Friction

Friction is examined explicitly in CBSE Class 11 Physics, Unit III "Laws of Motion," alongside Newton's three laws, momentum, equilibrium, and the dynamics of circular motion — none of that surrounding material has been rationalised out of the current 2025-26 prescription. The board expects you to state and apply the laws of static and kinetic friction, use the angle of repose, and combine friction with circular motion to find the maximum safe speed on level and banked roads. Work-energy questions (how much energy friction dissipates) belong to the next chapter, and torque or rolling-body dynamics belong to the System of Particles chapter — this article stays inside the Laws of Motion boundary.

When two surfaces press together, microscopic high points ("asperities") interlock and weld slightly at real points of contact, even though the surfaces look smooth to the eye. Any force that tries to slide one surface over the other must first break these tiny welds — that resistance is friction, and it always acts parallel to the contact surface, opposing relative motion or the tendency toward it. As long as the body stays at rest, static friction rises to exactly cancel the applied force along the surface, but it cannot rise without limit. Its ceiling is proportional to how hard the surfaces are pressed together, measured by the normal force $N$:

where: $f_s$ is the static friction force actually acting (in newtons), $\mu_s$ is the dimensionless coefficient of static friction (a property of the surface pair), and $N$ is the normal reaction. The physical reading: below the maximum $\mu_s N$, static friction is a response force — it takes whatever value is needed to prevent sliding, nothing more. Only exactly at the verge of slipping does $f_s$ equal $\mu_s N$. Once the body actually slides, the welds are continuously broken and reformed, and experiment shows the resisting force settles to a nearly constant value, independent of speed and of the (macroscopic) contact area:

where: $\mu_k$ is the coefficient of kinetic friction, always somewhat smaller than $\mu_s$ for the same pair of surfaces. That is the physical reason a stuck drawer needs a hard first tug and then slides free with less effort — you are crossing from the static ceiling $\mu_s N$ down to the lower kinetic value $\mu_k N$. A third, much smaller resistance opposes a body that rolls rather than slides — rolling friction — because a rolling surface deforms only slightly at the contact patch instead of scraping across it. Ranking the three: rolling friction is the smallest, kinetic friction next, and the maximum static friction is the largest, which is precisely why engineers put wheels and ball bearings under anything meant to move easily.

Friction is not only about sliding along a straight line — it is also the force that lets a vehicle turn. On a flat, unbanked curve, the only horizontal force available to bend the car's path toward the centre is friction between tyres and road, so friction itself must supply the entire centripetal force. On a banked curve, the road's own slope contributes part of that inward pull, so friction can supply less — or even none at all, at one special "optimum" speed.

where: $v_{\max}$ is the greatest speed at which the car can round the curve without skidding, $r$ is the radius of the circular path, $g$ is the acceleration due to gravity, $\mu_s$ is the coefficient of static friction between tyre and road, and $\theta$ is the banking angle. The physical reading: on the level road, a faster car demands more centripetal force $mv^2/r$, but friction cannot exceed $\mu_s N = \mu_s mg$, so there is a hard speed ceiling set purely by $\mu_s$, $r$, and $g$ — not by the car's mass. On the banked road, tilting the surface lets a component of the normal force point toward the centre, so at the one special speed $v = \sqrt{rg\tan\theta}$ no friction is needed at all; above or below that speed friction must again help out.

Board level — will the crate move, and how fast

A crate of mass $10\ \text{kg}$ sits on a level floor with $\mu_s = 0.40$ and $\mu_k = 0.30$. Take $g = 9.8\ \text{m/s}^2$. (a) Find the largest horizontal push the crate can withstand without moving. (b) Does a $30\ \text{N}$ push move it? (c) Once sliding under a $50\ \text{N}$ push, find its acceleration.

Step 1. Find the normal force. On the level floor the crate has no vertical acceleration, so $N = mg = 10 \times 9.8 = 98\ \text{N}$.

Step 2. Find the maximum static friction, which is the largest push the crate can resist without moving: $f_s^{\max} = \mu_s N = 0.40 \times 98 = 39.2\ \text{N}$.

Step 3. Compare the $30\ \text{N}$ push to this ceiling. Since $30\ \text{N} < 39.2\ \text{N}$, static friction simply rises to match the push — $f_s = 30\ \text{N}$, exactly, no more — and the crate stays put; it does not move.

Step 4. Analyse the $50\ \text{N}$ push, which exceeds $39.2\ \text{N}$, so the crate slides and kinetic friction takes over: $f_k = \mu_k N = 0.30 \times 98 = 29.4\ \text{N}$. The net force is $50 - 29.4 = 20.6\ \text{N}$, giving

The crate resists up to $39.2\ \text{N}$, stays still under $30\ \text{N}$, and accelerates at $2.06\ \text{m/s}^2$ under $50\ \text{N}$.

Competitive level — cornering speed, level versus banked

A car of mass $1200\ \text{kg}$ takes a curve of radius $r = 50\ \text{m}$ on a road with $\mu_s = 0.5$. Take $g = 10\ \text{m/s}^2$. (a) Find the maximum speed on the flat (unbanked) road. (b) Find the optimum (frictionless) banking angle for a design speed of $54\ \text{km/h}$. (c) With that same banking of part (b) and the same $\mu_s = 0.5$, is $90\ \text{km/h}$ safe?

Step 1. Level-road speed limit, from the derivation above: $v_{\max} = \sqrt{\mu_s r g} = \sqrt{0.5 \times 50 \times 10} = \sqrt{250} = 15.8\ \text{m/s}$, about $56.9\ \text{km/h}$. Notice the car's mass, $1200\ \text{kg}$, never entered the calculation — it cancelled out.

Step 2. Optimum banking angle for $v = 54\ \text{km/h} = 15\ \text{m/s}$: $\tan\theta = \dfrac{v^2}{rg} = \dfrac{15^2}{50 \times 10} = \dfrac{225}{500} = 0.45$, so $\theta = \tan^{-1}(0.45) \approx 24.2^\circ$. At exactly this speed on this banking, no friction is needed at all.

Step 3. Check $90\ \text{km/h} = 25\ \text{m/s}$ on that same banked, frictional road using $v_{\max} = \sqrt{\dfrac{rg(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta}}$. With $\mu_s = 0.5$ and $\tan\theta = 0.45$: numerator $= 50 \times 10 \times (0.5 + 0.45) = 500 \times 0.95 = 475$; denominator $= 1 - 0.5 \times 0.45 = 1 - 0.225 = 0.775$. So $v_{\max} = \sqrt{475/0.775} = \sqrt{613} \approx 24.8\ \text{m/s}$, i.e. about $89.2\ \text{km/h}$. A speed of $90\ \text{km/h}$ is just barely unsafe — it sits fractionally above the $\approx 89.2\ \text{km/h}$ maximum this banked, frictional road can hold. Timing note: budget under three minutes. Do the level-road case first as a sanity anchor (mass always cancels — do not waste time carrying it), then the optimum-banking angle by simple inversion of $\tan\theta = v^2/(rg)$, and only then the full banked-with-friction formula, which is worth memorising as a single block rather than re-deriving under time pressure. A quick giveaway that a "safe or not" question is close is when the computed $v_{\max}$ lands within a few percent of the test speed, as it does here — always double-check the arithmetic in $\tan\theta$ and $\mu_s\tan\theta$ before concluding.

• Treating $f_s = \mu_s N$ as a fact rather than a ceiling. Static friction equals only what is needed to prevent sliding; it is $\le \mu_s N$, and reaches $\mu_s N$ only right at the point of slipping.
• Using $\mu_s$ for a body that is already sliding. A moving body obeys $f_k = \mu_k N$ with $\mu_k < \mu_s$ — mixing the two coefficients gives a wrong (too-large) friction force during sliding.
• Forgetting the normal force changes on an incline. On a slope, $N = mg\cos\theta$, not $mg$, so friction there is $\mu N = \mu mg\cos\theta$, smaller than on flat ground.
• Adding a "centrifugal force" to a car turning a level road. In the inertial (ground) frame there is no outward centrifugal force to balance; friction alone provides the real, inward centripetal force. A centrifugal force appears only if you deliberately switch to the car's own rotating frame, which this chapter does not use.
• Forgetting mass cancels in $v_{\max} = \sqrt{\mu_s r g}$. Students sometimes plug in the vehicle's mass by habit; it never appears in the final level-road or optimum-banking formulas.
• Confusing angle of repose with angle of friction. They give the same number, $\tan\theta = \mu_s$, but the angle of repose describes a block on an incline about to slide, while the angle of friction $\lambda$ is defined from the resultant of $N$ and $f_s^{\max}$ — same formula, different physical picture.

• Distinguish static, kinetic, and rolling friction, and rank their sizes correctly: rolling smallest, then kinetic, then maximum static largest.
• Apply $f_s \le \mu_s N$ (an inequality, only equality at the point of slipping) and $f_k = \mu_k N$ (an equality once sliding), choosing the correct regime from the problem's wording.
• Decide whether an applied horizontal force will move a body at rest, and if it does, compute the resulting acceleration using kinetic friction.
• Relate the angle of repose (or angle of friction) to $\mu_s$ via $\tan\theta_r = \mu_s$, and recompute the normal force correctly on an incline.
• Derive and apply $v_{\max} = \sqrt{\mu_s r g}$ for a level circular road, and the banked-road formulas $\tan\theta = v^2/(rg)$ (frictionless optimum) and the full expression with friction, without adding a fictitious centrifugal force in the inertial frame.

Ask the tutor: if the static-versus-kinetic switch or the banked-road formula still feels shaky, open the AI tutor on this page and walk through one more cornering-speed problem step by step.