A metal wire of diameter D stretches by 0.100 mm when supporting a weight W. If the same-length wire is used to support a weight three times as heavy, what would its diameter have to be (in terms of D) so it still stretches only 0.100 mm?

Solution 21DQ Step 1: Initial diameter of the wire = D or 2R 2 Area of the wire A = R or2 D 4 Force applied by the weight F = W Step 2: Tensile stress = F A W Tensile stress = D2 4 4W Tensile stress = 2-------(1) D Step3: The weight is increased by three times, therefore F = 3W and the diameter is changed as D 1 F Tensile stress = A Tensile stress = 3W2 41 Tensile stress = 4*3W -----(2) D 12 Step 4: In both the cases the wire stretches by 0.10 mm. It means that the wire experienced the same tensile stress. Therefore Equation (1) = (2) 4W 4*3W D2 = D 2 1 2 2 D 1 = 3D D 1 = 3D or 1.732 D The new diameter of the wire when the weight increased three times as heavy is 1.732D