Question

A homogeneous rod of length $L$ is acted upon by two forces $F_1$ and $F_2$ applied to its ends and directed opposite to each other. With what force $F$ will the rod be stretched a the crosssection at a distance $l$ from the end where force $F_1$ is applied:

• A. $\frac{(F_2-F_1)}{2L}$
• B. $\frac{(F_2-F_1)l}{L}$
• C. $\frac{(F_2-F_1)l}{L}+F_1$
• D. $\frac{(F_2-F_1)l}{L}+F_2$

Answer 1

The correct option is C $\frac{(F_2-F_1)l}{L}+F_1$

Let $m$ be the mass per unit length and we assume $F_2<F_1$,
so the acceleration of the system in the left direction.
For part AP, $mla=F_1-F....\left(1\right)$
for part PB, $m(L-l)a=F-F_2....\left(2\right)$
$\left(2\right)/\left(1\right),(\frac{L}{l}-1)=\frac{F-F_2}{F_1-F}$
or, $(\frac{L}{l}-1)F_1-(\frac{L}{l}-1)F=F-F_2$
or, $F(\frac{L}{l}-1+1)=(\frac{L}{l}-1)F_1+F_2$
or, $F=(1-\frac{l}{L})F_1+F_2\left(\frac{l}{L}\right)$
or, $F=\frac{(F_2-F_1)l}{L}+F_1$