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Homework Help: Maximum Stress in a spider web

  1. Jan 28, 2009 #1
    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. jcsd
  3. Jan 28, 2009 #2

    LowlyPion

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    Homework Helper

    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?
     
  4. Jan 28, 2009 #3
    Re: stress

    Wouldn't it be Tension = sin theta * mg ???
     
  5. Jan 28, 2009 #4
    Re: stress

    I'm having a hard time visualizing the problem...
     
  6. Jan 28, 2009 #5

    LowlyPion

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    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.
     
  7. Jan 28, 2009 #6
    Re: stress

    I totally see it now thank you!
     
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