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A uniform metal bar of 70 cm length and 4 kg mass supported on two knife edges placed 10 cm from each end. A 6 kg load is suspended at 30 cm from one end. The reaction forces at the knife edges are $(g = 10 {ms}^{- 2})$

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30 N, 70 N

56 N, 44 N

50 N, 50 N

48 N, 52 N

answer is C.

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Detailed Solution

$R_{1} + R_{2} = (4 + 6) g$

About centre of mass of rod

$- R_{1} \times 0.25 + 0.05 W_{1} + 0.025 R_{2} = 0$

$\begin{array}{l} \Rightarrow R_{1} - R_{2} = 1.2 g \\ \Rightarrow R_{1} = 5.6 g a n d R_{2} = 4.4 g \end{array}$

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