Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 245: 24

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 $$F=29.5N$$
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