A rod of length `6m` has a mass of `12 kg`. It is hinged at one end of the rod at a distance of `3m` below the water surface.
a.What must be the weight of a block that is attached to the other end of the rod so that `5 m` of the rod's length is under water?
b. Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of rod is `0.5.`

Options

A. `17/3 kgf` in the downward direction
B. `8 kgf` in the downward direction
C. `4 kgf` in the downward direction
D. `5 kgf` in the downward direction

Answer

A

Explanation

Mass per unit length of rod is `2kg`.

Mass of the submerged portion of the rod is `10 kg` and its volume is `10//500 m^(3)`.
Buoyant force `F_(B)=10//500xx1000=20kgf`
Let `N` and `H` be the vertical and horizontal reactions of the hinge on the rod, respectively. considering the horizontal and vertical translational eqilibrium of the rod,
`N+12+W=20`
where `W` is the weight to be attached at the end.
`H=0`
Considering the rotational equilibrium of the rod about `A`
`-20xxgxx2.5costheta=12xxgxx3costheta+Wxx6costheta=0`
`implies 6W=50g-3g-36g=14g`
`impliesW=7/3g=7/3kgfimpliesN=8-7/3=17/3kgf`