Compiled from: NCERT Physics Part-1 (Laws of Motion) • H.C. Verma — Concepts of Physics Vol-1 (Ch. 5 & 6: Newton's Laws, Friction) • D.C. Pandey — Understanding Physics: Mechanics Part-1 (Laws of Motion, Friction, Circular Motion) • Pradeep's Fundamental Physics XI • S.L. Arora • Cengage (B.M. Sharma) Mechanics-I • Errorless Physics • I.E. Irodov (selected). Target exams: Boards • JEE Main • JEE Advanced • NEET • NDA.
Kinematics described motion; this chapter explains its cause — force. A force is a push or pull that can change the state of motion (or shape) of a body. Force is a vector; SI unit newton (N); dimension $[MLT^{-2}]$; 1 N = 10⁵ dyne; 1 kgf (kg-wt) = 9.8 N.
Aristotle claimed a force is needed to keep a body moving. This is wrong: a moving cart stops only because friction opposes it. If all opposing forces are removed, a body in motion keeps moving forever with the same velocity. Galileo established this with his inclined-plane thought experiment: a ball rolling down one incline rises to the same height on another; as the second incline is flattened, the ball travels farther and farther — on a frictionless horizontal plane it would never stop. Force is required to change motion, not to maintain it.
"State Aristotle's fallacy" / "Describe Galileo's experiment" are standard 2–3 mark questions (Pradeep & SL Arora highlight both). Key phrase for full marks: "an external force is required to change the state of motion, not to maintain uniform motion."
Statement: Every body continues in its state of rest or of uniform motion in a straight line unless compelled by an external unbalanced force to change that state. It is also called the law of inertia and it defines force qualitatively and defines inertial frames (frames where the first law holds; accelerating frames are non-inertial and need pseudo-forces).
Inertia = the inherent inability of a body to change its state by itself. Mass is the measure of inertia — heavier bodies resist change more.
| Type | Meaning | Classic Examples (asked verbatim in Boards) |
|---|---|---|
| Inertia of rest | Body at rest resists moving | Dust leaves a beaten carpet; passenger jerks backward when a bus starts suddenly; fruit falls when a branch is shaken; coin drops into glass when card is flicked |
| Inertia of motion | Moving body resists stopping | Passenger falls forward when a bus brakes; a runner runs some distance past the finish line; one jumps off a moving bus by running forward |
| Inertia of direction | Body resists change of direction | Passenger thrown outward at a sharp turn; mud from a spinning wheel flies off tangentially; sparks from a grinding wheel fly tangentially |
Vector, along velocity. SI unit kg·m/s; dimension $[MLT^{-1}]$. Relations with kinetic energy: $p = \sqrt{2mK}$ and $K = \dfrac{p^2}{2m}$ — for the same p, the lighter body has more KE; for the same K, the heavier body has more momentum (a NEET/NDA staple).
Statement: The rate of change of linear momentum of a body is directly proportional to the applied external force and takes place in the direction of the force.
$$\vec{F} \propto \frac{d\vec{p}}{dt} \;\Rightarrow\; \vec{F} = k\frac{d\vec{p}}{dt}$$Derivation. With constant mass:
$$\vec{F} = k\frac{d(m\vec{v})}{dt} = km\frac{d\vec{v}}{dt} = km\vec{a}$$The newton is defined so that k = 1 (1 N gives 1 kg an acceleration of 1 m/s²). Hence $\boxed{\vec{F} = m\vec{a} = \dfrac{d\vec{p}}{dt}}$. The second law is the real law of motion — the first law is contained in it (F = 0 ⟹ a = 0) and, with the third law, it gives momentum conservation.
Impulse = area under the F–t graph. Unit N·s = kg·m/s.
For a fixed Δp, force is large when time is short and small when time is long. Applications (Boards 3-markers): a cricketer lowers his hands while catching (increases Δt → smaller force); shockers in vehicles; jumping on sand vs concrete; china packed in straw; a person falls on a heap of hay unhurt. Conversely a karate chop shortens Δt to maximise force.
Q. A 150 g ball hits a wall normally at 20 m/s and rebounds at 20 m/s. Contact time is 0.01 s. Find the impulse and the average force on the ball.
Q. A force $F = (6t)\,$N acts on a 2 kg body at rest. Find its velocity at t = 2 s.
Statement: To every action there is always an equal and opposite reaction: $\vec{F}_{AB} = -\vec{F}_{BA}$. Forces always occur in pairs, are of the same nature, act simultaneously (no cause–effect delay), and — most importantly — act on different bodies, so they never cancel each other.
Statement: If the net external force on a system is zero, its total linear momentum remains constant.
Proof from Newton's laws. For two colliding bodies, during contact time Δt the third law gives $\vec{F}_{12} = -\vec{F}_{21}$. Impulses: $\Delta\vec{p}_1 = \vec{F}_{21}\Delta t$ and $\Delta\vec{p}_2 = \vec{F}_{12}\Delta t$. Adding:
$$\Delta\vec{p}_1 + \Delta\vec{p}_2 = 0 \;\Rightarrow\; m_1\vec{u}_1 + m_2\vec{u}_2 = m_1\vec{v}_1 + m_2\vec{v}_2$$Recoil speed of gun (mass M) firing a bullet (mass m, speed v); the − sign shows the backward direction. Rocket: exhaust gas thrown back pushes the rocket forward; thrust $= u\dfrac{dm}{dt}$ where u = exhaust speed relative to the rocket.
Q. A 5 kg gun fires a 20 g bullet at 500 m/s. Find the recoil speed of the gun.
| Force | Direction rule |
|---|---|
| Weight (mg) | Always vertically downward, at the centre of gravity |
| Normal reaction (N) | Perpendicular to the contact surface, pushing the body away from the surface (never pulls) |
| Tension (T) | Along the string, always pulling the body toward the string; same throughout a massless, frictionless string/pulley |
| Friction (f) | Along the surface, opposing relative motion (or its tendency) |
| Spring force | $F = -kx$ — opposes deformation; a spring can push or pull |
Equilibrium: $\sum\vec{F} = 0$ ⟹ a body at rest stays at rest, a moving body keeps constant velocity. For three concurrent forces in equilibrium, Lami's theorem: $\dfrac{F_1}{\sin\alpha} = \dfrac{F_2}{\sin\beta} = \dfrac{F_3}{\sin\gamma}$ (each force ÷ sine of the angle between the other two).
FBD gallery: block on table, hanging block, block on an incline
The three primitive FBDs of Section 6: every hard problem is assembled from these. On the incline, weight splits into mg sin θ (along) and mg cos θ (into the slope).
Q. A 6 kg lamp hangs from the ceiling by two strings making 30° and 60° with the ceiling. Find the tensions. (g = 10 m/s²)
The weighing machine (or the normal force N on the person) reads the apparent weight. Write Newton's second law on the person: taking up as positive, $N - mg = ma$.
| Lift's motion | Apparent weight N | Feeling |
|---|---|---|
| At rest / uniform velocity (up or down) | $N = mg$ | Normal — velocity does not matter, only acceleration! |
| Accelerating up (or decelerating while going down) with a | $N = m(g + a)$ | Heavier |
| Accelerating down (or decelerating while going up) with a | $N = m(g - a)$ | Lighter |
| Free fall (cable breaks, a = g) | $N = 0$ | Weightlessness (same physics as an orbiting satellite) |
| Downward acceleration a > g | Person leaves the floor | Contact only with the ceiling can push down |
Apparent weight in a lift — the three cases
Compare the green N arrows (drawn beside the passenger, clear of the walls): longer than mg going up, shorter coming down, zero in free fall.
Q. A 60 kg man stands on a weighing machine in a lift. Find the reading (in kgf) when the lift (a) moves up at constant 2 m/s, (b) accelerates up at 2 m/s², (c) accelerates down at 2 m/s². (g = 10 m/s²)
When several bodies move together with the same acceleration, first treat them as ONE system: $a = \dfrac{\text{net external force along motion}}{\text{total mass}}$. Then isolate a single body with an FBD to find internal forces (tension, contact force). Internal forces cancel in pairs for the system — never include them in the system equation.
Force F pushes block m₁ which pushes block m₂ on a smooth floor:
$$a = \frac{F}{m_1 + m_2}, \qquad \text{contact force on } m_2:\ N_{12} = m_2a = \frac{m_2F}{m_1+m_2}$$If instead F is applied on m₂ (from the other side), the contact force becomes $\dfrac{m_1F}{m_1+m_2}$ — the contact force always equals (mass being pushed ahead) × a. Same idea for three blocks: contact force between the 2nd and 3rd = $\dfrac{m_3F}{m_1+m_2+m_3}$.
F pulls m₁; a string connects m₁ to m₂ behind it: $a = \dfrac{F}{m_1+m_2}$ and string tension $T = m_2a = \dfrac{m_2F}{m_1+m_2}$ (tension drags only the trailing mass). For three masses in a chain pulled by F, the tension in each string equals (total mass behind that string) × a.
Setup: A uniform rope of mass M and length L is pulled on a smooth floor by force F at its front end. Acceleration of every part: $a = F/M$.
Tension at distance x from the REAR end: that cross-section drags the rear part of mass $\dfrac{M}{L}x$:
$$T(x) = \frac{M}{L}x \cdot \frac{F}{M} = \frac{F\,x}{L}$$So T = F at the pulled end, T = 0 at the free end, varying linearly. A hanging heavy rope similarly has $T(x) = \dfrac{M g x}{L}$ measured from the bottom. Only for a massless rope is tension the same everywhere — the assumption behind every "ideal string" problem.
Contact blocks, string-connected blocks, and the heavy rope with T(x) = Fx/L
Green tension arrows shrink toward the free end of the heavy rope; the inset graph shows the exact linear law.
Q. Blocks of 1 kg, 2 kg and 3 kg on a smooth floor are connected by strings and pulled by F = 12 N applied to the 3 kg block. Find the acceleration and both tensions (1 kg is at the rear).
Ideal pulley assumptions: pulley massless and frictionless, string massless and inextensible ⟹ tension is the same throughout the string and both connected bodies have the same magnitude of acceleration.
Setup: masses m₁ > m₂ hang on either side of a fixed frictionless pulley. m₁ goes down, m₂ goes up, both with acceleration a; tension T is common.
Newton's second law on each mass:
$$m_1:\quad m_1g - T = m_1a$$ $$m_2:\quad T - m_2g = m_2a$$Add (T cancels):
Force on the pulley's support (clamp) $= 2T = \dfrac{4m_1m_2g}{m_1+m_2}$ — less than $(m_1+m_2)g$ whenever the system accelerates (a slick JEE one-liner). Note T is the harmonic-mean-type combination: $T$ lies between $m_2g$ and $m_1g$.
Atwood machine with complete FBDs of both masses
Now the strings pay out and reel in with the blocks: the left string lengthens exactly as fast as the right one shortens, while the pulley turns through the matching arc — inextensibility made visible.
Setup: block m₁ on a smooth horizontal table, string over a light pulley at the edge, block m₂ hanging vertically.
$$m_2:\ m_2g - T = m_2a \qquad m_1:\ T = m_1a$$With table friction (coefficient μ): $a = \dfrac{(m_2 - \mu m_1)g}{m_1+m_2}$ (motion only if $m_2 > \mu m_1$). Force on the pulley $= \sqrt{T^2 + T^2} = T\sqrt2$ at 45° (the two string segments are perpendicular) — a classic JEE finish.
Table-edge pulley: M slides as m falls, and the pulley feels T√2
The horizontal string shortens on the table side exactly as the vertical segment lengthens below — one string, one tension, one acceleration. The T√2 construction sits in its own clear panel on the right.
Setup: m₁ on a smooth incline of angle θ, string over a pulley at the top, m₂ hanging vertically. If m₂g > m₁g sin θ, m₂ descends:
$$m_2g - T = m_2a \qquad T - m_1g\sin\theta = m_1a$$Equilibrium condition: $m_2 = m_1\sin\theta$. For a double incline (masses on two smooth slopes θ₁, θ₂ joined over the top pulley): $a = \dfrac{(m_1\sin\theta_1 - m_2\sin\theta_2)g}{m_1+m_2}$.
Incline–pulley system with the slope mass fully resolved
All four forces on m₁ are drawn at its centre; the boxed results now render as proper fractions.
Q. Masses 5 kg and 3 kg hang from a light frictionless pulley (g = 10 m/s²). Find the acceleration, the tension, and the force on the pulley's clamp.
Q. A monkey of mass 20 kg climbs a rope that can bear a maximum tension of 250 N. With what maximum acceleration can it climb up safely? What happens if it slides down with acceleration g? (g = 10 m/s²)
Friction is the tangential contact force opposing relative motion (or its tendency) between two surfaces. Cause: interlocking of surface irregularities and inter-molecular (adhesive) forces at contact points.
| Type | Nature | Formula |
|---|---|---|
| Static friction f_s | Self-adjusting: equals the applied force, up to a limit; acts before sliding starts | $f_s \le \mu_sN$ (inequality!) |
| Limiting friction | Maximum static friction, at the verge of sliding | $f_{s(max)} = \mu_sN$ |
| Kinetic (sliding) friction f_k | Constant once sliding starts; independent of speed (approximately) and of contact area | $f_k = \mu_kN$, with $\mu_k < \mu_s$ |
| Rolling friction | Opposes rolling; far smaller than sliding friction (that's why wheels and ball-bearings exist) | $f_r \ll f_k$ |
Plot friction f against applied force F: f rises along the line f = F (45°, self-adjusting static region), peaks at the limiting value μ_sN, then drops to the constant kinetic value μ_kN once sliding begins. The little drop at the peak is the signature — options without the drop are wrong. Also memorise: friction is independent of the apparent area of contact; it depends only on N and the nature of the surfaces.
Angle of friction (λ): the angle that the resultant of limiting friction and normal reaction makes with the normal:
$$\tan\lambda = \frac{f_{s(max)}}{N} = \mu_s$$Angle of repose (α): the maximum incline angle at which a block just rests without sliding. On the verge: $mg\sin\alpha = \mu_smg\cos\alpha$:
$$\boxed{\tan\alpha = \mu_s} \qquad\Rightarrow\qquad \alpha = \lambda$$The angle of repose equals the angle of friction — state and prove this for an easy 3-marker.
| Case | Friction direction | Result (a or F) |
|---|---|---|
| Sliding down (θ > α) | Up the slope | $a = g(\sin\theta - \mu_k\cos\theta)$ |
| Projected up the slope | Down the slope (both gravity & friction retard) | $a_{retard} = g(\sin\theta + \mu_k\cos\theta)$ |
| Force needed to just push up at constant speed | Down the slope | $F = mg(\sin\theta + \mu\cos\theta)$ |
| Force needed to just prevent sliding down | Up the slope | $F = mg(\sin\theta - \mu\cos\theta)$ |
| At rest with θ < α | Up the slope, f = mg sin θ (NOT μN!) | Static friction is only as large as needed |
If a block RESTS on a 20° incline, the friction on it is $mg \sin 20°$, not $\mu mg\cos20°$. Static friction takes only the value required for equilibrium. Writing μN for a resting body is the most common lost mark in this chapter.
Friction master figure: the f–F characteristic (traced live) and the angle of repose
The orange dot rides the 45° static line, snaps down at breakaway (μₖ < μₛ), then cruises on the flat kinetic branch — the whole friction story in one loop.
Q. A 10 kg block rests on a floor with μₛ = 0.4, μₖ = 0.3 (g = 10 m/s²). Find the friction force when the applied horizontal force is (a) 30 N, (b) 40 N, (c) 50 N, and the acceleration in case (c).
Q. A block slides down a 45° incline with μₖ = 0.5 (g = 10 m/s²). Find its acceleration. If instead it is projected up the same incline at 10 m/s, how far up does it travel?
Circular motion needs a net inward (centripetal) force $F_c = \dfrac{mv^2}{r} = m\omega^2r$. Centripetal force is not a new force — it is supplied by real forces: tension, friction, gravity, normal reaction, or their components. (Centrifugal force is only a pseudo-force felt in the rotating, non-inertial frame.)
Friction alone provides the centripetal force: $\dfrac{mv^2}{r} \le \mu_sN = \mu_smg$
$$\boxed{v_{max} = \sqrt{\mu_s r g}} \quad (\text{independent of the vehicle's mass!})$$Without friction: on a road banked at angle θ, resolve the normal N: vertical $N\cos\theta = mg$; horizontal $N\sin\theta = \dfrac{mv^2}{r}$. Divide:
$$\boxed{\tan\theta = \frac{v^{2}}{rg}} \qquad v_{ideal} = \sqrt{rg\tan\theta}$$With friction (JEE form): $$v_{max} = \sqrt{rg\,\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}}, \qquad v_{min} = \sqrt{rg\,\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}}$$ Between these speeds the car doesn't skid. A cyclist rounding a level curve leans inward by $\tan\theta = \dfrac{v^2}{rg}$ from the vertical (same formula, same derivation).
Banking of a road — full force resolution on the car
The vertical component of N supports the weight; the horizontal component supplies the centripetal force — giving tan θ = v²/rg at the design speed.
Q. (a) Find the maximum speed of a car on a level circular road of radius 45 m with μₛ = 0.2. (b) At what angle should the road be banked for an ideal speed of 30 m/s on a curve of radius 90 m? (g = 10 m/s²)
One worked model of each distinct question type asked from this chapter, tagged by exam. Attempt each yourself first, then check.
Q. Show that Newton's second law is the real law of motion (5 marks). / Prove the law of conservation of momentum from Newton's laws (3 marks). / Prove angle of repose = angle of friction (3 marks). / Derive the banking angle tan θ = v²/rg (5 marks).
Q. A 1000 kg car moving at 72 km/h is stopped in 50 m by brakes. Find the braking force.
Q. A 10 kg block on a rough floor (μ = 0.5) is pulled by F = 50 N at 37° above the horizontal (g = 10, sin 37° = 0.6, cos 37° = 0.8). Find the acceleration.
Q. In an Atwood machine with 6 kg and 4 kg, find a and T. If the string suddenly breaks, what is the acceleration of each mass? (g = 10 m/s²)
Q. A 5 kg block on a rough table (μ = 0.2) is connected over a light edge-pulley to a hanging 3 kg block (g = 10 m/s²). Find the acceleration and tension.
Q. A 2 kg block rests on a 8 kg block on a smooth floor; μ between the blocks is 0.4 (g = 10). What maximum horizontal force on the LOWER block lets both move together?
Q. A 20 g bullet moving at 400 m/s penetrates 10 cm into a fixed wooden block. Find the average resistive force.
Q. A 0.5 kg stone tied to a 1 m string is whirled in a horizontal circle. The string breaks at tension 50 N. Find the maximum speed and what happens to the stone on breaking.
Q. Sand falls at 2 kg/s onto a conveyor belt moving at 3 m/s. What extra force keeps the belt moving at constant speed, and what power is needed?
Q. Masses 4 kg (on a 30° smooth incline) and 2 kg (on a 60° smooth incline) are connected by a string over a pulley at the common apex (g = 10 m/s²). Find the acceleration and tension.
Read this the night before the exam:
| Concept | Formula | Condition / Note |
|---|---|---|
| Second law | $F = \frac{dp}{dt} = ma$ | Net external force; 1 N = 10⁵ dyne; 1 kgf = 9.8 N |
| Momentum–KE link | $p = \sqrt{2mK}$ | $K = p^2/2m$ |
| Impulse | $J = F\Delta t = \Delta p$ | Area under F–t graph; rebound: |Δp| = m(u+v) |
| Recoil of gun | $V = mv/M$ | Momentum conservation (backward) |
| Lift: up / down accel a | $N = m(g \pm a)$ | Free fall: N = 0; velocity irrelevant |
| Contact force (F on m₁) | $N_{12} = \frac{m_2F}{m_1+m_2}$ | = (mass ahead) × a |
| Heavy rope tension | $T(x) = Fx/L$ | x from the free end; massless rope → T uniform |
| Atwood machine | $a = \frac{(m_1-m_2)g}{m_1+m_2}$ | $T = \frac{2m_1m_2g}{m_1+m_2}$; clamp force = 2T |
| Table-edge pulley | $a = \frac{m_2g}{m_1+m_2}$ | Rough table: $a = \frac{(m_2-\mu m_1)g}{m_1+m_2}$; pulley force = T√2 |
| Incline + pulley | $a = \frac{(m_2 - m_1\sin\theta)g}{m_1+m_2}$ | Double incline: replace with $m\sin\theta$ on each side |
| Static friction | $f_s \le \mu_sN$ | Self-adjusting; limiting $= \mu_sN$ |
| Kinetic friction | $f_k = \mu_kN$ | $\mu_k < \mu_s$; independent of area & speed |
| Angle of repose | $\tan\alpha = \mu_s$ | = angle of friction λ |
| Rough incline (down) | $a = g(\sin\theta - \mu\cos\theta)$ | Up-projected: $g(\sin\theta + \mu\cos\theta)$ retardation |
| Min force to slide (angle λ) | $F_{min} = \frac{\mu mg}{\sqrt{1+\mu^2}}$ | Applied at the angle of friction |
| Level circular road | $v_{max} = \sqrt{\mu rg}$ | Mass-independent |
| Banking (ideal) | $\tan\theta = v^2/rg$ | Cyclist leans by the same angle |
| Banking + friction | $v_{max} = \sqrt{rg\frac{\mu+\tan\theta}{1-\mu\tan\theta}}$ | $v_{min}$: flip the signs of μ |
| Conical pendulum | $T_{period} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$ | $\tan\theta = v^2/rg$ |
| Belt / chain feeding | $F = v\,\frac{dm}{dt}$ | Rocket thrust $= u\,\frac{dm}{dt}$ |
How to use these notes: Day 1: Sections 1–5 (laws, momentum, impulse) + Fig 1; write all three law statements from memory. Day 2: Sections 6–8 (FBD method, lift, connected bodies) + Figs 2–3; practise drawing FBDs for five random setups. Day 3: Section 9 — derive all three pulley results from scratch + generate Figs 4–6 from the prompts. Day 4: Sections 10–11 (friction & circular dynamics) + Figs 7–8, then all ten Types of Section 12 without looking, followed by the mistakes checklist. Finish every session by writing the formula sheet from memory.
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