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.`

Explanation

Let `w` be the weight attached at the end `Q:X,Y` be the components of force exerted be hinge `P` on the rod: `B` be the buoyant force acting at the centre of byouancy `B_(0)` (mild point of submerged portion) `G` be the weight of rod and `theta` be the angle of the rod with vertical. Now balancing force is given by `B+Y=G+W `and `X=0`
balancing torque about `P` is given by
`Bsintheta(PB_(0))=Gsintheta(PG_(0))+Wsintheta(PQ)`
`implies Bsintheta(2.5)=Gsintheta(3)+Wsintheta(6)`
If area of cross section of the rod is `A`, then buoyant force `B`
`=5Arhog` then
`implies(5A)rhog(2.5)=(6A)dg(3)=W(6)`
`implies10^(3)(5xx1xx2.5)=10^(3)(6xx0.5xx3)=(6W)/(Ag)`
Mass of the rod `=12 Kg =6Ad=(6A)0.5xx10^(3)`
`implies A=(4m^(2))/1000`
`:. W=(3.5xx4xx9.9)/6=2.33xx9.8=22.83N` (or `2.33kg)`
`Y=G+W-B=(6Adg)+2.33g-(5Arhog)=-5.66g`
Therefore, the horizontal component of force by hinge is `X=0` and the vertical component is `5.66 kg` acting downwards.