Solution

If we consider a rod of mass $m$ as a single whole, then it will move with acceleration $$a = \frac{F}{m}$$ Since the rod is inextensible, the acceleration of all its parts is the same and equal to $a$

Let's consider a small section of the rod of length $dx$ and mass $dm$. Since the rod is homogeneous $$dm = m \frac{dx}{l}$$ Let's write down Newton's second law for this section. $$a\, dm = F(x+dx) - F(x)\tag{1}$$ Where $F(x+dx)$ and $F(x)$ are the interaction forces with neighbors

Let's integrate expression $(1)$ along the horizontal coordinate $x$: $$\int_x^l am\frac{dx}{l} = \int_x^l dF$$ $$F(x) = ma\frac{l-x}{l}$$ $$F(x) = \fbox{$F\left(1-\frac{x}{l}\right)$}$$