A bag of cement weighing 325 N hangs in equilibrium from three wires as suggeste

Question

A bag of cement weighing 325 N hangs in equilibrium from three wires as suggested in Figure P4.23. Two of the wires make angles theta1 = 60.0 degree and theta 2 = 40.0 degree with the horizontal. Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires. A bag of cement whose weight is Fg hangs in equilibrium from three wires as shown in Figure P4.23. Two of the wires make angles theta 1 and theta 2 with the horizontal. Assuming the system is in equilibrium, show that the tension in the left-hand wire is T1 = Fg cos theta 2/sin(theta 1 + theta 2) Figure P4.23 Problem 23 and 24.

Explanation / Answer

horizontal forces T1 cos theta1 = T2 cos theta2 T1 sin theta1 + T2 sin theta 2 = T3 = Fg T1 sin theta1 + T2 sin theta 2 == Fg then from first equation substitute T2 in terms of T1 we get T1 = Fg cos theta2 / ( sin theta1 +theta2 )

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