The steel I-beam in the drawing has a weight of 7.30 kN and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? (Assume α = 72.4°.)
Since the angles are similar for both cables, the
Tensions on each sling will be equal.
Summation of Forces vertical
∑Fv = 0
0 = Ty + Ty - 7.30kN
...= 2Ty - 7.30kN
0 = 2Tsin72.4º - 7.3kN
2T(0.953) = 7.3kN
T = 3.83 kN
The steel I-beam in the drawing has a weight of 7.30 kN and is being lifted...
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The answer with steps please
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please explain, thanks
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