Minimum Diameter for Steel Wire: Solve It Now

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Homework Help Overview

The problem involves determining the minimum diameter of a circular steel wire that must stretch a specified amount under a tensile force. The context is related to material properties and mechanical stress.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss attempts to apply formulas related to stress and area, with one participant suggesting the use of Young's Modulus. There are questions about the correctness of calculations and the need to re-check values.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the accuracy of their results. There is no consensus on the correct answer, and some guidance has been offered regarding relevant concepts.

Contextual Notes

Participants express uncertainty about their calculations and the assumptions made in applying the formulas. There is mention of a specific tensile force and stretch limit that must be adhered to.

asleight
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Homework Statement



A circular steel wire 1.91m long must stretch no more than 0.0024m when a tensile force of 450N is applied to each end of the wire.

What minimum diameter is required for the wire?

Homework Equations



[tex]p=\frac{F}{\Delta A}[/tex]

The Attempt at a Solution



I can't seem to solve for this at all. I've tried applying a volume to this exercise:

[tex]A=V/l[/tex], where l is the length, then [tex]dA=\sqrt{dV/l-Vdl/l^2}[/tex]...

[tex]p=\frac{F}{\Delta A}=\frac{F}{\sqrt{dV/l-Vdl/l^2}}[/tex]. This didn't work, as far as I remember...
 
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asleight said:

Homework Statement



A circular steel wire 1.91m long must stretch no more than 0.0024m when a tensile force of 450N is applied to each end of the wire.

What minimum diameter is required for the wire?

Homework Equations



[tex]p=\frac{F}{\Delta A}[/tex]

The Attempt at a Solution



I can't seem to solve for this at all. I've tried applying a volume to this exercise:

[tex]A=V/l[/tex], where l is the length, then [tex]dA=\sqrt{dV/l-Vdl/l^2}[/tex]...

[tex]p=\frac{F}{\Delta A}=\frac{F}{\sqrt{dV/l-Vdl/l^2}}[/tex]. This didn't work, as far as I remember...

Perhaps you want to look at Young's Modulus for the wire?

http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html#c2
 
I solved and got 2.1mm...

It's not correct.
 
asleight said:
I solved and got 2.1mm...

It's not correct.

It's not what I got either.

Maybe re-check your numbers?
 

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