Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 180: 36

$42.5N$

We know that $\Sigma F_x=-F+T_1cos\theta+T_2cos\theta=0$ As $T_1=T_2=mg$ $\implies \Sigma F_x=-F+mgcos30^{\circ}+mgcos30^{\circ}=0$ $\implies F=2mgcso30.0^{\circ}$ We plug in the known values to obtain: $F=2(2.50)(9.81)$ $F=42.5N$