force_cont - force_cont.nb 1 8AB 0.14 AC 0.06 CF 0.2 h 0.01...

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Force analysis via contour method 8 AB 0.14, AC 0.06, CF 0.2, h 0.01, d 0.01, hSlider 0.02, wSlider 0.05, ro 8000, g 9.807, Me 1000., phi @ t D 1.0472, phi @ t D 9.8696, phi @ t D 0 < Inertia forces and moments Link 1 m1 = ro AB h d = 0.112 @ kg D IC1 = m1 H AB^2 + h^2 12 = 0.000183867 @ kg m^2 D Fin1 = - m1 aC1 = 8 0.381844, 0.661373, 0 < @ N D F1 = Fin1 + G1 = 8 0.381844, - 0.437011, 0 < @ N D Min1 = M1 = - IC1 alpha10 = 8 0, 0, 0 < @ N m D Link 2 m2 = ro hSlider wSlider d = 0.08 @ kg D IC2 = m2 H hSlider^2 + wSlider^2 12 = 0.0000193333 @ kg m^2 D Fin2 = - m2 aC2 = 8 0.545491, 0.944818, 0 < @ N D F2 = Fin2 + G2 = 8 0.545491, 0.160258, 0 < @ N D Min2 = M2 = - IC2 alpha20 = 0 @ N m D Link 3 m3 = ro CF h d = 0.16 @ kg D IC3 = m3 H CF^2 + h^2 12 = 0.000534667 @ kg m^2 D Fin3 = - m3 aC3 = 8 3.30266, 1.02997, 0 < @ N D F3 = Fin3 + G3 = 8 3.30266, - 0.539153, 0 < @ N D Min3 = M3 = - IC3 alpha30 = 8 0, 0, - 0.0467673 < @ N m D M3e = - Sign @ omega2 D8 0, 0, Me < = 8 0, 0, - 1000. < @ N m
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This note was uploaded on 08/29/2011 for the course MECH 6420 taught by Professor Marghitu during the Summer '11 term at University of Florida.

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force_cont - force_cont.nb 1 8AB 0.14 AC 0.06 CF 0.2 h 0.01...

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