4.A uniform semicircular sign 1.00 m in diameter and a weight of 325 N is supported by two wires. What is the tension in each of wires supporting the sign?
the sum of the tension of both wires must add up to 325N if both wires are equal distance from the center of sign then the tension on each wire is 325/2=162.5 if the wires are at different distances you need to the distances to get the answer
Assuming the wires are vertical, connected evenly to the sign, and that nothing else is supporting the sign, then each wire carries half of the weight. T = 325 / 2 = 162.5N
