Answer
a) $r_{parent}=2.3m$
b) $r_{parent}=1.6m$
c) unchanged
Work Step by Step
(a) We know that
$\Sigma \tau=r_{parent}F_{parent}-r_{child}(m_{child}g)=0$
This can be rearranged as:
$r_{parent}=\frac{r_{child}{m_{child}g}}{F_{parent}}$
We plug in the known values to obtain:
$r_{parent}=\frac{(\frac{1}{2}\times 5.2)(19)(9.81)}{210}$
$r_{parent}=2.3m$
(b) $r_{parent}=\frac{r_{child}{m_{child}g}}{F_{parent}}$
We plug in the known values to obtain:
$r_{parent}=\frac{(\frac{1}{2}\times 5.2)(19)(9.81)}{310}$
$r_{parent}=1.6m$
(c) We know that answers to part (a) and part (b) will not change as the the mass of the teeter-totter is not included in the above calculations.
