Physics: Principles with Applications (7th Edition)

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 $tan(\theta)=\frac{.15m}{3.3m}\approx2.60^{\circ}$ $(\uparrow +) \sum \overrightarrow{F}_{y} =0$ $-19kg*9.8m/s^2+T*sin\theta+Tsin\theta=0$ $T=\frac{19kg*9.8m/s^2}{2*sin\theta}$ $T=\frac{19kg*9.8m/s^2}{2*sin2.60^{\circ}}$ $T\approx2100N$