[Physics] Forces on wheels in an accelerating vehicle

accelerationforcesmomentnewtonian-mechanics

Assume that a motorcycle of mass $m$ has two wheels that are equidistant from its centre i.e the force on each wheel is $mg/2$.

If the motorcycle accelerates forward, will the two forces on each wheel (measured instantaneously) remain the same? If not, how can one mathematically describe the change in forces measured on each wheel and will these forces oscillate before converging when the motorcycle reaches final velocity?

Now, if the weight of the motorcycle isn't uniformly distributed along it's length e.g force measured at one wheel is say $mg/3$ and $2mg/3$ at the other, how different would the vehicle dynamics (oscillations etc) be? Can one somehow infer the fact that the weight isn't uniformly distributed without actually measuring the forces at the wheels?

I know this question is very vague but any ideas are very welcome.

Best Answer

There is an old trick that turns a dynamics problem into a statics problem. Apply equal an opposite inertial forces on the center of mass.

Lets look at a free body diagram of a motorcycle accelerating with $\ddot{x}$.

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The balance of forces in the horizontal direction equate the tractive force $B_x$ to the acceleration $$ \left. m \ddot{x} - B_x = 0 \right\} B_x = m \ddot{x} $$

The balance of forces in vertical direction, together with the balance of moment about any point (I choose the rear contact point) give us the distribution of loads on the tires.

$$\left. \begin{align} A_y + B_y - m g & = 0 \\ -\ell A_y + c m g - h m \ddot{x} & = 0 \end{align} \right\} \begin{aligned} A_y &= \frac{c}{\ell} m g - \frac{h}{\ell} m \ddot{x} \\ B_y & = \left(1-\frac{c}{\ell}\right) m g + \frac{h}{\ell} m \ddot{x} \end{aligned} $$

ALSO, if you want to find the minimum acceleration for a wheelie then set $A_y=0$ to find $$\ddot{x} \ge \frac{c}{h} g$$

FINALLY, if traction is limited to $B_x \le \mu B_y$ then the equations above give us the maximum acceleration before traction is lost $$\ddot{x} \le \mu g \frac{\ell-c}{\ell-\mu h} $$

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