Min Steel Wire Diam for 390N: 0.296mm

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Homework Help Overview

The problem involves determining the minimum diameter of a steel wire that must stretch no more than a specified amount when subjected to a tensile force. The context is rooted in material properties, specifically Young's modulus, and the relationship between stress and strain in materials under load.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the relationship between stress, strain, and Young's modulus to find the required diameter. Some participants emphasize the importance of showing an initial attempt at a solution before receiving further assistance.

Discussion Status

The discussion has progressed with the original poster providing a mathematical approach to the problem. One participant acknowledges the improvement in the original poster's reasoning, indicating a positive direction in the exploration of the problem.

Contextual Notes

Participants note the necessity of adhering to forum rules regarding showing attempts at solutions before receiving help. There is also an emphasis on understanding the fundamental relationships involved in the problem, such as stress versus strain.

johntuan2009
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A steel wire of length 1.92 m with circular cross section must stretch no more than 0.210 cm when a tensile (stretching) force of 390 N is applied to each end of the wire.

What minimum diameter is required for the wire?

Express your answer in millimeters. Take Young's modulus for steel to be Y = 2.00×10^11 .
 
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Hello, johntuan! Welcome to PF! But per forum rules, before you receive help, you must show an attempt at a solution . Then we can provide assistance to help you obtain the solution, by seeing were you may have gone wrong. You should be familiar with the stress vs. strain or load versus elongation relationship for wires with given parameters of length, load, area, and Young's modulus.
 
E=Stress/Strain
E=(force/area)/(Change in length/original length)
E*(change in length/original length)= force/area
area=force/(E*(change in length/original length))
PiR^2=(390N)/((2.0*10^11)/(0.0021m/1.92m))
R^2=((390N)/((2.0*10^11)/(0.0021m/1.92m)))/Pi
R=sqrt{((390N)/((2.0*10^(11))/(0.0021m/1.92m)))/Pi
d=2R
then change into mm
 
Last edited:
Well, that is a lot better, and correct. Nice work!
 

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