A very stiff bar (AB) of negligible mass is suspended horizontally by two vertical rods as shown in figure. Length of the bar is 2.5L. The steel rod has length L and cross-sectional radius of r and the brass rod has length 2L and cross-sectional radius of 2r. A vertically downward force F is applied to the bar at a distance X from the steel rod and the bar remains horizontal. Find the value of X (in mm) if it is given that ratio of Young's modulus of steel and brass is $\frac{Y_{S}}{Y_{B}} = 2$ . [Take L=100cm]

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Detailed Solution

$\begin{array}{l} τ_{F} = 0 \\ T_{2} X = T_{1} (2.5 L - X) \\ \Rightarrow \frac{T_{2}}{T_{1}} = (2.5 \frac{L}{X} - 1) \\ 1 = (2.5 \frac{L}{X} - 1) \end{array}$

$\begin{array}{l} τ = 1.25 L \\ Y_{E} = σ \\ \Rightarrow T_{1} = Y_{B} π 4 r^{2} \frac{Δ L}{2 L} \\ \Rightarrow T_{2} = Y_{S} π r^{2} \frac{Δ l}{L} \\ = \frac{T_{1}}{T_{2}} = \frac{2 Y_{B}}{Y_{S}} = 1 \end{array}$