The bars AB and AC are joined by a pin at A and a horizontal cable. The vertical cable carrying the 200 kg mass is attached to the pin at A. Determine the tension in the horizontal cable. Neglect the weights of the bars. Verify if the solution and answer is acceptable and why.

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200 kg 3m 2m FACSin 60-FABSin 30 FAB = 3FAC FAC Cos60+ FAB Cos30+ Mg = 0 FAC 2 +3FA 3 +2009.81 = 0 AC 2 FAC = -981N FAC Cos30+T=0 T-FACCos30 = (-981)Cos30 - 849.57N 200 kg 3m 2m FACSin 60-FABSin 30 FAB = 3FAC FAC Cos60+ FAB Cos30+ Mg = 0 FAC 2 +3FA 3 +2009.81 = 0 AC 2 FAC = -981N FAC Cos30+T=0 T-FACCos30 = (-981)Cos30 - 849.57N

Answer rating: 100% (QA)

To verify if the calculated tension in the horizontal cable is correct and acceptable lets analyze t... View the full answer