Laws of Motion

A book rests on a table. Its weight is 10 N and the table pushes up with 10 N. Are these two a Newton's third law pair?

Most say yes. They are not.

A third-law pair always acts on two different bodies. Both of these act on the book, so they can never be a pair — they happen to be equal only because the book is in equilibrium.

The real partner of the book's weight is the book pulling the Earth up with 10 N.

The giveaway: put the book in an accelerating lift. The normal force changes; the weight does not. A third-law pair can never come apart like that.

Everything in this chapter is the same discipline — choose one body, draw every force acting on it, add them as vectors. Nearly every wrong answer is a force drawn that isn't there, one omitted that is, or one that belongs to a different body.

1. First law and inertial frames

First law — a body stays at rest or in uniform motion unless a net external force acts.

This is not a special case of the second law. It defines the frames in which the second law works: a frame where the first law holds is inertial.

Inertia — the property named. Mass — its quantitative measure.

A frame accelerating relative to an inertial one is non-inertial: bodies there appear to accelerate with no force acting. Section 10 handles those.

The Earth is inertial enough for every JEE problem.

2. Second law

${\vec{F}}_{net}=\frac{d\vec{p}}{dt}=\frac{d(m\vec{v})}{dt}$

Constant mass reduces this to:

${\vec{F}}_{net}=m\vec{a}$

The momentum form is the general one. Use it whenever mass changes — rockets, falling chains, conveyor belts.

It is a vector equation, so it holds independently along each axis. Resolve along two perpendicular directions, write two scalar equations.

Choose axes well. On an incline, take axes along and perpendicular to the slope — then the normal force never needs resolving.

Impulse $=\int \vec{F}dt=\Delta \vec{p}$. This is the tool for collisions and jerks, where the force itself is unknown.

Illustration 1

A 150 g ball strikes a wall at $20\text{m/s}$ and rebounds at $15\text{m/s}$, the contact lasting $0.02\text{s}$. Find the average force on the wall.

Momentum is a vector, so take the outward direction as positive and the incoming velocity as negative:

$\Delta p=m(v_f-v_i)=0.15(15-(-20))=0.15(35)=5.25\text{kg m/s}$

$F_{avg}=\frac{\Delta p}{\Delta t}=\frac{5.25}{0.02}=262.5\text{N}$

The trap is the sign. Using $15-20=-5$ gives 37.5 N, seven times too small. The reversal is most of the momentum change.

Scale check: the ball's own weight is only 1.5 N, so the wall delivers about 175 times its weight — which is why impact forces are treated separately from ordinary ones.

3. Third law

Third law — equal and opposite forces, along the same line, acting on different bodies.

The two never appear in the same free-body diagram. That is precisely why they never cancel.

The noun-swap test. "Earth pulls book" pairs with "book pulls Earth". If swapping the two nouns does not produce the other force, it is not a pair.

4. Conservation of linear momentum

Put the third law and the second law together. Internal forces in a system come in equal and opposite pairs, so they cancel in the total:

${\vec{F}}_{ext}=\frac{d{\vec{p}}_{total}}{dt}\Rightarrow {\vec{F}}_{ext}=0\text{ gives }{\vec{p}}_{total}=\text{constant}$

This is not a new law. It is the third law restated in a form you can compute with.

It is a vector statement, so it holds component by component. Momentum can be conserved horizontally while gravity destroys it vertically — which is exactly the situation in an explosion in flight.

Three standard applications:

For the rocket the constant-mass form $F=ma$ is simply wrong; you must go back to $\vec{F}=d\vec{p}/dt$. Gas leaving at speed $u$ relative to the rocket carries away momentum at the rate $u|dm/dt|$, and by the third law that rate is the forward thrust.

Illustration 2

A 4 kg rifle fires a 20 g bullet at $400\text{m/s}$. Find the recoil speed, and compare the kinetic energies.

No external horizontal force, so total momentum stays zero:

$0.02(400)=4v\Rightarrow v=2\text{m/s}$

Equal momentum, but 200 times the energy in the bullet. Since $K=p^2/2m$, at equal momentum the energy goes as $1/m$, and the mass ratio is exactly 200. That is why the recoil bruises a shoulder while the bullet penetrates a wall.

Illustration 3

A rocket of total mass 5000 kg burns fuel at $50\text{kg/s}$, ejecting it at $2\text{km/s}$ relative to itself. Find the thrust and the initial acceleration. $g=10$.

$\text{Thrust}=u\left|\frac{dm}{dt}\right|=2000(50)=100,000\text{N}$

The rocket must still support its own weight of $5000(10)=50,000$ N:

$a=\frac{100,000-50,000}{5000}=10{\text{m/s}}^2$

Read it back: thrust below $mg$ means the rocket cannot leave the pad at all, however much fuel it carries. And as fuel burns off, $m$ falls while the thrust holds steady, so the acceleration rises throughout the burn.

5. The common forces

"Just after release" questions turn on the last row. A string's tension can vanish instantly; a spring's force cannot change at all in that instant.

Illustration 4

Block A hangs from a spring fixed to the ceiling; block B, of the same mass $m$, hangs from A by a string. The string is cut. Find the acceleration of each block immediately afterwards.

Before the cut, the whole system is in equilibrium, so the spring carries both weights:

$F_{spring}=2mg$

Immediately after, apply the rule in the table above.

Block B — the string is gone, so only gravity acts, giving $a_B=g$ downward.

Block A — the spring's length has not had time to change, so it still pulls up with $2mg$, while A's own weight $mg$ pulls down:

$m{a}_A=2mg-mg\Rightarrow a_A=g\text{upward}$

The two blocks fly apart with equal accelerations in opposite directions. Assuming the spring force drops to $mg$ instantly gives $a_A=0$, and that is the standard wrong answer.

6. Equilibrium of concurrent forces

Concurrent forces all pass through a single point, so they produce no torque about it and only the force condition is needed:

$\sum \vec{F}=0⟺\sum F_x=0\text{ and }\sum F_y=0$

Drawn head to tail, forces in equilibrium form a closed polygon — three of them close into a triangle.

Lami's theorem turns that triangle into a formula. For exactly three concurrent forces in equilibrium:

$\frac{P}{\sin \alpha }=\frac{Q}{\sin \beta }=\frac{R}{\sin \gamma }$

where each angle is the one between the other two forces. It is the sine rule applied to the closed triangle, and the three angles must sum to $360°$ — a free check on your diagram before you compute anything.

Illustration 5

A 10 kg load hangs from a knot held by two ceiling strings, one at $30°$ and the other at $60°$ to the horizontal. Find both tensions. $g=10$.

By components. Horizontal: $T_1\cos 30°=T_2\cos 60°$, so $T_2=1.732T_1$.

Vertical: $T_1\sin 30°+T_2\sin 60°=100$

$0.5T_1+1.732T_1(0.866)=100\Rightarrow 2T_1=100$

$T_1=50\text{N},T_2=86.6\text{N}$

By Lami's theorem. The two strings are $90°$ apart; $T_2$ is $30°$ from the vertical, so the angle between $T_2$ and $mg$ is $150°$, and between $T_1$ and $mg$ it is $120°$. Those three sum to $360°$, as they must.

$\frac{100}{\sin 90°}=\frac{T_1}{\sin 150°}=\frac{T_2}{\sin 120°}$

$T_1=100(0.5)=50\text{N}$, $T_2=100(0.866)=86.6\text{N}$

Two independent routes, same answer. Note that the steeper string carries the larger tension.

7. Friction

$f_s\le {\mu }_{s}N,f_k={\mu }_{k}N$

Static friction adjusts itself — it takes whatever value prevents sliding, up to the limit. It is usually below maximum.

Trap. Never write $f={\mu }_{s}N$ unless the body is on the verge of sliding. This is the most common friction error.

Typically ${\mu }_k<{\mu }_s$, which is why a stationary object is harder to start than to keep moving.

Rolling friction is a third and much smaller resistance, arising from the slight deformation of the wheel and surface at the contact. It is typically one to two orders of magnitude below sliding friction — which is the entire reason wheels exist, and why a loaded trolley is easy to push but hard to drag.

Angle of repose — the incline angle at which sliding just begins:

$\tan {\theta }_{repose}={\mu }_s$

From balancing $mg\sin \theta$ against ${\mu }_{s}mg\cos \theta$. The mass cancels: the angle depends only on the surfaces.

Friction does not always oppose motion. The friction driving a car acts forward on the wheels — it opposes the relative slipping of tyre on road, not the car's motion.

Illustration 6

At what angle should a block be pulled so that the least force is needed to just start it sliding? Take ${\mu }_s=0.75$.

Pull at angle $\theta$ above the horizontal. The vertical component lightens the block, so the normal force is not $mg$:

$N=mg-F\sin \theta$

On the verge of sliding, $F\cos \theta =\mu N$:

$F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }$

$F$ is least when the denominator is greatest. Differentiating, $-\sin \theta +\mu \cos \theta =0$:

$\tan \theta =\mu \text{so}{F}_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$

With $\mu =0.75$: $\theta =36.9°$ and $F_{min}=\frac{0.75mg}{1.25}=0.6mg$.

Compare: pulling horizontally would need $0.75mg$. Angling the pull saves 20% of the effort — you are trading a little forward force for a large reduction in the normal force. The optimum angle is the angle of friction itself.

8. Constraints and connected bodies

Constraints supply the equations that close an under-determined system.

Single fixed pulley — inextensible string $\Rightarrow$ if one block descends $x$, the other rises $x$ $\Rightarrow$ equal acceleration magnitudes.

Movable pulley — supported by two segments $\Rightarrow$ its displacement is half the sum of the string ends' $\Rightarrow$ accelerations in a $2:1$ ratio.

General method, never fails: write the total string length as a sum of segments, set $\frac{dL}{dt}=0$, differentiate again. Safer than guessing the ratio.

For bodies in contact, the contact force is one unknown appearing in both diagrams with opposite signs — the third law doing useful work.

Illustration 7

3 kg and 5 kg hang over a frictionless pulley. $g=10$. Find $a$ and $T$.

Constraint: same $|a|$, opposite directions.

5 kg block: $50-T=5a$ 3 kg block: $T-30=3a$

Add, and $T$ cancels: $20=8a\Rightarrow a=2.5{\text{m/s}}^2$

Then $T=30+3(2.5)=37.5\text{N}$.

Check: $T$ lies between 30 N and 50 N, as it must — the heavy side falls, the light side rises.

Illustration 8

A rope runs from a load, up over a fixed pulley, down under a movable pulley and back up to a fixed point. The free end is pulled with acceleration $4{\text{m/s}}^2$. Find the load's acceleration.

Guessing the ratio is where marks are lost. Use the length method instead.

Let $x$ be the depth of the movable pulley below the fixed one and $y$ the length of free rope pulled in. Two rope segments span the gap, so

$L=2x+y+\text{constant}$

The rope is inextensible, so $\frac{dL}{dt}=0$, and differentiating twice:

$2\ddot{x}+\ddot{y}=0\Rightarrow |a_{load}|=\frac{1}{2}|a_{free}|=2{\text{m/s}}^2$

Read it back: the load moves half as far and half as fast, so the rope tension is half the load's — a movable pulley halves the force at the cost of doubling the distance. The energy bookkeeping is untouched, which is the sanity check on any pulley ratio.

9. Circular motion dynamics

$a_c=\frac{v^2}{r},F_c=\frac{m{v}^2}{r}$

Centripetal force is not a new force. It is the name for whatever real forces — tension, friction, gravity, normal — add up to point inward.

Flat road: friction supplies it. $\frac{m{v}^2}{r}\le \mu mg\Rightarrow v_{max}=\sqrt{\mu rg}$

Banked, frictionless: the horizontal component of $N$ supplies it. Dividing the two equations in the figure:

$\tan \theta =\frac{v^2}{rg}$

The mass cancels — the design speed of a banked curve is the same for a truck and a motorcycle.

Vertical circle: at the top, gravity alone can be the whole centripetal force, so the minimum speed there is $\sqrt{gr}$. Below that the string goes slack.

Conical pendulum: a bob swung in a horizontal circle on a string at angle $\theta$ to the vertical is the banked-road equations with $N$ replaced by $T$ — the same two components, the same $\tan \theta =v^2/rg$.

Illustration 9

A bob on a $1\text{m}$ string moves in a horizontal circle with the string at $30°$ to the vertical. Find the speed and the period. $g=10$.

The bob has no vertical acceleration, so the vertical components balance while the horizontal component turns it:

$T\cos 30°=mg,T\sin 30°=\frac{m{v}^2}{r}$

Dividing, $\tan 30°=\frac{v^2}{rg}$, with $r=L\sin 30°=0.5\text{m}$:

$v^2=0.577(0.5)(10)=2.89\Rightarrow v=1.70\text{m/s}$

Period $=\frac{2\pi r}{v}=\frac{\pi }{1.70}=1.85\text{s}$

Check by the standard form: $T_{period}=2\pi \sqrt{\frac{L\cos \theta }{g}}=2\pi \sqrt{0.0866}=1.85\text{s}$. The two routes agree.

Note: the string can never reach the horizontal. $T\cos \theta =mg$ would need infinite tension at $\theta =90°$.

Illustration 10

A car of mass 1000 kg rounds a curve of radius 50 m on a flat road, $\mu =0.4$, $g=10$. Find the maximum safe speed. Does it change for a 2000 kg truck?

$v_{max}=\sqrt{\mu rg}=\sqrt{0.4(50)(10)}=\sqrt{200}\approx 14.1\text{m/s}$

Mass does not appear. Same for the truck — the required force and the available friction both scale with $m$, so it cancels.

10. Pseudo forces

In a frame accelerating at ${\vec{a}}_0$, the second law still works if every body gets an extra force:

${\vec{F}}_{pseudo}=-m{\vec{a}}_0$

Opposite to the frame's acceleration, proportional to the body's mass.

It has no third-law partner. That is the formal sign it is bookkeeping, not physics.

Lift: accelerating up at $a$ gives apparent weight $m(g+a)$. In free fall it is zero — that is what weightlessness means.

Working in a non-inertial frame is a choice. Any problem can be done from the ground. Switch only when it makes a body stationary.

Illustration 11

A 2 kg block sits on the floor of a lift. $g=10$. Find the normal force when the lift accelerates (a) up at 2 m/s², (b) down at 2 m/s², (c) in free fall.

Ground frame, up positive: $N-mg=ma$

(a) $N=2(10+2)=24\text{N}$ (b) $N=2(10-2)=16\text{N}$ (c) $a=-g\Rightarrow N=0$ — the block floats.

In the lift's frame the same answers come from adding $-m{a}_0$ and setting the block in equilibrium. Different interpretation, identical prediction — always.

Summary

• Choose the body, draw every force on it, add as vectors. That is the whole method.
• The first law defines inertial frames; it is not a special case of the second.
• $\vec{F}=d\vec{p}/dt$ is the general law. $m\vec{a}$ only for constant mass.
• Third-law pairs act on different bodies, so they never appear in one FBD and never cancel. Use the noun-swap test.
• Zero external force conserves total momentum, component by component. Recoil, explosions and rockets are the three standard cases.
• Rocket thrust $=u|dm/dt|$, and the acceleration rises as fuel burns off.
• Three concurrent forces in equilibrium close into a triangle — that is all Lami's theorem says.
• $N=mg$ only on a horizontal surface with no vertical acceleration.
• A string's tension can change instantly; a spring's force cannot.
• $f_s\le {\mu }_{s}N$ — static friction is usually below maximum. Never assume equality without checking.
• $\tan {\theta }_{repose}={\mu }_s$, independent of mass.
• Rolling friction is one to two orders below sliding friction. That is why wheels exist.
• Least force to start a block sliding is at $\tan \theta =\mu$, giving $F_{min}=\mu mg/\sqrt{1+{\mu }^2}$.
• Constraints: write the string length, differentiate twice. Movable pulley gives $2:1$.
• Centripetal force names whatever real forces point inward. Banked: $\tan \theta =v^2/rg$, mass cancels. Vertical circle needs $\sqrt{gr}$ at the top.
• Pseudo force $-m{\vec{a}}_0$ in a non-inertial frame. No third-law partner.