A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?

Solution 25E Introduction We have to calculate the minimum diameter of the steel wire such that, when the applied force is 400 N, the wire does not stretch more than 0.25 cm. Step 1 The young’s modulus is given by FL Y = AL FL A = Y L The young’s modulus for the steel is 10 Y = 20 × 10 Pa We also have L = 2 m L = 0.25 cm = 0.25 × 10 2m F = 400 N Then we have A = (400 N)(2 m) = 1.6 × 10 6 m2 . (20×1010Pa)(0.25×102m)