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RA = 2 sin 60 = 1.732 kN Ans. RB = 2 sin 30 = 1 kN Ans. | 4-1 Chapter 4 d Scale of corner magnified f Chapter 4 51 4-2 Ra 2 sin 60 1.732 kN Ans. Rb 2 sin 30 1 kN Ans. S 0.6 m a tan 1 0.6 0.4 0.6 30.96 d Step 1 Find Ra and R e Ra 800 -- Ra 1100 N Ans. sin 135 sin 30.96-A Ro 800 -----0 ----------- Ro 377 N Ans. sin 14.04 sin 30.96 1.2 Ro --------- 2.078 kN Ans. O tan 30 1.2 Ra --------- 2.4 kN Ans. sin 30 4.5 h ------- 7.794m tan 30 2 ma 0 9Re - 7.794 400 cos 30 - 4.5 400sin30 0 Re 400 N Ans. J2 Fx 0 Rax 400 cos 30 0 Rax -346.4 N Fy 0 RAy 400 - 400sin30 0 RAy -200N Ra V346.42 2002 400 N Ans. 52 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Step 2 Find components of Re on link 4 and RD Me 0 400 4.5 - 7.794 - 1.9 Rd 0 Rd 305.4N Ans. Y F. 0 Rex 4 305.4 N Y Fy 0 Rey 4 400N Step 3 Find components of Re on link 2 Ef 0 Rex 2 305.4 346.4 0 Fy 0 Rey 2 200 N Rex 2 41N Ans. 4-3 a Mq 0 18 60 14 R2 8 30 4 40 0 R2 71.43 Ibf Y Fy 0 R1 40 30 71.43 60 0 R1 1.43 Ibf M1 1.43 4 5.72 Ibf in M2 5.72 41.43 4 171.44 Ibf in M3 171.44 11.43 6 240 Ibf in M4 240 60 4 0 checks
