Answer
$\frac{25}{24}L$
Work Step by Step
We can find the required distance as follows:
$d_1=\frac{L}{2}$
$d_2=\frac{1}{2}d_1$
The center of gravity of book 2 is given as
$d_2=\frac{1}{2}(\frac{L}{2})=\frac{L}{4}$
The center of gravity of book 3 is given as
$d_3=\frac{-m\frac{L}{2}+2m(0)}{2m+m}=\frac{-L}{6}$
The center of gravity of book 4 is given as
$d_4=\frac{3(0)m+m(-\frac{L}{2})}{3m+m}$
$d_4=-\frac{L}{8}$
Now $d=|d_1|+|d_2|+|d_3|+|d_4|$
We plug in the known values to obtain:
$d=\frac{L}{2}+\frac{L}{4}+\frac{L}{6}+\frac{L}{8}$
$\implies d=\frac{25}{24}L$
