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6.System of Particles and Rotational Motion
hard
Two light vertical springs with equal natural lengths and spring constants $k_1$ and $k_2$ are separated by a distance $l$. Their upper ends are fixed to the ceiling and their lower ends to the ends $A$ and $B$ of a light horizontal rod $AB$. $A$ vertical downwards force $F$ is applied at point $C$ on the rod. $AB$ will remain horizontal in equilibrium if the distance $AC$ is
A
$\frac{l}{2}$
B
$\frac{{{l}\,{k_1}}}{{{k_2} + {k_1}}}$
C
$\frac{lk_2}{k_1}$
D
$\frac{{{l}\,{k_2}}}{{{k_1} + {k_2}}}$
Solution
$\sum(\text { Moments about } C)=0$
$\therefore\left(k_{1} x\right) A C=\left(k_{2} x\right) B C$
$\therefore \frac{A C}{B C}=\frac{k_{2}}{k_{1}}$$…(i)$
$A C+B C=l \ldots(ii)$
Soving these two equation we get,
$A C=\left(\frac{k_{1}}{k_{1}+k_{2}}\right) l$
Standard 11
Physics
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