Question

A uniform rod( area of cross-section A, Young’s modulus Y, Length L, mass M) is pulled apart on a smooth horizontal surface as shown. Calculate the elongation of the rod after a long time.

Answer 1

Hint: The Young’s modulus of the material is defined as the ratio of the stress to the strain. The stress will reduce as it moves away from the point where the force is applied. Then find the force at the point of elongation. Then the stress can be calculated by the ratio of a new force to the area. Then substitute the force calculated and then the elongation can be calculated from the equation of young’s modulus.

Complete step-by-step solution
The Young’s modulus of the material is defined as the ratio of the stress to the strain. Here it is due to the longitudinal stress and longitudinal strain.
By the application of an external force the rod can accelerate by,
$a=\dfrac{F}{m}$
The stress will reduce as it moves away from the point where the force is applied. As a result, the stress decreases as it moves away from the end where force is applied.
To find out the elongation, we have to consider a small element for length dx at a distance x from the free end of the rod.
Then,
$\begin{align}
  & F'=\dfrac{Fx}{L} \\
 & \\
\end{align}$
Therefore stress can be calculated as,
$\sigma =\dfrac{F'}{A}$
$\Rightarrow \sigma =\dfrac{F}{A}\dfrac{x}{L}$
Thus the elongation produced is,
$d\delta =\dfrac{F}{YAL}xdx$
Then total elongation becomes,
$\delta =\dfrac{F}{YAL}\int\limits_{0}^{1}{xdx}$
$\Rightarrow \delta =\dfrac{F}{YAL}\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1}$
$\therefore \delta =\dfrac{1}{2}\dfrac{Fl}{YA}$

Note: The Young’s modulus is a mechanical property that defines the tensile property of a material. The Young’s modulus has the same unit of stress because the strain is a dimensionless quantity. When a force is applied to a solid material the material may either get elongated or compressed as the result of the applied force.