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Tensile Strength Question

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    In the course of the above research, they discovered that this spider silk has an ultimate strength #or
    tensile strength of 1850mPa (see 11.4 in University Physics). Tensile stress is a measure of the
    pressure in an object when it is stretched. Tensile strength is the maximum tensile stress that the
    object can endure before breaking. The equation for tensile stress is Tensile Stress=F/A

    where is the applied force F and A is the cross-sectional area of the object (in this case, the circle
    formed by a cross-section of a fibre).

    Ok, phew, so on to the question...

    If Spiderman has mass 63.2kg , what is the minimum diameter of fibre (of C. darwini spider silk) required to support his weight?

    2. Relevant equations

    Tensile strength=F/A

    3. The attempt at a solution

    I know that the area of a circle is 2Pi*r^2 but when I plug everything in I always get the wrong answer. I used force as mg, which didn't work. Could anyone please help?
    I usually end up with 0.412m.
    thanks alot.
  2. jcsd
  3. Nov 4, 2011 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    If you know the area of a circle is 2 pi r^2, you have another problem to solve besides this one.

    BTW, you are posting the same HW question in multiple threads.
  4. Nov 5, 2011 #3
    you might wanna check bridge building physics on this one. supporting a single weight differs on the distance to the anchor point. gravity will act on the "spider silk" as much as it acts on spiderman". The further he has to shoot his "line" the thicker it will have to be. You see that when spiderman shoots a line he swings on it. This emparts more forces on the line than just holiding a static weight. Silk is very thin so he'd have to use multiple strands of silk...as "spiderman" is in motion there is a friction element between the strands. All this has to be taken into consideration!

    Its an impossible question to answer without more information :)
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