you exceed that limit, you win — and the crate starts sliding.

Note

Static friction only exists when there is no relative motion between the surfaces. Once sliding begins, it's replaced by kinetic friction, which is usually weaker.

The Precise Statement

The static friction limit (also called limiting friction) is the maximum possible value of static friction that can act between two surfaces in contact before they start sliding relative to each other.

Mathematically:

f_s \leq \mu_s N

Where:

$f_s$ = static friction force (the actual value, which can be anything from 0 up to the limit)
$\mu_s$ = coefficient of static friction (a constant that depends on the two materials — rubber on concrete is high, ice on steel is low)
$N$ = normal reaction force (the force pressing the surfaces together, usually equal to weight on a horizontal surface)

The static friction limit is the equality case:

f_{s,\text{max}} = \mu_s N

This is the maximum static friction the surfaces can provide. Apply a force less than this, and the object stays put. Apply a force equal to this, and the object is on the verge of moving (impending motion). Apply a force greater than this, and the object accelerates.

A Concrete Example

A 10 kg block rests on a horizontal floor. $\mu_s = 0.4$ between the block and floor. Take $g = 10 \text{ m/s}^2$ .

Normal reaction: $N = mg = 10 \times 10 = 100 \text{ N}$

Static friction limit: $f_{s,\text{max}} = 0.4 \times 100 = 40 \text{ N}$

Now, what happens as you push?

Applied Force	Static Friction	Result
10 N	10 N (matches)	Block stays still
25 N	25 N (matches)	Block stays still
40 N	40 N (matches)	Block is just about to move
45 N	40 N (cannot exceed limit)	Block accelerates forward

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Static friction rises linearly with applied force until it reaches its maximum value $\mu_s N$ , then the block starts moving and kinetic friction (constant at $\mu_k N$ ) takes over. The graph shows a linear rise, a peak, then a horizontal line at a lower level.

The force $f$ from the rough surface is friction, and friction behaves differently depending on whether the block is stationary or moving. Understanding this distinction is the key to sketching the graph.

When you first apply a small horizontal force $F$ , the block doesn't move. Why? Because static friction automatically adjusts itself to exactly cancel whatever force you apply, keeping the net force zero. This self-adjusting property continues until static friction reaches its maximum possible value, $f_{\text{max}} = \mu_s N$ , where $\mu_s$ is the coefficient of static friction and $N$ is the normal force (equal to the block's weight on a horizontal surface).

Once $F$ exceeds $\mu_s N$ , static friction can no longer hold the block in place. The block begins to slide, and the nature of friction changes: kinetic friction takes over. Kinetic friction has a fixed magnitude $f_k = \mu_k N$ and doesn't depend on how hard you pull (as long as the block keeps moving). Typically, $\mu_k < \mu_s$ , so kinetic friction is weaker than the maximum static friction.

Let me walk through the regions of the graph:

Static regime ( $F < \mu_s N$ ): The block remains at rest. Static friction exactly balances the applied force, so $f = F$ . This gives a straight line through the origin with slope $1$ .
Threshold ( $F = \mu_s N$ ): Static friction reaches its maximum value. This is the peak of the graph, where $f = \mu_s N$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$30^{\circ}$
›Reveal solutionSolution
The block moves without acceleration at 30°, so the coefficient of friction equals tan(30°). When the incline is increased to 60°, the net force down the plane gives an acceleration of about 5.66 m s⁻², matching option (D).

The key idea is that “moves down without acceleration” means the net force along the incline is zero — the component of weight down the slope is exactly balanced by kinetic friction. That lets us find the coefficient of friction. Then, for the steeper incline, the same friction acts (still kinetic, since the block is moving), but now the downhill component of weight is larger, so there’s a net force and thus an acceleration.

First incline (30°) — equilibrium condition The forces along the incline are:

Downhill: $mg \sin 30^\circ$

Uphill (friction): $\mu_k N = \mu_k mg \cos 30^\circ$ Since acceleration is zero:

$mg \sin 30^\circ = \mu_k mg \cos 30^\circ$
Cancel $mg$ :
$\mu_k = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$

Second incline (30° + 30° = 60°) — net force and acceleration Now the angle is $60^\circ$ . The downhill component is $mg \sin 60^\circ$ , and friction is still $\mu_k mg \cos 60^\circ$ . Net force down the plane:

$F_{\text{net}} = mg \sin 60^\circ - \mu_k mg \cos 60^\circ$
Substitute $\mu_k = \tan 30^\circ$ :
$F_{\text{net}} = mg \left( \sin 60^\circ - \tan 30^\circ \cos 60^\circ \right)$
Using $\tan 30^\circ = \frac{\sin 30^\circ}{\cos 30^\circ}$ :
$F_{\text{net}} = mg \left( \sin 60^\circ - \frac{\sin 30^\circ}{\cos 30^\circ} \cos 60^\circ \right)$
Now plug exact values:

$\sin 60^\circ = \frac{\sqrt{3}}{2},\ \cos 60^\circ = \frac{1}{2},\ \sin 30^\circ = \frac{1}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2}$
$F_{\text{net}} = mg \left( \frac{\sqrt{3}}{2} - \frac{1/2}{\sqrt{3}/2} \cdot \frac{1}{2} \right) = mg \left( \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{3}} \cdot \frac{1}{2} \right)$ $= mg \left( \frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}} \right) …$
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COMEDK 2025Set 2025-A1 markMCQ
$\frac{\pi}{8}$
›Reveal solutionSolution
A body "just slides" down an incline when the coefficient of friction equals the tangent of the incline's angle: $\mu = \tan\theta$ . Reading the given slope as "the incline rises 1 m for every 2 m travelled along the slope" gives $\sin\theta = 1/2$ , so $\theta = 30^\circ = \pi/6$ . The correct option is (D).

Concept and Intuition

When a body "just slides" down a rough incline, it is on the verge of motion — the downhill component of its weight exactly balances the maximum force of static friction. At that condition, the coefficient of friction equals the tangent of the incline's angle: $\mu = \tan\theta$ . The mass of the body (125 kg) does not affect this result, since it cancels out of the balance equation.

Step-by-step reasoning

Find the angle of the incline.

"Rises 1 m in every 2 m" describes the incline's gradient as rise over the distance travelled along the slope, so $\sin\theta = \dfrac{1}{2}$ , giving $\theta = 30^\circ = \dfrac{\pi}{6}$ radians.

Apply the "just slides" condition.

For a body on the verge of sliding down a rough incline: …
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COMEDK 2024Set 2024-A1 markMCQ
$45^{\circ}$
›Reveal solutionSolution
The key idea is that the time ratio (with friction : without friction) is given as 3:1, and using kinematic equations for constant acceleration on an incline, we solve for the coefficient of friction, finding it to be 0.88.

Concept and Intuition

When a block slides down an incline without friction, its acceleration is $g \sin\theta$ . With friction, the net acceleration is reduced to $g(\sin\theta - \mu \cos\theta)$ . The problem states that the time taken with friction is three times the time taken without friction for the same distance. Since time is inversely related to the square root of acceleration (for constant acceleration from rest), we can set up a ratio and solve for $\mu$ .

Step-by-step solution

Define accelerations

For an incline of angle $\theta = 45^\circ$ :

Without friction: $a_0 = g \sin 45^\circ = \frac{g}{\sqrt{2}}$ .

With friction: $a_f = g(\sin 45^\circ - \mu \cos 45^\circ) = \frac{g}{\sqrt{2}}(1 - \mu)$ .

Relate time and acceleration for constant distance

Starting from rest, distance $s = \frac{1}{2} a t^2$ . For the same $s$ :

$\frac{1}{2} a_0 t_0^2 = \frac{1}{2} a_f t_f^2 \quad \Rightarrow \quad a_0 t_0^2 = a_f t_f^2.$

Use the given time ratio The problem says the block slides thrice the time with friction: $t_f = 3 t_0$ . Substitute:

$a_0 t_0^2 = a_f (3 t_0)^2 = 9 a_f t_0^2.$
Cancel $t_0^2$ (non-zero):
$a_0 = 9 a_f.$

Insert the expressions for acceleration
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KCET 2023Set A-31 markMCQ
$10\, \mathrm{kg}$
›Reveal solutionSolution
The applied force must first overcome the maximum static friction; once motion starts, kinetic friction opposes it. The net force gives an acceleration of about $1\, \mathrm{m/s^2}$ , so option (D) is correct.

The key here is that friction is not a fixed force — it adjusts to oppose relative motion up to a limit. When you push an object on a rough surface, the friction force that actually acts depends on whether the object is moving or not. For a stationary object, static friction can take any value from zero up to $\mu_s N$ (where $\mu_s$ is the coefficient of static friction). Once the object starts sliding, kinetic friction takes over at a constant value $\mu_k N$ .

In this problem, we are given the coefficient of kinetic friction ( $\mu_k = 0.5$ ), but we are not told the coefficient of static friction. However, a standard fact is that $\mu_s$ is generally slightly larger than $\mu_k$ for the same pair of surfaces. For typical exam problems, if only $\mu_k$ is given, we assume the applied force is enough to overcome static friction — but we must check that first.

Let’s work through it step by step.

Find the maximum static friction The normal reaction $N$ equals the weight of the body because the surface is horizontal:

$N = mg = 10 \times 9.8 = 98\, \mathrm{N}$
The maximum static friction force is $\mu_s N$ . Since $\mu_s$ is not given, but we know $\mu_s \ge \mu_k$ , the smallest possible maximum static friction is $\mu_k N = 0.5 \times 98 = 49\, \mathrm{N}$ . In reality, $\mu_s$ would be a bit larger, so the actual maximum static friction is at least $49\, \mathrm{N}$ .

Check if the applied force overcomes static friction

The applied horizontal force is $60\, \mathrm{N}$ . Since $60\, \mathrm{N} > 49\, \mathrm{N}$ , the applied force definitely exceeds the minimum possible maximum static friction. Even if $\mu_s$ were as high as, say, $0.6$ (giving $58.8\, \mathrm{N}$ ), $60\, \mathrm{N}$ still exceeds it. So the body will start moving.

Once moving, kinetic friction acts

Kinetic friction is constant:

$f_k = \mu_k N = 0.5 \times 98 = 49\, \mathrm{N}$
This opposes the motion.

Find the net force and acceleration …
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KCET 2021Set B-21 markMCQ
$4~\mathrm{cm}$
›Reveal solutionSolution

At the slipping threshold, limiting static friction equals the required centripetal force, giving $r \propto 1/\omega^2$ .

Step 1 — The physics of the coin on a turntable.

The only horizontal force available to keep the coin moving in a circle is static friction. The required centripetal force at radius $r$ and angular speed $\omega$ is
$F_c = m\omega^2 r.$
The most friction can supply is
$f_{\max} = \mu_s N = \mu_s m g.$
Step 2 — The "just slips" condition.

"Just slips" means the requirement has grown to exactly the maximum available:
$\mu_s m g = m\omega^2 r \quad\Longrightarrow\quad r = \frac{\mu_s g}{\omega^2}.$
The mass cancels — the answer does not depend on the coin. Note $\mu_s$ and $g$ are fixed, so
$r \propto \frac{1}{\omega^2} \quad\Longleftrightarrow\quad \omega^2 r = \text{constant}.$
Step 3 — Apply to the two cases.

$\omega_1^2 r_1 = \omega_2^2 r_2, \qquad r_1 = 4\ \mathrm{cm},\ \ \omega_2 = 2\omega_1.$ …
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KCET 2019Set A-11 markMCQ
$Q. A piston is performing S.H.M. in the vertical direction with a frequency of 0.5 Hz. A block of 10 kg is placed on the piston. The maximum amplitude of the system such that the block remains in contact with the piston is (A) 1 m (B) 0.5 m (C) 1.5 m (D) 0.1 m$
›Reveal solutionSolution

The block separates the instant the piston needs to pull it down faster than gravity can; the limit is $a_{\max} = \omega^2 A = g$ , so $A_{\max} = g/\omega^2$ .

Step 1 — The physics of "remains in contact"

At the highest point of the vertical SHM, the acceleration points downwards with magnitude $a_{\max} = \omega^2 A$ . Two forces act on the block: its weight $mg$ (down) and the normal reaction $N$ from the piston (up). Newton's second law, taking down as positive:
$mg - N = m a \quad\Longrightarrow\quad N = m(g - a) = m\left(g - \omega^2 A\right)$
A surface can only push, never pull, so we need $N \ge 0$ :
$g - \omega^2 A \ge 0 \quad\Longrightarrow\quad \boxed{\;\omega^2 A \le g\;}$
If the amplitude is any larger, the piston would have to pull the block down (impossible), so the block loses contact and flies off. The block's mass $m$ cancels out entirely — that is why the given "10 kg" is a deliberate red herring.

Step 2 — Find $\omega$
$\omega = 2\pi f = 2\pi (0.5) = \pi\ \mathrm{rad\,s^{-1}}$ $\omega^2 = \pi^2 \approx 9.87\ \mathrm{s^{-2}}$
Step 3 — Solve for the maximum amplitude

$A_{\max} = \frac{g}{\omega^2} = \frac{9.8}{\pi^2} = \frac{9.8}{9.87} \approx 0.993\ \mathrm{m}$ …
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Q.A block placed on a rough horizontal surface is pulled by a horizontal force $F$ . Let $f$ be the force applied by the rough surface on the block. Plot a graph of $f$ versus $F$ .

Concept understanding — Static Friction Limit

The Static Friction Limit: Why a Heavy Box Won't Move Until You Really Push

The Intuition: Friction as a "Smart" Force

The Precise Statement

A Concrete Example

Concept: Static friction self-adjusts up to a limit; kinetic friction is constant beyond it

Step 1 — The physics of "remains in contact"

Step 2 — Find $\omega$

Step 3 — Solve for the maximum amplitude

Q.A block placed on a rough horizontal surface is pulled by a horizontal force FFF. Let fff be the force applied by the rough surface on the block. Plot a graph of fff versus FFF.

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Q.A block placed on a rough horizontal surface is pulled by a horizontal force $F$ . Let $f$ be the force applied by the rough surface on the block. Plot a graph of $f$ versus $F$ .

Unlock everything free for 14 days