Answer
$1.72m$
Work Step by Step
We can find the required force in both cables and the required shortest length as follows:
We know that
$\Sigma F_y=0$
Since $AB$ and $AC$ are equal in length, therefore $T_{AC}=T_{AB}=T$
$\implies Tsin\theta+Tsin\theta=F$
$\implies 2Tsin\theta=F$
$2Tsin\theta=mg$
$\implies sin\theta=\frac{mg}{2T}$
We plug in the known values to obtain:
$sin\theta=\frac{500\times 9.81}{5\times 10^3}$
$\implies sin\theta=0.4905$
and $cos\theta=0.871$
Thus, the length of the cable is $\frac{1.5}{cos\theta}=\frac{1.5}{0.871}=1.72m$
