Unit 3: Laws of Motion & Friction

This unit is the mechanics workhorse of JEE. Free-body-diagram (FBD) and constraint problems show up everywhere — usually 1–2 direct questions in JEE Main, but feeding into nearly every multi-step JEE Advanced problem in mechanics. Expect pulley-string-incline systems, wedge constraints, and "is the block moving?" friction questions every year.

Typical question types:

• JEE Main: a block on an incline with friction (find acceleration / minimum force), Atwood machine (find acceleration & tension), two-block stacked problem (relative slipping).
• JEE Advanced: multi-pulley systems with constraint equations, wedge-block constraint, pseudo-force in accelerating elevator or accelerating wedge, impulsive tension problems.

Concept Map

• Newton's Laws
  • First law (inertia), second law ($\vec F=d\vec p/dt$), third law (action-reaction)
  • Inertial frames, non-inertial frames, pseudo forces
• Free-Body Diagrams
  • Identifying forces on each object: gravity, normal, tension, friction, spring, applied
  • Sign conventions, choosing axes
• Pulley & String Systems
  • Massless string assumption: tension uniform
  • Single fixed pulley, modified Atwood (one mass on table)
  • Multiple-pulley systems & constraint equations
  • Accelerating pulley (pseudo-force treatment)
  • Impulsive tension (sudden jerks)
• Inclined Plane
  • Smooth incline kinematics
  • Friction on incline (sliding vs static)
  • Minimum force to push/pull up; minimum to hold
• Wedge Problems
  • Block on smooth wedge, wedge free to move
  • Constraint between block and wedge motion
• Friction
  • Static & kinetic; coefficient $\mu_s, \mu_k$
  • Angle of friction $\lambda$, angle of repose $\theta_r$
  • Rolling friction (mention)
• Pseudo Forces
  • Linearly accelerating frames
  • Rotating frames hint (centrifugal, Coriolis ideas)

Topic 1: Newton's Laws of Motion

Sub-topic A: The three laws

First law (law of inertia). A body continues in its state of rest or uniform motion in a straight line unless acted on by a net external force. Defines the concept of an inertial frame.

Second law. Rate of change of linear momentum equals the net external force:

$\vec F_{net} = \frac{d\vec p}{dt}.$

For constant mass, $\vec F = m\vec a$.

Third law. Forces between two bodies are equal and opposite, acting on different bodies and along the same line.

Sub-topic B: Linear momentum and impulse

$\vec p = m\vec v$. Impulse $\vec J = \int \vec F\,dt = \Delta\vec p$. For a constant force, $\vec J=\vec F\cdot\Delta t$.

Sub-topic C: Inertial vs non-inertial frames

In an inertial frame, Newton's laws hold as stated. In a frame accelerating with $\vec a_0$, add a pseudo-force $-m\vec a_0$ to every body to recover $\vec F=m\vec a$.

Worked Examples (JEE Main level)

Example 1.1. A $2$ kg block on a smooth horizontal surface is pulled by $10$ N at $30^\circ$ above horizontal. Find acceleration and normal force ($g=10$).

Horizontal: $a = 10\cos30^\circ/2 = 5\sqrt 3/2\approx 4.33$ m/s².

Vertical: $N + 10\sin30^\circ = 20\Rightarrow N=15$ N.

Example 1.2. A ball of mass $0.2$ kg moving at $20$ m/s is stopped in $0.01$ s. Find avg. force.

Impulse $= 0.2\cdot 20 = 4$ N·s. $F = 4/0.01 = 400$ N.

Topic 2: Free-Body Diagrams (FBDs)

Sub-topic A: FBD methodology

1. Identify each body in the system.
2. List all forces acting on it: gravity ($mg$ down), normal from contact ($\perp$ surface), tension along string toward pulley, friction along surface (opposing relative motion or tendency), applied, spring ($-kx$).
3. Choose axes: if there's an inclined or constrained motion, align one axis with motion.
4. Apply $\sum F_x = ma_x$, $\sum F_y = ma_y$.
5. Use constraints (rope inextensibility, surface contact) to relate accelerations of different bodies.

Sub-topic B: Constraint equations

If a string of length $L$ connects two bodies via a pulley, $L = \text{const}$ gives an equation relating their positions. Differentiating twice gives the acceleration constraint.

Example: For a string over a fixed pulley with $m_1$ and $m_2$ hanging, if $m_1$ descends by $x$, $m_2$ rises by $x$. Hence $\vert a_1\vert =\vert a_2\vert$.

For a pulley itself moving with acceleration $a_p$, and a string of fixed length over it: if $a_1, a_2$ are accelerations of the two ends (in ground frame, positive down),

$a_1 + a_2 = 2 a_p.$

(Derivation: let $y_1, y_2$ be positions of the ends measured down from a fixed reference, $y_p$ position of pulley. String length is $(y_1 - y_p) + (y_2 - y_p) = L\Rightarrow y_1 + y_2 - 2 y_p = L\Rightarrow a_1+a_2 = 2a_p$.)

For a string passing over two pulleys, sum of speeds along the string is conserved.

Sub-topic C: Massless string and pulley

For a massless, inextensible string, tension is uniform along the string (and unchanged across an ideal pulley).

Worked Examples (JEE Main level)

Example 2.1. Two masses $m_1=3$ kg, $m_2=2$ kg connected by a string over a smooth pulley. Find $a$ and $T$.

For $m_1$ (heavier, going down): $m_1 g - T = m_1 a\Rightarrow 30 - T = 3a$.

For $m_2$ (going up): $T - m_2 g = m_2 a\Rightarrow T - 20 = 2a$.

Adding: $10 = 5a\Rightarrow a=2$ m/s². $T = 20 + 2\cdot2 = 24$ N.

Example 2.2. A block of $5$ kg on a smooth table is connected by a string over a pulley at the table edge to a hanging mass of $3$ kg. Find $a$ and $T$.

For hanging: $3g - T = 3a$.

For table block: $T = 5a$.

So $30 = 8a\Rightarrow a=3.75$ m/s², $T=18.75$ N.

Worked Examples (JEE Advanced level)

Example 2.A1. A mass $m_1$ hangs from one end of a string over a fixed pulley. The other end goes around a second pulley (movable, mass-less) and is tied to the ceiling. A mass $m_2$ hangs from the movable pulley. Find acceleration of $m_2$ in terms of $m_1, m_2, g$.

Let tension in string = $T$. The movable pulley has two string segments pulling up; for a massless pulley, $2T = m_2(a_2+0)$... but the pulley is massless, so net force on it is zero: $2T - T_{\text{rod}} - 0$ — instead, the rope tension is $T$ throughout (massless string), and the movable pulley is held by two rope-segments giving net upward force $2T$ minus weight of $m_2$ hanging from it. Force eqn on $m_2$: $m_2 g - 2T = m_2 a_2$.

For $m_1$: $m_1 g - T = m_1 a_1$.

Constraint: if $m_1$ goes down by $x_1$ and $m_2$ goes down by $x_2$, the lengths change as $\Delta x_1 + 2\Delta x_2 = 0\Rightarrow a_1 = -2 a_2$ (if $m_2$ descends, $m_1$ rises twice as fast). Sign convention: take downward positive for both. The string goes from $m_1$ up over the fixed pulley, down to and around the movable pulley, up to the ceiling. If movable pulley goes down by $x_2$, the rope on either side lengthens by $x_2$ each, total $2x_2$. The rope on the $m_1$ side must shorten by $2x_2$, so $m_1$ rises by $2x_2$, i.e. $x_1 = -2 x_2$. Hence $a_1 = -2 a_2$.

Substitute $a_1 = -2 a_2$ into $m_1 g - T = m_1 a_1 = -2 m_1 a_2\Rightarrow T = m_1 g + 2 m_1 a_2$.

Into the second equation: $m_2 g - 2(m_1 g + 2m_1 a_2) = m_2 a_2\Rightarrow (m_2 - 2m_1)g = (m_2 + 4 m_1) a_2$.

$a_2 = \frac{(m_2 - 2m_1)g}{m_2 + 4 m_1}.$

Example 2.A2 (Accelerating pulley — pseudo force). A pulley is moving up with acceleration $a_p$. Masses $m_1, m_2$ are connected over it. Find their accelerations in the ground frame.

In the pulley frame, add pseudo-force $m_i a_p$ down to each mass. Effective gravity in this frame is $g+a_p$. Acceleration in pulley frame: $a' = (m_1-m_2)(g+a_p)/(m_1+m_2)$. In ground frame: $a_1 = a_p + a'$, $a_2 = a_p - a'$ (for $m_1$ descending in pulley frame, $m_2$ ascending).

Topic 3: Inclined Plane

Sub-topic A: Smooth incline

A block of mass $m$ on a smooth incline of angle $\theta$. Resolve gravity:

• Along incline (down-slope): $mg\sin\theta$.
• Perpendicular (into incline): $mg\cos\theta$.

Normal force $N = mg\cos\theta$. Acceleration down the incline: $a = g\sin\theta$ (independent of mass).

Time to slide down length $L$ from rest: $L = \tfrac12 g\sin\theta\,t^2\Rightarrow t = \sqrt{2L/(g\sin\theta)}$. Speed at bottom: $v=\sqrt{2gL\sin\theta}=\sqrt{2gh}$ (consistent with energy conservation).

Sub-topic B: Friction on incline

If coefficient of static friction is $\mu_s$, the block remains at rest provided $\tan\theta\le\mu_s$. The critical angle, angle of repose, is

$\theta_r = \tan^{-1}\mu_s.$

If sliding occurs, kinetic friction acts up the incline: net acceleration

$a = g(\sin\theta - \mu_k\cos\theta)\quad(\text{down the incline}).$

If the block moves up under an external force, friction acts down:

$a = g(-\sin\theta - \mu_k\cos\theta) + F_{ext}/m.$

Sub-topic C: Minimum force on a block on incline

A block of mass $m$ on incline of angle $\theta$, $\mu$ between block and incline, is pulled up by force $F$ at angle $\phi$ above the incline. Minimum $F$ to just move:

Resolve: along incline, $F\cos\phi - mg\sin\theta - \mu N = 0$. Perpendicular: $N = mg\cos\theta - F\sin\phi$.

Combine: $F\cos\phi - mg\sin\theta - \mu(mg\cos\theta - F\sin\phi) = 0\Rightarrow F(\cos\phi+\mu\sin\phi) = mg(\sin\theta + \mu\cos\theta)$.

$F = \frac{mg(\sin\theta + \mu\cos\theta)}{\cos\phi + \mu\sin\phi}.$

Minimum w.r.t. $\phi$ (using $\mu=\tan\lambda$, $\lambda$ = angle of friction):

$\boxed{F_{\min} = mg\sin(\theta+\lambda)\quad\text{at}\quad \phi = \lambda.}$

(So pull at the angle of friction above the incline for minimum push.)

Worked Examples (JEE Main level)

Example 3.1. A block slides down a smooth incline of $30^\circ$ from rest. Speed after sliding $5$ m?

$a = g\sin30^\circ = 5$ m/s². $v=\sqrt{2\cdot5\cdot5}=\sqrt{50}\approx 7.07$ m/s.

Example 3.2. Same incline but $\mu_k = 0.2$. Acceleration?

$a = g(\sin30 - 0.2\cos30) = 10(0.5 - 0.2\cdot 0.866) = 10\cdot 0.327 = 3.27$ m/s².

Example 3.3. Angle of repose for $\mu_s=1/\sqrt 3$?

$\theta_r = \tan^{-1}(1/\sqrt 3) = 30^\circ$.

Worked Examples (JEE Advanced level)

Example 3.A1. A $1$ kg block is placed on a $30^\circ$ incline; $\mu = 0.5$. Find the minimum horizontal force (parallel to ground) required to hold the block stationary, and the minimum to push it up.

Resolve along and perp. to incline. Force $F$ horizontal; gravity $mg$ vertical.

Along incline (up positive): $F\cos\theta - mg\sin\theta + f = 0$ (to hold; $f$ can act up or down up to $\mu N$).

Perp.: $N = mg\cos\theta + F\sin\theta$.

For the block on the verge of sliding down, friction acts up the incline at max, $f = +\mu N$:

$F\cos\theta - mg\sin\theta + \mu(mg\cos\theta + F\sin\theta) = 0$.

$F(\cos\theta + \mu\sin\theta) = mg(\sin\theta - \mu\cos\theta)$.

$F_{\min,\text{hold}} = \dfrac{mg(\sin\theta-\mu\cos\theta)}{\cos\theta+\mu\sin\theta} = \dfrac{mg\tan(\theta-\lambda)}{1}$ (after dividing by $\cos\theta\cos\lambda$).

With $\theta=30^\circ$, $\mu=0.5\Rightarrow \lambda=\tan^{-1}0.5\approx 26.57^\circ$. $\theta-\lambda \approx 3.43^\circ$. $F_{\min,\text{hold}} \approx 1\cdot 10\cdot \tan 3.43^\circ \approx 0.6$ N.

For pushing up (verge of moving up), friction acts down:

$F_{\min,\text{up}} = mg\tan(\theta+\lambda) = 10\tan 56.57^\circ \approx 15.1$ N.

Example 3.A2 (Wedge problem). A block of mass $m$ rests on a smooth wedge of mass $M$, angle $\theta$. The wedge is free to move on a smooth floor. Find acceleration of the wedge.

Take ground frame. Let wedge accelerate with $A$ (to the left, say). Block has acceleration $\vec a_b$. Constraint: the block remains on the wedge surface — its acceleration relative to the wedge is along the incline.

Let $A$ = wedge acceleration (to the right positive), $a_{rel}$ = block's acceleration along incline (downhill positive in wedge frame).

In wedge frame (non-inertial): forces on block — gravity $mg$ down, normal $N$ perpendicular to incline, pseudo-force $-mA$ (to the left). Resolve perp. to incline:

$N = mg\cos\theta - mA\sin\theta$ (if $A$ is to the right, pseudo on block is to the left, component into incline is $-mA\sin\theta$... be careful with signs).

Actually easier: in ground frame, write FBDs. Block: gravity $-mg\hat\jmath$, normal $N$ at angle $\theta$ from vertical → components $-N\sin\theta\hat\imath$ (toward wedge, i.e. to the left if wedge is to the right of block? Set the wedge's incline rising to the right, block slides down-left. Let's place the wedge with its right-angle at the right; the incline rises from bottom-right to top-left. Then block on the incline slides down-right. Normal points up and to the right at angle $\theta$ from vertical. Components: $N_x = N\sin\theta$, $N_y = N\cos\theta$.

Wedge has gravity $Mg$, normal from floor $N'$, reaction from block $-N$ (down-left): horizontal component $-N\sin\theta$ on wedge (to the left).

Equation of motion of wedge: $-N\sin\theta = M A\Rightarrow A = -N\sin\theta/M$. (Wedge accelerates left.)

Block: $N\sin\theta = m a_{bx}$ and $N\cos\theta - mg = m a_{by}$.

Constraint (block stays on incline, which moves left with $A$): $(a_{bx} - A)/a_{by} = \cot\theta$... no, the slope of incline in wedge frame is $\tan\theta$ downhill to the right. So $(a_{by})/(a_{bx} - A) = -\tan\theta$ (block goes right and down relative to wedge).

Solving (skipping algebra):

$A = \frac{-m g\sin\theta\cos\theta}{M + m\sin^2\theta}\quad\text{(wedge accelerates opposite to block's horizontal motion)}.$

Magnitude: $\vert A\vert = \dfrac{m g\sin\theta\cos\theta}{M+m\sin^2\theta}$.

Block's horizontal acceleration:

$a_{bx} = \frac{Mg\sin\theta\cos\theta}{M+m\sin^2\theta},\quad a_{by}=-\frac{(M+m)g\sin^2\theta}{M+m\sin^2\theta}.$

(Standard JEE Advanced result.)

Topic 4: Friction

Sub-topic A: Static vs kinetic

• Static friction $f_s\le \mu_s N$ acts to prevent relative motion. It self-adjusts up to the maximum $\mu_s N$.
• Kinetic friction $f_k = \mu_k N$ acts when there is relative sliding, opposite to relative velocity.
• Generally $\mu_k < \mu_s$.

Sub-topic B: Angle of friction & angle of repose

Angle of friction $\lambda$: $\tan\lambda = \mu_s$. The resultant of $N$ and the maximum static friction makes angle $\lambda$ with $N$.

Angle of repose $\theta_r$: the maximum incline angle at which a block stays at rest under its own weight. $\tan\theta_r = \mu_s$. So $\theta_r = \lambda$.

Sub-topic C: Rolling friction

When a wheel rolls, the contact is not slipping but the wheel deforms slightly, producing a small friction $f_r\ll f_k$. Often ignored at JEE level except in conceptual questions.

Sub-topic D: Two-block friction problem

Block $A$ (mass $m_1$) on top of block $B$ (mass $m_2$) on smooth floor. $\mu$ between $A$ and $B$. Force $F$ applied to $B$ (horizontal). Find max $F$ for which they move together.

If they move together with acceleration $a$, $a=F/(m_1+m_2)$. The only horizontal force on $A$ is friction from $B$: $f = m_1 a\le \mu m_1 g\Rightarrow a\le \mu g\Rightarrow F\le \mu g(m_1+m_2)$.

If $F$ is applied to $A$ instead, similar analysis: friction on $B$ from $A$ provides $B$'s acceleration. They move together if $F/(m_1+m_2)\le \mu g$ requires $F\le \mu g(m_1+m_2) \cdot(m_2/m_1)$... reconsider: for $A$ pulled with $F$, sliding starts when friction between $A,B$ equals $\mu m_1 g$. Friction on $B$ = $\mu m_1 g$ at most, so $a_{B,\max} = \mu m_1 g/m_2$. For together, $a_A = F/(m_1+m_2)... actually for together (same$a$),$F - \mu m_1 g = m_1 a$,$\mu m_1 g = m_2 a$. Eliminate$a$: from 2nd,$a = \mu m_1 g/m_2$. Then$F = m_1\cdot\mu m_1 g/m_2 + \mu m_1 g = \mu m_1 g(m_1+m_2)/m_2$.

So max-without-slip $F = \mu m_1 g(m_1+m_2)/m_2$.

Worked Examples (JEE Main level)

Example 4.1. A $10$ kg block on a horizontal surface, $\mu_s = 0.4$. Min horizontal force to move?

$F_{\min} = \mu_s mg = 0.4\cdot10\cdot10 = 40$ N.

Example 4.2. A block on a $45^\circ$ incline just slides ($\mu_s$ unknown). Find $\mu_s$.

$\theta_r=45^\circ\Rightarrow \mu_s=\tan45^\circ=1$.

Example 4.3. A $2$ kg block on horizontal floor, $\mu=0.3$. A force $F=10$ N pulls at $30^\circ$ above horizontal. Friction force and acceleration?

$N = mg - F\sin30^\circ = 20 - 5 = 15$ N. $f_{\max} = 0.3\cdot 15 = 4.5$ N. Horizontal applied $= F\cos30^\circ \approx 8.66$ N $> 4.5$ N, so block slides; $f = 4.5$ N kinetic (assume $\mu_k=\mu_s$). $a = (8.66-4.5)/2 = 2.08$ m/s².

Worked Examples (JEE Advanced level)

Example 4.A1. A $5$ kg block on a horizontal floor ($\mu = 0.25$) is pulled by a string at angle $\theta$ with horizontal. Find $\theta$ for which the required force is minimum, and the value.

$F = \mu mg/(\cos\theta + \mu\sin\theta)$. Minimum when $\tan\theta = \mu = 0.25\Rightarrow \theta = \tan^{-1}(0.25)\approx 14^\circ$. $F_{\min} = \mu mg/\sqrt{1+\mu^2} = 0.25\cdot 50/\sqrt{1.0625}\approx 12.12$ N.

Example 4.A2. A book is pressed against a vertical wall with horizontal force $F$. The wall–book coefficient is $\mu_s$. Find min $F$ so the book doesn't slide.

Vertical equilibrium: $f_s = mg$. Also $f_s\le \mu_s N$ and $N = F$. So $F\ge mg/\mu_s$.

Topic 5: Pseudo Forces

Sub-topic A: In a linearly accelerating frame

If the frame accelerates with $\vec a_0$ (w.r.t. ground), in this frame Newton's law becomes

$\vec F_{\text{real}} + \vec F_{\text{pseudo}} = m\vec a',\quad \vec F_{\text{pseudo}} = -m\vec a_0.$

Common applications: man in elevator, block on accelerating wedge, train carriage.

Sub-topic B: Elevator problems

A person of mass $m$ in elevator accelerating up with $a$:

• Apparent weight (normal from floor) $N = m(g+a)$.
• If accelerating down with $a$: $N = m(g-a)$.
• Free-fall ($a=g$): weightless.

Sub-topic C: Rotating frame — hint

In a frame rotating with angular velocity $\vec\omega$, there are two pseudo-forces:

• Centrifugal: $-m\vec\omega\times(\vec\omega\times\vec r) = m\omega^2 r\,\hat r$ (outward, radial).
• Coriolis: $-2m\vec\omega\times\vec v'$ (depends on velocity in the rotating frame).

JEE Advanced occasionally tests centrifugal force conceptually; Coriolis is mostly Olympiad-level.

Worked Examples (JEE Main level)

Example 5.1. A man stands on a balance in a lift. The reading is $720$ N when the lift accelerates up at $2$ m/s². His mass?

$720 = m(10+2)\Rightarrow m = 60$ kg.

Example 5.2. A pendulum hangs in a car accelerating at $a$ horizontally. Find the angle of inclination.

$\tan\theta = a/g$.

Worked Examples (JEE Advanced level)

Example 5.A1. A block of mass $m$ rests on a smooth wedge of angle $\theta$. What horizontal acceleration $a$ of the wedge (toward the incline base) will keep the block stationary on the wedge?

In wedge frame, pseudo $-ma$ horizontal. Block in equilibrium under gravity, normal, and pseudo. Along the incline: $-mg\sin\theta + ma\cos\theta = 0\Rightarrow a = g\tan\theta$.

Example 5.A2. A bead is at the bottom of a smooth hemispherical bowl of radius $R$ rotating about its vertical axis with angular speed $\omega$. Find the height $h$ at which the bead settles.

In rotating frame, centrifugal force $m\omega^2 r$ outward (r = horizontal distance from axis). For equilibrium on bowl: tangent to bowl makes angle $\theta$ with horizontal where $\tan\theta = m\omega^2 r/(mg)$. At polar angle $\phi$ from bottom, $r=R\sin\phi$, and the tangent angle equals $\phi$ (geometry of sphere). So $\tan\phi = \omega^2 R\sin\phi/g\Rightarrow \cos\phi = g/(\omega^2 R)$. Height above the bottom: $h = R(1-\cos\phi) = R - g/\omega^2$. Valid when $\omega^2 R\ge g$.

Topic 6: Spring & Impulsive Tension Problems

Sub-topic A: Spring forces

Spring force on mass attached at one end: $F = -k(\text{extension or compression})$. A spring with natural length $\ell_0$ stretched to $\ell$ has force $k(\ell-\ell_0)$ on each end, directed along the spring.

Sub-topic B: Impulsive (sudden) tension

When a string is suddenly jerked (e.g. tightens from slack), tension can be impulsive (infinite force in zero time, finite impulse). Use impulse-momentum: $J = T\,\Delta t$ remains finite. Common in problems where one mass falls and yanks another via a string.

Sub-topic C: When a string breaks

Just after a string in a system breaks, the system's accelerations change discontinuously, but velocities are continuous. To find accelerations just after, write the new FBD with $T=0$ on the broken side.

When a spring is cut (instead of string), the spring force vanishes immediately too, but if the spring was previously stretched, the mass connected on the other side keeps the same instantaneous force until the spring relaxes — actually since spring forces depend on length, if you cut a spring, force on both attached objects becomes zero instantly. (Trap: in a problem with mass connected to ceiling by a spring and a string from above, when the string is cut, the spring still pulls — different from cutting the spring.)

Worked Examples (JEE Main level)

Example 6.1. A mass $m$ hangs in equilibrium from a spring of constant $k$. Find extension.

$kx = mg\Rightarrow x = mg/k$.

Example 6.2. Two blocks $m_1, m_2$ on a smooth table connected by a spring. $F$ applied to $m_2$. In steady state (when both have same acceleration), find spring force.

$a = F/(m_1+m_2)$. Spring force on $m_1$: $F_s = m_1 a = m_1 F/(m_1+m_2)$.

Worked Examples (JEE Advanced level)

Example 6.A1. A block of mass $m$ hangs at rest from a spring (constant $k$). A second mass $m$ is suddenly attached to the first. Describe the subsequent motion.

Before: spring extension $x_0 = mg/k$. After attaching, equilibrium extension would be $2mg/k$. The combined mass starts at $x_0$ with zero velocity; it executes SHM about the new equilibrium with amplitude $\vert x_0 - 2mg/k\vert = mg/k$. Period $T = 2\pi\sqrt{2m/k}$.

Example 6.A2. Two equal masses are connected by a string over a smooth pulley. One mass also has a spring (constant $k$) attached below it. Initially, the spring is unstretched and the system is held. Find acceleration of the masses just after release and the maximum extension.

Just after release: spring force zero, system behaves like normal Atwood with equal masses $\Rightarrow a = 0$. Hmm — actually need details. If the spring is attached only to the lower mass (anchored to floor?), spring exerts no force initially, system has zero net force. So the masses sit still.

(This problem is under-specified; in JEE Advanced it would have additional geometry. The lesson: read the problem carefully for spring connections.)

Problem-Solving Heuristics

1. Draw clear FBDs: one per body. Label every force with its source.
2. Pick a single, sensible coordinate system for each body — usually aligned with the surface it lies on.
3. Constraint equations are essential for connected bodies — never assume two bodies have the same acceleration unless rigidly connected.
4. Use the ground frame by default; switch to a non-inertial frame only when it makes the problem (e.g. accelerating wedge) easier — and remember to add pseudo-forces.
5. For friction: first check if the body is moving (kinetic) or on the verge (static at max). Don't blindly apply $\mu N$ as the actual friction — that's only the maximum.
6. For incline + friction problems, the angle of friction $\lambda$ makes formulas elegant: $F_{\min,\text{up}}=mg\sin(\theta+\lambda)$.
7. Impulse simplifies sudden-force problems: $\int F\,dt = \Delta p$.
8. Just after release / just after a string is cut: velocity is continuous; acceleration usually jumps. Spring force is continuous (depends on length); tension is not (depends on whether string is taut).
9. Pulleys with mass require torque equation; in JEE Main/Adv unless told otherwise, assume massless.

Common Traps & Mistakes

• Assuming friction is always at $\mu N$: only true when on the verge or sliding.
• Forgetting that normal force changes when an external force has a vertical component (e.g. pulling at angle reduces $N$ — and hence friction).
• Pseudo-force direction: it's $-m\vec a_0$, opposite to the frame's acceleration.
• In a pulley problem with the pulley accelerating, $a_1 + a_2 = 2 a_p$ (signs!), not $a_1 = a_2$.
• For two-block stacked problems: when force is applied to top vs bottom, the max no-slip force differs.
• Sudden cutting of string vs spring: string force vanishes instantly; spring force is continuous in time (since length is continuous).
• Angle of friction $\lambda$ vs angle of repose $\theta_r$: numerically equal for the same $\mu_s$, but conceptually different — $\lambda$ is about the contact, $\theta_r$ is the maximum tilt before sliding.
• Confusing action-reaction with two forces on the same body. Newton's third law involves two different bodies.
• In Atwood, the tension is NOT $(m_1+m_2)g/2$ — that's only when the system is massless/static. Use both FBDs.

Quick Revision Card

• $\vec F = m\vec a$; constraints relate accelerations.
• Atwood ($m_1>m_2$): $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2 g/(m_1+m_2)$.
• Smooth incline: $a = g\sin\theta$, $N = mg\cos\theta$.
• Rough incline (sliding down): $a = g(\sin\theta - \mu\cos\theta)$.
• Min force to push up incline: $F_{\min} = mg\sin(\theta+\lambda)$ at angle $\lambda$ above incline.
• Angle of repose: $\theta_r = \tan^{-1}\mu_s$.
• Pseudo force: $-m\vec a_0$ in a frame accelerating with $\vec a_0$.
• Wedge-block: $A_{\text{wedge}} = -m g\sin\theta\cos\theta/(M+m\sin^2\theta)$.
• Pulley accelerating up with $a_p$: effective $g_{eff} = g+a_p$.

Formula Sheet