Answer
The tension in the rope is $~~7765~N$
Work Step by Step
We can find the angle $\theta$ that the rope sags below the horizontal:
$sin~\theta = \frac{0.350~m}{1.72~m}$
$\theta = sin^{-1}~(\frac{0.350~m}{1.72~m})$
$\theta = 11.7^{\circ}$
The vertical component of the tension in each side of the rope pulls up on the weight and keeps the system in equilibrium.
We can use the sum of the vertical forces to find the tension $F_T$ in the rope:
$F_{net, y} = 0$
$2~F_T~sin~\theta-mg = 0$
$2~F_T~sin~\theta = mg$
$F_T = \frac{mg}{2~sin~\theta}$
$F_T = \frac{3160~N}{2~sin~11.7^{\circ}}$
$F_T = 7765~N$
The tension in the rope is $~~7765~N$.
