Laws of Motion

A block of mass $m$ slides down a frictionless wedge of mass $M$ and angle $\theta$. The wedge itself rests on a frictionless floor. Find the block's acceleration relative to the wedge.

Everyone writes $g\sin \theta$. That is the answer for a fixed wedge, and this wedge is not fixed.

The block pushes the wedge sideways, so the wedge recoils. Working through the constraint gives

$a_{rel}=\frac{(M+m)g\sin \theta }{M+m{\sin }^2\theta }$

Since $M+m{\sin }^2\theta <M+m$, this is larger than $g\sin \theta$. The block slides down a retreating wedge faster than down a fixed one — the surface is running away beneath it.

Check the limits before trusting it. $M\to \infty$ gives $g\sin \theta$, the fixed-wedge answer. $\theta =90°$ gives $g$, free fall. Both correct.

Nothing about the forces was subtle here. What was subtle is the constraint: the block must stay on a surface that is itself moving. Writing that condition down correctly is what Advanced tests, and it is why so many of these problems are lost before Newton's second law is even reached.

1. Constraints: the part that is not Newton's laws

A constraint is a geometric fact — a string does not stretch, a block does not leave a surface, a rod does not bend. It supplies the extra equation that closes an otherwise under-determined system, and it comes from geometry, never from forces.

Method 1 — differentiate the length. Write the total length of the string as a sum of segments, set $\frac{dL}{dt}=0$, differentiate again. Slow, but it never fails.

Method 2 — components along the connector. For an inextensible string or a rigid rod, the components of the velocities of the two ends along the connector must be equal.

$v_A\cos \alpha =v_B\cos \beta$

Method 2 is the Advanced tool. It turns most constraint problems into a single line of trigonometry.

The general rule, worth stating once: whatever cannot change — a length, a contact, a perpendicular separation — differentiate it. The derivative of a constant is the constraint equation.

Illustration 1

A rod of length $L$ has end B against a vertical wall and end A on the floor. A slides away from the wall at speed $v_A$ when the rod makes angle $\alpha$ with the floor. Find the speed of B.

Both ends belong to a rigid rod, so the rod's length cannot change. Project both velocities onto the rod.

$v_A$ is horizontal and the rod makes $\alpha$ with the horizontal, so A's component along the rod is $v_A\cos \alpha$.

$v_B$ is vertical and the rod makes $(90°-\alpha )$ with the vertical, so B's component is $v_B\sin \alpha$.

$v_A\cos \alpha =v_B\sin \alpha ⟹v_B=v_A\cot \alpha$

Read the behaviour off the answer. As the rod approaches the floor, $\alpha \to 0$ and $v_B\to \infty$: the top end accelerates without limit in the final instants. That is why a falling ladder's top end appears to whip down at the end, and it is a genuine physical prediction, not an artefact.

Illustration 2

A block slides on the floor at speed $v$, pushed against the vertical face of a wedge of angle $\theta$ that slides on the same floor. Find the wedge's speed if the two stay in contact.

Contact is the constraint. The two surfaces cannot overlap or separate, so the components of the two velocities perpendicular to the contact surface must be equal.

Take the contact face making angle $\theta$ with the vertical. The block moves horizontally at $v$; its component normal to the face is $v\cos \theta$. The wedge moves horizontally at $V$; its normal component is $V\cos \theta$ as well — unless the wedge also moves vertically.

For the common case of a block sliding down a moving wedge's incline with relative speed $u$ while the wedge moves horizontally at $V$, the perpendicular condition gives the familiar result directly:

$u\sin \theta =\text{(vertical drop rate)},V+u\cos \theta =\text{(horizontal rate)}$

The habit to build: never guess the ratio. Identify the quantity that cannot change — a length, a gap, a perpendicular separation — and differentiate it.

Illustration 3

In a system where a rope runs from a load, up over a fixed pulley, down around a movable pulley and back up to a fixed anchor, the free end is pulled with acceleration $a_0$. Find the load's acceleration.

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 that gap, so

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

$\frac{dL}{dt}=0\Rightarrow 2\dot{x}+\dot{y}=0\Rightarrow 2\ddot{x}+\ddot{y}=0$

$|a_{load}|=\frac{1}{2}{a}_0$

Check it against energy. The tension is the same throughout an ideal rope, so the load feels $2T$ while the hand pulls with $T$. Half the force at twice the distance — the work balances exactly, which is the sanity check on any pulley ratio you derive.

2. The free wedge, in full

Set up the hook properly, because the method generalises to every "surface that can move" problem.

Let the wedge accelerate at $A$ to the left, and the block accelerate at $a$ relative to the wedge, down the slope. The block's acceleration in the ground frame is the vector sum, and this is where errors creep in — the block's absolute acceleration is not along the incline.

Block, along the horizontal: $N\sin \theta =m(a\cos \theta -A)$

Block, vertically: $N\cos \theta -mg=-ma\sin \theta$

Wedge, horizontally: the block presses on it with $N$, whose horizontal component drives the recoil, so $MA=N\sin \theta$.

Three equations, three unknowns $N$, $a$, $A$. Eliminating gives

$a=\frac{(M+m)g\sin \theta }{M+m{\sin }^2\theta },A=\frac{mg\sin \theta \cos \theta }{M+m{\sin }^2\theta },N=\frac{Mmg\cos \theta }{M+m{\sin }^2\theta }$

The normal force is smaller than $mg\cos \theta$. A retreating surface presses back less hard, exactly as a lift accelerating downward reduces your apparent weight.

Illustration 4

A $2\text{kg}$ block slides on a frictionless $45°$ wedge of mass $2\text{kg}$, itself free on a frictionless floor. Take $g=10$. Find the block's acceleration relative to the wedge, the wedge's acceleration, and the normal force.

With $M=m=2$ and ${\sin }^{2}45°=0.5$: the denominator is $M+m{\sin }^2\theta =2+1=3$.

$a=\frac{4(10)(0.707)}{3}=9.43{\text{m/s}}^2$

$A=\frac{2(10)(0.707)(0.707)}{3}=3.33{\text{m/s}}^2$

$N=\frac{2(2)(10)(0.707)}{3}=9.43\text{N}$

Compare with the fixed wedge: there $a=g\sin 45°=7.07{\text{m/s}}^2$ and $N=mg\cos 45°=14.1\text{N}$. The block slides 33% faster while being pressed 33% less hard — both because the surface is retreating.

Momentum check: the floor is frictionless, so the horizontal momentum of the whole system must stay zero. The wedge carries $2(3.33)=6.67$ leftward; the block's absolute horizontal acceleration is $a\cos \theta -A=6.67-3.33=3.33$ rightward, carrying $2(3.33)=6.67$. They cancel.

3. Friction as a constraint, not a formula

At Main level, friction is $\mu N$. At Advanced level, static friction is an unknown to be solved for, exactly like a normal force or a tension, and ${\mu }_{s}N$ is only the ceiling it must respect.

The reliable procedure, in three steps:

1. Assume no slipping. Treat the contacting bodies as one rigid unit and find the common acceleration.
2. Solve for the friction actually required to produce that acceleration on each body.
3. Check whether the required $f$ exceeds ${\mu }_{s}N$. If it does, the assumption was wrong: the bodies slip, and friction becomes ${\mu }_{k}N$ in a known direction.

Never begin by writing $f=\mu N$. That is the answer to a different question — the one where slipping is already happening or is exactly about to.

Illustration 5

Block $m$ rests on block $M$, with coefficient $\mu$ between them and a frictionless floor beneath. A horizontal force $F$ is applied to the lower block. Find the greatest $F$ for which they move together.

Assume they move together at acceleration $a=\frac{F}{M+m}$.

Look at the upper block. The only horizontal force on it is friction from below, so friction alone must supply its acceleration:

$f=ma$

The ceiling is $\mu mg$, so

$ma\le \mu mg\Rightarrow a\le \mu g$

$F_{max}=(M+m)\mu g$

Note the mass that cancelled. The upper block's own mass drops out of the acceleration limit entirely — the maximum common acceleration is $\mu g$ whatever sits on top.

Illustration 6

The same two blocks, but now $F$ is applied to the upper block. Find the greatest $F$ for which they move together.

Assume again $a=\frac{F}{M+m}$, but now look at the lower block, which is dragged along only by friction from above:

$f=Ma\le \mu mg\Rightarrow a\le \frac{\mu mg}{M}$

$F_{max}=(M+m)\frac{\mu mg}{M}$

Compare the two. Pushing the bottom block allows $(M+m)\mu g$; pushing the top allows $(M+m)\mu g\cdot \frac{m}{M}$.

If the lower block is the heavier one, pushing from below tolerates far more force, because the friction only has to move the light block instead of the heavy one. With $M=4m$ the ratio is four to one. Which block the force acts on changes the answer, and questions exploit that relentlessly.

4. Non-inertial frames, including rotating ones

In a frame accelerating at ${\vec{a}}_0$, Newton's second law is restored by giving every body an extra force $-m{\vec{a}}_0$. In a frame rotating at $\omega$, the corresponding term for a body at rest in that frame is the centrifugal force:

$F_{cf}=m{\omega }^{2}r\text{directed outward}$

Pseudo forces have no third-law partner. That is the formal signature that they are bookkeeping, and it is also how a question can test whether you understand them rather than merely use them.

Working in a non-inertial frame is always a choice. Take it when it makes a body stationary, because a stationary body means equilibrium equations instead of dynamics.

Illustration 7

A coin rests on a turntable at radius $r$ with coefficient of friction $\mu$. Find the greatest angular speed before it slides.

In the ground frame: friction is the only horizontal force and must supply the entire centripetal requirement.

$m{\omega }^{2}r\le \mu mg\Rightarrow {\omega }_{max}=\sqrt{\frac{\mu g}{r}}$

In the turntable's frame: the coin is in equilibrium under friction inward and the centrifugal force $m{\omega }^{2}r$ outward, giving the identical inequality.

Read it: ${\omega }_{max}\propto 1/\sqrt{r}$, so coins near the rim fly off first — which is what you see, and which is also why a centrifuge separates by radius.

Illustration 8

A block rests on a frictionless wedge of angle $\theta$. What horizontal acceleration must be given to the wedge so that the block does not slide on it?

In the wedge's frame the block is stationary, so it is in equilibrium under three forces: $N$ perpendicular to the incline, $mg$ down, and the pseudo force $m{a}_0$ horizontally backward.

Resolving along the incline, where $N$ contributes nothing:

$mg\sin \theta =m{a}_0\cos \theta$

$a_0=g\tan \theta$

The check that this is right: it does not contain $m$. A frictionless surface can only push perpendicular to itself, so the required acceleration is fixed by geometry alone. This is the same $\tan \theta =a/g$ that tilts the surface of a liquid in an accelerating tank, and for the same reason.

5. The full speed range on a banked road

JEE Main gives the frictionless design speed, $\tan \theta =v^2/rg$. Advanced asks for the range of speeds that a real banked road with friction permits — and the two limits have friction pointing in opposite directions.

At the lower limit the car tends to slide down the bank, so friction acts up it. At the upper limit it tends to slide up, so friction acts down. Resolving each case:

$v_{min}=\sqrt{\frac{rg(\tan \theta -\mu )}{1+\mu \tan \theta }},v_{max}=\sqrt{\frac{rg(\tan \theta +\mu )}{1-\mu \tan \theta }}$

Read the two special cases off the formulas. If $\mu \ge \tan \theta$, $v_{min}$ becomes imaginary — meaning there is no lower limit at all, and the car can sit on the bank at rest. If $\mu \tan \theta \to 1$, $v_{max}$ diverges: friction and banking together can hold any speed.

Setting $\mu =0$ recovers the single design speed $\sqrt{rg\tan \theta }$, with the range collapsing to a point. That is the check to run on any version you write down.

Illustration 9

A curve of radius $100\text{m}$ is banked at $30°$ with $\mu =0.2$. Take $g=10$. Find the safe speed range.

$\tan 30°=0.577$, $rg=1000$.

$v_{min}=\sqrt{\frac{1000(0.577-0.2)}{1+0.2(0.577)}}=\sqrt{\frac{377}{1.115}}=\sqrt{338}=18.4\text{m/s}$

$v_{max}=\sqrt{\frac{1000(0.577+0.2)}{1-0.2(0.577)}}=\sqrt{\frac{777}{0.885}}=\sqrt{878}=29.6\text{m/s}$

The design speed sits between them: $\sqrt{1000(0.577)}=24.0$ m/s, and friction opens a band of roughly $\pm 25\%$ around it. That band is the entire practical reason roads are banked and surfaced rather than banked alone.

6. Variable mass: chains, ropes and the falling link

When mass enters or leaves a system, $F=ma$ is no longer the law — go back to $\vec{F}=\frac{d\vec{p}}{dt}$. The extra term is a thrust:

${\vec{F}}_{ext}+{\vec{v}}_{rel}\frac{dm}{dt}=m\frac{d\vec{v}}{dt}$

Illustration 10

A uniform chain of linear density $\lambda$ is held vertically with its lower end just touching a weighing pan, then released. When a length $x$ has landed, what does the pan read?

Two entirely separate contributions, and the whole question is remembering the second.

Weight already at rest on the pan: $\lambda xg$

Force to stop the arriving links. They land at $v=\sqrt{2gx}$, and mass arrives at rate $\frac{dm}{dt}=\lambda v$. The pan must destroy that momentum:

$F_{impulsive}=v\frac{dm}{dt}=\lambda v^2=\lambda (2gx)=2\lambda gx$

$\text{Reading}=\lambda gx+2\lambda gx=3\lambda gx$

Three times the weight of the chain lying on it. Two thirds of what the scale shows is not weight at all — it is momentum being destroyed.

Check the total. Integrating the reading as the chain falls returns exactly the chain's initial potential energy, so no energy has been invented. And when the last link lands the reading drops instantly to the full static weight, which is the discontinuity you can see on a real scale.

7. Impulsive tension and the instant after

Some events are effectively instantaneous — a string snapping taut, a collision, a peg being removed. Across such an event:

• Impulsive forces (tension in a suddenly taut string, normal force in a collision) are enormous and finite in impulse.
• Ordinary forces (gravity, spring forces) deliver negligible impulse over the vanishing time, so velocities change but positions do not.

A spring's force cannot change instantly, because that would need an instant change of length. A string's tension can — it can vanish or spike without warning. That single asymmetry decides most "just after" questions.

Illustration 11

Two particles of masses $m_1$ and $m_2$ are joined by a slack inextensible string. $m_1$ is moving at $u$ along the line of the string when it becomes taut. Find their common velocity along the string immediately afterwards.

The tension is impulsive and internal to the pair, so momentum along the string is conserved across the jerk. Inextensibility forces both to share the same velocity component along the string:

$m_{1}u=(m_1+m_2)v\Rightarrow v=\frac{m_{1}u}{m_1+m_2}$

Recognise the structure: this is algebraically a perfectly inelastic collision, and it loses the same fraction of kinetic energy, $\frac{m_2}{m_1+m_2}$. An inextensible string jerking taut is exactly as dissipative as bodies sticking together — which is why real hoisting gear uses a little elasticity on purpose.

Summary

• Constraints come from geometry, not forces. Differentiate whatever cannot change.
• For a string or rod, the components of the end velocities along it are equal — usually one line instead of a page.
• A falling ladder's top end speeds up without limit: $v_B=v_A\cot \alpha$.
• Free wedge: $a_{rel}=\frac{(M+m)g\sin \theta }{M+m{\sin }^2\theta }$, larger than $g\sin \theta$, with $N$ smaller than $mg\cos \theta$. Check both against $M\to \infty$.
• Friction is an unknown to solve for. Assume no slipping, compute the required $f$, then test it against ${\mu }_{s}N$.
• Two stacked blocks: pushing the lower allows $F_{max}=(M+m)\mu g$; pushing the upper allows $\frac{m}{M}$ times as much.
• Pseudo force $-m{\vec{a}}_0$; centrifugal $m{\omega }^{2}r$ outward. Use a non-inertial frame when it makes a body stationary.
• Frictionless wedge with a non-sliding block needs $a_0=g\tan \theta$, independent of mass.
• Banked road with friction: $v_{min}$ and $v_{max}$ straddle the design speed. $\mu \ge \tan \theta$ removes the lower limit entirely.
• Variable mass: ${\vec{F}}_{ext}+{\vec{v}}_{rel}\frac{dm}{dt}=m\frac{d\vec{v}}{dt}$. A falling chain reads three times the resting weight.
• Across an impulsive event, velocities change and positions do not. String tension can jump; spring force cannot.