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)
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
Since the solution to 25E from 11 chapter was answered, more than 527 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “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?” is broken down into a number of easy to follow steps, and 37 words. This full solution covers the following key subjects: Wire, minimum, diameter, end, Force. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 25E from chapter: 11 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460.