Tensile Forces

  • Thread starter asleight
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  • #1
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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...
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,097
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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...

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

http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html#c2
 
  • #4
152
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I solved and got 2.1mm...

It's not correct.
 
  • #5
LowlyPion
Homework Helper
3,097
5
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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