I cannot get this question i don't know what i'm doing wrong.?

Question

A steel I-beam has a weight of 8.40 kN and is being lifted at a constant velocity by two cables attached to its ends. The angle between the cables and the beam is θ=74.9°. What is the tension in each cable? /_\

I have the x components as

T1 -cos(105.1)

T2 cos(74.9)

y components

T1 -sin(105.1)

T2 -sin(74.9)

My book isn't helping.

Answer 1

I don't have the diagram so I might be wrong, but I'm going to start by making a few assumptions:

(1) There are two cables, one attached to one of the beam, and another attached to the other end of the beam

(2) Both cables form 74.9° angles with the beam

(3) The beam is not moving horizontally, and since the vertical velocity of the beam is constant, the net horizontal and vertical forces equal zero. But if that's the case, then the tensions in both cables must be same, (i.e. T1 = T2). The reason being that their horizontal components cancel out, (it's the only way the net horizontal force equals zero), and they both form the same angles with the beam.  

That means that in terms of the vertical components of the forces acting on the beam, let upwards be positive,

2Ty - 8.40 = 0

=> 2Ty = 8.40

=> Ty = 4.20 kN

Therefore, the vertical component of the tension in each cable, Ty = 4.20 kN

Each cable forms a 74.9° with the beam, so if you draw the diagram of the cables and the beam, you should find that:

Ty = T * sin(74.9°), where T is the tension in (each) cable

=> T = 4.20 / sin(74.9°) = 4.35 kN

Therefore, the tension in each cable is 4.35 kN.