A bag of cement weighing 350 N hangs in equilibrium from three wires as suggested in the figure below. Two of the wires make angles θ1 = 61.0° and θ2 = 41.0° with the horizontal. Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires.T1 = T2 = T3 =
The tension in the third wire is equal to the weight of the bag of cement. The sum of the vertical components of the tension in each of the other wires is equal to the tension in the third wire. So, the sum of the vertical components of the tension in each of the other wires is equal the weight of the bag of cement. Vertical component = T * sin θ For the left wire, vertical component = T1 * sin 61 For the right wire, vertical component = T2 * sin 41 Eq#1: T1 * sin 61 + T2 * sin 41 = 350 Since the system is in equilibrium, the magnitude of the horizontal components of the tension in each of the other wires is equal. Horizontal component = T * cos θ Eq#2: T1 * cos 61 = T2 * cos 41 T1 = T2 * cos 41/cos 61 Substitute (T2 * cos 41/cos 61) for T1 in Eq#1, and solve for T2. T2 * cos 41/cos 61 * sin 61 + T2 * sin 41 = 350 T2 * (cos 41/cos 61 * sin 61 + sin 41) = 350 T2 = 350 ÷ (cos 41/cos 61 * sin 61 + sin 41) = 173.4741944 N This is approximately 173.5 N Substitute 173.4741944 for T2 in Eq#1 and solve for T1. Eq#1: T1 * sin 61 + 173.4741944 * sin 41 = 350 T1 = (350 – 173.4741944 * sin 41) ÷ sin 61 = 270.0495844 N This is approximately 270 N. To check this answer, substitute both answers in Eq#2: T1 * cos 61 = T2 * cos 41 270 * cos 61 = 173.5 * cos 41 131 = 131 This proves that the tensions are correct! I hope that my work has helped you understand how to solve this type of problem.
