Answer
$F=\frac{m(g+a_B)\sqrt{4y^2+d^2}}{4y}$
Work Step by Step
We can determine the magnitude of the force as follows:
$\Sigma F_y=ma_y$
$\implies 2Fcos\theta-mg=ma_B$
But from the given figure,
$cos\theta=\frac{y}{\sqrt{y^2+(d/2)^2}}$
Now the above equation becomes
$2F(\frac{y}{\sqrt{y^2+(d/2)^2}})-mg=ma_B$
$\implies F=\frac{mg+ma_B}{2(\frac{y}{\sqrt{y^2+(d/2)^2}})}$
$\implies F=\frac{m(g+a_B)}{4(\frac{y}{\sqrt{4y^2+d^2}})}$
This simplifies to:
$F=\frac{m(g+a_B)\sqrt{4y^2+d^2}}{4y}$
