Question

For the mechanism shown in the schematics, determine the magnitudes, directions, and locations of the inertia forces acting on links 2,3, and 4. Draw the given acceleration polygon to scale for use in determi ning unknown acceleration. 2. 200 ft/s2 (60.9 m/s2) 0% SA 100 ft/s2 (30.5 m/s2) 8 in. (203 mm) 5 in. (127 mm) 83 10 in. (254 mm) W2-W 6 lb (27 N) W3 10 lb (44.5 N) 3 -0.009 lb s ft (0.012 kg m2

Answer 1

nW 8 44.5N 3012 Kg m2. 51-34 Acceleration Diagram 203 mm 60:9m/s2 26-603 38 66 264mm 30-5 ms 60多 osing concept oF Acceleration image AG3 127 2542 08V6o9 + 3062 60:92 +3052 42 +2022 AG3-26, 603 m/s2 Applying Cosine rule in Acceleration diagram, O6092+ 26.6032-2(609) (26603) cos 26 603) 043 38 978 m/s? 60.92 t 38.97 82-26.6032 17. 7950. case -р 2x 60.9x 38978 Free- Body dtagram of F2j Free-Body dtagram of links:ey Nesoton's Lano As net Force is in direction oF acceleration & Inertia Torce is opposite to direction oF acceleration 161 614 N x 609 21 210 9 81 Similarly F4 9.81 9.81

3i m2 1ayl = m3 local 445x 38 978176.811 N 7.793 9.8 6-17-796 net a, 3 For inertial torque we need the angular acceterathion OF linK 3 Angular acceleration calculation requi res acceleration of any point on linka mrt mass centre 26603- 3866 From velocity diagram acceleration diagram s4 l 26603 m/s2 S3 %, x AG3 26-603 x Sin(12051) X3= x 0 127 26.603 x sin 12:051-_ 0127 aslangential Con about s component Tnhertial Torque = 0.012×43.7ラら 0 529 N-m [Cw sense about so ]