Answer
$F=29.5N$
Work Step by Step
We take the axis of rotation to be at the point A where the wheel touches the step. There are 3 forces that produce torque as portrayed in the figure below:
- Force F creates a clockwise torque $-F(r-h)$
- The wheel's weight $W$ creates a counterclockwise torque $Wl_w=W\sqrt{r^2-(r-h)^2}$
- The normal force from the ground $G$ creates a clockwise torque $-Gl_g=-G\sqrt{r^2-(r-h)^2}$
At first, the bicycle wheel is in equilibrium, so the net torque is zero: $$W\sqrt{r^2-(r-h)^2}-G\sqrt{r^2-(r-h)^2}-F(r-h)=0$$ $$F(r-h)=(W-G)\sqrt{r^2-(r-h)^2}$$ $$F=\frac{(W-G)\sqrt{r^2-(r-h)^2}}{r-h}$$
When the wheel starts to lose contact with the ground, there is no longer $G$ anymore, which means $$F=\frac{W\sqrt{r^2-(r-h)^2}}{r-h}$$
We have $W=25N$, $r=0.34m$ and $h=0.12m$, so $$F=29.5N$$
