Question

A uniform elastic plank moves over a smooth horizontal plane due to a constant force $F_0$ distributed uniformly over the end face. The surface of the end face is equal to $S$ and Young's modulus of the material is $E$. If the compressive strain of the plank in the direction of the acting force is $\sigma =\frac{F_0}{xSE}$, find the value of '$x$'.

Answer 1

Let us consider an element of the rod at a distance $x$ from the free end (figure shown below). For the considered element, 'T' and 'T+dT' are internal restoring forces which produce elongation and dT provide the acceleration to the element. For the element from Newton's law:

$dT=\left(dm\right)w=\left(\frac{m}{l}dx\right)\frac{F_0}{m}=\frac{F_0}{l}dx$

As free end has zero tension, on integrating the above expression,

we get, ${\int }_0^{T}dT=\frac{F_0}{l}{\int }_0^{x}dx$ or $T=\frac{F_0}{l}x$

Elongation in the considered element of length $dx$:

$\partial \xi =\frac{\sigma }{E}\left(x\right)dx=\frac{T}{SE}dx=\frac{F_{x0}xdx}{SEL}$

Thus total elongation $\xi =\frac{F_0}{SEl}{\int }_x^{l}xdx=\frac{F_{0}l}{2SE}$

Hence the sought strain $\sigma =\frac{\xi }{l}=\frac{F_0}{2SE}$