A wooden rod weighing `25N` is mounted on a hinge below the free surface of water as shown. The rod is `3m` long and uniform in cross section and the support is 1.6m below the free surface. At what angle `alpha` rod is in equilibrium? The cross-section of the rod is `9.5 xx10^(-4)m^(2)` in area. Density of water is `1000kg//m^(3)`. Assume buoyancy to act at centre of immersion. `g=9.8m//s^(2)`. Also find reaction on the hinge in this position.
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Answer

A, C, D

Explanation

Let `G` be the mid-point of `AB` and `E` the mid point of `AC`(i.e. the centre of buoyancy)

`AC=1.6 cosec alpha`
Volume of `AC`
`(1.6xx9.5xx10^(-4)) cosec (alpha)`
Weight of water displaced by `AC`
`=(1.6xx9.5xx10^(-4))cosec alpha`
`=14.896 cosec alpha`
Hence, the buoyant force is `14.896 cosed alpha` vertically upwards at` E`. While the weight of the rod is `25N` acting vertically downwards at `G`. Taking moments about A.
`(14.896 cosec alpha)(AE cos alpha)=(25)(AG cos alpha)`
or, `(14.896 cosec alpha)((1.6 cosec alpha)/(2))=25xx(3)/(2)`
or, `sin^(2)alpha =0.32`
`:. sinalpha=0.56`
or. `alpha=34.3^(@)`
Further. Let` F` be the reaction at hinge in vertically downward direction. Then, considering the tranlatory equilibrium of in vertical direction we have,
`F+`weight of the rod=upthrust
`F=`upthrust -weight of the rod
`=14.896 cosec(34.3^(@))-25`
`=26.6-25`
` :. F=1.6N`.