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. If $F_2>F_1$, then with what force $F$ will the rod be stretched at the across section at a distance $l$ from the end where $F_1$ is applied?

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

Answer 1

Answer: (B)

$\frac{(F_2-F_1)l}{L}+F_1$

mass of both parts of rods are :

$m_1=(\frac{M}{L})l,m_2=(\frac{M}{L}\left)\right(L-l)$
Also $F_2-F=m_{2}a\dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$
$F-F_1=m_{1}a\dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(2\right)$
and $a=\frac{F_2-F_1}{M}$
Acceleration of rod $=\frac{F_2-F_1}{M}$
This will be acceleration of 'l' part
$F_1+\frac{Ml}{L}\times \frac{F_2-F_1}{M}=F$

$\Rightarrow F=F_1+\frac{(F_2-F_1)l}{L}$