Physics Applied to Aristotle's Wheel

Let $f_{1}$ be the friction force on the rim and let $f_{2}$ be the friction force on the tire. Since the center of the tire moves without acceleration, the net force on the tire is 0. Therefore, the vertical forces on the wheel must add to 0. Likewise, the horizontal forces on the wheel must add to 0. Thus, we have the equations

$N_1 + N_2 - mg = 0 \quad \text{ and } \quad F + f_1 + f_2 = 0 .$

The equality of the normal forces gives

$N_1 = \dfrac{mg}2 ,$

and the second equation above gives

$F = -f_1 - f_2 .$

Let $\tau_{1}$ and $\tau_{2}$ be the torques produced by $f_{1}$ and $f_{2}$ , respectively, about the center of the wheel. Let $\tau_\text{net}$ be the net torque on the wheel. There is no angular acceleration, so we must have $\tau_\text{net}=0$ . Since the only torques on the wheel are those produced by $f_{1}$ and $f_{2}$ , we have

$\tau_\text{net} = \tau_1 + \tau_2 = 0 .$

Hence,

$\tau_1 = -\tau_2 .$

However,

$| \tau_1 | = |f_1| r$ and $|\tau_2 | = |f_2| R$ .

Therefore,

$|f_2 | R = |f_1| r .$

Since the center of the wheel covers a distance equal to the circumference of the tire, the tire rolls without slipping. However, the center of the rim moves a distance greater than the circumference of the rim and the rim must therefore slip on the surface it contacts. This means that the friction $f_{1}$ is kinetic friction. Furthermore, $f_{1}$ is to the left in the figure above. Letting the direction to the left be negative, we have, from the net torque equation above,

$\tau_2 =f_2 R = -\tau_1 = -f_1 r = \mu_{k_1} N_1 r .$

Thus,

$f_2 = \mu_{k_1} N_1 \dfrac rR,$

and consequently,

$F = \mu_{k_1} N_1 \left(1 - \dfrac rR \right) = 0.2 \cdot \dfrac{mg}2 \left(1 - \dfrac rR \right) = \boxed{9.8} \text{ Newtons}.$