At equilibrium, the sum of forces and moments acting at the point where
the tree cables meet are zero.
- The tension in the 30° inclined cable is 75.0 N
- The tension in the 60° inclined cable is approximately 129.9 N
- The tension in the bird feeder tension cable 150 N
Reasons:
Required;
The tension in each cable
Solution:
Let T₁ represent the tension in the cable attached to the feeder, let T₂
represent the tension in the cable inclined 60°, and let T₃ represent the
tension in the 30° cable.
At equilibrium, we have;
The tension in cable T₁ is the weight = 150 N
T₁ = T₂·sin(60°) + T₃·sin(30°)
T₂·cos(60°) = T₃·cos(30°)
By evaluating, we have;
\(150 = \dfrac{\sqrt{3} }{2} \cdot T_2 + \dfrac{1 }{2} \cdot T_3...(1)\)
\(\dfrac{1 }{2} \cdot T_2 = \dfrac{\sqrt{3} }{2} \cdot T_3 ...(2)\)
T₂ = √3·T₃
Which gives;
\(150 = \dfrac{\sqrt{3} }{2} \cdot \sqrt{3} \cdot T_3 + \dfrac{1 }{2} \cdot T_3 = 2 \cdot T_3\)
- \(T_3 = \dfrac{150}{2} = 75.0\)
The tension in the 30° inclined cable, T₃ = 75.0 N
T₂ = √3·T₃
- T₂ = √3 × 75.0 N ≈ 129.9 N
The tension in the 60° inclined cable, T₂ ≈ 129.9 N
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