Answer to Question #124893 in Mechanics | Relativity for titi

Question

A fisherman uses a fishing rod to pull a fish of weight 6.5 kg out of a river. The fish is then suspended vertically on the end of a fishing line of uniform circular cross section. The line has a breaking stress of 65 MPa.
Calculate:
(a)
The minimum diameter of the fishing line required if the line does not break.
(b)
The extension per metre of this line under these conditions.

Answer 1

(a)

$\sigma=\frac{F}{S}=\frac{4F}{\pi d^2}\to d=\sqrt{\frac{4F}{\pi\sigma}}=\sqrt{\frac{4\cdot 6.5\cdot9.81}{3.14\cdot 65\cdot10^6}}=0.0011m$

(b)

$\sigma=E\frac{\Delta l}{l}$ if $l=1 m$ then $\Delta l=\frac{\sigma}{E}$

Assume that $E=1.4\cdot10^9Pa$

So, we get

$\Delta l=\frac{\sigma}{E}=\frac{65\cdot10^6}{1.4\cdot10^9}=0.046 m$