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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Concept and data

Maximum force for common motion

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