Question

A uniform elastic plank moves over a smooth horizontal plane due to a constant force $F _{0}$ distributed uniformly over the end force. The surface area of the end face is equal to $A$ and Young's modulus of the material is $Y$. If the compressive strain of the plank along the direction of the force is $\frac{ F _{0}}{ CAY }$. Find $C$.

Answer 1

The force at any section is due to the inertia behind the section. Hence, the stress increases from zero to maximum at the end where force is applied.
Consider a small element of length $dx$ at a distance $x$ from the free end as shown below.

Now, $F _{ x }=ma$
$=\left(\frac{ M }{ L } x \right) \times \frac{ F _{0}}{ M }=\frac{ F _{0} x }{ L }$
Elongation of the element $d l=\frac{ F _{ x }( dx )}{ AY }$
$=\frac{\left(\frac{F_{0} x}{L}\right) d x}{A Y}$
$\therefore $ Total elongation,
$l=\int\limits_{0}^{ L } \frac{ F _{0}}{ ALY } xdx $
$=\frac{ F _{0}}{2 ALY } L ^{2}=\frac{ F _{0} L }{2 AY }$
$\therefore $ Compressive strain $=\frac{l}{ L }=\frac{ F _{0}}{2 AY }$
$\Rightarrow C =2$