NCERT Ch 5

Laws of Motion & Friction

Why things move the way they do · Class 11 Physics

1From “how” to “why”

Kinematics told us how an object moves once we know its acceleration. This chapter answers the deeper question: what decides the acceleration in the first place? The answer, in one word, is force — and the three laws Isaac Newton set down in 1687 still carry essentially every mechanics problem you will meet in this exam.

The whole subject rests on one habit of mind. Before writing a single equation you must ask: which object am I talking about, and what pushes or pulls on it? Get that right and the algebra is easy. Get it muddled — mixing forces on different bodies into one equation — and no amount of formula-hunting will save you. That habit has a name, the free-body diagram, and we build up to it deliberately.

2Newton's first law — inertia

A body continues in its state of rest, or of uniform motion in a straight line, unless a net external force acts on it. Read it carefully: it does not say a moving body needs a force to keep moving. It says exactly the opposite — a force is needed only to change the motion. A puck sliding on frictionless ice would glide forever; it slows on a real floor only because friction, a real force, acts on it.

This built-in reluctance to change velocity is called inertia, and its measure is mass. A loaded truck is harder to get going — and harder to stop — than a bicycle, because it has more mass and therefore more inertia. The first law is really the definition of what a force does: it is the thing that overcomes inertia. Note that “no net force” allows many forces that cancel — a book resting on a table has gravity pulling down and the table's normal force pushing up in perfect balance, so it stays at rest.

3Newton's second law — force, momentum, impulse

The second law makes the first quantitative. Newton framed it not with acceleration but with linear momentum, the product of mass and velocity:

Linear momentump = m v

Momentum is a vector — it points the way the velocity points — and it captures “how hard it is to stop” something better than speed alone: a slow lorry and a fast pebble can carry the same punch. The law states that the rate of change of momentum equals the net force:

Newton's second lawF = d pd t = m a

When the mass is constant, d p / d t = m (d v / d t) = m a, which recovers the familiar F = m a. The unit of force follows directly: one newton is the force that gives a 1 kg mass an acceleration of 1 m s−2, so 1 N = 1 kg m s−2. Force and acceleration always point the same way; mass is just the constant of proportionality between them.

Impulse — force acting over time

Rearrange the law over a short interval Δt. The quantity F Δt is called the impulse, and it equals the change in momentum:

Impulse–momentum theoremJ = F Δt = Δp = m v − m u

This one line explains a surprising amount of everyday physics. To stop a moving object you must remove a fixed amount of momentum, Δp. If you stop it slowly — large Δt — the force is gentle; if you stop it abruptly — tiny Δt — the force is huge. A cricketer “gives” with the ball, pulling the hands back on the catch to stretch out Δt and shrink the sting. Airbags, crumple zones and a high-jumper's foam landing all work the same way: same momentum change, longer time, smaller force.

4Newton's third law — action & reaction

To every action there is an equal and opposite reaction. If body A pushes on body B with a force, then B pushes back on A with a force of the same size and the opposite direction. But there is a subtlety students constantly miss, and it is worth putting in a box.

Remember this

The action and reaction act on two different bodies — never on the same one. That is precisely why they do not cancel out. A pair of forces can only cancel if they act on the same object.

When you walk, your foot pushes backward on the ground; the ground pushes forward on you, and that reaction is what drives you along. A rocket throws gas downward; the gas throws the rocket upward. A gun pushes the bullet forward; the bullet pushes the gun back — the recoil. In each case the two forces sit on different objects, so each object responds to the single force it feels. If you ever find yourself adding an action and its reaction to “cancel,” stop — you have almost certainly put them on the wrong body.

5Conservation of momentum

The third law hands us one of the most powerful tools in physics, almost for free. Consider two bodies that interact only with each other — a bullet and a gun, two colliding carts, an exploding shell. During the interaction body 1 feels a force F12 from body 2, and body 2 feels F21 from body 1. By the third law these are equal and opposite: F12 = −F21.

Apply the second law to each and add. The internal forces cancel exactly, so the total momentum cannot change:

dd t(p1 + p2) = F12 + F21 = 0

Therefore, if no external force acts on a system, its total momentum stays constant. This is conservation of linear momentum, and it holds even when the internal forces are messy, brief and impossible to measure — which is exactly the situation in every collision and explosion. A rifle of mass M firing a bullet of mass m at speed v recoils at V = m v / M, found in one line by setting the total momentum before (zero) equal to the total after (m v − M V). No knowledge of the explosion inside is needed.

6The free-body diagram — the master method

Almost every problem in this chapter is solved by the same disciplined recipe. It is worth doing consciously until it becomes automatic:

  1. Choose one body and mentally cut it free of everything touching it.
  2. Draw every force acting on it — weight mg down, the normal N perpendicular to the contact surface, tension, applied pushes, and friction. Nothing else, and nothing that acts on other bodies.
  3. Pick sensible axes (for an incline, tilt them along and perpendicular to the slope) and split each force into components.
  4. Apply F = m a along each axis and solve.

The figure below is the free-body diagram for a block resting on a rough incline — the case we analyse next. Study which way each arrow points before reading on.

θ mg N mg sin θ f (friction)
Free-body diagram of a block on a rough incline. Weight mg acts straight down; the surface pushes back with the normal N; along the slope, gravity's component mg sin θ pulls the block down while friction f resists up-slope.

7Friction — static vs kinetic

Friction is the sideways contact force between two surfaces, and it always opposes the relative sliding (or the tendency to slide). It comes in two flavours you must keep apart.

Static friction acts when the surfaces are not yet sliding. It is a self-adjusting force: it grows to match whatever you apply, keeping the object still, right up to a maximum. Push a heavy crate gently and it does not move — static friction exactly balances your push. Push harder and it still balances, until you exceed the limit and it lets go. That limit is

Static friction (a range, not a fixed value)fs ≤ μs N

The coefficient μs depends on the two materials, and N is the normal force pressing them together. The crucial point is the “≤”: static friction is anything from zero up to μs N, whatever is needed to prevent motion.

Kinetic friction takes over once the surfaces are actually sliding. Now the force is essentially constant, no longer self-adjusting:

Kinetic friction (fixed while sliding)fk = μk N

Why is μs > μk? At rest, the microscopic bumps of the two surfaces settle into each other and tiny cold-welds form at the points of true contact — it takes a larger force to break them all at once. Once sliding begins the surfaces skate over the tips, giving the welds no time to set, so a smaller force keeps the motion going. This is the familiar lurch when a heavy box suddenly “breaks free” and then slides more easily: you had to beat the larger μs N to start it, but only the smaller μk N to keep it going.

Remember this

Friction does not depend on the area of contact or on the speed of sliding — only on the nature of the surfaces (μ) and how hard they are pressed together (N). And static friction is a range capped at μs N, whereas kinetic friction is a single fixed value μk N.

8Block on an incline

Now we cash in the free-body diagram. Put a block of mass m on a slope of angle θ. Tilt the axes along and across the slope so that only gravity needs splitting. The weight mg resolves into a part along the slope, mg sin θ (pulling the block down the incline), and a part into the slope, mg cos θ (pressing it against the surface).

Perpendicular to the slope there is no acceleration, so the normal force balances the pressing component:

N = m g cos θ

Will it stay or slide? The block sits still as long as static friction can match the down-slope pull, that is while mg sin θ ≤ μs N = μs mg cos θ. Cancelling mg cos θ gives the clean condition

Stays put while…tan θ ≤ μs

The special angle at which it is just about to slip, tan θ = μs, is called the angle of repose — a neat way to measure μs by simply tilting a surface until things start to slide.

Once it slides, kinetic friction μk mg cos θ acts up the slope while mg sin θ pulls down. Newton's second law along the slope gives mg sin θ − μk mg cos θ = m a. The mass cancels, leaving the acceleration down the incline:

Acceleration of a sliding blocka = g (sin θ − μ cos θ)

Two sanity checks worth internalising. With no friction (μ = 0) this reduces to a = g sin θ, the frictionless slide. And the block only accelerates if sin θ > μ cos θ, i.e. tan θ > μ — exactly the slipping condition from above. The two results are one idea seen from two sides.

◆ Activity — forces on an incline

Drag the two sliders. The red arrow is the down-slope pull mg sin θ; the blue arrow is friction resisting up-slope. Watch friction grow to match the pull — until the slope is too steep or too slippery, and the block breaks free. When it slides, the readout gives a = g (sin θ − μ cos θ).

9Circular motion & banking of roads

Recall from kinematics that a body moving in a circle of radius r at speed v is accelerating toward the centre, with magnitude v2/r. By Newton's second law that acceleration needs a real, inward (centripetal) force:

Centripetal forceFc = m v2r

“Centripetal” is not a new kind of force — it is the role played by whatever real force points to the centre: tension for a whirled stone, gravity for an orbiting satellite, and friction for a car on a flat bend. On a flat road, only sideways friction bends the car, and if you go too fast for the available friction you skid straight on.

Engineers solve this by banking the road — tilting it inward through an angle θ so the surface itself helps turn the car. Consider the ideal case with no friction. Two forces act: gravity mg down, and the normal force N perpendicular to the tilted road. The normal now leans inward, so it has a horizontal part.

θ mg N N sin θ N cos θ
On a banked road the normal force N tilts inward. Its horizontal part N sin θ supplies the centripetal force; its vertical part N cos θ holds up the weight.

Write Newton's second law in the two directions. Vertically there is no acceleration, so the upward component balances gravity; horizontally the inward component supplies the centripetal force:

N cos θ = m g   ·   N sin θ = m v2r

Divide the second equation by the first. The N and the m both cancel, leaving the banking angle in a strikingly simple form:

Banking of roadstan θ = v2r g

Notice the mass has vanished — the correct bank angle is the same for a scooter and a lorry. It is fixed for one design speed v; below it you lean on friction to stop sliding down, above it friction must stop you sliding out. This is why race tracks and highway curves are steeply banked while a slow city corner is nearly flat.

Worked example

A 2 kg block is placed on a slope inclined at 30°. The coefficients of friction are μs = 0.5 and μk = 0.3. Take g = 10 m s−2. Does it slide, and if so with what acceleration?

Step 1 — will it move? Compare tan θ with μs. Here tan 30° = 0.577, which is greater than μs = 0.5, so the down-slope pull beats the most static friction can offer — the block slides.

Step 2 — the forces along the slope. Down-slope: mg sin θ = 2 × 10 × 0.5 = 10 N. Normal: N = mg cos θ = 2 × 10 × 0.866 = 17.3 N. Kinetic friction (up-slope): fk = μk N = 0.3 × 17.3 = 5.2 N.

Step 3 — Newton's second law. Net down-slope force = 10 − 5.2 = 4.8 N, so a = 4.82 = 2.4 m s−2 down the incline.

Check with the formula: a = g(sin θ − μk cos θ) = 10(0.5 − 0.3 × 0.866) = 10 × 0.24 = 2.4 m s−2 ✓. The mass never entered the acceleration — only the “stays or slides” test needs the actual forces.

10Common confusions to clear up

11Check yourself

Class 11 Physics · Laws of Motion & Friction · aligned to NCERT Chapter 5 · SmartStudy.School