Question

A steel rope has length $L$, area of cross-section $A$, Young's modulus $Y$ and density as $d$. It is pulled on a horizontal frictionless floor with a constant horizontal force $F=\frac{dALg}{2}$ applied at one end. Find the strain at the midpoint.

• A. $\frac{dgL}{2Y}$
• B. $\frac{dgL}{4Y}$
• C. $\frac{dgL}{6Y}$
• D. $\frac{dgL}{8Y}$

Answer 1

The correct option is B $\frac{dgL}{4Y}$

Since steel rope has mass therefore, tension in it will not remain constant and it will proportional to the length of it from the end with which it is tied up.

$T\propto l$

Therefore tension at $l=L/2$ will be half.

$T_{l=L/2}=T/2$

$T_{l=L/2}=\frac{dALg}{4}$

We know that,
$stress=\frac{F}{A}=\frac{T}{A}$

$stress=\frac{dALg}{4A}$

$strain=\frac{stress}{Y}$

$strain=\frac{dgL}{4Y}$