# Tension and Angles

by Heat
Tags: angles, tension
 P: 273 1. The problem statement, all variables and given/known data A frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Neglect any friction between the wall and the picture frame.) 2. Relevant equations F=ma w=mg 3. The attempt at a solution I think it looks something like this: It says .75 weight of frame, so I did the following to convert to N. w=.75(9.8) = 7.35N That is 7.35 on each side, and this is the part that get's me "if the tension in each wire is equal to 0.75 of the weight of the frame?" I am lost, here. but for angle I think it should be something like this arc cos (14.7/7.35) well at least i tried :D
 Mentor P: 41,568 You are not given the weight of the frame so don't assume any particular value. Instead, just call that weight W. You are given the tension in the wires in terms of that weight. The tension equals 0.75*weight, so T = 0.75*W. Now apply the condition for equilibrium, namely that vertical forces acting on the frame must add to zero. (What are the vertical components of the wire tension?)
 P: 273 so the tension on each wire is .75w, where w is unknown. "condition for equilibrium, namely that vertical forces acting on the frame must add to zero." vertical must add up to w, for them to cancel out, right? if so where do I go from there. the angle would not be:::: arc cosine (w/.75w)?
 Mentor P: 41,568 Tension and Angles Almost! Realize that there are two wires pulling up on the frame.
 P: 273 so that means they are sharing the weight, so it should be half of w. ?
Mentor
P: 41,568
 Quote by Heat so that means they are sharing the weight, so it should be half of w. ?
You can certainly think of it that way.
 P: 273 so it would be arc cosine of .5w/.75w = arc cosine .66 = 48.19 ?
 Mentor P: 41,568 Yep. Looks good.
 P: 273 thank you :)

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