# Example 3.3: Vehicle accelerating on level ground

A vehicle of mass $m$ accelerates on level ground with acceleration $a$. Its center of mass $C$ is at a height $h$ above the ground, at a distance $a_1$ from the front axle and $a_2$ from the rear one (wheelbase $L = a_1 + a_2$). What is the vertical reaction force at each wheel? Apply it to $m = 1500$ kg, $a = 3$ m/s², $a_1 = 1.2$ m, $a_2 = 1.5$ m and $h = 0.55$ m.

<img src="figs/cocheAcelerandoHorizontal.png" alt="Accelerating vehicle" width="560px">

The vehicle is assumed to be symmetric, so both wheels of an axle carry the same load: $2F_{z1}$ at the front axle and $2F_{z2}$ at the rear one.

:::{dropdown} Solution
We study the whole vehicle as a single rigid body (the wheels rotate, but their mass is small compared with the rest, and we neglect their rotational inertia).

**Inertia loads.** The vehicle translates with acceleration $a$ forwards ($+x$): by D'Alembert's principle, we add an inertia force $-m a$ **backwards**, applied at the center of mass $C$.

**Equilibrium.** The forces on the vehicle are: the weight $mg$ at $C$, the vertical reactions $2F_{z1}$ and $2F_{z2}$ and the horizontal ones $2F_{x1}$ and $2F_{x2}$ at the wheel contacts, and the inertia force. The equations are:

$$
\begin{aligned}
\sum F_x = 0: &\quad 2F_{x1} + 2F_{x2} - m a = 0 \\
\sum F_z = 0: &\quad 2F_{z1} + 2F_{z2} - m g = 0 \\
\sum M_{\text{rear wheel}} = 0: &\quad 2F_{z1}\, (a_1 + a_2) - m g\, a_2 + m a\, h = 0
\end{aligned}
$$

In the moment equation, taken about the contact point of the rear wheel (counterclockwise positive), neither the rear wheel forces nor any horizontal force on the ground appear; the inertia force, with arm $h$, does have a moment. From the third and second equations:

$$
F_{z1} = \frac{mg}{2}\frac{a_2}{L} - \frac{ma}{2}\frac{h}{L}, \qquad
F_{z2} = \frac{mg}{2}\frac{a_1}{L} + \frac{ma}{2}\frac{h}{L}
$$

The first term is the static weight distribution; the second one is the **load transfer** due to the acceleration: when accelerating, the front axle is unloaded and the rear one is loaded (the opposite happens when braking).

**Numerical application.** With $L = 2.7$ m:

$$
\begin{aligned}
F_{z1} &= \frac{1500 \cdot 9.81}{2}\frac{1.5}{2.7} - \frac{1500 \cdot 3}{2}\frac{0.55}{2.7} = 4087.5 - 458.3 = 3629~\text{N} \\
F_{z2} &= \frac{1500 \cdot 9.81}{2}\frac{1.2}{2.7} + \frac{1500 \cdot 3}{2}\frac{0.55}{2.7} = 3270.0 + 458.3 = 3728~\text{N}
\end{aligned}
$$

The front wheels would lift off ($F_{z1} = 0$) at an acceleration $a = g\,a_2/h = 26.8$ m/s², far above what a passenger car can reach. On a motorcycle, with a large $h$ and a short $L$, this condition is reached (a "wheelie").

**Horizontal forces.** We only know that $2F_{x1} + 2F_{x2} = m a$: how they are shared between the axles is **indeterminate** with the equations of statics (a hyperstatic problem). It depends on the drivetrain: in a front-wheel drive vehicle, $F_{x2} \approx 0$ (except for rolling resistance), and in a rear-wheel drive one, $F_{x1} \approx 0$.
:::