< Mentor Note -- thread moved to HH from the New Member Introduction forum, so no HH Template is shown > Hi there, I found a thread from 2004 that perfectly answered my homework question here, but I don't understand how the person who responded came up with the answer. I would have just replied to the thread and resurrected it, but it says the thread is no longer open for replies. I'll paraphrase it here: Original question: Arlene is to walk across a high wire strung horizontally between two buildings 10 m apart. The sag in the rope when she is at the mid-point should not exceed 10 degrees. If her mass is 50 kg what must be the tension in the rope? Original answer: The sum of forces in the vertical direction would look like this: 0 = 2Tsin(10) - 490.5, T = 1412.3384 N I actually had the exact same problem with the exact same numbers, but I have no idea why 1412.33 N is the right answer. I know the equation where ΣFy = FN - mg I also get that the ΣFy is 0 if there is no movement in the y direction. 0 = FN - mg. mg = 50kg X 9.8m/s2 = 490 N 0 = FN - 490 N... How do you know that FN = 2Tsin(θ)? I can't find that equation anywhere in the dynamics chapter in my book, my notes, or my teacher's powerpoint. This is supposed to be homework from the dynamics chapter, and I would have been totally unable to solve this problem if I hadn't found this website. I also tried googling the formula and didn't find anything helpful. Do I just need to take this for granted or is there somewhere I can see how the 2Tsin(θ) was derived?