Example 4 As shown in Figure 3, in the rectangular coordinate system, the coordinates of vertex $B$ of rectangle $O A B C$ are $(15,6)$, and the line $y=\frac{1}{3} x+b$ exactly divides rectangle $O A B C$ into two parts of equal area. Therefore, $b=$ $\qquad$ (2000, National Junior High School Mathematics Competition)
Solution: By the symmetry of the rectangle, the line that bisects its area must pass through its center (i.e., the intersection of the diagonals).
Let the center be point $O^{\prime}$.
Given $B(15,6)$, then $O^{\prime}\left(\frac{15}{2}, 3\right)$.
Substituting into $y=\frac{1}{3} x+b$, we get $3=\frac{1}{3} \times \frac{15}{2}+b$. Solving for $b$ yields $b=\frac{1}{2}$.