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Tension and Angle Analysis for Traction in a Broken Leg
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[QUOTE="defaultusername, post: 5487308, member: 595548"] [B][I]b)[/I][/B] Find the proper angle for the upper rope: First I try to find the amount of force on the boot. using the same equation w = mg = (4.0kg)(9.8) = 39.2 N of force acting on the boot Then I take the sine to find the correct angle, right? So, 58.8 sin(θ) = 39.2 + 58.8 sin(15°) θ = 54.42° ----additional calculations---- 58.8sin(θ) = 54.42° sin(θ) = 54.42 / 58.8 sin(θ) = 0.9255 sin[SUP]-1[/SUP](0.9255) = 67.74° That is [I]close [/I]to 67.8°...but I am not sure if I did it correctly or not. From there: [B][I]c) [/I][/B]Tcos(15°)+Tcos(67.7°) = 1.345382 = (1.345382)(58.8°) = 79.12N (rounded to 79N) net traction force pulling on the leg I am shooting in the dark here and am not sure if I am even doing it correctly. Thanks in advance [/QUOTE]
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Tension and Angle Analysis for Traction in a Broken Leg
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