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Stretching of a Steel wire

  • Thread starter GiovanniG
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  • #1
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Homework Statement


A steel wire 2.9mm in diameter stretches by 0.038{\rm \\%} when a mass is suspended from it. The elastic modulus for steel is 2.0×1011N/m2

*that {\rm \\%} value is in my homework, I assumed it was a mistake and was meant to be mm

Homework Equations


young's module for elasticity F/A=E(Δl/l)

The Attempt at a Solution


So to answer this I think I need to find F because that equals mg, so I'll just divide F by 9.8.

However I'm not given l so inserting the numbers into the equation I get
F/ [(.5*.0029)^2*π]=2.0*10^11{[(l+.00038)-l]/l}
There are two unknowns and I don't know about other equations to answer this problem, usually original length is given so I'm very confused.
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


A steel wire 2.9mm in diameter stretches by 0.038{\rm \\%} when a mass is suspended from it. The elastic modulus for steel is 2.0×1011N/m2

*that {\rm \\%} value is in my homework, I assumed it was a mistake and was meant to be mm

Homework Equations


young's module for elasticity F/A=E(Δl/l)

The Attempt at a Solution


So to answer this I think I need to find F because that equals mg, so I'll just divide F by 9.8.

However I'm not given l so inserting the numbers into the equation I get
F/ [(.5*.0029)^2*π]=2.0*10^11{[(l+.00038)-l]/l}
There are two unknowns and I don't know about other equations to answer this problem, usually original length is given so I'm very confused.
It's not clear what you are supposed to find here, F or the length of the wire.
 
  • #3
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The exact question is "How large is the mass?"
 
  • #4
SteamKing
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The exact question is "How large is the mass?"
What if instead of the wire stretching by 0.038 mm, it stretched by 0.038% of its original length?
 
  • #5
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What if instead of the wire stretching by 0.038 mm, it stretched by 0.038% of its original length?
That makes so much more sense and I was able to solve it, thank you so much I was working on this problem forever :)
 

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