# Homework Help: Maximum Stress in a spider web

1. Jan 28, 2009

### dimpledur

1. The problem statement, all variables and given/known data
Spider Silk has a Young Modulus of 4.0x10^9 N/m^2 and can withstand stresses of up to 1.4x10^9 N/m^2. A single webstrand has a cross sectional area of 1.0x10^-11 m^2, and a web is made up of 50 radial strands. A bug lands in the centre of a horizontal web. With what mass should the bug be to exert this maximum stress?

2. Relevant equations

F/A = 1.4x10^9 N/m^2

3. The attempt at a solution

F/A = 1.4x10^9 N/m^2
mg / A = 1.4x10^9 N/m^2

A = 50 strands * 1.0x10^-10 N/m^2 = 5.0x10^-10

so m = (1.4x10^9 N/m^2)(5.0x10^-10) / 9.8
m = 71.4g

HOWEVER, the answer is 48 grams. I am unsure of what to do with the area. Like, I'm pretty sure that I am messing that portion up.

2. Jan 28, 2009

### LowlyPion

Re: stress

Since the web is horizontal, won't the strands stretch?

It seems like the angle of that stretch will be Cosθ = L/(L +ΔL).

Then it's a matter of tension, with the vertical m*g being the Tmax*Sinθ isn't it?

Now if only there was some way to figure the stretch at Tmax?

3. Jan 28, 2009

### dimpledur

Re: stress

Wouldn't it be Tension = sin theta * mg ???

4. Jan 28, 2009

### dimpledur

Re: stress

I'm having a hard time visualizing the problem...

5. Jan 28, 2009

### LowlyPion

Re: stress

Draw a diagram.

The web is horizontal. Looking in cross-section it will look like a weight supported by 2 lines in Tension sagging down. (Ignore for a moment the other 24 pairs of strands.) The angle θ it makes with the horizontal can be used to express the vertical component of the tension which I think you should see is T*sinθ = m*g.

6. Jan 28, 2009

### dimpledur

Re: stress

I totally see it now thank you!