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Thermal Expansion

A homogeneous...

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 cross section at a distance $l$ from the end where $F_{1}$ is applied?

A

\frac{(F_{2} - F_{1}) l}{L}

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B

\frac{(F_{2} - F_{1}) l}{L} + F_{1}

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C

\frac{(F_{2} + F_{1}) l}{2 L}

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D

\frac{({F^{2}}_{2} + F_{1}^{2}) l}{F_{1} L}

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Solution

The correct option is C $\frac{(F_{2} - F_{1}) l}{L} + F_{1}$
consider, left side of the cross-section is a block of mass $m_{1} = \frac{M l}{L}$ and right side of the cross-section as a block of mass $m_{2} = \frac{M (L - l)}{L}$
From FBD , we get
for block $m_{1}$ , $m_{1} a = F - F_{1} . . . . . . . . . . (1)$
for block $m_{2}$ , $m_{2} a = F_{2} - F . . . . . . . . . . (2)$
from (1), $a = \frac{F - F_{1}}{m_{1}}$
now (2) becomes, $\frac{m_{2}}{m_{1}} (F - F_{1}) = F_{2} - F$
$(\frac{m_{2}}{m_{1}} + 1) F = F_{2} + F_{1} (\frac{m_{2}}{m_{1}})$
put $\frac{m_{2}}{m_{1}} = \frac{L}{l} - 1$
$\frac{L}{l} F = F_{2} + (\frac{L}{l} - 1) F_{1}$
$F = \frac{(F_{2} - F_{1}) l}{L} + F_{1}$

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