Young's modulus of spider thread

AI Thread Summary
The discussion centers on calculating the fractional increase in length of a spider's thread when a spider hangs from it, using Young's modulus. The spider's thread has a Young's modulus of 4.7 x 10^9 N/m² and a radius of 9.8 x 10^-6 m, leading to a calculated fractional increase of approximately 1.799 x 10^-3 L₀. Additionally, the problem involves determining the required radius of a nylon rope, with a Young's modulus of 3.7 x 10^9 N/m², to achieve the same fractional increase in length when a 76 kg person hangs from it. The solution indicates that the radius needed for the nylon rope is approximately 5.97 x 10^-3 m, and clarifies that the initial lengths of the spider's thread and the rope do not need to be equal for the calculations of fractional increase to hold true.
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Homework Statement



Q 1. A spider with a mass of 0.26 g hangs vertically by one of its threads. The thread has a Young’s modulus of 4.7 x 10^9 N/m2 and a radius of 9.8 x 10^- ^6 m.
a) What is the fractional increase in the thread’s length caused by the spider?
b) Suppose a 76 kg person hangs vertically from a nylon rope (Young ’s modulus of 3.7 x 10^9 N/m2). What radius must the rope have if its fractional increase in length is to be the same as that of the spider’s thread?

Homework Equations



F = Y (\frac{A_o}{L_o}) \Delta L

The Attempt at a Solution



This is the only equation that i know for this. I realize that the area that i need to solve this problem using this equation is the cross sectional area so, is it possible to work this out with this information as i don't have the cross sectional area and i can't use A = \pi r^2 because the radius that I've been given is for the length of the string itself.
 
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faoltaem said:

Homework Statement



Q 1. A spider with a mass of 0.26 g hangs vertically by one of its threads. The thread has a Young’s modulus of 4.7 x 10^9 N/m2 and a radius of 9.8 x 10^- ^6 m.
a) What is the fractional increase in the thread’s length caused by the spider?
b) Suppose a 76 kg person hangs vertically from a nylon rope (Young ’s modulus of 3.7 x 10^9 N/m2). What radius must the rope have if its fractional increase in length is to be the same as that of the spider’s thread?

Homework Equations



F = Y (\frac{A_o}{L_o}) \Delta L

The Attempt at a Solution



This is the only equation that i know for this. I realize that the area that i need to solve this problem using this equation is the cross sectional area so, is it possible to work this out with this information as i don't have the cross sectional area and i can't use A = \pi r^2 because the radius that I've been given is for the length of the string itself.
The spider is hanging vertically by a single thread. The radiius of the thread is given, and from this, you can determine the cross sectional area and solve for the length change using the equation you have correctly noted. I think you might be assuming that the given radius is the radius of the web? This is not the case.
 
Why can't you use the radius you are given?
Young's modulus is just stress/strain.
You know the stress is just force (weight of spider ) / area and you are solving for strain.
Just use the initial diameter of the thread.
 
Yeah, i was taking the radius as the radius of the web.
so I've done this:
nb: s=spider p=person

a) What is the fractional increase in the thread’s length caused by the spider?
Y_s = 4.7x10^9 N/m^2
m = 0.26g = 2.6x10^-^4 kg
r_s = 9.8x10^-^6 m

F = Y(\frac{A_o}{L_o})\Delta L => mg = Y_s(\frac{\pi r_s^2}{L_o})\Delta L =>\frac{\Delta L}{L_o} = \frac{mg}{Y_s \pi r_s^2}

\frac{\Delta L}{L_o} = \frac{2.6\times10^4 \times 9.81}{4.7\times10^9 \times \pi \times (9.8\times10^-^6)^2}
\Delta L = 1.799\times10^-^3 Lo

b) Suppose a 76 kg person hangs vertically from a nylon rope (Young ’s modulus of 3.7 x 10 N/m2). What radius must the rope have if its fractional increase in length is to be the same as that of the spider’s thread?
Y_s = 3.7x10^9 N/m^2
m = 76kg
\Delta L_p = \Delta L_s = 1.799\times^-^3 L_o

assume: Lo_p = Lo_s

r^2 = \frac{mgL_o}{Y_p \Delta L \pi}

r^2 = \frac{76\times9.81\times L_o}{3.7\times10^9 \times 1.799\times10^-^3 \times L_o \times \pi} = 3.567x10^-^5

r = \sqrt{3.457\times10^-^5} = 5.97x10^-^3 m



Is this correct? Is it possible to find L_o in part (a)? And is it right to assume that Lo_p = Lo_s
 
faoltaem said:
Is this correct? Is it possible to find L_o in part (a)? And is it right to assume that Lo_p = Lo_s
I didn't check all your numbers, but your solution method and approach looks good to me. You are asked to find a fractional or percentage increase in length, which is the 'strain' value that mgb_phys alluded to above. You cannot find the Lo value without additional data, but it makes no difference what it is when you are just looking at fractional increases. Also, you need not assume that Lo_p = Lo_s, the result for the fractional increase (strain) is the same regardless of length.
 
thanks for all your help
 
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