Question

The crossed four-bar linkage has a constant crank angular velocity 2=10 rad/s ccw. The dimensions and the results of the kinematic analysis are given below for the position shown. (a) Draw the free-body diagrams of all the links of the mechanism, (b) Find the reaction forces at all the pins and the driving torque at link 2. The gravity centers of link 2 and link 4 are located at O2 and O4 respectively.

y A O2A-6 in, O204-18 in, AB-18 in, O4B-6 in, AC-24 in, AG,-12 in, w,-41b, G2-o-0.063 lb - in-s, IG3 0.497 lb-in-s o-1.43 rad/s2, cw, -11.43 rad/s cw aas (25.92i+24.58jft/s 03-04-84.8 rad/s2 ccw Fc 30 lb 60° 261.8° 3218 04 4

Answer 1

Given that for link (3) Sigma F_x = 0 Sigma F_y = 0 Sigma M_0 = 0 rightarrow F_2, 3 cos 38.2 - f_4,3 cos 38.2 = 0 -(1) F_2,3 sin 363 - F_4,3 sin 38.2 + 30 = 0 -(2) F_4,3 AB os 51.8 degree +F_C A_C cos 31.8 = 0 -(3) 0.78 F_2,3 - 0.78 f_4, 3 - (4) 0.618 F_2,3 - 0.628 F_4,3 + 30 = 0 -(5) 19.84 F_C -11.13 F_4,3 = 0 -(6) 11.13 f_4,3 = 14.84 times 30 lb/4 F_4,3 = 40 lb F_4.13^x = -F_3,4^x = 4.0 lb rightarrow 0.76 F_2,3 = 0.70 F_4,3 F_2,3 = F_4,3^x = 40 lb \r\n for link (2) Sigma F_X = 0 Sigma F_y = 0 Sigma m_02 = 0 Rightarrow F_12 cos 60 degree - F_3,2 cos 60 degree = 0 F_32 sin 60 degree - F_1,2 sin 60 degree = 0 F_3,2 AO_2 sin 60 degree + F_3,2 AO_2 cos 60 degree + F_12 = 0 - (-34.64) times 6 + (-20) 6 + T/R = 0 T_12 = -87.8416 in T_12 = F_3,2 cos 60 degree/cos 60 degree F_1,2 = f_3,2 F_1,2 = -40