## Engineering Mechanics: Statics & Dynamics (14th Edition)

$T=30.5lb$ $\theta=20^{\circ}$
First we will sum the forces along rope D $(\searrow +) \sum \overrightarrow{F}_{x} =0$ $60\cos10^{\circ}-T-T\cos\theta =0$ Now we will sum the forces perpendicularly to rope D $(\nearrow +) \sum \overrightarrow{F}_{y} =0$ $T\sin\theta-60\sin10^{\circ}$ Solving the system of two equations, we obtain : $T=30.5lb$ $\theta=20^{\circ}$