The top acceleration is 16.884 m/s2. The front wheels can go around at will. Leave exclude the wheels' mass from your calculations. 1.72 is the coefficient of friction.
By condition of equilibrium, we can find the Load at each wheel as
Rear load Wr= 224lb=224*4.45=996.8N
Front-load Wf= 261lb=261*4.45=1161.1N
CG of the entire mass will be located at (2.30, 1.62).
CG location Can be found by
x=sum(Wi*xi) /sum(Wi)
Load transfer due to acceleration is given by
DW=m*a*(h/L)
Where h= height of CG= 1.62ft=0.486m
L= wheel base=5ft=1.5m
m= total mass in kg=(335+150) *0.45=212.25kg
For the front wheel to be lifted,
dW=Wf
solving we get the acceleration value as
a= 16.884m/s² of acceleration will be the max for tipping of the vehicle.
We know that,
Tractive force = ma
Gives us tractive force as T=212.25*16.88=3582.78N
This will be accomplished when the rear end's friction force is at its maximum tractive state. Static friction force equals trajectories.
T= Friction force= (coefficient of friction) *N
Where N= Wf+Wr for tipping condition
Coefficient of friction= 3582.78/(Wf+Wr) =1.72
We require a coefficient of friction greater than or equal to 1.72 for a vehicle to tip.
Since when tipping the load will be transferred to the back wheel, I have assumed the total weight in the coefficient of friction equation.
Since both are inversely proportional, the rider should hunch down to boost acceleration. This will lower the height of the center of gravity (CG) above the ground.
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