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

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Homework Help Overview

The problem involves determining the mass of a bug that would maximize the stress on a spider web, given specific material properties of the spider silk and the geometry of the web. The context includes the Young's modulus and maximum stress that the silk can withstand, as well as the arrangement of the web strands.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between force, area, and stress, with one participant attempting to calculate the mass based on the maximum stress formula. Others question the implications of the web being horizontal and the resulting tension in the strands. There are also considerations about the geometry of the web and the angles involved in the tension calculations.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some have provided diagrams to aid understanding, and there is a recognition of the complexities involved in visualizing the forces at play. However, no consensus has been reached regarding the correct approach or final answer.

Contextual Notes

Participants note potential confusion regarding the area used in calculations and the effects of the web's horizontal orientation on the tension and stress experienced by the strands. There is also mention of the need to consider the stretch of the strands under load.

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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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