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A homogeneous rod of length L is acted upon by the 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 at the cross section at a distance l from the end where force ${{F}_{1}}$ is applied? (A) $\dfrac{{{F}_{2}}-{{F}_{1}}}{2L}$(B) $\dfrac{({{F}_{2}}-{{F}_{1}})l}{L}$(C) $\dfrac{({{F}_{2}}-{{F}_{1}})l}{L}+{{F}_{1}}$(D) $\dfrac{({{F}_{2}}-{{F}_{1}})l}{L}+{{F}_{2}}$

Last updated date: 20th Jun 2024
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Hint: We are given in the question that the rod is homogeneous which means mass is uniformly throughout and two forces act. We are not given which force is greater in magnitude. We need to find the magnitude

Let m be mass per unit length and since it is not given which force is greater in magnitude, assuming ${{F}_{2}}<{{F}_{1}}$
So, the system will move towards the left. Let a be the acceleration of the system
For AP:
Length=l
Mass of this length is ml
Applying Newton’s second law $(ml)a={{F}_{1}}-F$-------(1)
For PB:
$\Rightarrow m(L-1)a=F-{{F}_{2}}$-----(2)
Divide eq (2) by (1) we get, $\dfrac{F-{{F}_{2}}}{{{F}_{1}}-F}=\dfrac{L-1}{l}$
Cross multiplying, we get,
\begin{align} &\Rightarrow (\dfrac{L}{l}-1){{F}_{1}}-(\dfrac{L}{l}-1)F=F-{{F}_{2}} \\ &\Rightarrow F=(1-\dfrac{l}{L}){{F}_{1}}+{{F}_{2}}(\dfrac{l}{L}) \\ &\therefore F=\dfrac{({{F}_{2}}-{{F}_{1}})l}{L}+{{F}_{1}} \\ \end{align}

Thus, the correct option is (C).