What Mass Can a Bug Have to Maximize Stress on a Spider Web?

AI Thread Summary
To maximize stress on a spider web, the bug's mass must be calculated based on the web's material properties and geometry. The spider silk has a Young's modulus of 4.0x10^9 N/m^2 and can withstand a maximum stress of 1.4x10^9 N/m^2. The total cross-sectional area of the web, made up of 50 strands, is 5.0x10^-10 m^2. The correct mass for the bug to exert this maximum stress is determined to be 48 grams, accounting for the tension and angle of the strands when the web is horizontal. Visualizing the problem through a diagram helps clarify the relationship between tension, angle, and the weight of the bug.
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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.
 
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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?
 


LowlyPion said:
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?

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


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


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.
 


LowlyPion said:
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!
 
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