Answer
Part A) The tension in the rope is 230N
Part B) The tension in the rope is 2100N
Work Step by Step
Part A) We will sum the forces vertically to find the tension in the rope, T. First we will find the angle of depression of the rope.
$tan(\theta)=\frac{1.5m}{3.3m}\approx24.4^{\circ}$
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$-19kg*9.8m/s^2+T*sin\theta+Tsin\theta=0$
$T=\frac{19kg*9.8m/s^2}{2*sin\theta}$
$T=\frac{19kg*9.8m/s^2}{2*sin24.4^{\circ}}$
$T\approx230N$
Part B) Same as part A but use .15m instead of 1.5m
$tan(\theta)=\frac{.15m}{3.3m}\approx2.60^{\circ}$
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$-19kg*9.8m/s^2+T*sin\theta+Tsin\theta=0$
$T=\frac{19kg*9.8m/s^2}{2*sin\theta}$
$T=\frac{19kg*9.8m/s^2}{2*sin2.60^{\circ}}$
$T\approx2100N$
