Answer
$948N$
Work Step by Step
We can find the required tension as follows:
$\Sigma F_y=2Tsin\theta -mg=0$
This can be rearranged as:
$T=\frac{mg}{2sin\theta}$
We plug in the known values to obtain:
$T=\frac{(50.0)(9.81)}{2sin15.0^{\circ}}$
$T=948N$
