## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$T_x = 281~N$ $T_y = 351~N$
We can find the angle $\theta$ that the chains make with the vertical. $cos(\theta) = \frac{2.5~m-0.55~m}{2.5~m}$ $\theta = arccos(\frac{2.5~m-0.55~m}{2.5~m})$ $\theta = 38.7^{\circ}$ We can find the horizontal component of the tension. $T_x = T~sin(\theta)$ $T_x = (450~N)~sin(38.7^{\circ})$ $T_x = 281~N$ We can find the vertical component of the tension. $T_y = T~cos(\theta)$ $T_y = (450~N)~cos(38.7^{\circ})$ $T_y = 351~N$