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force_half_final_cont - force_half_final_cont.nb 1!AB 0.14...

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Force analysis via contour method ! AB ! 0.14, AC ! 0.06,CF ! 0.25,h ! 0.01,d ! 0.01,hSlider ! 0.02,wSlider ! 0.05, ro ! 8000, g ! 9.807,Me ! 1000.,phi " t # ! 1.0472, phi " " t # ! 52.3599,phi "" " t # ! 0 $ Inertia forces and moments Link 1 m1 # ro AB h d # 0.112 " kg # IC1 # m1 % AB^2 $ h^2 &’ 12 # 0.000183867 " kg m^2 # Fin1 # % m1 aC1 # ! 10.7469,18.6142,0 $ " N # F1 # Fin1 $ G1 # ! 10.7469,17.5158,0 $ " N # Min1 # M1 # % IC1 alpha10 # ! 0,0,0 $ " N m # Link 2 m2 # ro hSlider wSlider d # 0.08 " kg # IC2 # m2 % hSlider^2 $ wSlider^2 &’ 12 # 0.0000193333 " kg m^2 # Fin2 # % m2 aC2 # ! 15.3527,26.5917,0 $ " N # F2 # Fin2 $ G2 # ! 15.3527,25.8071,0 $ " N # Min2 # M2 # % IC2 alpha20 # ! 0,0, % 0.0475952 $ " N m # Link 3 m3 # ro CF h d # 0.2 " kg # IC3 # m3 % CF^2 $ h^2 &’ 12 # 0.00104333 " kg m^2 # Fin3 # % m3 aC3 # ! 145.238,45.2941,0 $ " N # F3 # Fin3 $ G3 # ! 145.238,43.3327,0 $ " N # Min3 # M3 # % IC3 alpha30 # ! 0,0, % 2.5685 $ " N m # M3e # % Sign " omega2 #! 0, 0, Me $ # ! 0,0, % 1000. $ " N m # Joint reactions Joint C_R F03 % 3 % B_T % 2 % B_R B_T: Sum F for 3 upon rCB direction: % F3 $ F03 & .rCB # 0 0.07 % 145.238 $ F03x & $ 0.0612436 % 43.3327 $ F03y & ## 0 % 1 & B_R: Sum M wrt B for 3 & 2: rBC x F03 $ rBC3 x F3 $ M3 $ M3e $ M2 # 0 % 1004.63 $ 0.0612436F03x % 0.07F03y ## 0 % 2 & force_half_final_cont.nb 1
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From Eqs. % 1 &% 2 & #& F03x, F03y F03 # ! F03x, F03y, 0 $ # ! 7008.6, % 8220.01,0 $ " N # Joint B_T I: F23 % 3 % C_R % 0 II: F32 % 2 % B_R % 1 F23 perpendicular to BC: F32.BC # 0 #& % 0.07F23x % 0.0612436F23y ## 0 % 3 & point Q is on BC: % yB % yC &’% xB % xC & # % yQ % yC &’% xQ % xC & #& 0.874908 ## % 0.06 $ yQ ’’’’’’’’’’’’’’’’ ’’’’’’’’’’ xQ % 4 & I: C_R: Sum M wrt C for 3 rCQ x F23 $ rCC3 x F3 $ M3e $ M3 # 0 % 1010.45 $ 0.06F23x $ F23yxQ % F23xyQ ## 0 % 5 & II: B_R: Sum M wrt B for 2 rBQ x F32 $ M2 # 0 % 0.0475952 % 0.121244F23x $ 0.07F23y % F23yxQ $ F23xyQ ## 0 % 6 & From Eqs. % 3 & % % 6 & #& F23x, F23y, xQ, yQ
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