Laws of Motion · Part 2: Friction

Even polished surfaces are rough on a microscopic scale. They touch only at a few high points, and the real contact area is a tiny fraction of the apparent area. At these points the atoms are so close that they bond (a kind of "cold welding"). Friction is the force needed to shear these bonds and to plough the high points through each other.

The laws of limiting friction follow from this picture:

1. Limiting friction acts along the surface, opposite to the direction in which the body is about to slip.
2. It is proportional to the normal force: \(f_{s,\max}=\mu_sN\). A larger N squeezes more high points into contact.
3. It does not depend on the apparent area of contact. Spreading the same N over a larger area lowers the pressure, so each high point flattens less. The real contact area stays about the same.
4. It depends on the nature and condition of the two surfaces, which is what \(\mu_s\) describes.

Kinetic friction obeys the same laws with \(\mu_k\). Usually \(\mu_k<\mu_s\), because sliding surfaces do not get time to form as many bonds. This is why it is harder to start a heavy box moving than to keep it moving.

Push a block with a horizontal force F that grows slowly from zero (see the figure).

• Static region: the block stays at rest, so friction exactly balances the push, f = F. The graph is a straight line through the origin with slope 1 (45° when both axes use the same scale).
• Limiting point: at F = μₛN the block is just about to move. This is the peak of the graph.
• Kinetic region: once the block slides, friction drops to μₖN and stays almost constant, whatever F is. The extra force F − μₖN now produces acceleration.

If a question gives the graph and the mass, read μₛ from the peak and μₖ from the flat level, each divided by N = mg.

The surface exerts one contact force on the block. Its two components are N (perpendicular) and f (along the surface). Its magnitude is \(\sqrt{N^2+f^2}\).

On a horizontal surface with only a horizontal push, N = mg while f can be anything from 0 to μₛmg. So the contact force lies between mg (no friction needed) and mg√(1+μₛ²) (limiting friction). It is never less than mg.

At the limit, the contact force tilts away from the normal by an angle λ with \(\tan\lambda=f_{\max}/N=\mu_s\). This λ is the angle of friction. It returns in the incline section as the angle of repose.

A body sliding on a level floor with no other horizontal force slows down only because of kinetic friction.

1. Friction = μₖmg, so the deceleration is a = μₖg. The mass cancels.
2. From v² = u² − 2as with v = 0: stopping distance s = u²/(2μₖg).
3. From v = u − at: stopping time t = u/(μₖg).
4. The same result comes from the work–energy theorem: friction work −μₖmg·s removes the whole kinetic energy ½mu².

Proportional reasoning: s ∝ u² and t ∝ u. Doubling the speed makes the stopping distance four times larger but the time only twice as long. For the same kinetic energy K, s = K/(μₖmg), so the lighter body slides farther.

Kinetic friction opposes the relative sliding of the two surfaces. On a fixed floor that is the same as opposing the body's motion, so its work is negative. If the other surface moves, look at each body in the ground frame:

• A box dropped on a moving conveyor belt slides backward relative to the belt. Friction on the box points forward and does positive work on the box, speeding it up.
• Friction on the belt points backward and does negative work on the belt. The motor must supply that energy.
• The two works add to −f × (relative sliding distance). This net negative amount is the heat produced at the contact.

1. Tilt a plane slowly until the block is just about to slide. Call this angle θᵣ, the angle of repose.
2. Along the plane: friction balances the downhill pull, f = mg sinθᵣ. Perpendicular: N = mg cosθᵣ.
3. At the limit f = μₛN, so mg sinθᵣ = μₛ mg cosθᵣ, giving tanθᵣ = μₛ.
4. Only two forces now act: the weight and the total contact force. So the contact force must be vertical. It makes angle θᵣ with the normal to the plane.
5. But the angle between the limiting contact force and the normal is, by definition, the angle of friction λ. Hence θᵣ = λ, and tanθᵣ = tanλ = μₛ.

Below θᵣ the block stays at rest and friction is only mg sinθ. Above θᵣ it slides with a = g(sinθ − μₖcosθ).

A force F pushes the block along the incline (parallel to the surface). Friction always points opposite to the way the block would slip.

• To start moving it up (or move it up at constant speed with μₖ): friction acts down, so F = mg(sinθ + μcosθ).
• To just stop it sliding down (only needed when tanθ > μ): friction acts up, so F = mg(sinθ − μcosθ).
• For any F between these two values the block stays at rest, and static friction takes whatever value and direction is needed.

Going up: gravity and friction both act down the slope, so the deceleration is g(sinθ + μₖcosθ). Coming down (if tanθ > μₛ): a = g(sinθ − μₖcosθ). The trip down is slower, so it takes longer than the trip up over the same distance: t_down/t_up = √(a_up/a_down).

Rough vs smooth: if sliding down a rough incline takes n times as long as down the same smooth incline (same length L from rest), then L = ½ a t² gives a_smooth/a_rough = n². So sinθ = n²(sinθ − μcosθ), which rearranges to μ = tanθ(1 − 1/n²). At 45° this is simply μ = 1 − 1/n².

1. Pull with P at angle θ above the horizontal (figure). Vertical balance: N = mg − P sinθ.
2. At the point of sliding: P cosθ = μN = μ(mg − P sinθ).
3. So \(P=\dfrac{\mu mg}{\cos\theta+\mu\sin\theta}\).
4. P is least when the denominator is greatest. Write cosθ + μ sinθ = √(1 + μ²) cos(θ − λ), where tanλ = μ. Its maximum, √(1 + μ²), occurs at θ = λ.
5. So the best angle is the angle of friction, tanθ = μ, and \(P_{\min}=\dfrac{\mu mg}{\sqrt{1+\mu^2}}=mg\sin\lambda\).

Physical picture: tilting the pull upward does two things. Its horizontal part shrinks (bad), but it lifts some weight off the floor, cutting friction (good). The best balance is at θ = λ.

A push directed downward at θ below the horizontal presses the block into the floor: N = mg + P sinθ. At the point of sliding P cosθ = μ(mg + P sinθ), so

\[P_{\rm push}=\frac{\mu mg}{\cos\theta-\mu\sin\theta}.\]

The denominator is smaller than for a pull, so pushing always needs more force. If cosθ ≤ μ sinθ (that is, tanθ ≥ 1/μ), no push however large can start the block: every extra newton of push adds at least as much friction as driving force. This is why a lawn roller is pulled rather than pushed.

A block m₁ lies on a rough table and is tied over a light pulley to a hanging block m₂.

1. Will it move? The pull available is m₂g. The most static friction can supply is μₛm₁g. Motion starts only if m₂g > μₛm₁g, i.e. m₂ > μₛm₁.
2. If it moves: treat the two blocks as one system along the string. Driving force m₂g, opposing force μₖm₁g, total mass m₁ + m₂: \[a=\frac{(m_2-\mu_km_1)g}{m_1+m_2}\]
3. Tension: from the hanging block, T = m₂(g − a).
4. If it does not move: a = 0, T = m₂g, and friction = m₂g (not μₛm₁g).

On a road banked at θ, friction can act down the slope (car tending to skid outward at high speed) or up the slope (car tending to slide inward at low speed). Resolving forces gives the safe range of speeds:

\[v_{\max}=\sqrt{rg\,\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}},\qquad v_{\min}=\sqrt{rg\,\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}}\]

Special cases: θ = 0 gives v_max = √(μₛrg), the level-road result. μₛ = 0 gives v = √(rg tanθ), the design speed of a frictionless bank. If μₛ ≥ tanθ, v_min = 0 and a car can even stand still on the bank.

Driving and braking forces on a car come from static friction between tyres and road. If all wheels are driven or braked, the largest acceleration or deceleration is μₛg. A box on a truck floor can only be accelerated by friction too, so the truck must not accelerate or brake harder than μₛg (box–floor) or the box slides.

1. Assume no slip. Common acceleration a = F/(m + M) (smooth floor).
2. Isolate the block without F. Friction is the only horizontal force on it, so the friction needed is its mass times a.
3. Compare with the limit μₛmg.
   • If needed ≤ limit: they move together, and friction equals the needed value (not μₛmg).
   • If needed > limit: they slip. Use kinetic friction μₖmg on each block, in opposite directions.

Force F on the lower block. The upper block is carried by friction: m a ≤ μₛmg, so a ≤ μₛg. The pair then needs F = (m + M)a, giving F_max = μₛ(m + M)g.

Force F on the upper block. Now the lower block is carried by friction: M a ≤ μₛmg, so a ≤ μₛmg/M. Hence F_max = (m + M)μₛmg/M.

Below either limit, the friction on the carried block is a fraction of F: f = mF/(m + M) in the first case and f = MF/(m + M) in the second.

Take F on the lower block, smooth floor, F above the limit.

• Upper block: only kinetic friction acts, forward: a_m = μₖg (independent of F).
• Lower block: a_M = (F − μₖmg)/M.
• Relative acceleration a_rel = a_M − a_m. The upper block slides backward relative to the lower one (but still moves forward relative to the ground).
• If it must slide a length L to fall off, starting from rest relative to M: t = √(2L/a_rel).

With a rough floor (coefficient μ₂), add the floor friction μ₂(m + M)g on the lower block. Its normal force from the floor is the weight of both blocks.

• Lubrication: oil or grease puts a thin fluid layer between the surfaces. Solid–solid friction is replaced by the much smaller fluid (viscous) friction.
• Ball and roller bearings: replace sliding by rolling.
• Air cushion: a hovercraft or air-track glider floats on a thin layer of air.
• Polishing: reduces the high points, but only up to a point. Very highly polished flat surfaces can have more friction, because more atoms come close enough to bond (adhesion).
• Streamlining: reduces fluid friction (drag) on vehicles and aircraft.

• Walking: your foot pushes backward on the ground. The ground’s static friction on the foot points forward and drives you ahead.
• Bicycle being pedalled: the rear wheel is driven, its bottom tends to slip backward, so friction on the rear wheel acts forward. The front wheel is pushed along, so friction on it acts backward and simply makes it turn.
• Bicycle freewheeling (no pedalling): friction on both wheels acts backward.
• Car tyres have treads so water is pushed out, keeping contact and friction on wet roads.
• Friction is also why nails and screws hold, why brakes work, and why belts drive pulleys.