Young's Modulus Formula / Steel Cable

[Miguel Velasquez]

A steel cable 3.00 cm 2 in cross-sectional area has a mass of
2.40 kg per meter of length. If 500 m of the cable is hung
over a vertical cliff, how much does the cable stretch under
its own weight? Take Y steel ϭ 2.00 ϫ 10 11 N/m 2 .Y=([L][/o]F)/(A*delta_L)

My attempt of solution: http://docdro.id/Fft5plu

This problem is driving me crazy, the textbook says the correct answer is 0.0490m. Can anyone tell me where i went wrong?NOTE: This problem was taken from the textbook "Physics for Scientists and Engineers with modern physics. 7h ed, page 360, problem 56"

 

[RUber]

I can't see your work, the link is no good for me.
Here is the process:
You are given $2\times10^{11}\frac{N}{m^2} = \frac{500*F}{A*\Delta L }.$
(note that area must be changed to square meters)
Rewriting to solve for Delta L:
$\Delta L = \frac{500*F}{A*(2\times10^{11}) }.$
Force acts linearly on the cable, so it can be averaged to get
$F= 500/2 * 2.4*9.8(N).$
So
$\Delta L = \frac{500*(F)}{A*(2\times10^{11}) }.$

 

[Miguel Velasquez]

When i plug in the data into your eq. i get 0.016333333 m which is not the right answer. The link seems to works for me, can anyone else test the link i gave? Thank you Ruber for trying.

 

[RUber]

Are you using .0003 for area or .0001?

 

[Miguel Velasquez]

Wow! you were right, i plugged in a wrong value, thank you RUber! Could you tell me why the force must be averaged? What i did is use F=M_tot*g=(2.4Kg/m)(500m)(9.8m/s2), but this seems to be wrong.

 

[Chestermiller]

To get the average stress, you need to use half the weight. The local tension in the cable is $$T=\gamma gz_0$$ where z₀ is the (unstretched) distance measured up from the bottom. The local stress in the cable is $$\sigma=\frac{\gamma g z_0}{A}$$. The local strain in the cable is $$\epsilon=\frac{\gamma g z_0}{YA}$$ The total stretched length of the cable is $$L=\int_0^{L_0}\left(1+\frac{\gamma g z_0}{YA}\right)dz_0$$