The tension in both cables are calculated as; T₁ = 127.63 N and T₂ = 185.48 N
How to find the Tension in Suspended Cables?
Resolving forces in the y direction gives us;
T₁ sin 29° + T₂ sin 53° = 210 ------(eq 1)
Resolving forces in the horizontal direction gives;
T₁ cos 29° = T₂ cos 53° ----(eq 2)
T₁ = T₂ cos 53°/cos 29°
Thus;
(T₂ cos 53°/cos 29°)sin 29° + T₂ sin 53° = 210
0.3336T₂ + 0.7986T₂ = 210
T₂ = 210/1.1322
T₂ = 185.48 N
Thus;
T₁ = 185.48 cos 53°/cos 29°
T₁ = 127.63 N
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