Answer
The maximum value of $~~a_4~~$ is $~~\frac{L}{8}$
Work Step by Step
In part (a), we found that $a_1 = \frac{L}{2}$
In part (b), we found that $a_2 = \frac{L}{4}$
In part (c), we found that $a_3 = \frac{L}{6}$
The center of mass of the four bricks must be directly above part of the bottom surface. If the center of mass of the four bricks is above only air, the bricks will fall.
We can find the center of mass of the four bricks as a distance from the right side of the diagram:
$\frac{(m)(L/2)+(m)(L)+(m)(5L/4)+(m)(11L/12+L/2)}{4m} = \frac{50mL/12}{4m} = \frac{25L}{24}$
We can find the maximum value of $a_4$:
$\frac{L}{2}+\frac{L}{4}+\frac{L}{6}+a_4 = \frac{25L}{24}$
$a_4 = \frac{25L}{24}-\frac{L}{2}-\frac{L}{4}-\frac{L}{6}$
$a_4 = \frac{L}{8}$
The maximum value of $~~a_4~~$ is $~~\frac{L}{8}$
