You might have heard, during your first driving lessons, that it is not recommended to brake heavily during cornering. And this is correct, it’s better to brake before the corner, to slow down, and only steer for the whole length of the corner.

To understand what’s happening, let’s have a look at the **forces acting on a braked wheel**.

where:

*v _{w}* [m/s] – wheel linear speed

*M*[Nm] – braking torque (moment)

_{B}*G*[N] – vehicle weight acting on the wheel

_{w}*F*[N] – road reaction (normal) force acting on the wheel

_{z,w}*F*[N] – vehicle body longitudinal force pushing the wheel

_{x,w}*X*[N] – wheel friction force trying to stop the wheel

_{w}When the driver brakes, the vehicle inertia is trying to still push the wheel forward. On the other side, the friction between the tire and the road is trying to keep the wheel in place. If the friction force is greater than the vehicle push force, the wheel will slow down and come to a stop.

The **friction force** between the wheel and the road depends on two things:

- normal load (vehicle weight)
- coefficient of friction (
*μ*)

According to **Newton’s third law**, for every action force, there is an equal and opposite reaction force. This means that the normal reaction force in the contact patch is equal to the vehicle weight acting on the wheel.

Now, we can calculate the **friction (braking) force** in the contact patch, function of vehicle weight.

Bear in mind that this is the **maximum friction force** that can be applied. If the driver brakes the braking force can be anywhere between *0* and *X _{w}*.

Let’s calculate what is the the maximum friction force or a vehicle mass of *2000* kg. The friction coefficient is not constant, it depends on the wheel slip and road type. In order to simplify the calculation, we are going to assume a constant **static friction coefficient**, with the value of *0.9*.

X_w &= \mu \cdot G_w = \mu \cdot \frac{m}{4} \cdot g\\

X_w &= 0.9 \cdot \frac{2000}{4} \cdot 9.81\\

X_w &= 4414 \text{ N}

\end{split} \end{equation*} \]

We also assumed that the vehicle mass acting on the wheel is a quarter (1/4) of the total mass. In reality it’s a bit different, because during braking, the vehicle rotates slightly around the lateral, y-axis, and more weight is transferred on the front axle. During braking the vehicle weight is not equally distributed between front and rear axle.

For a better understanding we are going to draw a circle around the wheel. The circle is named **Kamm circle** and represents the maximum total friction force that can applied in the contact patch.

If the driver only brakes, in straight line, all the available friction force will be used in longitudinal direction *X _{w}*, to slow the vehicle down. If the braking force exceeds this limit, the wheel will lock and slip.

Let’s see what happens when the driver only steers, during a corner, without any braking.

As you can see, all the available friction force is used by the lateral force to steer the vehicle. The lateral force *Y _{w}* is counteracting the centrifugal force and the vehicle changes direction. The wheel will not slip as long as the lateral force will be withing the Kamm circle.

**What happens if the driver steers and brakes in the same time ?**

As you can see in the image above, the **total friction force at the wheel ( F_{w})** is now split between the longitudinal component (

*X*) and lateral component (

_{w}*Y*).

_{w}Keep in mind that* X _{w}* and

*Y*are the maximum longitudinal and lateral forces that can be transmitted.

_{w}In this situation, **the vehicle can skid** if one of the below conditions is met:

- the wheel brake force is higher than the maximum longitudinal friction force (
*X*)_{w} - the vehicle centrifugal force (which increases with speed) is higher than the sum of the lateral wheel forces (
*Y*)_{w}

Since both longitudinal and lateral forces have to share the total maximum friction force (*F _{w}*), wheel slip limit can be reached easier. If one of the two conditions is met, the driver can lose the control of the vehicle.

Fortunately most of the modern vehicles have **electronic stability control systems** which react to these situation by redistributing the wheel forces and thus preventing vehicle skid.

## Jose Menendez

Here is a physics clarification. you write “According to Newton’s third law, for every action force, there is an equal and opposite reaction force. This means that the normal reaction force in the contact patch is equal to the vehicle weight acting on the wheel.”

These two forces are indeed equal, but it is not because of Newton’s third law. The weight of the vehicle is a force from the entire Earth on the vehicle. By Newton’s third law, this force is equal to the force with which the vehicle attracts the Earth. The “normal reaction force” is a force from the road on the car. By Newton’s third law, the car presses down on the road with a force equal to the normal reaction force.

The are two vertical forces acting on the vehicle: the weight and the normal reaction force. But since we know that the acceleration of the car in the vertical direction is zero (in fact, the vehicle is not moving at all in the vertical direction), then the net force in the vertical direction has to be zero by Newton’s second law, and this implies that weight is equal to the normal reaction force.