How Much Does a Steel Wire Stretch When Supporting Moving Masses?

AI Thread Summary
The discussion focuses on calculating the stretch of a steel wire supporting two moving masses over a pulley. A 1.80-m long wire with a diameter of 0.80 mm is used, with one end attached to a 23.0-kg mass and the other to a 14.0-kg mass. The tension in the wire is calculated to be 171 N, leading to a tensile stress of approximately 340,163 N/m². Participants clarify that this stress should be used in conjunction with Young's modulus for steel (210 GN/m²) to determine the wire's stretch. The conversation emphasizes the importance of correctly applying the formulas for stress and strain to find the final stretch value.
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A 1.80-m long round steel wire with a cross-sectional diameter of 0.80 mm is placed over a light frictionless pulley, with one end of the wire connected to a 23.0-kg mass and the other end connected to a 14.0-kg mass. By how much does the wire stretch when the masses are in motion? S= F/A
e= ΔL/L°
S= Ye (where Y is Young's modulus for steel= 210 GN/m^2)

I figured the F i needed in the first equation would be the tension in the wire. I got 171N for that answer and then I applied it to S= F/A. I found my S to be about 340163 N/m^2. I don't know whether to use that S for S=Ye or not. Where do I go from here?
 
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I haven't check your numbers, but yes, the tensile stress is the tension force divided by the cross-sectional area, and this is the stress that goes into your equation involving young's modulus.
 
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