Answer
$ 1.4\times 10^2\ N.m$
Work Step by Step
Given:
mass of cyclist $m =70\ kg$
Diameter of cross section $d =0.40\ m$
radius $r =\frac{d}{2} = \frac{0.40\ m}{2}=0.2\ m$
The magnitude of the maximum torque is
$\tau =F\times r$
$\tau =mg\times r$
$\tau =(70\ kg )(9.8\ m/s^2)\times 0.2 = 1.37\times 10^2\ Nm$
$\tau = 1.4\times 10^2\ N.m$
