# Young's modulus of spider thread

1. Oct 21, 2007

### faoltaem

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. The attempt at a solution

This is the only equation that i know for this. I realise 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.

2. Oct 21, 2007

### PhanthomJay

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.

3. Oct 21, 2007

### mgb_phys

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.

4. Oct 22, 2007

### faoltaem

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 $$Lo_p = Lo_s$$

5. Oct 22, 2007

### PhanthomJay

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.

6. Oct 22, 2007