Question

the steel I-beam in thedrawing has a weight of 8.00 kN and is being lifted at a constantvelocity. what is the tension in each cable attached to it'sends?

Answer 1

Since the two angles are equal, it makes the problemconsiderably easier!The lift velocity is constant, so acceleration is zero, so netforce in the vertical is zero.ΣF_y = ΣF_T + (-mg) = 0=2F_Tsin70 - 8000 N = 0∴2F_T sin70 = 8000 NF_T= 8000 / (2 sin70)= 4256N=4.26 kN [3sig figs]This solution readilygives the tension in each cablesince I already let 2F_Tsinθ be the net force in thevertical.