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

Answer: 2

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 $dl=\frac{F_x(dx)}{AY}$
$=\frac{\left(\frac{F_{0}x}{L}\right)dx}{AY}$
$\therefore$ Total elongation,
$l=\underset{0}{\overset{L}{\int }}\frac{F_0}{ALY}xdx$
$=\frac{F_0}{2ALY}{L}^2=\frac{F_{0}L}{2AY}$
$\therefore$ Compressive strain $=\frac{l}{L}=\frac{F_0}{2AY}$
$\Rightarrow C=2$