A rigid uniform horizontal bar of mass m1 80.0 kg and length L 5.40 m is supported by two vertical massless stringsString A is attached at a distance d 1.60 m from the left end of the bar and is connected to the top plateString B is attached to the left end of the bar and is connected to the floorAn object of mass m2 2000 kg is supported by the crane at a distance x 5.20 m from the left end of the bar.Throughout this problem, positive torque is counterclockwise and use 9.80 m/s2 for the magnitude of the acceleration due to gravity.Find TA-Tension at string A Find TB- Tension at string B THANK YOU :)
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Sum the moments about string BΣM 0 Ta1.60m - 80.0kg9.8m/s?2.70m - 2000kg9.8m/s?5.20m so Ta 65 023 N Now sum the vertical forces: 65023N - (80.0 + 2000)kg 9.8m/s? - Tb 0 Tb 44 639 N If you find this helpful, please award Best Answer!
