Young's modulus of spider thread

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Homework Help Overview

The discussion revolves around a physics problem involving Young's modulus, specifically related to a spider hanging from its thread and a comparison with a nylon rope. The problem requires participants to explore the fractional increase in length of the thread due to the weight of the spider and to determine the necessary radius of a nylon rope to achieve the same fractional increase in length.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the formula for Young's modulus and the relationship between force, area, and length change. There is a focus on understanding how to calculate the cross-sectional area from the given radius and the implications of using the radius of the thread versus the web.

Discussion Status

Some participants have provided guidance on the use of the formula and the interpretation of the radius. There is acknowledgment that the fractional increase in length can be calculated without needing the original length, and that assumptions about the lengths of the spider's thread and the nylon rope may not be necessary for the calculations being performed.

Contextual Notes

Participants note that the problem does not provide the original length of the thread, which raises questions about how to approach the calculations without this information. There is also discussion about the assumption that the original lengths of the spider's thread and the nylon rope are equal.

faoltaem
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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[tex]^9[/tex] N/m2 and a radius of 9.8 x 10[tex]^- ^6[/tex] 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[tex]^9[/tex] 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



[tex]F = Y (\frac{A_o}{L_o}) \Delta L[/tex]

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 [tex]A = \pi r^2[/tex] 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[tex]^9[/tex] N/m2 and a radius of 9.8 x 10[tex]^- ^6[/tex] 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[tex]^9[/tex] 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



[tex]F = Y (\frac{A_o}{L_o}) \Delta L[/tex]

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 [tex]A = \pi r^2[/tex] 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?
[tex]Y_s[/tex] = 4.7x10[tex]^9[/tex] N/m[tex]^2[/tex]
m = 0.26g = 2.6x10[tex]^-^4[/tex] kg
[tex]r_s[/tex] = 9.8x10[tex]^-^6[/tex] m

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

[tex]\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}[/tex]
[tex]\Delta L = 1.799\times10^-^3 Lo[/tex]

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?
[tex]Y_s[/tex] = 3.7x10[tex]^9[/tex] N/m[tex]^2[/tex]
m = 76kg
[tex]\Delta L_p = \Delta L_s = 1.799\times^-^3 L_o[/tex]

assume: [tex]Lo_p = Lo_s[/tex]

[tex]r^2 = \frac{mgL_o}{Y_p \Delta L \pi}[/tex]

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

r = [tex]\sqrt{3.457\times10^-^5}[/tex] = 5.97x10[tex]^-^3[/tex] m



Is this correct? Is it possible to find [tex]L_o[/tex] in part (a)? And is it right to assume that [tex]Lo_p = Lo_s[/tex]
 
faoltaem said:
Is this correct? Is it possible to find [tex]L_o[/tex] in part (a)? And is it right to assume that [tex]Lo_p = Lo_s[/tex]
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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