Finding the mass of an object with max stress

[walnuts]

Homework Statement

A spider is caught in the midpoint of a spider thread. The thread breaks under a stress of 8.2E8 N/m^2 and a strain of 2.00. Initially, it was horizontal and had a length of 2.00 cm and a cross-sectional area of 8.E-12 m^2. As the thread was stretched under the weight of the insect, its volume remained constant. If the weight of the insect puts the thread on the verge of breaking, what is the insect’s mass?

Homework Equations

stress=F/A
A=cross sectional area

The Attempt at a Solution

Stress*A/g=w

I know I'm missing a key point here because I do not take in account the length of the wire, but how does the length of wire affect this problem?

 

[tiny-tim]

Welcome to PF!

Hi walnuts! Welcome to PF!

(try using the X² tag just above the Reply box )

A spider is caught in the midpoint of a spider thread. The thread breaks under a stress of 8.2E8 N/m^2 and a strain of 2.00. Initially, it was horizontal and had a length of 2.00 cm and a cross-sectional area of 8.E-12 m^2. As the thread was stretched under the weight of the insect, its volume remained constant. If the weight of the insect puts the thread on the verge of breaking, what is the insect’s mass?
…
I know I'm missing a key point here because I do not take in account the length of the wire, but how does the length of wire affect this problem?

The (wire?) thread is horizontal, so the length of the thread will affect how far down the weight of the spider moves it.

 

[walnuts]

I turned it in earlier, here's what I did. I'm known for just making stuff up, so correct me if I'm wrong.

stress/strain=(F/A)/(delta L/ L)
delta L/L=2 and since the volume didn't change A=.5A so,

2*stress/(strain*.5)=mg

4*stress/(strain*g)=m