Maximum Stress in a spider web

[dimpledur]

Homework Statement

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?

Homework Equations

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

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.

 

[LowlyPion]

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 T_max*Sinθ isn't it?

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

 

[dimpledur]

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 T_max*Sinθ isn't it?

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

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

 

[dimpledur]

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

 

[LowlyPion]

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.

 

[dimpledur]

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.

I totally see it now thank you!