11-8 [100 pts]

Use 0.3 lbf/in3 for the specific weight of steel. Show your work (e.g., include a printout from FOURBAR or MathCAD) for your complete kinematics analysis. θ3=10.1°, Ω3=-41.6 rad/sec, α3=-336 rad/sec2 θ4=113°, Ω4=26.3 rad/sec, α4=2964 rad/sec2 For link 3, use: m3 = 0.038 slug, RCG3 = 3.977 in, ΔCG3 = 38.199°, IG3 = 0.467 slug in2. You are not required to show work for these quantities. aG3x = -8636 in/sec2, aG3y = -12210 in/sec2 If using MathCAD (which is what I recommend), the most proper way to deal with units in matrix equation 11.9, is to divide each element in rows 3, 6, and 9 of the matrix equation (both sides) by your length unit (e.g., "in") because these rows represent torque equations that have different units (in-lbf) than the other equations which involve force (with units of lbf). Then, after solving the matrix equation, you would have to multiply the 9th element in your results vector by the same unit (in) to get the desired quantity (T12). T12 = 23 ft lbf Calculate the resultant force in each joint. F12 = 58.0 lbf, F32 = 60.7 lbf, F43 = 105 lbf, F14 = 109 lbf