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What angles produce a net torque of 1250 Nm on an angled arm?
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[QUOTE="draupe, post: 6142860, member: 658226"] [h2]Homework Statement [/h2] [ATTACH=full]239770[/ATTACH]My teacher gave us a solution of + 1250 Nm Where CCW = positive torque I know that the torque of the 600N + 300N forces + 1250Nm = the torque of the 500N force. I can't figure out what angles work with the forces at the end of the pipe. [h2]Homework Equations[/h2] Σ T = 1250 Nm T =Fr sinΘ [h2]The Attempt at a Solution[/h2] ΣT = (500N x5.5m) - (600N x 1m) - (300N x 5.5m x sin 45°) ΣT = (2750Nm) - (600Nm) - (1166.7 Nm) ΣT = +983.3 Nm This is the answer I get when I attempted the problem. I've also tried setting those forces with the hypotenuse as the r for the 500 and 300 N forces. The triangle would be 4.58m base, 1.58m height. Pythagoras would say this triangle's last side( hypotenuse) would be 4.84m. Also I tried sin of 135° turns out it is equivalent to sin 45° ΣT = (500N x 4.84m) - (600N x 1m) - (300N x 4.84 x sin 45°) ΣT = 2420Nm - 600Nm - 1026.7 Nm ΣT = 793.3 Nm [/QUOTE]
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What angles produce a net torque of 1250 Nm on an angled arm?
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