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Question 1 (1 point) Im k = 600 N/m 1.5 m 1.5 m - UB The rod supports a cylinder of mass 74 kg and is pinned at its end A. If

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1840+(725.94 x 1.5coso)-(1800 sin Ox 3 coso) =0 840 + 1088.91coso - 5400 Sis O coso se 840 +1088.91 Coso - 2700 sis 20=0 | 10Using free body diagram and equilibrium equations to get the result.

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