Min Steel Wire Diam for 390N: 0.296mm

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To determine the minimum diameter of a steel wire that must stretch no more than 0.210 cm under a tensile force of 390 N, the relationship between stress, strain, and Young's modulus is applied. Using the formula E = Stress/Strain, the calculations involve rearranging to find the required area and radius of the wire. The area is derived from the force divided by the product of Young's modulus and the strain. The radius is then calculated from the area, and finally, the diameter is obtained by doubling the radius and converting it to millimeters. The discussion emphasizes the importance of showing an attempt at a solution before receiving assistance.
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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
 
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Well, that is a lot better, and correct. Nice work!
 
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