Question

The steel I-beam in the drawing has a weight of 8.80 kN and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? (Assume α = 69.3°.)

Answer 1

First, each cable is identical, and we're probably to assume that the I-beam is of constant density all along its length, so that if you flip the I-beam around, the situation is exactly the same. That means that whatever the tension we find in one cable, the tension in the other is the same.

That means that 8.80 kN /2 = 4.4kN is held by each cable.

The cables aren't hanging straight down. This means that some fraction of the tension doesn't go into lifting the I-beam, but (in this case) into applying a compressional force along the I-beam. This can be neglected (presumably the beam isn't so weak that it's going to crumple).

So, what is the component of the tension that actually lifts the beam? It's T*sin(69.3) ~ 0.93T. Each cable provides this lift, so we have

2*0.93*T = 8.8 kN
==> T = 4.73 kN.