A small block A of mass $m/2$ rests on top of a larger block B of mass $m$, and B rests on a horizontal floor. The coefficient of friction between the floor and block B is $0.1$, and the coefficient of friction between the two blocks A and B is $0.2$. A horizontal force $\mathbf{F}$ is applied to the lower block B (pointing to the right). Which of the following statements are true? (Note: more than one of the given options may be correct.) (a) The bodies will move together if $F = 0.25\,mg$. (b) The body A will slip with respect to B if $F = 0.5\,mg$. (c) The bodies will move together if $F = 0.5\,mg$. (d) The bodies will be at rest if $F = 0.1\,mg$. (e) The maximum value of $F$ for which the two bodies will move together is $0.45\,mg$.

> [!TLDR] > Block B is pushed and it drags A along only through the friction acting between them. The floor friction sets the threshold to start moving; the limited friction on A sets the largest acceleration the pair can share. Crunching the numbers gives a maximum common force of $0.45\,mg$, so (A), (B), (D) and (E) are true and (C) is false. ### Concept and data - Masses: $m_A=m/2$, $m_B=m$, total $=\tfrac{3m}{2}$. - Floor friction on B: $\mu_{\text{floor}}=0.1$, normal $=\tfrac{3m}{2}g$, so $f_{\text{floor}}^{\max}=0.1\times\tfrac{3m}{2}g=0.15\,mg$. - Friction between A and B: $\mu=0.2$, normal on A $=\tfrac{m}{2}g$, so $f_{AB}^{\max}=0.2\times\tfrac{m}{2}g=0.1\,mg$. ### Maximum force for common motion Treating both as one body: $F-f_{\text{floor}}=\tfrac{3m}{2}a$. Block A is accelerated **only** by the friction from B, so $f_{AB}=\tfrac{m}{2}a$. Its ceiling is $0.1\,mg$, giving $$a_{\max}=\frac{0.1\,mg}{m/2}=0.2\,g.$$ The largest force that still moves them together is $$F_{\max}=f_{\text{floor}}^{\max}+\tfrac{3m}{2}a_{\max}=0.15\,mg+\tfrac{3m}{2}(0.2g)=0.15\,mg+0.30\,mg=0.45\,mg.$$ This confirms **(E)**. ### Testing each statement …

A small block A of mass m/2 rests on top of a larger block B of mass m, and B rests on a horizontal floor. The

NCERT Exemplar · Q11

m/2

(a) The bodies will move together if

F = 0.25\,mg

(b) The body A will slip with respect to B if

F = 0.5\,mg

F = 0.5\,mg

(d) The bodies will be at rest if

F = 0.1\,mg

(e) The maximum value of

F

for which the two bodies will move together is

0.45\,mg

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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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Block B is pushed and it drags A along only through the friction acting between them. The floor friction sets the threshold to start moving; the limited friction on A sets the largest acceleration the pair can share. Crunching the numbers gives a maximum common force of $0.45\,mg$ , so (A), (B), (D) and (E) are true and (C) is false.

Concept and data

Masses: $m_A=m/2$ , $m_B=m$ , total $=\tfrac{3m}{2}$ .
Floor friction on B: $\mu_{\text{floor}}=0.1$ , normal $=\tfrac{3m}{2}g$ , so $f_{\text{floor}}^{\max}=0.1\times\tfrac{3m}{2}g=0.15\,mg$ .
Friction between A and B: $\mu=0.2$ , normal on A $=\tfrac{m}{2}g$ , so $f_{AB}^{\max}=0.2\times\tfrac{m}{2}g=0.1\,mg$ .

Maximum force for common motion

Treating both as one body: $F-f_{\text{floor}}=\tfrac{3m}{2}a$ . Block A is accelerated only by the friction from B, so $f_{AB}=\tfrac{m}{2}a$ . Its ceiling is $0.1\,mg$ , giving

a_{\max}=\frac{0.1\,mg}{m/2}=0.2\,g.

The largest force that still moves them together is

F_{\max}=f_{\text{floor}}^{\max}+\tfrac{3m}{2}a_{\max}=0.15\,mg+\tfrac{3m}{2}(0.2g)=0.15\,mg+0.30\,mg=0.45\,mg.

This confirms (E).

Testing each statement …

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

Showing the 12 most recent of 27 on this concept.

AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$[g = 10\ ms^{-2}]$
›Reveal solutionSolution
This tests Newton's second law with kinetic friction opposing an applied force. The answer is (B), $1\,ms^{-2}$ .

Concept and Intuition

When a block is pulled along a rough horizontal surface, friction opposes the applied force. The net force driving the acceleration is the applied force minus the friction force (which depends on the normal reaction, here equal to weight since the surface is horizontal and the pull is horizontal).

Step-by-Step Solution

Normal reaction $N=mg=5\times10=50\,N$ (horizontal pull, horizontal surface).

Maximum (kinetic) friction force $f=\mu N=0.2\times50=10\,N$ .

Applied force is $15\,N$ , which exceeds friction, so the block does move and kinetic friction of $10\,N$ opposes it.

Net force $=15-10=5\,N$ .

Acceleration $a=\dfrac{F_{net}}{m}=\dfrac{5}{5}=1\,ms^{-2}$ .

Common Mistakes …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$45°$
›Reveal solutionSolution
This tests force analysis on a wedge system: with the block frictionlessly sliding, the minimum ground friction coefficient needed to keep a wedge stationary comes from balancing the horizontal component of the block's normal reaction against the wedge's total normal reaction from the ground.

Concept and Intuition

Since there's no friction between block and wedge, and we want the minimum $\mu$ for the wedge to just barely stay put, the wedge is exactly in equilibrium (on the verge of sliding) while the block slides freely down the incline. Because the wedge doesn't move, the block's acceleration is purely along the incline surface (it cannot leave the surface), so the usual incline result $N = mg\cos\theta$ holds for the normal contact force between block and wedge. This normal force pushes back on the wedge (Newton's third law) with a horizontal component that friction from the ground must resist, and a vertical component that adds to the wedge's own weight to determine the ground's normal reaction (and hence the maximum available friction).

Step-by-Step Solution

Block on frictionless incline (wedge fixed): along the incline, $mg\sin\theta = ma$ ; perpendicular to the incline (no perpendicular acceleration since motion is confined to the surface), $N = mg\cos\theta$ .

By Newton's third law, the block pushes on the wedge with force $N$ directed into the incline surface. Its horizontal component is $N\sin\theta = mg\sin\theta\cos\theta$ , and its vertical (downward) component is $N\cos\theta = mg\cos^2\theta$ .

Vertical equilibrium of the wedge (mass $2m$ ): ground normal reaction $N_g = 2mg + mg\cos^2\theta$ . …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$\sqrt{3}\ \mathrm{kg}$
›Reveal solutionSolution
This tests force resolution for a push applied at an angle below the horizontal, which increases the normal force and hence the maximum friction available; the answer is (D).

Concept and Intuition

When a force is applied downward-and-forward (pushing into the surface) rather than upward-and-forward (pulling away from it), its vertical component adds to the weight, increasing the normal force and therefore the maximum static friction the surface can supply. This means a larger horizontal force can be resisted before slipping begins — unlike a pulling force, which would reduce the normal force.

Step-by-Step Solution

Resolve $F$ (applied at $60^\circ$ to horizontal, pushing down into the block) into horizontal component $F\cos60^\circ$ and vertical (downward) component $F\sin60^\circ$ .

Vertical equilibrium: normal force $N=mg+F\sin60^\circ$ (since the push adds to the weight pressing the block down).

At the verge of sliding, horizontal driving force equals maximum static friction: $F\cos60^\circ=\mu_s N=\mu_s(mg+F\sin60^\circ)$ .

Substitute values: $m=\sqrt3$ kg, $g=10\ \mathrm{m/s^2}$ , $\mu_s=\dfrac{1}{2\sqrt3}$ , $\cos60^\circ=\frac12$ , $\sin60^\circ=\dfrac{\sqrt3}{2}$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$0.94\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
Since the applied horizontal force (4 N) is smaller than the maximum available static friction (6 N), the block stays at rest and its acceleration is zero.

Concept and Intuition

Static friction is a self-adjusting force — it exactly balances any applied force up to a maximum value $\mu_s N$ ( $N$ = normal reaction). Only once the applied force exceeds this maximum does the body start to slide (and then kinetic friction, generally a bit smaller, takes over).

Step-by-Step Solution

Normal reaction on a horizontal surface: $N = mg = 2 \times 10 = 20$ N.

Maximum static friction available: $f_{s,max} = \mu_s N = 0.3 \times 20 = 6$ N.

Applied force $F = 4$ N $< f_{s,max} = 6$ N.

Since the applied force does not exceed the maximum static friction, static friction simply equals the applied force (4 N) in the opposite direction, keeping the block in equilibrium. …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$7\ \mathrm{m\,s^{-2}}$
›Reveal solutionSolution
This tests Newton's second law with kinetic friction opposing an applied horizontal force: the friction force is whatever is left after accounting for the net force that actually produced the observed acceleration.

Concept and Intuition

When a block accelerates under an applied force with friction present, the friction force isn't found from a formula alone first — it's found by balancing forces: (applied force) − (friction) = (mass)×(acceleration). Once friction's actual magnitude is known, the coefficient of kinetic friction follows from $f=\mu N$ , with $N=mg$ on a horizontal surface.

Step-by-Step Solution

Required net force for the given acceleration: $F_{net}=ma=2\times7=14$ N.

Applied force is $20$ N, so friction opposing the motion is $f=F_{applied}-F_{net}=20-14=6$ N.

Normal force on a horizontal surface: $N=mg=2\times10=20$ N. …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$g = 10\ \mathrm{ms}^{-2}$
›Reveal solutionSolution
The key is to check whether the applied force exceeds the maximum static friction between the blocks. If it does, the blocks slip relative to each other and the upper block accelerates under the net force of the applied force minus kinetic friction. Here, the maximum static friction is 40 N, but F = 250 N, so slipping occurs; the acceleration of block A is 21 m/s², corresponding to option (D).

Concept and Intuition: The Static Friction Limit

When you push the top block, the only horizontal force that can make the bottom block move is friction at the contact surface between the blocks. But friction has a maximum value — the static friction limit — given by $f_{\text{max}} = \mu_s N$ , where $N$ is the normal force between the blocks.

If the applied force $F$ is less than or equal to this limit, both blocks move together as one unit (no slipping).

If $F$ exceeds this limit, the top block slips over the bottom block, and the friction becomes kinetic (which is usually slightly less, but here we treat it as equal to the maximum static friction for simplicity, since the problem gives only $\mu_s$ ).

So the first step is always: Find the maximum static friction and compare it to the applied force.

Step-by-step solution

Find the normal force between the blocks Block A weighs 100 N, so the normal force from block B on block A is equal to its weight (since there is no vertical acceleration):

$N = 100\ \text{N}$

Calculate the maximum static friction

$f_{\text{max}} = \mu_s N = 0.4 \times 100 = 40\ \text{N}$
This is the largest horizontal force that friction can exert on block A (to the left) and on block B (to the right) without slipping.

Compare applied force to $f_{\text{max}}$

The applied force $F = 250\ \text{N}$ is much larger than 40 N. Therefore, the static friction limit is exceeded — the blocks slip relative to each other.

Determine the net force on block A after slipping

Once slipping occurs, the friction force becomes kinetic, but here we assume it remains at the maximum value (since $\mu_k$ is not given, we use $\mu_s$ as the sliding friction).

Applied force to the right: $+250\ \text{N}$

Friction force to the left (opposing motion): $-40\ \text{N}$ …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$\mu = 0.2$
›Reveal solutionSolution
This tests the maximum acceleration a surface can impart to a resting block purely through static friction, before the block starts to slip.

Concept and Intuition

The block has no other horizontal force acting on it except friction from the plank. As the plank accelerates, friction must act on the block to accelerate it along with the plank (Newton's third law pair on the plank). Friction can only supply a force up to its maximum static value $\mu mg$ . Beyond that acceleration, friction cannot keep up, and the block starts sliding backward relative to the plank — it is no longer "in equilibrium" (moving together) with the plank.

Step-by-Step Solution

For the block (mass $m$ ) to accelerate along with the plank at acceleration $a$ , the net horizontal force needed is $ma$ .

The only horizontal force available is static friction, bounded by $f_{max}=\mu m g$ .

At the maximum possible common acceleration, $ma_{max}=\mu m g \Rightarrow a_{max}=\mu g$ . …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$A = 300\text{N}, B = 100\text{N}$
›Reveal solutionSolution
The least horizontal force $P$ that will start motion of any part of the system is determined by comparing the friction limits at each interface; the smallest force that overcomes any one of these limits is $P = 60\,\text{N}$ , which corresponds to slipping between blocks A and B. The correct option is (A).

Concept and Intuition: The Static Friction Limit

When you push block A with a gradually increasing force $P$ , nothing moves until the static friction at some interface is overcome. The key idea is that motion will begin at the interface with the smallest friction limit — because that is the "weakest link." We do not need to consider all blocks moving together; we only need the first slip. So we compute the maximum static friction force at each contact surface (between A & B, B & C, and C & ground) and see which one is smallest. That smallest value is the force $P$ required to start motion somewhere.

Watch out

A common mistake is to assume that all blocks move together, or to add up all friction forces. But the question asks for motion of any part — so only the weakest interface matters.

Step-by-Step Solution

1. Identify the normal forces at each interface.

Since the blocks are stacked vertically, the normal force at each interface equals the total weight above it.

Between A and B: Only block A sits above.

$N_{AB} = W_A = 300\,\text{N}$

Between B and C: Blocks A and B are above.

$N_{BC} = W_A + W_B = 300 + 100 = 400\,\text{N}$

Between C and ground: All three blocks are above.

$N_{Cg} = W_A + W_B + W_C = 300 + 100 + 200 = 600\,\text{N}$
2. Compute the maximum static friction force at each interface.

Use $f_{\text{max}} = \mu N$ .

Between A and B ( $\mu = 0.3$ ):

$f_{AB}^{\text{max}} = 0.3 \times 300 = 90\,\text{N}$

Between B and C ( $\mu = 0.2$ ):

$f_{BC}^{\text{max}} = 0.2 \times 400 = 80\,\text{N}$

Between C and ground ( $\mu = 0.1$ ):

$f_{Cg}^{\text{max}} = 0.1 \times 600 = 60\,\text{N}$
3. Interpret the meaning of each friction limit.

If $P$ exceeds $90\,\text{N}$ , block A will slip relative to B (since the friction between them can no longer hold A).

If $P$ exceeds $80\,\text{N}$ , block B will slip relative to C (but note: this requires that the force is transmitted through A and B).

If $P$ exceeds $60\,\text{N}$ , block C will slip relative to the ground (the entire stack slides as one). …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$10\ \mathrm{ms^{-1}}$
›Reveal solutionSolution
This tests friction on an incline. The constant-velocity descent fixes $\mu=\tan\theta$ ; going up then combines gravity and friction, both decelerating the body, giving a stopping distance of 20 m.

Concept and Intuition

When a body slides down an incline at CONSTANT velocity, the net force is zero, so kinetic friction exactly balances the component of gravity along the incline: $\mu mg\cos\theta = mg\sin\theta \Rightarrow \mu=\tan\theta$ . When the same body is instead pushed UP the incline, both gravity's component and friction act down the slope (opposing the upward motion), so the deceleration is the SUM of the two, not their difference.

Step-by-Step Solution

From constant-velocity descent: $\mu = \tan30° = \dfrac{1}{\sqrt3}$ .

Deceleration while moving up: $a = g\sin\theta + \mu g\cos\theta$ .

Since $\mu g\cos\theta = (\tan\theta)(g\cos\theta) = g\sin\theta$ , we get $a = g\sin\theta+g\sin\theta = 2g\sin\theta$ . …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$R$
›Reveal solutionSolution
The insect can climb until the tangential pull of gravity exceeds the maximum static friction the curved surface can supply; that critical angle fixes the maximum height. Answer: $R\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]$ .

Concept and Intuition

On a curved surface, gravity naturally splits into a component along the surface (trying to slide the insect down) and a component perpendicular to it (balanced by the normal reaction). Friction can only resist up to $\mu N$ ; the highest point reachable is where the insect is on the verge of slipping — friction is exactly maxed out.

Step-by-Step Solution

Let $\theta$ be the angle between the radius to the insect's position and the vertical (radius to the bottom of the bowl). Height above the bottom: $h=R(1-\cos\theta)$ .

Resolve gravity: radial component $mg\cos\theta$ (balanced by normal force, so $N=mg\cos\theta$ ), tangential component $mg\sin\theta$ (tends to slide the insect down, resisted by friction). …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$\mu = 0.05$
›Reveal solutionSolution
The body stays with the oscillating board only if friction can supply the needed force at the point of maximum acceleration; equating $\mu mg$ to $m\omega^2A$ gives $\mu \approx 0.075$ .

Concept and Intuition

For the heavy body to move together with the board (no slipping), the only horizontal force available to accelerate it is friction. Since SHM acceleration is largest at the extreme positions ( $a = \omega^2 x$ , maximum at $x = A$ ), that is the critical instant: if friction is enough to supply $m\omega^2 A$ there, it is enough everywhere else in the cycle too.

Step-by-Step Solution

Angular frequency: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\ \text{rad s}^{-1}$ .

Maximum acceleration of the board (and hence of the body, if it doesn't slip): $a_{max} = \omega^2 A = \left(\dfrac{\pi}{2}\right)^2 \times 0.3 = 2.467 \times 0.3 \approx 0.740\ \text{m/s}^2$ .

The force needed to give the body this acceleration is provided entirely by friction: $f = ma_{max}$ .

For no slipping, this must not exceed the maximum static friction $\mu mg$ : …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$Q. The following is not the method of reducing friction (A) using ball bearings (B) applying grease (C) applying paint (D) forming a thin air cushion$
›Reveal solutionSolution
Ball bearings, grease, and air cushions are all standard, taught methods of reducing friction; applying paint is not — it is done mainly for protection, not friction reduction.

Concept and Intuition

Friction between two surfaces can be reduced by (i) making the surfaces smoother/lubricating them, (ii) replacing sliding friction with rolling friction (ball bearings), or (iii) separating the surfaces entirely with a fluid layer (air cushion, as in hovercraft/air-track). Painting a surface serves a different purpose — corrosion resistance and finish — and is not among the standard physics methods for reducing friction.

Step-by-Step Solution

(A) Using ball bearings: converts sliding friction into much smaller rolling friction — a genuine friction-reduction method.

(B) Applying grease: lubrication reduces the coefficient of friction between contacting surfaces — a genuine method.

(D) Forming a thin air cushion: keeps the surfaces from direct contact, nearly eliminating friction (used in air-cushion vehicles) — a genuine method. …
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Concept and data

Maximum force for common motion

Testing each statement …

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