Question

A unniform elastic rod of cross section area A, natural length L and Youngs modulus Y is placed on a smooth horizontal surface. Now two horizontal forces (of magnitude F and 3F) directed along the length of rod and in opposite direction act at two of its ends as shown. After the rod has acquired steady state, the extension of the rod will be

• A. $\frac{2F}{YA}L$
• B. $\frac{4F}{YA}L$
• C. $\frac{F}{YA}L$
• D. $\frac{3F}{2YA}L$

Answer 1

The correct option is A $\frac{2F}{YA}L$

$\begin{array}{c}\text{Let, in the element of length}dx\\ \text{ds be the elongation.}\\ Y=\frac{T}{A\left(ds/dx\right)}⟶\left(1\right)\text{.}\end{array}$

$\begin{array}{c}\text{Where}T\text{is the tension at a length}x\\ \text{causing the eleongation.}\end{array}$

$\begin{array}{c}\text{The Teusion is varying linearly along the length of the rod, Hence}⟶\\ T\text{(at position}x)=3F-\frac{2F}{L}x\to (2)\\ \text{subsituting (2)}in\text{(1) and then integrating}\end{array}$

$\begin{array}{cc}\Rightarrow \int ds& ={\int }_0^L\frac{F}{AY}(3-\frac{2x}{L})du\\ \Rightarrow s& =\frac{F}{AY}\times 2L=\frac{2FL}{AY}\\ & \therefore Ans:\left(A\right)\end{array}$