Originally Posted by **PeachBlossoms** The steel I-beam in the drawing has a weight of 7.40 kN and is being lifted at a constant velocity. What is the tension in each cable attached to its ends?
The answer is in Newtons.
I know that it is at an equilibrium which means that the sum of the forces equals 0. But other than that I'm lost. |

That's a good start.

I would sketch a free body diagram, not of the I-beam, but the point where the three cords meet and another at the CM of the beam. For both diagrams I will choose a coordinate system such that +x is horizontal and to the right, and where +y is straight upward.

First do the FBD on the I-beam. There are three forces here: the weight w, and two tensions T which have the same magnitude both acting along the cords: one is directed at 70 degrees above the positive x axis and the other is directed at 70 degrees above the negative x axis. (I'll let you decide how to prove that these two tensions are equal in magnitude.) So applying Newton's 2nd in the +y direction:

$\displaystyle \sum F_y = T~sin(70) + T~sin(70) - mg = 0$

(The 0 N on the right hand side is due to the acceleration being 0. Else this is ma.) So you can now find the magnitude of these tensions.

For the second FBD we have one tension R straight upward and two tensions T of equal magnitude: one directed at 70 degrees below the positive x axis and one directed at 70 degrees below the negative x axis. Again using Newton's 2nd in the y direction:

$\displaystyle \sum F_y = -T~sin(70) - T~sin(70) + R = 0$

(Here we have not only a 0 acceleration, but the FBD is on the point of intersection of three cables and this point has no mass (for an ideal cable.)) The T's are equal to the same T as in the first diagram, via Newton's 3rd.

-Dan