Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 137: 47

$a).$ The tension depends on (3) both the tree separation and the sag. $b).$ $613N$

a). The tension depends on (3) both the tree separation and the sag. b). Vertically, $2Tsin\theta-mg=0$ $T=\frac{mg}{2sin\theta}$ Horizontally, $Tcos\theta-Tcos\theta=0$ $\theta=tan^{-1}\frac{0.2}{5}=2.29^{\circ}$ So, $T=\frac{mg}{2sin\theta}=\frac{5\times9.8}{2sin2.29^{\circ}}=613N$