Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw \({\rm{0}}{\rm{.160}}\;{\rm{m}}\) to the left of the pivot (on the side of the lighter child) and assuming a mass of \({\rm{12}}{\rm{.0}}\;{\rm{kg}}\) for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium

For a system, the linear momentum remains the same when no external force is acting).

The mass is \({{\rm{m}}_{\rm{1}}}{\rm{ = 26}}\;{\rm{kg}}\).

The distance from the hinge is \({{\rm{r}}_{\rm{1}}}{\rm{ = 1}}{\rm{.60}}\;{\rm{m}}\).

The mass of the other is\({{\rm{m}}_{\rm{2}}}{\rm{ = 32}}\;{\rm{kg}}\)

The mass of the see-saw is \({{\rm{m}}_{\rm{s}}}{\rm{ = 12}}\;{\rm{kg}}\) .

The torque is zero about the pivot point. So, we can write,

\(\begin{array}{c}{{\rm{m}}_{\rm{1}}}{\rm{g}}{{\rm{r}}_{\rm{1}}}{\rm{ + }}{{\rm{m}}_{\rm{s}}}{\rm{g}}{{\rm{r}}_{\rm{s}}}{\rm{ = }}{{\rm{m}}_{\rm{2}}}{\rm{g}}{{\rm{r}}_{\rm{2}}}\\{\rm{26 \times 1}}{\rm{.6 + 12 \times 0}}{\rm{.16 = 32 \times }}{{\rm{r}}_{\rm{2}}}\\{{\rm{r}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{26 \times 1}}{\rm{.6 + 12 \times 0}}{\rm{.16}}}}{{{\rm{32}}}}\\{{\rm{r}}_{\rm{2}}}{\rm{ = 1}}{\rm{.36}}\;{\rm{m}}\end{array}\)

Hence, the distance is \({{\rm{r}}_{\rm{2}}}{\rm{ = 1}}{\rm{.36}}\;{\rm{m}}\).