The question in the above link describes a dragster (1000kg) of unlimited power towing a cart (500kg), and a static coefficient of friction between the wheels and ground of 0.6. Given the vertical (0.5m) and horizontal (1m) distances between the COG and the rear wheel base, along with the wheelbase (6m), what is the maximum acceleration of the body?
I have been told that the answer is in the form of an equation resembling one below, however it eludes me as to how it is derived. $$ a = \frac{\mu g (l-b)}{b-\mu h} $$ where $l$ and $h$ are the horizontal and vertical COG distances from the rear wheel base respectively, and $b$ is the length of the wheel base
A similar question is given in the link below: Similar_Q
This probably belongs in a physics forum, as the math involved is quite simple, but I’ll point you toward a solution by solving the somewhat simpler second problem that you linked in your question.
The thrust that produces forward acceleration is applied at the tire-road interface (the “contact patch”). This is not in line with the vehicle’s CG, so it generates a moment that tends to pitch the nose upwards. This tendency is resisted by the ground pushing back on the front and rear tires, with the net effect of transferring load from the front to the rear.
The steady-state load under acceleration can be computed using a quasi-static model: the vertical forces balance, the resultant horizontal force produces the given acceleration, and the net moment is zero. Taking the positive $x$-direction to be forward and the positive $y$-direction to be downward (so that the vehicle’s weight is positive), we compute the net moment: $$M=-F_{thrust}h+F_{rear,y}l_2-F_{front,y}l_1=0,$$ where $l_1$ is the distance from the CG to the front wheels and $l_2$ the distance to the rear wheels, so that $l_1+l_2=l$. The sum of the front and rear loads equals the vehicle’s weight, so we can substitute and rearrange to get $$F_{front,y}=mg(l_2/l)-F_{thrust}(h/l)\tag{1}$$ for the load on the front (driven) wheels. The first term is just the static load on the front axle, so this equation says that under acceleration the load on the front axle is reduced by an amount that depends on the applied thrust and the vehicle’s geometry.
Wheelspin begins at the point that the applied thrust equals the static frictional force at the contact patch. For this problem, we assume that the coefficient of friction is constant (in real life it depends nonlinearly on tire load, among other things), from which we get $$F_{thrust, max}=\mu_sF_{front,y}=\mu_smg(l_2/l)-\mu_sF_{thrust,max}(h/l).$$ Solving for the maximum thrust yields $$F_{thrust,max}={\mu_smg(l_2/l)\over1+\mu_s(h/l)}$$ so the maximum acceleration is $$a_{max}={\mu_s(l_2/l)\over1+\mu_s(h/l)}g.\tag{2}$$ A quick dimensional analysis verifies that this solution is plausible. Plugging in the values from the problem and using $g=9.8\,m/s^2$ we get $$a_{max}={0.3\times1/3\over1+0.3\times1\,m/3.0\,m}\times9.8\,m/s^2\approx0.89\,m/s^2.$$
The other problem is similar. However, since the driven wheels are at the rear, the acceleration will instead increase the load on the driven wheels, and so increase the maximum acceleration. You will also have to take into account the extra loading and moments created by the trailer, but the calculations involved are similar.
It might not come up, but another factor limiting maximum acceleration is wheelstand: increasing thrust beyond the point at which the front load is zero could flip the vehicle over. This limit is easy to compute using the information presented above.
Acknowledgement: The above derivation follows in general outline that presented in chapters 16 and 17 of Beikmann’s Physics For Gearheads.