Physics · Mechanics · Grade 7

Car braking

Why is it essential to slow down on a wet road?

The situation

You are driving along a city street. Suddenly a mannequin steps out onto the pedestrian crossing ahead. Even before the driver sees the danger and presses the brake, the car keeps going down the road. The road surface, the speed and attention — which of them will save a life?

Topic

Force of friction. Braking of a car: dry, wet and icy roads

Key ideas

Force of friction
Newton's second law
Uniformly decelerated motion

Formulas

s_{r} = v_{0} t_{r}

Reaction distance: until the driver presses the brake, the car moves uniformly at speed $v_{0}$ .

F_{f r} = μ m g \Rightarrow a = \frac{F_{f r}}{m} = μ g

Newton's second law: the deceleration does not depend on mass.

s_{t} = \frac{v_{0}^{2}}{2 μ g}, S = v_{0} t_{r} + \frac{v_{0}^{2}}{2 μ g}

The braking distance is proportional to the square of the speed; $S$ is the total stopping distance.

v_{z} = \sqrt{v_{0}^{2} - 2 μ g (D - s_{r})}

If the car cannot stop within distance $D$ — the impact speed.

F_{t} \leq μ_{0} N, F_{s} = μ N, F_{d} = μ_{d} N, μ_{d} ≪ μ < μ_{0}

Static, sliding and rolling friction. In the laboratory, with uniform pulling, $μ = F / N$ .

Theory

Two stages

Stopping has two parts. During the reaction time ( $t_{r}$ ) the driver sees the danger and moves a foot to the brake — the car is not yet slowing down. Then comes braking: the friction force between the wheels and the road stops the car.

Why doesn't mass matter?

In a heavy car the friction force is also large: $F = μ m g$ . But it is just as hard to stop the car: $a = F / m = μ g$ . As a result, in the ideal case a light car and a heavy car stop in the same distance. In real life lorries take longer to stop — their brakes overheat and their tyres are different.

The square of the speed

The braking distance is proportional to $v_{0}^{2}$ : if the speed increases 2 times, the braking distance becomes 4 times longer. Going from 60 km/h to 90 km/h increases the braking distance 2.25 times.

What does ABS do?

A locked wheel slides on the road and sliding friction acts. ABS stops the wheel from locking and holds the tyre at the limit of slipping — the coefficient of static friction is larger, and the driver also keeps control of the car.

Types of friction: lab work

Friction arises from the roughness of surfaces and from the mutual attraction of the molecules where they touch. Static friction is equal to the pulling force and opposite to it: in “Measuring friction” mode the force-meter reading grows while the sledge stays in place. When the force reaches $μ_{0} N$ the sledge starts to move and sliding friction $μ N$ acts — the peak and the flat part of the $F (t)$ graph. In uniform pulling $F_{meter} = F_{f r}$ , so $μ = F / N$ ; if you change $N$ by putting discs on the sledge, $F (N)$ is a straight line through the origin. $μ$ depends on the material and roughness of the surfaces and hardly at all on the area of contact. When a wheel rolls, $μ_{r} \approx 0.01$ – $0.02$ — tens of times smaller than for sliding: this is why we need the wheel and the bearing. In engineering friction is both a friend (brakes, tyre tread, sand spread on the road in winter) and an enemy (wear, heating). Viscous friction in a liquid depends on speed — see the “Viscosity and flow” experiment.

Newton's laws on the road

The main task of dynamics is to find the motion of a body from the forces acting on it (or the force from the motion). Here, from the friction force we get $a = μ g$ , and from that the braking distance. Several forces act on a car: its weight $m g$ , the normal reaction of the road $N$ (vertically $N = m g$ ), friction and air resistance. First law (inertia): during the reaction time the driver holds the speed — the resultant force is zero and the car moves uniformly. A frame of reference fixed to the Earth is approximately inertial; a braking car is non-inertial: a passenger without a seat belt continues forward by inertia, and inside the car it seems as if an inertial force $- m \vec{a}$ directed forward acts on them. Second law: $a = F / m$ ; if passengers sit in the car, the masses add (mass is additive), but $a = μ g$ does not change. Third law: the tyre pushes the road forward and the road pushes the tyre backward with the same force; whatever force the sledge pulls the force meter with, the force meter pulls the sledge with the same force.

Did you know?

An impact at 50 km/h is equal to a fall from a height $h = v^{2} / 2 g \approx 9.8$ m — roughly from the roof of a three-storey house.
On wet asphalt the friction coefficient can be 1.5–2 times smaller than on dry asphalt, and on ice 10 times smaller.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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