Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 369: 42

$0.25m$

We know that $\Sigma \tau=r_{cm}m_{board}g-r_{cat}m_{cat}g=0$ This can be rearranged as: $r_{cat}=\frac{m_{board}}{m_{cat}}r_{cm}$ We plug in the known values to obtain: $r_{cat}=\frac{7.00}{2.8}(0.50)$ Now $d_{cat}=1.50-1.25=0.25m$