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  1. What is the minimum cross-sectional area required of a vertical steel cable from which is suspended a 270-kg chandelier? Assume a safety factor of 7.0.
  2. If the cable is 7.5 m long, how much does it elongate?

Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.
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Quick Answer: 
  1. $3.7 \times 10^{-5} \textrm{ m}$
  2. $2.7 \textrm{ mm}$

Giancoli 7th Edition, Chapter 9, Problem 53


Chapter 9, Problem 53 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. The tensile stress equals the force divided by area and that equals allowable tensile stress in this case and allowable tensile stress is the maximum tensile stress the steel material can tolerate divided by the safety factor. And we'll solve this for area and we get A equals safety factor times F divided by maximum tensile stress. So that's 7 times—the weight of the chandelier—270 kilograms times 9.8 newtons per kilogram divided by 500 times 10 to the 6 newtons per square meter— the maximum tensile stress for steel— and that gives 3.7 times 10 to the minus 5 meters. And if you wanna figure out how much this cable stretches, we know that stress equals elastic modulus times the change in length divided by the original length. So multiply both sides by l naught and divide by E and you get the Δl is F times l naught over E times A. So that's 270 kilograms times 9.8 newtons per kilogram times the original length of 7.5 meters divided by 200 times 10 to the 9 newtons per square meter— Young's modulus for steel— times 3.704 times 10 to the minus 5 square meters which we found out before is the area of this cable and that gives about 2.7 millimeters that it will stretch.