Two light vertical springs with equal natural | vedclass

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(D) For the rod $AB$ to remain horizontal,the extension $x$ in both springs must be equal. Let the extension be $x$.The forces exerted by the springs

Vedclass · Accepted Answer

$\frac{l k_2}{k_1 + k_2}$

Vedclass · Answer

(D) For the rod $AB$ to remain horizontal,the extension $x$ in both springs must be equal. Let the extension be $x$.The forces exerted by the springs at ends $A$ and $B$ are $F_A = k_1 x$ and $F_B = k_2 x$ respectively.For rotational equilibrium about point $C$,the sum of moments must be zero:$\sum 	au_C = 0$$(k_1 x) \cdot AC = (k_2 x) \cdot BC$Canceling $x$ from both sides:$k_1 \cdot AC = k_2 \cdot BC$$\frac{AC}{BC} = \frac{k_2}{k_1} \dots (i)$We know that $AC + BC = l$,so $BC = l - AC$. Substituting this into equation $(i)$:$\frac{AC}{l - AC} = \frac{k_2}{k_1}$$k_1 \cdot AC = k_2 (l - AC)$$k_1 \cdot AC = k_2 l - k_2 \cdot AC$$AC (k_1 + k_2) = k_2 l$$AC = \frac{l k_2}{k_1 + k_2}$

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