How Thick Should Spider Silk Be to Hold Spiderman?

[mattyc33]

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?

I know that the area of a circle is Pi*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?

 

[SteamKing]

Check your units. Your problem statement did not indicate what units it wanted the diameter of the fibre expressed in.

 

[mattyc33]

I have verified that the question is needed in meters. If anyone could help me with this I would appreciate it greatly!

 

[SteamKing]

Is the tensile strength of the webbing 1850 milli pascals (mPa) or 1850 mega pascals (MPa)? It makes a difference. Do you know the derived units for a pascal?

 

[asteriatic]

I can provide a response to this question by using the formula for tensile stress and considering the given information. First, let's convert Spiderman's mass from kilograms to Newtons, as force is measured in Newtons. This gives us a force of approximately 620 N (using the conversion factor 1 kg = 9.8 N).

Next, we need to determine the cross-sectional area of the spider silk fiber. We can do this by rearranging the equation for tensile stress to solve for A: A = F/Tensile Stress. Plugging in the values of 620 N for F and 1850mPa (or 1.85 GPa) for tensile strength, we get a cross-sectional area of approximately 0.000335 m^2 or 335 mm^2.

Finally, to find the minimum diameter of the fiber, we can use the formula for the area of a circle (A = Pi*r^2) and solve for r. Plugging in the calculated area of 335 mm^2, we get a minimum diameter of approximately 20.6 mm. Therefore, the minimum diameter of the spider silk fiber needed to support Spiderman's weight would be approximately 20.6 mm.

I would also like to mention that this calculation assumes that the spider silk fiber is able to support Spiderman's weight without breaking, which may not be accurate in real-world scenarios. Other factors such as the strength and structure of the anchor points and the distribution of weight on the fiber would also need to be considered in order to accurately determine the minimum diameter of the fiber needed.