## Answers

FBD Four AC => sigma M_c = 0 110 times 70 - F_BD cos theta times 19.01 = -F_BD sin theta times 40.70 7700 F_BD times 0.995 times 19.01 = -F_BD times 0.090 times 40.78 7700 = F_BD (+ 15.26) F_BD = + 504.495 N force in BD sigma F_x = 0 F_BD cos theta = 110 cos 65 + R_cx R_cx = 455.96 N so R_cx = Q Q = 455.96 N

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