Answer
The 4.0-kg cat should stand a distance of 1.5 meters to the left of the pivot.
Work Step by Step
Let $m_1$ be the mass of the bowl.
Let $m_2$ be the mass of the 4.0-kg cat.
Let $m_3$ be the mass of the 5.0-kg cat.
To keep the seesaw balanced, the net torque must be zero.
$\sum \tau = 0$
$(r_1\times m_1~g)+(d\times m_2~g)+(r_3\times m_3~g) = 0$
$d~m_2~g = r_3~m_3~g-r_1~m_1~g$
$d = \frac{r_3~m_3-r_1~m_1}{m_2}$
$d = \frac{(2.0~m)(5.0~kg)-(2.0~m)(2.0~kg)}{4.0~kg}$
$d = 1.5~m$
The 4.0-kg cat should stand a distance of 1.5 meters to the left of the pivot.
