A very stiff bar AB of negligible mass is suspended horizontally by two vertical rods as shown. The length of the bar is 2.5L the steel rod has length L and cross-sectional radius r and the brass rod has length 2L and cross-sectional radius 2r. A vertically downward force F is applied to the bar at a distance "x" from the steel rod and the bar remains horizontal. Ratio of young 's modulus of steel to brass is 2.

x = 2 L

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x = 1.25 L

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x = 5/3L

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Solution

The correct option is B x = 5/3L
$\begin{matrix} Applying torque at A . \begin{matrix} F_{2} x = F_{1} (2.5 L - x) ⟶ (1) \sum F_{y} = 0 \to F = F_{1} + F_{2} ⟶ (2) \end{matrix} \end{matrix}$

$\begin{matrix} Using (1) and (2) F_{1} = \frac{F x}{2 \cdot 5 L}, F_{2} = F {1 \frac{x}{2 \cdot 5 L}} For the vertical rods : - \frac{F_{1}}{A_{B}} = Y_{B} \frac{Δ l_{1}}{2 L} \Rightarrow Δ l_{1} = \frac{2 L F_{1}}{A_{B} Y_{B}} \end{matrix}$

$\begin{matrix} r_{B} & = 2 r r_{S} & = r Y_{B} & = 1 \times 10^{11} N / m^{2} Y_{S} & = 2 \times 10^{11} N / m^{2} \end{matrix}$

$\begin{matrix} similarly Δ l_{2} = \frac{F_{2} L}{4_{s} y_{s}} \frac{Δ l_{1}}{Δ l_{2}} = \frac{2 F_{1} L}{A_{B} y_{B}} \div \frac{F_{2} L}{A_{s} y_{s}} = \frac{2 F_{1} (A_{s} y_{s})}{A_{B} Y_{B}} F_{2} \end{matrix}$

$\begin{matrix} = \frac{2 F_{1}}{F_{2}} \frac{1 \times 2}{4} = \frac{F_{1}}{F_{2}} = 2 \Rightarrow \frac{x / 2.5 L}{1 - x / 2.5 L} = 2 \Rightarrow \frac{x}{2.5 L - x} = 2 \Rightarrow x = 5 L - 2 x \Rightarrow x = 5 L / 3 \end{matrix}$

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