# Stretching of a Steel wire

## 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.

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SteamKing
Staff Emeritus
Homework Helper

## 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.

The exact question is "How large is the mass?"

SteamKing
Staff Emeritus