1. Why friction exists
To the eye, two polished steel plates pressed together look smooth, but under a microscope both surfaces are jagged ranges of asperities touching at only a few peaks. Where those peaks meet, atoms get close enough to form weak adhesive bonds — cold welds. Sliding one surface over the other means continually shearing those bonds and reforming new ones further along. The net horizontal force you must overcome is what we call friction.
Two book-keeping facts follow. Friction is a component of the contact force — the same interaction that supplies the normal reaction \(N\) also supplies a tangential piece \(f\). And that tangential piece opposes the relative motion of the two surfaces in contact, not necessarily the body's motion through space.
2. Static friction
Put a block of mass \(m\) on a horizontal table. Gravity \(mg\) is balanced by the normal reaction \(N\). Push horizontally with a small force \(F\); the block does not move, so a frictional force \(f_s\) must oppose \(F\) exactly. This is static friction, with three properties NEET tests repeatedly.
(i) It is self-adjusting. Push with 5 N and \(f_s = 5\) N. Push with 12 N (still no motion) and \(f_s = 12\) N. Friction takes whatever value enforces equilibrium. (ii) It has a maximum: \((f_s)_{\max} = \mu_s N\). Beyond that the block slides. (iii) The law of static friction is therefore an inequality:
\[ f_s \;\le\; \mu_s N. \]
The dimensionless \(\mu_s\) is the coefficient of static friction, depending only on the surface pair — not on mass, area or velocity. The law is empirical: it summarises experiments rather than deriving from a fundamental force.
3. Kinetic (sliding) friction
Push hard enough and the block breaks free. The instant relative sliding begins, the frictional force drops to a smaller, roughly constant value — kinetic (or sliding) friction:
\[ f_k = \mu_k N. \]
\(\mu_k\) is dimensionless, depends only on the surface pair, and experimentally satisfies \(\mu_k < \mu_s\). To elementary precision, \(f_k\) is also independent of the relative velocity. The step-down from \(\mu_s\) to \(\mu_k\) explains why a stuck drawer jolts forward when you yank it, and why ABS keeps the wheel on the verge of skidding rather than letting it skid — maximum braking comes from \(\mu_s\), not \(\mu_k\).
4. Static vs kinetic — the graph NEET loves
Plot frictional force \(f\) (y-axis) against applied force \(F\) (x-axis). For small \(F\), \(f = F\) — a 45° line through the origin — until \(f\) hits the static ceiling \(\mu_s N\). At that point the block slides and \(f\) drops to \(\mu_k N\), staying flat as \(F\) grows. The little step at breakaway is the graphical signature of \(\mu_k < \mu_s\).
| Region | Condition | Friction | Block |
|---|---|---|---|
| 1. Below threshold | \(F < \mu_s N\) | \(f_s = F\) (self-adjusting) | At rest |
| 2. At threshold | \(F = \mu_s N\) | \(f_s = (f_s)_{\max} = \mu_s N\) | About to slide |
| 3. Above threshold | \(F > \mu_s N\) | \(f_k = \mu_k N\) (constant) | Sliding, possibly accelerating |
5. Why the coefficient of friction is dimensionless
From \(f = \mu N\), both sides are forces:
\[ [\mu] = \frac{[f]}{[N]} = \frac{[M L T^{-2}]}{[M L T^{-2}]} = [M^0 L^0 T^0]. \]
\(\mu\) is a pure number with no SI unit — exactly the answer to NEET 2018 Q.15, where the distractor "coefficient of sliding friction has dimensions of length" is the wrong statement. Related: the maximum acceleration that friction can produce on a body on a horizontal surface is \(a_{\max} = \mu_s g\) — mass-independent, a recurring NEET theme.
6. Why friction is independent of contact area
Lay a brick flat on a table — wide face down — and measure the minimum force to slide it. Stand the brick on a narrow edge so the apparent contact area is much smaller. Repeat. To NEET-level accuracy, the answer is the same. Friction depends on \(N\) and the surface pair, not on the macroscopic footprint.
The microscopic reason: the true contact area — the sum of asperity patches — is itself proportional to \(N\), not to the apparent footprint. Spread the same weight over more area and each asperity carries less load; the true contact area stays the same.
7. Angle of repose — derivation
Place a block on a plank and slowly tilt. At first the block sits still — static friction \(f_s\) up the slope balances the down-slope pull of gravity. At a critical angle \(\theta_{\max}\) the block just begins to slide. This is the angle of repose — the cleanest way to measure \(\mu_s\) experimentally.
Resolve \(mg\): down the incline \(mg\sin\theta\), perpendicular \(mg\cos\theta\), so \(N = mg\cos\theta\). At the verge of sliding, static friction is maximal:
\[ f_s = mg \sin\theta_{\max} = \mu_s N = \mu_s\, mg \cos\theta_{\max}. \]
\[ \boxed{\;\tan\theta_{\max} = \mu_s\;} \]
Mass cancels — angle of repose is mass-independent. A heavy block and a light block of the same material slide at the same angle. The angle of repose equals the angle of friction \(\varphi\), defined by \(\tan\varphi = \mu_s\) — same number, two names.
Friction on an inclined plane is the warm-up for banking of roads — where \(\mu_s\) sets the maximum safe speed on a curve.
8. Rolling friction — and why the wheel is a civilisational hack
A rigid sphere rolling without slipping has only one point of contact at each instant, with zero velocity relative to the surface. With zero relative velocity, sliding friction cannot act — in principle a wheel should experience no friction at all.
In practice, both surfaces deform slightly under load. The contact point spreads into a small patch, and asymmetric deformation at its leading and trailing edges produces a small horizontal force opposing motion — rolling friction. It is typically two to three orders of magnitude smaller than sliding friction. That factor of 100–1000 is the engineering case for the wheel and for ball bearings, which replace sliding axle contact with rolling balls in a race. Air cushions and magnetic levitation eliminate contact altogether.
9. Friction we need vs friction we waste
We need static friction between foot and ground (walking), tyre and road (driving), brake pad and disc (stopping), fingertips and object (gripping). We waste it inside machines, where moving parts dissipate kinetic energy as heat. Mitigations attack \(\mu_k\) or \(N\): lubricants (oil, grease, graphite) interpose a thin layer between asperities; ball bearings replace sliding with rolling; air cushions eliminate contact. Where you need grip, you go the other way — rubber-soled shoes, treaded tyres, rosin on a bow.
10. Worked examples
A box of mass \(m\) lies on the floor of a train. The coefficient of static friction between the box and the floor is \(\mu_s = 0.15\). What is the largest acceleration of the train for which the box will not slide on the floor? Take \(g = 10\) m s\(^{-2}\).
Setup. Look at the box in the ground frame. The only horizontal force on it is static friction from the floor, which must supply the acceleration \(a\) of the train so the box keeps up. From Newton's second law, \(f_s = m a\).
Limit. The largest \(a\) is reached when \(f_s\) hits its ceiling, \(f_s = \mu_s N = \mu_s m g\). Setting the two expressions equal: \(\mu_s m g = m a_{\max}\), so \(a_{\max} = \mu_s g\).
Number. \(a_{\max} = 0.15 \times 10 = \mathbf{1.5}\) m s\(^{-2}\). The mass cancels, so the answer is the same whether the box weighs 1 kg or 100 kg.
A 4 kg block rests on a flat plane that is gradually tilted. The block just begins to slide when the angle of the plane with the horizontal reaches \(\theta_{\max} = 15^\circ\). Find the coefficient of static friction between the block and the plane.
Setup. Resolve the weight along and perpendicular to the plane. Down the slope: \(mg \sin\theta\). Into the plane: \(mg \cos\theta\). The normal reaction is \(N = mg \cos\theta\), and at the angle of repose static friction is at its maximum, \(f_s = \mu_s N\), directed up the slope.
Equilibrium. Along the slope, \(mg \sin\theta_{\max} = \mu_s mg \cos\theta_{\max}\), giving \(\mu_s = \tan\theta_{\max}\).
Number. \(\mu_s = \tan 15^\circ \approx \mathbf{0.27}\). The mass 4 kg never enters the answer — angle of repose is mass-independent.
A trolley of mass 20 kg sits on a horizontal surface and is connected by a light string over a frictionless pulley to a hanging block of mass 3 kg. The coefficient of kinetic friction between the trolley and the surface is \(\mu_k = 0.04\). Find the acceleration of the system and the tension in the string after release. Take \(g = 10\) m s\(^{-2}\).
FBD of the hanging block. Weight \(3g = 30\) N down, tension \(T\) up. Newton's second law (taking downward as positive for this block): \(30 - T = 3a\).
FBD of the trolley. Tension \(T\) forward, kinetic friction \(f_k\) backward. The normal reaction is \(N = 20g = 200\) N, so \(f_k = \mu_k N = 0.04 \times 200 = 8\) N. Newton's second law: \(T - 8 = 20 a\).
Solve. From the second equation \(T = 20 a + 8\). Substitute into the first: \(30 - (20a + 8) = 3a \Rightarrow 22 = 23 a \Rightarrow a \approx \mathbf{0.96}\) m s\(^{-2}\). Tension: \(T = 30 - 3(0.96) \approx \mathbf{27.1}\) N.
Friction — the seven facts NEET tests
- Static friction: \(f_s \le \mu_s N\); self-adjusting; reaches \(\mu_s N\) only at impending motion.
- Kinetic friction: \(f_k = \mu_k N\); \(\mu_k < \mu_s\); approximately velocity-independent.
- Both \(\mu_s\) and \(\mu_k\) are dimensionless properties of the surface pair.
- Friction is independent of apparent contact area.
- Angle of repose: \(\tan\theta_{\max} = \mu_s\); independent of mass.
- Rolling friction is 2–3 orders of magnitude smaller than sliding friction.
- Friction opposes relative motion, not motion of the body in space.