Question

n the 4- cable set-up shown, find the tension in each cable to maintain equilibrium of the system.

Cable AD is horizontal

Angle ADE=120°

Angle BDE=105°

Angle CED=105°

Answer 1

So,

$-AD+BD\cos45°+DE\cos60°=0$

$BD\sin45°-DE\sin60°=1320$

$-DE\cos60°+EC\cos15°=0$

$DE\sin60°+EC\sin15°=1760$

$-AD+BD\cdot0.7071+DE\cdot0.5=0$

$BD\cdot0.7071-DE\cdot0.866=1320$

$-DE\cdot0.5+EC\cdot0.9659=0$

$DE\cdot0.866+EC\cdot0.2588=1760\to$

$AD=3724\ (N)$

$BD=4022\ (N)$

$DE=1760\ (N)$

$EC=911\ (N)$ . Answer