Why Highway Curves Are Built Tilted Instead of Flat

Without friction you could not walk, a car could not turn, and a knot would slip open. With too much, machines wear out and waste energy as heat.

Friction also decides how fast a vehicle can safely take a curve, which is why highway bends are tilted.

This part covers the types and laws of friction, lubrication, circular motion on level and banked roads, and solving problems with several bodies.

**Static friction adjusts to prevent sliding up to a maximum $f_{s,max}={\mu }_{s}N$ (limiting friction); kinetic friction $f_k={\mu }_{k}N$ acts during sliding with ${\mu }_k<{\mu }_s$; rolling friction is much smaller than both; and the angle of repose satisfies $\tan {\theta }_r={\mu }_s$.

Laws of friction: limiting friction is proportional to the normal force, is independent of the area of contact over a wide range, and depends on the nature of the surfaces.

Worked example.** A $10$ kg box on a floor has ${\mu }_s=0.5$ and ${\mu }_k=0.4$; take $g=10$ m/s${}^2$, so $N=100$ N.

- Limiting friction $=0.5\times 100=50$ N
- **Push of $30$ N**: static friction is $30$ N, and the box stays put
- **Push of $60$ N**: the box slides, friction drops to $0.4\times 100=40$ N, and

$$a=\frac{60-40}{10}=2{\text{ m/s}}^2$$

Angle of repose. On an incline, a block just starts to slide when $mg\sin \theta ={\mu }_{s}mg\cos \theta$, so

$$\tan {\theta }_r={\mu }_s=0.5\Rightarrow {\theta }_r\approx 26.6°$$

An everyday example. A heavy steel almirah is hard to get moving, but once it slides it needs a smaller push.

The substance. **Static friction is not always ${\mu }_{s}N$** — it equals the applied force until that limit is reached.

A lubricant fills the tiny hills and valleys of two surfaces with a thin fluid layer, so the solids slide over the fluid instead of gripping each other; ball bearings go further by replacing sliding with rolling.

Worked example. Suppose a $200$ N load in a machine slides with $\mu =0.3$, and bearings bring the effective coefficient down to $0.02$.

$$f_{\text{before}}=0.3\times 200=60\text{ N}{f}_{\text{after}}=0.02\times 200=4\text{ N}$$

Friction is useful for walking, braking, tyres gripping the road, holding a nail in wood, and writing with a pen.

Friction is wasteful in engines, gears and axles, where it causes wear and turns useful energy into heat.

An everyday example. Oiling a squeaky bicycle chain makes pedalling easier and stops the metal links grinding each other down.

The substance. Friction can be reduced but not removed completely — and tyres are given treads precisely to keep enough of it on wet roads.

**On a level road only friction supplies the centripetal force, giving $v_{max}=\sqrt{{\mu }_{s}rg}$; on a road banked at $\theta$, part of the normal force also points towards the centre, giving $v_{max}=\sqrt{\frac{rg({\mu }_s+\tan \theta )}{1-{\mu }_s\tan \theta }}$.

Level road.** $\frac{m{v}^2}{r}\le {\mu }_{s}mg$, so $v_{max}=\sqrt{{\mu }_{s}rg}$.

Banked road. Resolving $N$ and $f={\mu }_{s}N$ (down the slope at top speed):

$$N\cos \theta =mg+f\sin \theta N\sin \theta +f\cos \theta =\frac{m{v}^2}{r}$$

Dividing gives the formula above. With no friction, the optimum speed is $v_o=\sqrt{rg\tan \theta }$.

Worked example. A curve has $r=50$ m, ${\mu }_s=0.4$ and $g=10$ m/s${}^2$.

$$\text{level: }{v}_{max}=\sqrt{0.4\times 50\times 10}=\sqrt{200}\approx 14.1\text{ m/s}\approx 51\text{ km/h}$$

Banked at $15°$, with $\tan 15°\approx 0.268$:

$$v_o=\sqrt{50\times 10\times 0.268}\approx 11.6\text{ m/s}{v}_{max}=\sqrt{\frac{500(0.4+0.268)}{1-0.4\times 0.268}}\approx 19.3\text{ m/s}\approx 70\text{ km/h}$$

An everyday example. A cyclist leans into a turn for the same reason a highway bend is tilted — so the ground's push has an inward part.

The substance. At the optimum speed, no friction is needed at all, which is why banking reduces tyre wear.

Draw a separate free-body diagram for each body, choose axes along each body's motion, write Newton's second law for each, and solve the equations together; treat the whole system as one body to find the common acceleration quickly.

Take $g=10$ m/s${}^2$.

Worked example 1 — blocks in contact. A $20$ N push acts on a $4$ kg block touching a $6$ kg block on a smooth floor.

$$a=\frac{20}{4+6}=2{\text{ m/s}}^2\text{contact force on }6\text{ kg}=6\times 2=12\text{ N}$$

Worked example 2 — Atwood machine. Masses of $3$ kg and $2$ kg hang over a light pulley.

$$a=\frac{(3-2)(10)}{5}=2{\text{ m/s}}^{2}T=2(10+2)=24\text{ N}$$

Worked example 3 — with friction. A $5$ kg block on a table (${\mu }_k=0.2$) is pulled by a hanging $3$ kg block over a pulley.

$$a=\frac{30-0.2\times 50}{8}=2.5{\text{ m/s}}^{2}T=3(10-2.5)=22.5\text{ N}$$

An everyday example. A tractor pulling a loaded trolley is a connected-body system linked by the tow bar.

The substance. Internal forces such as tension cancel for the whole system — isolate one body to find them.