Posted by Haley on Sunday, January 9, 2011 at 1:24pm.
A tightly stretched "high wire" is 44 m long. It sags 3.8 m when a 56 kg tightrope walker stands at its center. What is the tension in the wire?
1 NIs it possible to increase the tension in the wire so that there is no sag?
so far i know you do the inverse sin of (3.8/22) to find theta
then (56x9.8)/2 to get 274
however, now im stuck at sin(9.95)/sin(9.95)= (274)/(3.8x9.8)
i can't figure out if i did the wrong equation or just why i am stuck.

physics  drwls, Sunday, January 9, 2011 at 1:33pm
arcsin 3.8/22 = 9.95 degrees is the sag angle.
So far, so good.
2 T sin 9.95 = M g = 549 N
That is the vertical force balance equation. T is the tension on either side of the tightrope walker.
T = 549/[2*0.173) = 1590 N
<<Is it possible to increase the tension in the wire so that there is no sag? >>
No

physics  Wjones, Friday, October 26, 2012 at 11:41am
I believe you would use arctan. The rope is 44m long and 1/2 that becomes the adjacent leg. The distance sagged is the opposite leg so the angle should be arctan(3.8/22) or 9.79degrees. Plugging that in results in 1613N or approximately 1600N
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