Question

A rope has length 3.72 m, cross sectional area of 0.154 x 10-4 m^2, and Young's modulusof 19.1 x10^10 N•m^-2. A load of mass 286 kg hangs from the rope.A) What is the increase in length (in mm) of the rope?

Answer 1

Given data:

* The original length of the rope is L = 3.72 m.

* The area of cross-section of rope is,

$A=0.154\times10^{-4}m^2$

* The Young's modulus of rope is,

$Y=19.1\times10^{10}Nm^{-2}$

* The mass of the load is,

$m=286\text{ kg}$

Solution:

The weight of the load is,

$\begin{gathered} W=mg \\ W=286\times9.8 \\ W=2802.8\text{ N} \end{gathered}$

Young's modulus of the rope in terms of the weight, area, and the original length of the rope is,

$Y=\frac{WL}{Al}$

where l is the change in the length,

Substituting the known values,

$\begin{gathered} 19.1\times10^{10}=\frac{2802.8\times3.72}{0.154\times10^{-4}\times l} \\ l=\frac{2802.8\times3.72}{0.154\times10^{-4}\times19.1\times10^{10}} \end{gathered}$

By simplifying,

$\begin{gathered} l=\frac{10426.416}{2.9414\times10^6} \\ l=3544.7\times10^{-6}\text{ m} \\ l=3.54\times10^{-3}\text{ m} \\ l=3.54\text{ mm} \end{gathered}$

Thus, the increase in the length of the rope is 3.54 mm.